Help me with Concatenating strings

[c]

Hi everybody.

I'm working on converting a program wriiten on perl to C, and facing a
problem with concatenate strings.
Now here is a small program that descripe the problem, if you help me
to solve in this small code, I can solve it on my own program...you
don't want to have head-ache :-)

So, the problem excatly is, I built an array..just like this one.

char *my_array [10];
my_array[0] = "Sally\n";
my_array[1] = "Ndiya\n";
my_array[2] = "Samantha\n";
my_array[3] = "Sara\n";
my_array[4] = "Cadillac\n";
my_array[5] = "GM For Ever\n";
my_array[6] = "SlacWare\n";
my_array[7] = "Google\n";
my_array[8] = "Google more and more\n";
my_array[9] = "You\n";
my_array[10] = "Computers\n";

my program will recieve an argument -char *argv[]- from the user.

int main(int argc,char *argv[])
I want to concatenate the given argument to at the first of each line
of my array.
so, if we imagine the array after the concatenating, it would look like
this (suppose the given argument is "ILove").

my_array[0] = "ILoveSally\n";
my_array[1] = "ILoveNdiya\n";
my_array[2] = "ILovesamantha\n";
my_array[3] = "ILoveSara\n";
my_array[4] = "ILoveCadillac\n";
my_array[5] = "ILoveGM For Ever\n";
my_array[6] = "ILoveSlacWare\n";
my_array[7] = "ILoveGoogle\n";
my_array[8] = "ILoveGoogle again\n";
my_array[9] = "ILoveSAMI\n";
my_array[10] = "ILoveComputers\n";*/

I tried strcat() -strings.h- but the result wasn't what I'm looking
for.

read the code..and the comments will help to understand the
problem..here is the whole code of my problem example.

#include <stdio.h>

int main(int argc,char *argv[])
{int i=-1;
char *my_array [10];

my_array[0] = "Sally\n";
my_array[1] = "Ndiya\n";
my_array[2] = "Samantha\n";
my_array[3] = "Sara\n";
my_array[4] = "Cadillac\n";
my_array[5] = "GM For Ever\n";
my_array[6] = "SlacWare\n";
my_array[7] = "Google\n";
my_array[8] = "Google more and more\n";
my_array[9] = "You\n";
my_array[10] = "Computers\n";

while(i++ < 10)
{
printf(argv[1]); //Don't do it like this..read my comments please.
printf(my_array[i]);
}

// I need your help to fix below loop, I want to cancatenate the given
data -*argv[]- to each line of my array.
// For Example, if the user enetered
// root@localhost# test ILove
// the result would be:
// ILoveSally
// ILoveNdiya
// ..etc
//
//But don't do it by printting the given data followed by my_array
lines(I mean by printting it twoice), because I need the given data to
be concatenated at the first of each line of my array.
//So, my_array would look like this
// my_array[0] = "ILoveSally\n";
// my_array[1] = "ILoveNdiya\n";
// my_array[2] = "ILovesamantha\n";
// my_array[3] = "ILoveSara\n";
// my_array[4] = "ILoveCadillac\n";
// my_array[5] = "ILoveGM For Ever\n";
// my_array[6] = "ILoveSlacWare\n";
// my_array[7] = "ILoveGoogle\n";
// my_array[8] = "ILoveGoogle again\n";
// my_array[9] = "ILoveSAMI\n";
// my_array[10] = "ILoveComputers\n";*/

while (i++ < 10)
{
// type your code here..Thanks
}
return 0;
}
thank you for your time.

[Ancient_Hacker]

you can't change anything in that array, as those are all read-only
constants.

So you either need another array, either fixed size strings, or get the
memory dynamically, either thru malloc() or out of your own array. I
recommend something aloong the lines of the latter for simplicity. Not
tested, but generally on track below code is. Add some obvious error
checks for extra credit. Use snprintf for extra safety.
char BigPool[ 1000000 ]; // or whatever size you expect;

char Strs[ STRINGS_IN_LIST ] = { "foo", "bar", ........ };
int Free = 0;

int i, Len;

for( i = 0; i < STRINGS_IN_LIST; i++ )
Free += sprintf( BigPool[ Free ], "%s%s", argv[ whatever ], Strs[
i ] ) + 1;

