I have looked this program up and down and I don't see what's wrong with
it. But it always breaks and gives me an error "mode error" no matter which
mode binary or text I choose. This simple program is supposed to take as
argv[1] a "b" or "t" for binary or text. It's not taking anything.
#include <stdio.h>
int main(int argc, char **argv) {
char *b;
int a;
FILE *ifp,*ofp;
if (argc!=4) {
fprintf(stderr,"usage error\n");
return -1;
}
if (argv[1]=="b") {
b="rb";
}
if (argv[1]=="t") {
b="rt";
}
if (argv[1]!="t"||argv[1]!="b") {
fprintf(stderr,"mode error\n");
return -1;
}
if ((ifp=fopen(argv[2],b))==0) {
fprintf(stderr,"open error i\n");
return -1;
}
if ((ofp=fopen(argv[3],b))==0) {
fprintf(stderr,"open error o\n");
return -1;
}
while(a!=EOF)
a=fgetc(ifp);
fputc(a,ofp);
printf("done\n");
return 0;}
Is anyone good enough to glance at this and see what's wrong?
Bill
35  1179 
Well I now see I left out fclose I will correct that but I don't think
that's my problem.
Bill
On May 2, 5:40 pm, "Bill Cunningham" <nos...@nspam.comwrote:
I have looked this program up and down and I don't see what's wrong with
it. But it always breaks and gives me an error "mode error" no matter which
mode binary or text I choose. This simple program is supposed to take as
argv[1] a "b" or "t" for binary or text. It's not taking anything.
#include <stdio.h>
int main(int argc, char **argv) {
char *b;
int a;
FILE *ifp,*ofp;
if (argc!=4) {
fprintf(stderr,"usage error\n");
return -1;
Not a portable return value from main.
}
if (argv[1]=="b") {
This isn't doing what you think it is. This is comparing the address
of first program argument to the address of the string literal "b"
which will always be different. Either use (strcmp(argv[1], "b") ==
0) or say:
if (argv[1][0] == 'b') {
which will compare the first letter of the first argument to the
character constant b.
b="rb";
}
if (argv[1]=="t") {
b="rt";
}
if (argv[1]!="t"||argv[1]!="b") {
fprintf(stderr,"mode error\n");
return -1;
}
if ((ifp=fopen(argv[2],b))==0) {
fprintf(stderr,"open error i\n");
return -1;
}
if ((ofp=fopen(argv[3],b))==0) {
fprintf(stderr,"open error o\n");
return -1;
}
while(a!=EOF)
a is not initialized, referencing its value invokes undefined
behavior.
a=fgetc(ifp);
fputc(a,ofp);
I think you want braces around these statements, yes?
Try (not tested):
while ( (a = fgetc(ifp)) != EOF && fputc(a, ofp) != EOF )
;
printf("done\n");
puts("done"); would work equally well.
return 0;}
Is anyone good enough to glance at this and see what's wrong?
Bill
--
Robert Gamble
On May 2, 5:53 pm, "Bill Cunningham" <nos...@nspam.comwrote:
Well I now see I left out fclose I will correct that but I don't think
that's my problem.
Bill
There were a number of problems with the program but as long as the
program was terminating normally the fclose() wasn't one of them, all
files are properly closed during normal program termination (although
it is usually a good idea to explicitly close files yourself).
--
Robert Gamble
"Robert Gamble" <rg*******@gmail.comwrote in message
news:2d**********************************@8g2000hs e.googlegroups.com...
>a is not initialized, referencing its value invokes undefined
behavior.
> a=fgetc(ifp);
fputc(a,ofp);
In the beginning a was declared. Is it not enough in this case to simply
declare an int? Or should I have done this int a=0; at the beginning of the
program?
Bill
"Robert Gamble" <rg*******@gmail.comwrote in message
news:2d**********************************@8g2000hs e.googlegroups.com...
Not a portable return value from main.
