Hi All,
I have one question regarding the code.
#include<stdio. h>
char *f1(void);
char *f1(void)
{
char *abc ="Hello";
return abc;
}
int main(void )
{
char *p;
p = f1();
printf("%s\n", p);
return 0;
}
While compiling the above program as mentioned below I am getting the
meaning
$ gcc -W -Wall -ansi -pedantic -Wformat-nonliteral -Wcast-align -
Wpointer-arith -Wbad-function-cast -Wmissing-prototypes -Wstrict-
prototypes -Wmissing-declarations -Winline -Wundef -Wnested-externs -
Wcast-qual -Wshadow -Wconversion -Wwrite-strings -ffloat-store -O2
jj.c
jj.c: In function `f1':
jj.c:5: warning: initialization discards qualifiers from pointer
target type
I have two question
1) What is the meaning of the warning ?
2) is it safe to return the local pointer value (i.e return abc ;) is
correct ?
My understanding is we should not return address of local
variable .So above code may not be working always .
Regards,
Somenath
Dec 14 '07
20  2019 
On Fri, 14 Dec 2007 17:48:06 -0500, CBFalconer wrote:
Harald van D?k wrote:
>>
... snip ...
>>
(However, even without -Wwrite-strings, gcc has a bug in which the
equally strictly conforming
int main(void) {
if (0) ""[0] = 'x';
}
is unconditionally rejected.)
Not a bug. "" is a constant string, stored in (possibly) constant
memory, and thus is not writable. You may be complaining that gcc
hasn't bother to notice that the statment won't be executed, and thus
should suppress the message. However, compilers are allowed to emit all
the messages they wish.
I am complaining not that GCC warns about this program (which would be
conforming), but that it issues an error message and refuses to compile
it at all, in all non-conforming, mostly-conforming, and conforming
modes. I posted quite clearly that the program was "unconditionall y
rejected". If it got compiled, with or without warnings, it would have
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