[Ancient_Hacker]

oops, I forgot to log the address of each created string, try this:
char BigPool[ 1000000 ]; // or whatever size you expect;

char Strs[ STRINGS_IN_LIST ] = { "foo", "bar", ........ };

char * OutStrs[ STRINGS_IN_LIST ];
int Free = 0;

int i, Len;

for( i = 0; i < STRINGS_IN_LIST; i++ )
Free += sprintf( OutStrs[ i ] = BigPool[ Free ], "%s%s", argv[
whatever ], Strs[ i ] ) + 1;

[Frederick Gotham]

c posted:

char *my_array [10];

This array contains ten elements.

my_array[0] = "Sally\n";
my_array[1] = "Ndiya\n";
my_array[2] = "Samantha\n";
my_array[3] = "Sara\n";
my_array[4] = "Cadillac\n";
my_array[5] = "GM For Ever\n";
my_array[6] = "SlacWare\n";
my_array[7] = "Google\n";
my_array[8] = "Google more and more\n";
my_array[9] = "You\n";
my_array[10] = "Computers\n";

Here, you assign to eleven separate elements.

#include <stdio.h>

int main(int argc,char *argv[])
{int i=-1;
char *my_array [10];

my_array[0] = "Sally\n";
my_array[1] = "Ndiya\n";
my_array[2] = "Samantha\n";
my_array[3] = "Sara\n";
my_array[4] = "Cadillac\n";
my_array[5] = "GM For Ever\n";
my_array[6] = "SlacWare\n";
my_array[7] = "Google\n";
my_array[8] = "Google more and more\n";
my_array[9] = "You\n";
my_array[10] = "Computers\n";

Again here, you assign to eleven elements.

while(i++ < 10)
{
printf(argv[1]); //Don't do it like this..read my comments please.
printf(my_array[i]);
}

Why resort to that? Looping 0 through 10 can be much simpler:

unsigned i = 0;

do {

}while(++i != 11);

Or, if you'd prefer the less efficient simplicity of:

for(unsigned i = 0; i != 11; ++i)
{

}

You're still pretending that you're array has eleven elements though...

--

Frederick Gotham

[c]

Thank you, but this is just a peace of code i don't care about it that
much.

Frederick Gotham wrote:

c posted:

char *my_array [10];

This array contains ten elements.

my_array[0] = "Sally\n";
my_array[1] = "Ndiya\n";
my_array[2] = "Samantha\n";
my_array[3] = "Sara\n";
my_array[4] = "Cadillac\n";
my_array[5] = "GM For Ever\n";
my_array[6] = "SlacWare\n";
my_array[7] = "Google\n";
my_array[8] = "Google more and more\n";
my_array[9] = "You\n";
my_array[10] = "Computers\n";

Here, you assign to eleven separate elements.

#include <stdio.h>

int main(int argc,char *argv[])
{int i=-1;
char *my_array [10];

my_array[0] = "Sally\n";
my_array[1] = "Ndiya\n";
my_array[2] = "Samantha\n";
my_array[3] = "Sara\n";
my_array[4] = "Cadillac\n";
my_array[5] = "GM For Ever\n";
my_array[6] = "SlacWare\n";
my_array[7] = "Google\n";
my_array[8] = "Google more and more\n";
my_array[9] = "You\n";
my_array[10] = "Computers\n";

Again here, you assign to eleven elements.

while(i++ < 10)
{
printf(argv[1]); //Don't do it like this..read my comments please.
printf(my_array[i]);
}

Why resort to that? Looping 0 through 10 can be much simpler:

unsigned i = 0;

do {

}while(++i != 11);

Or, if you'd prefer the less efficient simplicity of:

for(unsigned i = 0; i != 11; ++i)
{

}

You're still pretending that you're array has eleven elements though...

--

Frederick Gotham

[c]

Thank you for helping me, I'm going to try and reply again if I have
some Questions.

Ancient_Hacker wrote:

oops, I forgot to log the address of each created string, try this:
char BigPool[ 1000000 ]; // or whatever size you expect;

char Strs[ STRINGS_IN_LIST ] = { "foo", "bar", ........ };

char * OutStrs[ STRINGS_IN_LIST ];
int Free = 0;

int i, Len;

for( i = 0; i < STRINGS_IN_LIST; i++ )
Free += sprintf( OutStrs[ i ] = BigPool[ Free ], "%s%s", argv[
whatever ], Strs[ i ] ) + 1;

.