> }
if (argv[1]=="b") {
This isn't doing what you think it is. This is comparing the address
of first program argument to the address of the string literal "b"
Was declaring b as a pointer to char proper? Or should a simple b of
type char all that's needed?
which will always be different. Either use (strcmp(argv[1], "b") ==
0) or say:
if (argv[1][0] == 'b') {
which will compare the first letter of the first argument to the
character constant b.
I do not understand the above but it looks like a good way to write
code. Why is a zero used as the second dimension of this array?
Bill
Bill Cunningham wrote:
I have looked this program up and down and I don't see what's wrong with
it. But it always breaks and gives me an error "mode error" no matter which
mode binary or text I choose. This simple program is supposed to take as
argv[1] a "b" or "t" for binary or text. It's not taking anything.
#include <stdio.h>
int main(int argc, char **argv) {
char *b;
int a;
FILE *ifp,*ofp;
if (argc!=4) {
fprintf(stderr,"usage error\n");
return -1;
}
if (argv[1]=="b") {
b="rb";
}
if (argv[1]=="t") {
b="rt";
}
if (argv[1]!="t"||argv[1]!="b") {
fprintf(stderr,"mode error\n");
return -1;
}
if ((ifp=fopen(argv[2],b))==0) {
fprintf(stderr,"open error i\n");
return -1;
}
if ((ofp=fopen(argv[3],b))==0) {
fprintf(stderr,"open error o\n");
return -1;
}
while(a!=EOF)
a=fgetc(ifp);
fputc(a,ofp);
printf("done\n");
return 0;}
Is anyone good enough to glance at this and see what's wrong?
Bill
You can't use == to compare strings. That's what strcmp() is for. This
is the first error I find, not the only one.
--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---
On May 2, 6:16 pm, "Bill Cunningham" <nos...@nspam.comwrote:
"Robert Gamble" <rgambl...@gmail.comwrote in message
news:2d**********************************@8g2000hs e.googlegroups.com...
a is not initialized, referencing its value invokes undefined
behavior.
a=fgetc(ifp);
fputc(a,ofp);
In the beginning a was declared. Is it not enough in this case to simply
declare an int? Or should I have done this int a=0; at the beginning of the
program?
You need to store a value into a variable before you use the value of
that variable, either through initialization (implicit or explicit) or
assignment. If you don't do this then the variable may contain
garbage ("indeterminate value" in standards parlance), that garbage
may be a trap value which will invoke UB when read. Note that the
alternative I provided does not have this problem because "a" is
assigned before its value is used.
--
Robert Gamble
Bill Cunningham wrote:
I have looked this program up and down and I don't see what's wrong with
it. But it always breaks and gives me an error "mode error" no matter which
mode binary or text I choose. This simple program is supposed to take as
argv[1] a "b" or "t" for binary or text. It's not taking anything.
#include <stdio.h>
int main(int argc, char **argv) {
char *b;
int a;
FILE *ifp,*ofp;
if (argc!=4) {
fprintf(stderr,"usage error\n");
return -1;
}
if (argv[1]=="b") {
b="rb";
}
if (argv[1]=="t") {
b="rt";
}
if (argv[1]!="t"||argv[1]!="b") {
fprintf(stderr,"mode error\n");
return -1;
}
if ((ifp=fopen(argv[2],b))==0) {
fprintf(stderr,"open error i\n");
return -1;
}
if ((ofp=fopen(argv[3],b))==0) {
fprintf(stderr,"open error o\n");
return -1;
}
while(a!=EOF)
a=fgetc(ifp);
fputc(a,ofp);
printf("done\n");
return 0;}
Is anyone good enough to glance at this and see what's wrong?
Someone stole the whitespace?
Why are you comparing characters with string literals? Surly your
compiler gave you some warnings?
--
Ian Collins.
"Ian Collins" <ia******@hotmail.comwrote in message
news:68**************@mid.individual.net...
Why are you comparing characters with string literals? Surly your
compiler gave you some warnings?
