char pointers?

[mdh]

In trying to understand the issue, I wrote this;

#include <stdio.h>
void f_output(char arg1[6], int limit);
int main () {

f();

return 0;
}

void f(void) {
char matrix[2][6] ={ {'h','e','l','l ','o','\0'},
{'w','o','r','l ','d','\0'}};
f_output(matrix , 10);
}
void f_output(char *s, int limit) {
int x, i;
for (i=0; i < limit; i++)
printf("%s\n", s++);

}

Desired output: "Hello world"
Actual output:

hello
ello
llo
lo
o

world
orld
rld
ld
My naive way of looking at this was to assume that the declaration

"void f_output(char arg1[6], int limit)"

would "tell" the pointer that the type was an array of 6 chars, hence s
++ would increment to matrix[2]. I understand this is a totally
useless and pointless function, except in trying to further my
understanding.. .so please go easy!!! :-)

[Ian Collins]

Eric Sosman wrote:

Ian Collins wrote On 04/11/07 17:37,:

>>mdh wrote:

>>>In trying to understand the issue, I wrote this;

#include <stdio.h>
void f_output(char arg1[6], int limit);
[...]

void f_output(char *s, int limit) {

Head the warning your compiler should have given you here - this doesn't
match the prototype.

As far as I can see, the definition and declaration
match perfectly. Yet both you and Walter Roberson say
they disagree ... In light of 6.3.5.7p7

A declaration of a parameter as "array of /type/"
shall be adjusted to "qualified pointer to /type/"
[...]

.... could one or both of you explain the disagreement?

My interpretation was that this applied where the parameter is type[]
and not where it is type[6]. The former being a generic array, the
latter and array of 6 type. It looks like I was mistaken.

--
Ian Collins.

[mdh]

Thanks to all. I know everyone approaches C differently, but I
certainly often learn more from trying...as long as this does not
exasperate all of you...but then I am sure, some would let me
know!!! :-)

[Richard Heathfield]

mdh said:

Thanks to all. I know everyone approaches C differently, but I
certainly often learn more from trying...as long as this does not
exasperate all of you...but then I am sure, some would let me
know!!! :-)

Um, any article posted here is bound to exasperate some people. Your
article will exasperate those who don't like non-technical back-chat,
and this reply will exasperate those who don't like protracted
non-technical back-chat.

So don't worry /too/ much about that - as long as you stick to the
subject of programming in the C language and behave reasonably
cluefully (as you appear to have done so far), very few people will
have any reasonable cause to be genuinely exasperated with you.

Not that it will stop some of us trying! :-)

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

[Chris Torek]

[given]

>char matrix[2][6] = ...
f_output(matri x, 10);

In article <11************ **********@q75g 2000hsh.googleg roups.com>
Peter Nilsson <ai***@acay.com .auwrote:

>... matrix will decay to a pointer to its first element. Thus it is
equivalent to &matrix[0] which has type char[6].

Actually char (*)[6], which is a very odd-looking type (and the
parentheses are in fact required). To declare a local variable
with such a type, the name goes after the "*", before the close
parenthesis, e.g.:

char (*p)[6] = &matrix[0];

To turn a declaration into a type-name, one simply removes the
variable name from the declaration, hence the odd-looking type.

The declaration itself is also odd-looking, but at least here
there is an "obvious" reason for the parentheses: without them
the brackets would bind more tightly than the "*". That is:

char *q[6];

"means":

char *(q[6]);

which declares q as a variable of type "array 6 of pointer to
char". We want "pointer to array 6 of char", so we need to bind
the "*" first, even when the variable name is omitted.
--
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.

[Peter Nilsson]

Chris Torek <nos...@torek.n etwrote:

char matrix[2][6] = ...
f_output(matrix , 10);

Peter Nilsson <[e-mail]. auwrote:

... matrix will decay to a pointer to its first element. Thus
it is equivalent to &matrix[0] which has type char[6].

Actually char (*)[6],

Yes. Thanks for the correction.