[Eric Sosman]

c wrote:

Hi everybody.

I'm working on converting a program wriiten on perl to C, and facing a
problem with concatenate strings.
Now here is a small program that descripe the problem, if you help me
to solve in this small code, I can solve it on my own program...you
don't want to have head-ache :-)

So, the problem excatly is, I built an array..just like this one.

char *my_array [10];
my_array[0] = "Sally\n";
my_array[1] = "Ndiya\n";
my_array[2] = "Samantha\n";
my_array[3] = "Sara\n";
my_array[4] = "Cadillac\n";
my_array[5] = "GM For Ever\n";
my_array[6] = "SlacWare\n";
my_array[7] = "Google\n";
my_array[8] = "Google more and more\n";
my_array[9] = "You\n";
my_array[10] = "Computers\n";

"Just like" this one? Including the attempt to store a
value in the non-existent eleventh element of a ten-element
array? In C an array of N elements has indices that run from
[0] through [N-1], inclusive, but does not have an [N] element.

By the way, C offers a more convenient syntax for array
initialization:

char *my_array[10] = { "Sally\n", "Ndiya\n", "Samantha\n",
"Sara\n", "Cadillac\n", "GM For Ever\n", "SlacWare\n",
"Google\n", "Google more and more\n", "You\n",
"Computers\n", };

A few points about this: First, since you've declared the array
to have ten elements but actually provided eleven initializers,
the compiler will complain and point out your mistake. Second,
you could replace [10] with [] and let the compiler figure out
the array size by counting the initializers itself instead of
making you do it (sizeof my_array / sizeof my_array[0] will then
tell you how big the compiler decided the array should be). And
last, the comma after the final initializer is optional -- it's
sometimes convenient if the code is being generated by a program
or if you anticipate adding more initializers later, but you can
use it or omit it according to your own taste.

my program will recieve an argument -char *argv[]- from the user.

int main(int argc,char *argv[])

I want to concatenate the given argument to at the first of each line
of my array. [...]

You need to confront the simplicity of C's memory management.
I mean "simplicity" in the sense that C doesn't do very much about
memory management on its own; you must attend to it for yourself.
This is both a weakness and a strength of C: It's a weakness because
it forces you to fret about the details, and a strength because it
doesn't burden you with the machinery of automatic managers whose
style may not suit your needs.

In this case, C has already done some minimal memory management
for you: It has set aside space for the array my_array and filled
its elements with pointers to strings, and it has set aside space
for those strings and filled them with their intended contents. But
now you need something different: You need to create new strings in
memory someplace, and change the array elements to point to those
new strings instead of the old ones.

Here's a quickie solution, with explanation to follow:

for (i = 0; i < sizeof my_array / sizeof my_array[0]; ++i) {
char *new;
new = malloc(strlen(argv[1]) + strlen(my_array[i]) + 1);
if (new == NULL)
exit (EXIT_FAILURE);
strcpy (new, argv[1]);
strcat (new, my_array[i]);
my_array[i] = new;
}

The first line steps the variable i through all the valid index
values of the array my_array, executing the body of the loop once
for each i value. The second line declares a temporary variable called
new that will point to the memory for one of the new strings you're
about to create -- you can't just obliterate my_array[i] yet, because
you still need to work with the original string it's still pointing to.

The third line requests some memory to be allocated. How much?
You'll need enough for all the characters of the argv[1] string, plus
enough for all the characters of the my_array[i] string, plus one more
for the '\0' that must be present at the end of any string. (It is
astonishing how frequently people forget about that "plus one;" be
on guard against making that all-too-commmon mistake.)

The fourth and fifth lines are important: You call malloc to
*request* a certain amount of memory, but the request might not be
granted. Perhaps the system is running low on memory, perhaps it
has plenty available but it's all scattered around in dribs and
drabs none of which are big enough, perhaps the phase of the Moon
is wrong. *Always* check for a NULL return from malloc, and take
corrective action if you get one -- in this case, the "corrective
action" is simply to terminate the program unsuccessfully, but in
other programs you might do something more elaborate.