Not as whisper. As it is typed as is.
Bill
On May 2, 6:28 pm, Ian Collins <ian-n...@hotmail.comwrote:
Bill Cunningham wrote:
I have looked this program up and down and I don't see what's wrong with
it. But it always breaks and gives me an error "mode error" no matter which
mode binary or text I choose. This simple program is supposed to take as
argv[1] a "b" or "t" for binary or text. It's not taking anything.
#include <stdio.h>
int main(int argc, char **argv) {
char *b;
int a;
FILE *ifp,*ofp;
if (argc!=4) {
fprintf(stderr,"usage error\n");
return -1;
}
if (argv[1]=="b") {
b="rb";
}
if (argv[1]=="t") {
b="rt";
}
if (argv[1]!="t"||argv[1]!="b") {
fprintf(stderr,"mode error\n");
return -1;
}
if ((ifp=fopen(argv[2],b))==0) {
fprintf(stderr,"open error i\n");
return -1;
}
if ((ofp=fopen(argv[3],b))==0) {
fprintf(stderr,"open error o\n");
return -1;
}
while(a!=EOF)
a=fgetc(ifp);
fputc(a,ofp);
printf("done\n");
return 0;}
Is anyone good enough to glance at this and see what's wrong?
Someone stole the whitespace?
Why are you comparing characters with string literals? Surly your
compiler gave you some warnings?
My compiler didn't provide any warnings for this code, did yours?
--
Robert Gamble
"Ian Collins" <ia******@hotmail.comwrote in message
news:68**************@mid.individual.net...
Someone stole the whitespace?
I was always told there were several choices in C indentation. These two
and maybe a third.
function { /*notice the one space after the first brace*/
} closing brace on a line by itself.
This is the style I use and another
function
{
body
}
With this style all braces are on lines by themselves.
Bill
Ian Collins wrote:
>
.... snip ...
>
Someone stole the whitespace?
Why are you comparing characters with string literals? Surly
your compiler gave you some warnings?
Surly compilers have that fault. :-)
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
** Posted from http://www.teranews.com **
Bill Cunningham wrote:
"Robert Gamble" <rg*******@gmail.comwrote in message
>a is not initialized, referencing its value invokes undefined
behavior.
a=fgetc(ifp);
fputc(a,ofp);
In the beginning a was declared. Is it not enough in this case to
simply declare an int? Or should I have done this int a=0; at the
beginning of the program?
You are using it before loading it. You want:
while (EOF != (a = fgetc(ifp))) fputc(a, ofp);
Note the use of blanks. As I recall there are other errors also.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
** Posted from http://www.teranews.com **
Robert Gamble wrote:
On May 2, 6:28 pm, Ian Collins <ian-n...@hotmail.comwrote:
>>
Why are you comparing characters with string literals? Surly your
compiler gave you some warnings?
My compiler didn't provide any warnings for this code, did yours?
Oops, my bad, I spotted the comparisons an didn't spot they were with
argv[n].
--
Ian Collins.
Ian Collins wrote:
Robert Gamble wrote:
>On May 2, 6:28 pm, Ian Collins <ian-n...@hotmail.comwrote:
>>Why are you comparing characters with string literals? Surly your
compiler gave you some warnings?
My compiler didn't provide any warnings for this code, did yours?
Oops, my bad, I spotted the comparisons an didn't spot they were with
argv[n].
So what?
if (argv[1] == "b")
is not an error the compiler need diagnose. Both arguments are of type
(char *) and the chances the two pointer values are equal is nil but the
expression is completely valid.
--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---
Joe Wright wrote:
Ian Collins wrote:
>Robert Gamble wrote:
>>On May 2, 6:28 pm, Ian Collins <ian-n...@hotmail.comwrote:
Why are you comparing characters with string literals? Surly your
compiler gave you some warnings?
My compiler didn't provide any warnings for this code, did yours?