--
Peter

[rasztasd]

On Apr 11, 11:26 pm, "mdh" <m...@comcast.n etwrote:

In trying to understand the issue, I wrote this;

#include <stdio.h>
void f_output(char arg1[6], int limit);
int main () {

f();

return 0;

}

void f(void) {
char matrix[2][6] ={ {'h','e','l','l ','o','\0'},
{'w','o','r','l ','d','\0'}};
f_output(matrix , 10);

}

void f_output(char *s, int limit) {
int x, i;
for (i=0; i < limit; i++)
printf("%s\n", s++);

}

Desired output: "Hello world"
Actual output:

hello
ello
llo
lo
o

world
orld
rld
ld

My naive way of looking at this was to assume that the declaration

"void f_output(char arg1[6], int limit)"

would "tell" the pointer that the type was an array of 6 chars, hence s
++ would increment to matrix[2]. I understand this is a totally
useless and pointless function, except in trying to further my
understanding.. .so please go easy!!! :-)

Hi.
I will explain u.
First, ur code won't compile, u have to modify a little like this:
#include <stdio.h>
void f_output(char arg1[6], int limit);
void f(void) {
char matrix[2][6] ={ {'h','e','l','l ','o','\0'},
{'w','o','r','l ','d','\0'}};
f_output(matrix , 10);

}
int main () {

f();
getch();
return 0;

}

void f_output(char *s, int limit) {
int x, i; //int x is't used
for (i=0; i < limit; i++)
printf("%s\n", s++);

}
The only difference is i put the void f(void) function's declaration
before it's called.
In C u can overload a function. It means u declare a function with the
same name but with not the same paramteres.
Anyway, u call the void f_output(char *s, int limit) function, this
void f_output(char arg1[6], int limit) does do nothing.
Ur decrlared string Hello World looks like this in the memory: 'h' 'e'
'l' 'l' 'o' '\0' 'w' 'o' 'r' 'l' 'd' '\0' and these characters are one
after the other.
When u call the function void f_output(char *s, int limit) s will b
the memory address of h. U call the function printf with s. So Hello
will b printed. Then u raise the memory address. s contains the
addresss of e. So when u call printf with the second time it will
print ello and so on. there is an empty line between Hello and World
and at the end, this is because u call the printf when s contains the
address of a null character, so it dosn't print anything.

[Default User]

ra******@gmail. com wrote:

I will explain u.

Please don't use shorthand code like "u". Your post, frankly, is hard
enough to read without that sort of thing.

First, ur code won't compile, u have to modify a little like this:
#include <stdio.h>
void f_output(char arg1[6], int limit);
void f(void) {
char matrix[2][6] ={ {'h','e','l','l ','o','\0'},
{'w','o','r','l ','d','\0'}};
f_output(matrix , 10);

}
int main () {

f();
getch();

There's no such C function as getch(), and you didn't define one.

return 0;

}

void f_output(char *s, int limit) {
int x, i; //int x is't used
for (i=0; i < limit; i++)
printf("%s\n", s++);

}
The only difference is i put the void f(void) function's declaration
before it's called.

That wasn't the main problem, so your fix didn't help much.

In C u can overload a function. It means u declare a function with the
same name but with not the same paramteres.

That's not true. You are perhaps thinking of C++.

Anyway, u call the void f_output(char *s, int limit) function, this
void f_output(char arg1[6], int limit) does do nothing.

This make no sense, and I suspect you think that the two declarations
are different. They aren't.

The program has pretty been analyzed by the people here, and correct
given.


Brian

[Flash Gordon]

ra******@gmail. com wrote, On 13/04/07 19:57:

<snip>

In C u can overload a function. It means u declare a function with the
same name but with not the same paramteres.

No you can't. C++ may well have such a facility, but that is a different
language.

Also, please do not use contractions like "u" for "you" and "ur" for
"your", it makes it far harder to read. In fact, it made it so hard for
me I gave up, it was just luck I spotted the word "overload" and so
realised you were saying things that are wrong.

<snip>
--
Flash Gordon