The sixth and seventh lines build the new string from the two
pieces. Line six copies the argv[1] string into the new memory area,
and line seven appends the my_array[i] string to it. Note carefully
that line six uses strcpy and seven uses strcat; it would be wrong
to interchange them or to use the same function (either one) in both
places. You need to copy the first string into the uninitialized
brand-new memory area, and then append the second.

Finally, line eight changes my_array[i] so it now points to the
newly-built string.

At the beginning of the program, you'll need to #include the
<string.hheader (to declare the strlen, strcpy, and strcat functions)
and the <stdlib.hheader (for malloc, exit, and EXIT_FAILURE). You
may need other headers like <stdio.hif you intend to display any
results. Also, it would be a good idea for your program to check
that it actually received an argument string (check for argc>1)
before running blindly ahead and trying to do things with a string
that doesn't even exist.

--
Eric Sosman
es*****@acm-dot-org.invalid

[c]

THANK YOU VERY VERY MUCH, I'm going to try it and reply if I have
another questions..you just wrote an article for me, it must be printed
:-)

Eric Sosman wrote:

c wrote:

Hi everybody.

I'm working on converting a program wriiten on perl to C, and facing a
problem with concatenate strings.
Now here is a small program that descripe the problem, if you help me
to solve in this small code, I can solve it on my own program...you
don't want to have head-ache :-)

So, the problem excatly is, I built an array..just like this one.

char *my_array [10];
my_array[0] = "Sally\n";
my_array[1] = "Ndiya\n";
my_array[2] = "Samantha\n";
my_array[3] = "Sara\n";
my_array[4] = "Cadillac\n";
my_array[5] = "GM For Ever\n";
my_array[6] = "SlacWare\n";
my_array[7] = "Google\n";
my_array[8] = "Google more and more\n";
my_array[9] = "You\n";
my_array[10] = "Computers\n";

"Just like" this one? Including the attempt to store a
value in the non-existent eleventh element of a ten-element
array? In C an array of N elements has indices that run from
[0] through [N-1], inclusive, but does not have an [N] element.

By the way, C offers a more convenient syntax for array
initialization:

char *my_array[10] = { "Sally\n", "Ndiya\n", "Samantha\n",
"Sara\n", "Cadillac\n", "GM For Ever\n", "SlacWare\n",
"Google\n", "Google more and more\n", "You\n",
"Computers\n", };

A few points about this: First, since you've declared the array
to have ten elements but actually provided eleven initializers,
the compiler will complain and point out your mistake. Second,
you could replace [10] with [] and let the compiler figure out
the array size by counting the initializers itself instead of
making you do it (sizeof my_array / sizeof my_array[0] will then
tell you how big the compiler decided the array should be). And
last, the comma after the final initializer is optional -- it's
sometimes convenient if the code is being generated by a program
or if you anticipate adding more initializers later, but you can
use it or omit it according to your own taste.

my program will recieve an argument -char *argv[]- from the user.

int main(int argc,char *argv[])

I want to concatenate the given argument to at the first of each line
of my array. [...]

You need to confront the simplicity of C's memory management.
I mean "simplicity" in the sense that C doesn't do very much about
memory management on its own; you must attend to it for yourself.
This is both a weakness and a strength of C: It's a weakness because
it forces you to fret about the details, and a strength because it
doesn't burden you with the machinery of automatic managers whose
style may not suit your needs.

In this case, C has already done some minimal memory management
for you: It has set aside space for the array my_array and filled
its elements with pointers to strings, and it has set aside space
for those strings and filled them with their intended contents. But
now you need something different: You need to create new strings in
memory someplace, and change the array elements to point to those
new strings instead of the old ones.

Here's a quickie solution, with explanation to follow:

for (i = 0; i < sizeof my_array / sizeof my_array[0]; ++i) {
char *new;
new = malloc(strlen(argv[1]) + strlen(my_array[i]) + 1);
if (new == NULL)
exit (EXIT_FAILURE);
strcpy (new, argv[1]);
strcat (new, my_array[i]);
my_array[i] = new;
}

The first line steps the variable i through all the valid index
values of the array my_array, executing the body of the loop once
for each i value. The second line declares a temporary variable called
new that will point to the memory for one of the new strings you're
about to create -- you can't just obliterate my_array[i] yet, because
you still need to work with the original string it's still pointing to.