Oops, my bad, I spotted the comparisons an didn't spot they were with
argv[n].
So what?
if (argv[1] == "b")
is not an error the compiler need diagnose. Both arguments are of type
(char *) and the chances the two pointer values are equal is nil but the
expression is completely valid.
I know, that's why I said I didn't spot the comparison was with argv[n].
If I had, I wouldn't have mentioned compiler warnings.
--
Ian Collins.
On May 2, 2:40*pm, "Bill Cunningham" <nos...@nspam.comwrote:
* * I have looked this program up and down and I don't see what's wrong with
it. But it always breaks and gives me an error "mode error" no matter which
mode binary or text I choose. This simple program is supposed to take as
argv[1] a "b" or "t" for binary or text. It's not taking anything.
#include <stdio.h>
int main(int argc, char **argv) {
*char *b;
*int a;
*FILE *ifp,*ofp;
*if (argc!=4) {
*fprintf(stderr,"usage error\n");
*return -1;
*}
*if (argv[1]=="b") {
*b="rb";
*}
*if (argv[1]=="t") {
*b="rt";
*}
*if (argv[1]!="t"||argv[1]!="b") {
*fprintf(stderr,"mode error\n");
*return -1;
*}
*if ((ifp=fopen(argv[2],b))==0) {
*fprintf(stderr,"open error i\n");
*return -1;
*}
*if ((ofp=fopen(argv[3],b))==0) {
*fprintf(stderr,"open error o\n");
*return -1;
*}
*while(a!=EOF)
*a=fgetc(ifp);
*fputc(a,ofp);
*printf("done\n");
*return 0;}
* * Is anyone good enough to glance at this and see what's wrong?
Too many things for a glance. It requires a perusal.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char **argv)
{
char *imode = 0;
char *omode = 0;
char option;
int ch = 0;
FILE *ifp,
*ofp;
if (argc != 4) {
fprintf(stderr, "usage error\n");
return EXIT_FAILURE;
}
option = *argv[1];
if (option == 'b') {
imode = "rb";
omode = "wb";
}
if (option == 't') {
/* This may or may not do something useful on your system */
/* The 't' in the mode is not portable, but is a fairly */
/* popular extension for text mode. */
imode = "rt";
omode = "wt";
}
if (option != 't' && option != 'b') {
fprintf(stderr, "mode error\n");
return EXIT_FAILURE;
}
if ((ifp = fopen(argv[2], imode)) == 0) {
fprintf(stderr, "open error i\n");
return EXIT_FAILURE;
}
if ((ofp = fopen(argv[3], omode)) == 0) {
fprintf(stderr, "open error o\n");
return EXIT_FAILURE;
}
/* I guess that you want to write what you read, hence the {} */
do {
ch = fgetc(ifp);
if (ch != EOF)
fputc(ch, ofp);
} while (ch != EOF);
fclose(ifp);
fclose(ofp);
printf("done\n");
return 0;
}
"Bill Cunningham" <no****@nspam.comwrites:
"Ian Collins" <ia******@hotmail.comwrote in message
news:68**************@mid.individual.net...
>Someone stole the whitespace?
I was always told there were several choices in C indentation.
[...]
Yes, there are serveral choices. No indentation at all is not one of
them.
If you want help with your code, indent it. It doesn't matter much
which indentation style you choose, but pick *something* that reflects
the structure of your code. (To be clear, indentation refers to
whitespace at the beginning of each line.)
--
Keith Thompson (The_Other_Keith) <ks***@mib.org>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
Keith Thompson said:
"Bill Cunningham" <no****@nspam.comwrites:
>"Ian Collins" <ia******@hotmail.comwrote in message
news:68**************@mid.individual.net...
>>Someone stole the whitespace?
I was always told there were several choices in C indentation.
[...]
Yes, there are serveral choices. No indentation at all is not one of
them.
