I am not a GCC developer, just a regular user of C. But I have some
comments below:
On 6/9/2019 3:21 PM, L A Walsh wrote:
If I have a function returning NULL on error (including EOF).
So the program calls exit if the function doesn't return
a non-zero value (func() || exit(1)).
I have:
--/tmp/ex.c--
main() { char * buf[512];
while (gets(buf) || exit(0)) puts(buf);
}
-- compile w/:
gcc -fpermissive --no-warnings -o /tmp/ex /tmp/ex.c
Got:
/tmp/ex.c: In function ‘main’:
/tmp/ex.c:2:22: error: void value not ignored as it ought to be
while (gets(buf) || exit(0) ) puts(buf);
^~~~~~~
I understand that the while is testing the value of exit
"when it returns", but since it doesn't, why flag an error
(if exit is part of C-standard?) but *especially*, why
not a warning, like a type mismatch or such?i
There is a slight misunderstanding of operators here. This is not while
testing the value of exit(0). This is the "||" operator. It needs a
value on *both* sides of || so it can evaluate them both and return a
boolean result. Your code has no value on the right-hand side so it's
meaningless.
But this:
--/tmp/ex2.c--
main(){ char * buf[512];
while (1) {
fgets(buf) || exit(1);
fputs(buf);
}
}
---compile + output:
gcc -fpermissive --no-warnings -o /tmp/ex2 /tmp/ex2.c
/tmp/ex2.c: In function ‘main’:
/tmp/ex2.c:4:18: error: void value not ignored as it ought to be
fgets(buf) || exit(1);
^~~~~~~
Ultra confusing -- how is exit value used? Isn't it thrown away?
Same problem here.
I haven't tested it but you might be able to make this work with a comma
operator like "while (gets(buf) || (exit(0), 1)) puts(buf);"
Oh, and "gets" is no longer a function. It's not just deprecated. It's
*gone*.
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Study Hard -- Be Evil