The third line requests some memory to be allocated. How much?
You'll need enough for all the characters of the argv[1] string, plus
enough for all the characters of the my_array[i] string, plus one more
for the '\0' that must be present at the end of any string. (It is
astonishing how frequently people forget about that "plus one;" be
on guard against making that all-too-commmon mistake.)

The fourth and fifth lines are important: You call malloc to
*request* a certain amount of memory, but the request might not be
granted. Perhaps the system is running low on memory, perhaps it
has plenty available but it's all scattered around in dribs and
drabs none of which are big enough, perhaps the phase of the Moon
is wrong. *Always* check for a NULL return from malloc, and take
corrective action if you get one -- in this case, the "corrective
action" is simply to terminate the program unsuccessfully, but in
other programs you might do something more elaborate.

The sixth and seventh lines build the new string from the two
pieces. Line six copies the argv[1] string into the new memory area,
and line seven appends the my_array[i] string to it. Note carefully
that line six uses strcpy and seven uses strcat; it would be wrong
to interchange them or to use the same function (either one) in both
places. You need to copy the first string into the uninitialized
brand-new memory area, and then append the second.

Finally, line eight changes my_array[i] so it now points to the
newly-built string.

At the beginning of the program, you'll need to #include the
<string.hheader (to declare the strlen, strcpy, and strcat functions)
and the <stdlib.hheader (for malloc, exit, and EXIT_FAILURE). You
may need other headers like <stdio.hif you intend to display any
results. Also, it would be a good idea for your program to check
that it actually received an argument string (check for argc>1)
before running blindly ahead and trying to do things with a string
that doesn't even exist.

--
Eric Sosman
es*****@acm-dot-org.invalid

[CBFalconer]

c wrote:

>
I'm working on converting a program wriiten on perl to C, and
facing a problem with concatenate strings.

Try strlcat and strlcpy. You can get source and documentation for
these from:

<http://cbfalconer.home.att.net/download/>

--
Some informative links:
<news:news.announce.newusers
<http://www.geocities.com/nnqweb/>
<http://www.catb.org/~esr/faqs/smart-questions.html>
<http://www.caliburn.nl/topposting.html>
<http://www.netmeister.org/news/learn2quote.html>
<http://cfaj.freeshell.org/google/>

[Joe Estock]

Frederick Gotham wrote:

c posted:

>char *my_array [10];

This array contains ten elements.

> my_array[0] = "Sally\n";
my_array[1] = "Ndiya\n";
my_array[2] = "Samantha\n";
my_array[3] = "Sara\n";
my_array[4] = "Cadillac\n";
my_array[5] = "GM For Ever\n";
my_array[6] = "SlacWare\n";
my_array[7] = "Google\n";
my_array[8] = "Google more and more\n";
my_array[9] = "You\n";
my_array[10] = "Computers\n";

Here, you assign to eleven separate elements.

To someone who is new at C, the above code would not be as easy to
notice something wrong in.

In c array indexing starts at 0. When you declare char *my_array[10];
you are saying the following: "Give me an array of 10 pointers to type
char". You can now safely assign to 0 - 9 of those without worrying.
Assigning to anything beyond index 9 would invoke UB (Undefined Behaviour).

It is also important for you to note that in your declaration you are
(in laymen terms) saying: "Give me an array of 10 pointers to type char
that I may not modify after initialization".

For example:

char *foo;

/* this is fine. you are initializing foo */
foo = "some string";

/* this is fine too. you are declaring and initializing foo */
char *foo = "some string";

/* this is *not* fine. you are writing to memory that you shouldn't */
foo[3] = '\0';

char foo[100];

/* the below is fine. we are allowed to write to foo provided that we do
not go beyond element 99. */
strcpy(foo, "testing");

foo[4] = '\0';

>

>#include <stdio.h>

int main(int argc,char *argv[])
{int i=-1;
char *my_array [10];

my_array[0] = "Sally\n";
my_array[1] = "Ndiya\n";
my_array[2] = "Samantha\n";
my_array[3] = "Sara\n";
my_array[4] = "Cadillac\n";
my_array[5] = "GM For Ever\n";
my_array[6] = "SlacWare\n";
my_array[7] = "Google\n";
my_array[8] = "Google more and more\n";
my_array[9] = "You\n";
my_array[10] = "Computers\n";

Again here, you assign to eleven elements.