Wrong. No indentation at all *is* a valid choice. The OP is under no
obligation whatsoever to write his code in a way that encourages people to
help him.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
"user923005" <dc*****@connx.comwrote in message
news:6f**********************************@i76g2000 hsf.googlegroups.com...
#include <stdio.h>
int main(int argc, char **argv) {
char *b;
int a;
FILE *ifp,*ofp;
if (argc!=4) {
fprintf(stderr,"usage error\n");
return -1;
}
if (argv[1]=="b") {
b="rb";
}
if (argv[1]=="t") {
b="rt";
}
if (argv[1]!="t"||argv[1]!="b") {
fprintf(stderr,"mode error\n");
return -1;
}
if ((ifp=fopen(argv[2],b))==0) {
fprintf(stderr,"open error i\n");
return -1;
}
if ((ofp=fopen(argv[3],b))==0) {
fprintf(stderr,"open error o\n");
return -1;
}
while(a!=EOF)
a=fgetc(ifp);
fputc(a,ofp);
printf("done\n");
return 0;}
Is anyone good enough to glance at this and see what's wrong?
Too many things for a glance. It requires a perusal.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char **argv)
{
char *imode = 0;
char *omode = 0;
char option;
int ch = 0;
FILE *ifp,
*ofp;
if (argc != 4) {
fprintf(stderr, "usage error\n");
return EXIT_FAILURE;
}
option = *argv[1];
if (option == 'b') {
imode = "rb";
omode = "wb";
}
if (option == 't') {
/* This may or may not do something useful on your system */
/* The 't' in the mode is not portable, but is a fairly */
/* popular extension for text mode. */
imode = "rt";
omode = "wt";
}
if (option != 't' && option != 'b') {
fprintf(stderr, "mode error\n");
return EXIT_FAILURE;
}
if ((ifp = fopen(argv[2], imode)) == 0) {
fprintf(stderr, "open error i\n");
return EXIT_FAILURE;
}
if ((ofp = fopen(argv[3], omode)) == 0) {
fprintf(stderr, "open error o\n");
return EXIT_FAILURE;
}
/* I guess that you want to write what you read, hence the {} */
do {
ch = fgetc(ifp);
if (ch != EOF)
fputc(ch, ofp);
} while (ch != EOF);
fclose(ifp);
fclose(ofp);
printf("done\n");
return 0;
}
I will definately save this code and study it. I really do no use do so
maybe this will open my eyes to it some.
Bill
On May 3, 6:52 pm, "Bill Cunningham" <nos...@nspam.comwrote:
"user923005" <dcor...@connx.comwrote in message
news:6f**********************************@i76g2000 hsf.googlegroups.com...
#include <stdio.h>
int main(int argc, char **argv) {
char *b;
int a;
FILE *ifp,*ofp;
if (argc!=4) {
fprintf(stderr,"usage error\n");
return -1;
}
if (argv[1]=="b") {
b="rb";
}
if (argv[1]=="t") {
b="rt";
}
if (argv[1]!="t"||argv[1]!="b") {
fprintf(stderr,"mode error\n");
return -1;
}
if ((ifp=fopen(argv[2],b))==0) {
fprintf(stderr,"open error i\n");
return -1;
}
if ((ofp=fopen(argv[3],b))==0) {
fprintf(stderr,"open error o\n");
return -1;
}
while(a!=EOF)
a=fgetc(ifp);
fputc(a,ofp);
printf("done\n");
return 0;}
Is anyone good enough to glance at this and see what's wrong?
Too many things for a glance. It requires a perusal.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char **argv)
{
char *imode = 0;
char *omode = 0;
char option;
int ch = 0;
FILE *ifp,
*ofp;
if (argc != 4) {
fprintf(stderr, "usage error\n");
return EXIT_FAILURE;
}
option = *argv[1];
if (option == 'b') {
imode = "rb";
omode = "wb";
}
if (option == 't') {
/* This may or may not do something useful on your system */
/* The 't' in the mode is not portable, but is a fairly */
/* popular extension for text mode. */
imode = "rt";
omode = "wt";
Hahah, what else will the troll come up with? You're entertaining, but
it's sad that posters take you seriously.
