CODEWITHSUNDEEP
c++cgcccompiler-warningsimplicit-conversion
c0dehunter
c0dehunter

Reputation: 6140

Resolving conversion warnings with compound assignment operators

At our company we have a policy to compile with -Wconversion which produces some conversion warnings. While I do agree this extra checking prevents bugs, it is annoying to see warnings on shorthand operators such as in the following case:

uint8_t byte;
byte += 8; // conversion to 'uint8_t' from 'int' may alter its value [-Wconversion]

Now this can be solved by rewriting it as byte = (uint8_t)(byte+8) which in turn reduces code readability.

Is there any better way to do this?

Upvotes: 10

Views: 1055

Answers (2)

Lundin
Lundin

Reputation: 213276

Consider the reason why you get the warning, namely that the integer constant 8 is of type int. That everything in C has to be promoted to (signed) int is a well-known design flaw of the language.

Suppose you had byte += 256; or byte += -1; or byte += function_that_returns_int();. All of them are potentially severe bugs, so the warning certainly makes sense to enable.

There's really no other work-around than to cast the result of the operation to the intended type, uint8_t. Which isn't necessarily a bad thing, as it creates self-documenting code saying "yes, I have actually considered which types that are used in this calculation so there should be no overflows".

Upvotes: 3

the kamilz
the kamilz

Reputation: 1988

This may not exactly solve your problem but at least gives you a hint that there is a solution for almost everything.

#include <stdio.h>
#include <stdint.h>

#define SAFE_ADD(a,b) ((a) = (typeof(a))((a)+(b)))

int main(void)
{
        uint8_t byte = 0;
        SAFE_ADD(byte, 8);

        fprintf(stderr, "byte = %d \n", byte);
        return 0;
}

Compiled w/o warnings with gcc 4.8.4 (gcc -Wall -Wconversion byte.c)
Hope that helps.

Upvotes: -1

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