>while(i++ < 10)
{
printf(argv[1]); //Don't do it like this..read my comments please.
printf(my_array[i]);
}

Why resort to that? Looping 0 through 10 can be much simpler:

unsigned i = 0;

do {

}while(++i != 11);

This isn't safe, especially if you're suggesting that someone who is
obviously new to C to use it. What happens if OP increments i within the
while loop? Consider the following:

i = 0;

do
{
i += 20;
} while(++i != 11);

There you have an infinite loop and the OP may not know what is wrong.
The code below would give clearer meaning of your intent.

i = 0;

do
{
/* code */
} while (++i < 11);

>
Or, if you'd prefer the less efficient simplicity of:

for(unsigned i = 0; i != 11; ++i)

Not valid C89 or C90 syntax. Variable declarations must be at the
beginning of a block. Again you insist on checking i to see if it is
*exactly* 11.

{

}

You're still pretending that you're array has eleven elements though...

It would be more helpful if you could point out to the OP *why* (see my
reply towards the top).

[sjdevnull]

Frederick Gotham wrote:

c posted:

while(i++ < 10)
{
printf(argv[1]); //Don't do it like this..read my comments please.
printf(my_array[i]);
}

Why resort to that? Looping 0 through 10 can be much simpler:

unsigned i = 0;

do {

}while(++i != 11);

These aren't equivalent. His has _2_ off-by-one errors; it loops from
1 to 10. Yours goes from 0 to 10; you've inadvertantly fixed one
problem.

To the OP: You want to loop from 0 to 9. One very common way to do
that is:

for(i=0; i<10; i++) {
/* Do stuff */
}

[Frederick Gotham]

posted:

Yours goes from 0 to 10; you've inadvertantly fixed one
problem.

Which is why I mentioned that he was pretending the array had eleven
elements.

--

Frederick Gotham

[Jack Klein]

On 24 Sep 2006 06:20:12 -0700, "Ancient_Hacker" <gr**@comcast.net>
wrote in comp.lang.c:

Kindly quote at least some context.

you can't change anything in that array, as those are all read-only
constants.

In what array? In the OP's code, which you failed to quote:

char *my_array [10];
my_array[0] = "Sally\n";
my_array[1] = "Ndiya\n";
my_array[2] = "Samantha\n";
my_array[3] = "Sara\n";
my_array[4] = "Cadillac\n";
my_array[5] = "GM For Ever\n";
my_array[6] = "SlacWare\n";
my_array[7] = "Google\n";
my_array[8] = "Google more and more\n";
my_array[9] = "You\n";
my_array[10] = "Computers\n";

The array contains pointers to char. The pointers are not constant at
all. Nor are the character arrays they point to constant. They are
string literals, and the type of string literals is "array of char",
most specifically not "array of const char".

Attempting to modify a string literal produces undefined behavior in C
because the C standard specifically states that it does, not because
the string literal itself has the type "array of const char".

And there is neither requirement nor guarantee in the C standard that
string literals are "read-only". The point of having an undefined
behavior category is that there are no requirements or guarantees on
the result. On some platforms, writing to string literals changes
them.

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://c-faq.com/
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.learn.c-c++
http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html

[Nick Keighley]

Frederick Gotham wrote:

c posted:

<snip>

while(i++ < 10)
{
printf(argv[1]); //Don't do it like this..read my comments please.
printf(my_array[i]);
}

Why resort to that? Looping 0 through 10 can be much simpler:

unsigned i = 0;

do {

}while(++i != 11);

I don't see why your version is simpler than the original. In fact I'd
say
while is safer than do just in case the length of the array is
specified as
zero.

Or, if you'd prefer the less efficient simplicity of:

for(unsigned i = 0; i != 11; ++i)
{

}

why is this less effiecient? Do you have benchmarks?

<snip>

--
Nick Keighley

[Frederick Gotham]

Nick Keighley posted:

Frederick Gotham wrote:

>c posted:

<snip>

while(i++ < 10)
{
printf(argv[1]); //Don't do it like this..read my comments

please.

printf(my_array[i]);
}

Why resort to that? Looping 0 through 10 can be much simpler:

unsigned i = 0;

do {

}while(++i != 11);

I don't see why your version is simpler than the original. In fact I'd
say
while is safer than do just in case the length of the array is
specified as
zero.