"Robert Gamble" <rg*******@gmail.comwrote in message
news:2d**********************************@8g2000hs e.googlegroups.com...
[snip]
I think you want braces around these statements, yes?
I don't know. I've ran code before and seen a lot of people use while
and statements after it and the code works. No braces. So I usually use it
like that. It's not a typo I don't usually use braces with while.
Try (not tested):
while ( (a = fgetc(ifp)) != EOF && fputc(a, ofp) != EOF )
;
> printf("done\n");
puts("done"); would work equally well.
> return 0;}
Is anyone good enough to glance at this and see what's wrong?
Bill
--
Robert Gamble
"Bill Cunningham" <no****@nspam.comwrites:
"Robert Gamble" <rg*******@gmail.comwrote in message
news:2d**********************************@8g2000hs e.googlegroups.com...
[snip]
>I think you want braces around these statements, yes?
I don't know. I've ran code before and seen a lot of people use while
and statements after it and the code works. No braces. So I usually use it
like that. It's not a typo I don't usually use braces with while.
[...]
Do you understand what the braces mean?
Here's the tail end of the code Robert was commenting on:
[...]
while(a!=EOF)
a=fgetc(ifp);
fputc(a,ofp);
printf("done\n");
return 0;}
Putting the closing brace immediately after the semicolon like that is
horribly bad style. Putting everything at the same indentation level
is also horribly bad style. The compiler doesn't care, but anyone
trying to read your code does.
The syntax of a while statement is:
while ( expression ) statement
Note that that's a *single* statement. If you want the body to be
multiple statements, you need to wrap them in braces to create a
compound statement, which is itself as single statement. (The same
thing applies to if statements; you correctly used braces for those,
so I don't know why the while loop was such a problem.)
The code I quoted is equivalent to this:
[...]
while (a != EOF)
a = fgetc(ifp);
fputc(a, ofp);
printf("done\n");
return 0;
}
With proper indentation, you can see that the while controls only a
single statement, ``a = fgetc(ifp);''. The fputc call will occur
exactly once, after the while loop terminates. I can't be certain,
but I *think* you wanted both the fgetc and fputc calls to be in the
body of the loop. To do that, you need braces:
[...]
while (a != EOF) {
a = fgetc(ifp);
fputc(a, ofp);
}
printf("done\n");
return 0;
}
In my opinion, it's a good habit always to use braces even if you only
want only a single statement in the body. It makes the code more
consistent and easier to maintain if you later want to add a second
statement. (Others will disagree.)
Judicious use of whitespace, especially around operators and after
commas, also makes the code much easier to read.
From now on, if you post a chunk of nearly illegible code like what
you posted upthread, with no indentation and virtually no whitespace,
I will ask you to format it properly before I'll even consider helping
you fix any other problems.
--
Keith Thompson (The_Other_Keith) <ks***@mib.org>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
"Keith Thompson" <ks***@mib.orgwrote in message
news:87************@kvetch.smov.org...
The syntax of a while statement is:
while ( expression ) statement
Note that that's a *single* statement. If you want the body to be
multiple statements, you need to wrap them in braces to create a
compound statement,
I see.
which is itself as single statement. (The same
thing applies to if statements; you correctly used braces for those,
I think braces are required with if.
so I don't know why the while loop was such a problem.)
The code I quoted is equivalent to this:
[...]
while (a != EOF)
a = fgetc(ifp);
fputc(a, ofp);
printf("done\n");
return 0;
}
With proper indentation, you can see that the while controls only a
single statement, ``a = fgetc(ifp);''. The fputc call will occur
exactly once, after the while loop terminates. I can't be certain,
but I *think* you wanted both the fgetc and fputc calls to be in the
body of the loop. To do that, you need braces:
[...]
while (a != EOF) {
a = fgetc(ifp);
fputc(a, ofp);
}
printf("done\n");
return 0;
}
In my opinion, it's a good habit always to use braces even if you only
want only a single statement in the body. It makes the code more
consistent and easier to maintain if you later want to add a second
statement. (Others will disagree.)
