How to make gcc/g++ warn on comparing signed and unsigned char - gcc

Question

I have the following piece of code:
#include <stdlib.h>
#include <stdio.h>
void test(unsigned char * arg) { }
int main() {
char *pc = (char *) malloc(1);
unsigned char *pcu = (unsigned char *) malloc(1);
*pcu = *pc = -1; /* line 10 */
if (*pc == *pcu) puts("equal"); else puts("not equal"); /* line 12 */
pcu = pc; /* line 14 */
if (pcu == pc) { /* line 16 */
test(pc); /* line 18 */
}
return 0;
}
If I compile it with gcc version 4.6.3 (Ubuntu/Linaro 4.6.3-1ubuntu5) (but it is not limited to this particular version) with options
gcc a.c -pedantic -Wall -Wextra -Wsign-conversion -Wno-unused-parameter; ./a.out
I get the following warnings
test.c: In function ‘main’:
test.c:10:21: warning: conversion to ‘unsigned char’ from ‘char’ may change the sign of the result [-Wsign-conversion]
test.c:14:13: warning: pointer targets in assignment differ in signedness [-Wpointer-sign]
test.c:16:17: warning: comparison of distinct pointer types lacks a cast [enabled by default]
test.c:18:17: warning: pointer targets in passing argument 1 of ‘test’ differ in signedness [-Wpointer-sign]
test.c:4:6: note: expected ‘unsigned char *’ but argument is of type ‘char *’
not equal
g++ warnings/errors are similar. I hope I understand why the comparison on line 12 is evaluated to false, but is there any way to get a warning also in such cases? If not, is there some principial difference between line 12 and the lines which cause warnings? Is there any specific reason why comparison of char and unsigned char shouldn't deserve its warning? Because at least at first glance, line 12 seems to me more "dangerous" than e.g. line 16.
A short "story behind": I have to put together pieces of code from various sources. Some of them use char and some of them use unsigned char. -funsigned-char would work fine, but I am forced to avoid it and rather to add proper type conversions. That's why such a warning would be useful for me, because now, if I forget to add a type conversion in such a case, the program silently fails.
Thanks in advance, P.

Answer 1

I believe this is caused by integer promotion.
When you deal with char or short, what C actually does (and this is defined by the standard, not the implementation) is promote those types to int before doing any operations. The theory, I think, is that int is supposed to be the natural size used by the underlying machine, and therefore the fastest, most efficient size; in fact, most architectures will do this conversion on loading a byte without being asked.
Since both signed char and unsigned char will fit happily within the range of a signed int, the compiler uses that for both, and the comparison becomes a pure signed comparison.
When you have a mismatched type on the left-hand-side of the expression (lines 10 and 14) then it needs to convert that back to the smaller type, but it can't, so you get a warning.
When you compared the mismatching pointers (line 16) and passed the mismatching pointer (line 18), the integer promotion is not in play because you never actually dereference the pointers, and so no integers are ever compared (char is an integer type also, of course).