>Or, if you'd prefer the less efficient simplicity of:

for(unsigned i = 0; i != 11; ++i)
{

}

why is this less effiecient? Do you have benchmarks?

It is less efficient (conceptually) because it performs a redundant
equality operation. I don't have benchmarks at the moment, but I would
hazard a guess that the following statement:

i != 11;

takes longer to execute than the following statement:

;

--

Frederick Gotham

[sjdevnull]

Frederick Gotham wrote:

posted:

Yours goes from 0 to 10; you've inadvertantly fixed one
problem.

Which is why I mentioned that he was pretending the array had eleven
elements.

Please disregard my earlier message, I missed his initialization to -1.
Sorry for the confusion.

[Nick Keighley]

Frederick Gotham wrote:

Nick Keighley posted:

Frederick Gotham wrote:

c posted:

while(i++ < 10)
{
printf(argv[1]); //Don't do it like this..read my comments please.
printf(my_array[i]);
}
Why resort to that? Looping 0 through 10 can be much simpler:

unsigned i = 0;

do {

}while(++i != 11);

I don't see why your version is simpler than the original. In fact I'd
say while is safer than do just in case the length of the array is
specified as zero.

Or, if you'd prefer the less efficient simplicity of:

for(unsigned i = 0; i != 11; ++i)
{

}

why is this less effiecient? Do you have benchmarks?

It is less efficient (conceptually) because it performs a redundant
equality operation. I don't have benchmarks at the moment, but I would
hazard a guess that the following statement:

i != 11;

takes longer to execute than the following statement:

;

both versions have to perform the test. I wouldn't be surprised if they

generate identical code. Don't micro-optimse.
--
Nick Keighley

[Frederick Gotham]

Nick Keighley posted:

both versions have to perform the test. I wouldn't be surprised if they

generate identical code. Don't micro-optimse.

Don't rely on an optimiser.

If you tell John to drive your car past the neighbouring town to drop it
into the mechanic, don't just sit back and expect him to take the shorter
route up past the old cemetary -- explicitly instruct him to do so.

When you write:

for(size_t i = 0; max != i; ++i) /* Body */

you're *explicitly* telling the compiler that the conditional must be
evaluated before the first iteration. Of course, in accordance with the "as
if" principle, the compiler may elide the first evaluation if it realises
that it's utterly redundant (just like John _might_ take the quicker
route).

If you _don't_ want the conditional to be evaluated for the first
iteration, then simply TELL the compiler:

size_t i = p;

do /* Body */
while(max != ++i);

There's micro-optimisation, and then there's wanton sabotage of the
efficiency of one's code by being lax and saying "Hey compiler, just get
the job done however you like, I'm not too pushed whether it's efficient or
not."

--

Frederick Gotham

[Nick Keighley]

Frederick Gotham wrote:

Nick Keighley posted:

my reply isn't really addressed to Frederick Gotham as he has
demonstrated before that once he has an idea fixed in his head it's
hard to shift. I'm more expressing what I believe to be the balance
of opinion amongst experienced programmers for the benefit of the
lurkers.

both versions have to perform the test. I wouldn't be surprised if they
generate identical code.

to be honest, I'd be astonished if they generated different code.

Don't micro-optimse.

"premature optimistation is the root of all evil"

Don't rely on an optimiser.

***DO*** rely on the optimser. Changing your optimisation level might
take
a few minutes to recompile. Analysing and optimising an application may

take days, or weeks.

If you tell John to drive your car past the neighbouring town to drop it
into the mechanic, don't just sit back and expect him to take the shorter
route up past the old cemetary -- explicitly instruct him to do so.

bad analogy. John is not an optimising compiler.

When you write:

for(size_t i = 0; max != i; ++i) /* Body */

you're *explicitly* telling the compiler that the conditional must be
evaluated before the first iteration.

which, of course is what I want... You snipped the bit where I said I
wouldn't normally replace while-loop with a do-loop. It's not 1957
anymore and we don't program in FORTRAN.

Of course, in accordance with the "as
if" principle, the compiler may elide the first evaluation if it realises
that it's utterly redundant (just like John _might_ take the quicker
route).