Judicious use of whitespace, especially around operators and after
commas, also makes the code much easier to read.
From now on, if you post a chunk of nearly illegible code like what
you posted upthread, with no indentation and virtually no whitespace,
I will ask you to format it properly before I'll even consider helping
you fix any other problems.
OK
Bill
On May 4, 1:58 am, "Bill Cunningham" <nos...@nspam.comwrote:
"Keith Thompson" <ks...@mib.orgwrote in message
news:87************@kvetch.smov.org...
The syntax of a while statement is:
while ( expression ) statement
Note that that's a *single* statement. If you want the body to be
multiple statements, you need to wrap them in braces to create a
compound statement,
I see.
which is itself as single statement. (The same
thing applies to if statements; you correctly used braces for those,
I think braces are required with if.
No they are not. You know it, you could try it, you could read about
it.
*OR* you could just read what mr Keith said; *the* *same* *thing*
*applies* *to* *if* *statements*.
Just once more, just in case:
*the* *same* *thing* *applies* *to* *if* *statements*.
Now, every day you wake up, you should run this commant in your
terminal: 'yes the same thing applies to if statements', until you
memorize it. Then you can chant it quietly, until you understand it.
Mr Keith (and anyone else who actually cared to reply to this troll)
I'm really tired of seeing clc replying to him.
Assuming he is actually not able to learn after so many years, and
assuming he can't understand what he reads in usenet messages (as
shown just now), you are simply not helping him. Just how stupid does
one have to be to say "I think B isn't" when you say "A is, same
applies for B?"
And what's up with all the horrible indentation he comes up with every
single time, the horrible questions "is anyone good enough to [...]"
in his original message, which could simply be answered with
"yes"/"no", the repeating mistakes, the tendency to reply without
quoting, taking quotes out of context (notice his reply to Mr Gamble)
Honestly, I'm tired of these threads. I also think I make a fool of
myself, since such replies most likely entertain him and the other
trolls...
"Bill Cunningham" <no****@nspam.comwrites:
"Keith Thompson" <ks***@mib.orgwrote in message
news:87************@kvetch.smov.org...
>The syntax of a while statement is:
while ( expression ) statement
Note that that's a *single* statement. If you want the body to be
multiple statements, you need to wrap them in braces to create a
compound statement,
I see.
which is itself as single statement. (The same
>thing applies to if statements; you correctly used braces for those,
I think braces are required with if.
[...]
You're guessing. Don't guess. Look it up.
--
Keith Thompson (The_Other_Keith) <ks***@mib.org>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister" vi******@gmail.com wrote:
>
.... snip ...
>
Mr Keith (and anyone else who actually cared to reply to this
troll) I'm really tired of seeing clc replying to him. Assuming
he is actually not able to learn after so many years, and assuming
he can't understand what he reads in usenet messages (as shown
just now), you are simply not helping him. Just how stupid does
one have to be to say "I think B isn't" when you say "A is, same
applies for B?"
....
Suit yourself. If you don't like it, it takes a few seconds to
plonk Cunningham, and possibly to threadplonk any threads he
starts. Then you won't be bothered. Some of us admire his
stick-to-it-ness. We don't really expect him to learn, but he has
at times. The problem is that he then forgets it all.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
** Posted from http://www.teranews.com **
Keith Thompson wrote:
From now on, if you post a chunk of nearly illegible code like what
you posted upthread, with no indentation and virtually no whitespace,
I will ask you to format it properly before I'll even consider helping
you fix any other problems.
It will feel good once you stop beating your head against the wall.
