maerics maerics - 1 year ago 66
C Question

Why doesn't GCC produce a warning when assigning a signed literal to an unsigned type?

Several questions on this website reveal pitfalls when mixing signed and unsigned types and most compilers seem to do a good job about generating warnings of this type. However, GCC doesn't seem to care when assigning a signed constant to an unsigned type! Consider the following program:

/* foo.c */
#include <stdio.h>
int main(void)
unsigned int x=20, y=-30;
if (x > y) {
printf("%d > %d\n", x, y);
} else {
printf("%d <= %d\n", x, y);
return 0;

Compilation with GCC 4.2.1 as below produces no output on the console:

gcc -Werror -Wall -Wextra -pedantic foo.c -o foo

The resulting executable generates the following output:

$ ./foo
20 <= -30

Is there some reason that GCC doesn't generate any warning or error message when assigning the signed value
to the unsigned integer variable

Answer Source

Use -Wconversion:

~/src> gcc -Wconversion -Werror -Wall -Wextra -pedantic -o signwarn signwarn.c
cc1: warnings being treated as errors
signwarn.c: In function 'main':
signwarn.c:5: error: negative integer implicitly converted to unsigned type

I guess the thing here is that gcc is actually pretty good at generating warnings, but it defaults to not doing so for (sometimes unexpected) cases. It's a good idea to browse through the available warnings and choose a set of options that generate those you feel would help. Or just all of them, and polish that code until it shines! :)

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