If you _don't_ want the conditional to be evaluated for the first
iteration, then simply TELL the compiler:

size_t i = p;

do /* Body */
while(max != ++i);

There's micro-optimisation, and then there's wanton sabotage of the
efficiency of one's code by being lax and saying "Hey compiler, just get
the job done however you like, I'm not too pushed whether it's efficient or
not."

yes, and this is *not* an example. If I use bubble sort instead of
quicksort I can hardly expect the compiler to fix my error.

Choose good algorithms, write clear simple code. If it is too slow
then use a profiler and find out where it is slow. Only then start
applying optimsation tricks.

Your pessimisation saves one test at loop setup and is slightly more
insecure.
--
Nick Keighley

"There are two ways of constructing a software design: One way is to
make it so simple that there are obviously no deficiencies, and the
other way is to make it so complicated that there are no obvious
deficiencies. The first method is far more difficult."
-- C.A.R. Hoare

[Thomas J. Gritzan]

Frederick Gotham schrieb:

Nick Keighley posted:

>both versions have to perform the test. I wouldn't be surprised if they
generate identical code. Don't micro-optimse.

Don't rely on an optimiser.

Do you know this - not directly C or C++ related - article?
http://www.flounder.com/optimization.htm

When you write:

for(size_t i = 0; max != i; ++i) /* Body */

you're *explicitly* telling the compiler that the conditional must be
evaluated before the first iteration. Of course, in accordance with the "as
if" principle, the compiler may elide the first evaluation if it realises
that it's utterly redundant (just like John _might_ take the quicker
route).

If you _don't_ want the conditional to be evaluated for the first
iteration, then simply TELL the compiler:

size_t i = p;

do /* Body */
while(max != ++i);

If you _know_, that max never ever may fall to zero, you can do it.
However, a simple compare of two variables that are already stored in
registers won't do any damage. A buffer overflow caused by a superfluous
and redundant check, sacrificed for a few microseconds, will do damage.

--
Thomas
http://www.netmeister.org/news/learn2quote.html

[Frederick Gotham]

Thomas J. Gritzan posted:

However, a simple compare of two variables that are already stored in
registers won't do any damage.

Of course it will, several million femtoseconds in fact.

A buffer overflow caused by a superfluous and redundant check,
sacrificed for a few microseconds, will do damage.

Indeed, which is why on would be wise to destory the possibility.

--

Frederick Gotham

[Chris Torek]

In article <7b*******************@news.indigo.ie>
Frederick Gotham <fg*******@SPAM.comwrote:

>If you tell John to drive your car past the neighbouring town to drop it
into the mechanic, don't just sit back and expect him to take the shorter
route up past the old cemetary -- explicitly instruct him to do so.

Well, there are shortcuts, and then there is "make sure you tell
John not to go from your house to your neighbor's on the west by
going 24,000 miles east"... :-)

>When you write:

for(size_t i = 0; max != i; ++i) /* Body */

you're *explicitly* telling the compiler that the conditional must be
evaluated before the first iteration. Of course, in accordance with the "as
if" principle, the compiler may elide the first evaluation if it realises
that it's utterly redundant (just like John _might_ take the quicker
route).

There are really only three "interesting" cases for this code:

a) max is a constant.
b) max is a variable, and might be zero, and if it is zero the
loop must not run.
c) max is a variable, and guaranteed not to be zero.

In case (a), a compiler that fails to optimize away a "test at
top of loop" is the moral equivalent of the guy who goes 24,000
miles east so as to wind up half a block west.

In case (b), the test at the top is needed. Omitting it makes the
code incorrect.

In general, then, only case (c) is worth worrying about -- and if
that test at the top is truly significant, the loop may well need
to be special-cased and/or unrolled or otherwise handled in a
fancier fashion than this:

>If you _don't_ want the conditional to be evaluated for the first
iteration, then simply TELL the compiler:

size_t i = p;

do /* Body */
while(max != ++i);

[I assume this was meant to read "size_t i = 0;".]

The do-while form is *most* appropriate (not "only", but "most") when
the loop has to run at least once even if whatever condition is to be
tested is true initially.

>There's micro-optimisation, and then there's wanton sabotage of the
efficiency of one's code by being lax and saying "Hey compiler, just get
the job done however you like, I'm not too pushed whether it's efficient or
not."

And then there's "cluttering up the source code with repetitive
instructions not to drive 24,000 miles the wrong way"... :-)
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) [phone]
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.