question about gcc option..

[herrcho]

here is the code

#include <stdio.h>
int main()
{
char *str1="Borland International";
printf("%d\n", strlen(str1));
return 0;
}

i learned when i use strlen i have to include <string.h>
but the above runs o.k without <string.h>
i compiled it with gcc -Wall -ansi -pedantic -o

Could anyone explain about this ?

[Emmanuel Delahaye]

In 'comp.lang.c', "herrcho" <he*********@kornet.net> wrote:

here is the code

#include <stdio.h>
Missing
#include <string.h>
int main()
{
char *str1="Borland International";
printf("%d\n", strlen(str1));
printf ("%d\n", (int) strlen(str1));

or better

printf ("%lu\n", (unsigned long) strlen(str1));
return 0;
}

i learned when i use strlen i have to include <string.h>
True enough.
but the above runs o.k without <string.h>
i compiled it with gcc -Wall -ansi -pedantic -o
What was the warning messages?
Could anyone explain about this ?

It runs by chance only. It's a UB, and UB are tricky. Don't do that.

--
-ed- em**********@noos.fr [remove YOURBRA before answering me]
The C-language FAQ: http://www.eskimo.com/~scs/C-faq/top.html
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[herrcho]

Emmanuel Delahaye wrote:

In 'comp.lang.c', "herrcho" <he*********@kornet.net> wrote:

here is the code

#include <stdio.h>

Missing
#include <string.h>

int main()
{
char *str1="Borland International";
printf("%d\n", strlen(str1));

printf ("%d\n", (int) strlen(str1));

or better

printf ("%lu\n", (unsigned long) strlen(str1));

return 0;
}

i learned when i use strlen i have to include <string.h>

True enough.

but the above runs o.k without <string.h>
i compiled it with gcc -Wall -ansi -pedantic -o

What was the warning messages?

Could anyone explain about this ?

It runs by chance only. It's a UB, and UB are tricky. Don't do that.

there was no warning message..

by the way.. what is UB ?? i'm new to c programming..

[Emmanuel Delahaye]

In 'comp.lang.c', "herrcho" <he*********@kornet.net> wrote:

> i compiled it with gcc -Wall -ansi -pedantic -o
What was the warning messages?

> Could anyone explain about this ?

It runs by chance only. It's a UB, and UB are tricky. Don't do that.

there was no warning message..

Try again with

gcc -W -Wall -ansi -pedantic -o
by the way.. what is UB ?? i'm new to c programming..

Sorry about that. Undefined Behaviour. anything can happen, including 'good'
behaviour. This is why it's so a dangerous bug.

--
-ed- em**********@noos.fr [remove YOURBRA before answering me]
The C-language FAQ: http://www.eskimo.com/~scs/C-faq/top.html
<blank line>
FAQ de f.c.l.c : http://www.isty-info.uvsq.fr/~rumeau/fclc/

[Emmanuel Delahaye]

In 'comp.lang.c', Emmanuel Delahaye <em**********@noos.fr> wrote:

Try again with

gcc -W -Wall -ansi -pedantic -o

I meant -O (or -O2, -O3 etc.)

gcc -W -Wall -ansi -pedantic -O2

--
-ed- em**********@noos.fr [remove YOURBRA before answering me]
The C-language FAQ: http://www.eskimo.com/~scs/C-faq/top.html
<blank line>
FAQ de f.c.l.c : http://www.isty-info.uvsq.fr/~rumeau/fclc/

[bd]

On Thu, 07 Aug 2003 22:05:33 +0900, herrcho wrote:

here is the code

#include <stdio.h>
int main()
{
char *str1="Borland International";
printf("%d\n", strlen(str1));
return 0;
}

i learned when i use strlen i have to include <string.h> but the above
runs o.k without <string.h> i compiled it with gcc -Wall -ansi -pedantic
-o

Could anyone explain about this ?

If you don't #include it, it assumes it's a function that returns int, and
has whatever you gave it as arguments. strlen()'s prototype is:

size_t strlen(const char *)

Since size_t isn't the same as int, this is Undefined Behavior, so
anything can happen. In this case it apparently worked, but you shouldn't
rely on it.

Also, "%d" is incorrect - if you have a C99 compiler, you can use "%zu",
otherwise, cast to the proper type (for "%d", int)

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