Brian
"Default User" <de***********@yahoo.comwrites:
Keith Thompson wrote:
>From now on, if you post a chunk of nearly illegible code like what
you posted upthread, with no indentation and virtually no whitespace,
I will ask you to format it properly before I'll even consider helping
you fix any other problems.
It will feel good once you stop beating your head against the wall.
Thanks for the advice.
Again.
--
Keith Thompson (The_Other_Keith) <ks***@mib.org>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
"Bill Cunningham" <no****@nspam.comwrites:
"Keith Thompson" <ks***@mib.orgwrote in message
news:87************@kvetch.smov.org...
>The syntax of a while statement is:
while ( expression ) statement
Note that that's a *single* statement. If you want the body to be
multiple statements, you need to wrap them in braces to create a
compound statement,
I see.
which is itself as single statement. (The same
>thing applies to if statements; you correctly used braces for those,
I think braces are required with if.
Did you ever try to read a C book? It seems you do not have much of an
idea of anything or what braces are for.
Keith Thompson wrote:
"Default User" <de***********@yahoo.comwrites:
It will feel good once you stop beating your head against the wall.
Thanks for the advice.
Happy to help!
Brian
"Eligiusz Narutowicz" <el*************@hotmail.comwrote in message
news:fv**********@registered.motzarella.org...
Did you ever try to read a C book? It seems you do not have much of an
idea of anything or what braces are for.
k and r 2 can sometimes go over my head. So I don't use it much but I do
have a little handy dandy little book called "C pocket reference". It's what
I do most of my reading from. It's a small reference and not a big tutorial
but it's pretty easy for me and on my level much more than some texts.
Bill
On Sun, 04 May 2008 21:25:27 GMT, "Bill Cunningham" <no****@nspam.com>
wrote:
>
"Eligiusz Narutowicz" <el*************@hotmail.comwrote in message
news:fv**********@registered.motzarella.org...
>Did you ever try to read a C book? It seems you do not have much of an
idea of anything or what braces are for.
k and r 2 can sometimes go over my head. So I don't use it much but I do
have a little handy dandy little book called "C pocket reference". It's what
I do most of my reading from. It's a small reference and not a big tutorial
but it's pretty easy for me and on my level much more than some texts.
Is this something you found on the web or did you buy it. Can you
provide a reference so we can check to see if it is the source of
misinformation you seem to be suffering from?
Remove del for email
"Barry Schwarz" <sc******@dqel.comwrote in message
news:hc********************************@4ax.com...
Is this something you found on the web or did you buy it. Can you
provide a reference so we can check to see if it is the source of
misinformation you seem to be suffering from?
It's a very small paper back book. Like a reference. From Oreilly books.
Bill
On 2 May, 22:40, "Bill Cunningham" <nos...@nspam.comwrote:
* * I have looked this program up and down and I don't see what's wrong with
it. But it always breaks and gives me an error "mode error" no matter which
mode binary or text I choose. This simple program is supposed to take as
argv[1] a "b" or "t" for binary or text. It's not taking anything.
#include <stdio.h>
int main(int argc, char **argv) {
*char *b;
*int a;
*FILE *ifp,*ofp;
*if (argc!=4) {
*fprintf(stderr,"usage error\n");
*return -1;
*}
*if (argv[1]=="b") {
*b="rb";
*}
*if (argv[1]=="t") {
*b="rt";
*}
*if (argv[1]!="t"||argv[1]!="b") {
*fprintf(stderr,"mode error\n");
*return -1;
*}
*if ((ifp=fopen(argv[2],b))==0) {
*fprintf(stderr,"open error i\n");
*return -1;
*}
*if ((ofp=fopen(argv[3],b))==0) {
*fprintf(stderr,"open error o\n");
*return -1;
*}
*while(a!=EOF)
*a=fgetc(ifp);
*fputc(a,ofp);
*printf("done\n");
*return 0;}
* * Is anyone good enough to glance at this and see what's wrong?
your space key is broken.
You can't be bothered to indent your code.
I can't be bothered to read your unindented code.
--
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