Question

This is an example:

#include <stdbool.h>

void foo(bool b){};
void bar(bool b) {foo(b);}

int main() {
    bar(false);
}

I compile with:

gcc -Wtraditional-conversion test.c

I get these warnings:

test.c: In function 'bar':
test.c:4: warning: passing argument 1 of 'foo' with different width due to prototype
test.c: In function 'main':
test.c:7: warning: passing argument 1 of 'bar' with different width due to prototype

Why do these warnings occur? As far as I can see the arguments are all the same type, so should be the same width. What is -Wtraditional-conversion doing to cause these warnings in this very simple piece of code?

I started to get these errors when I switched from using my own typedef of bool to the stdbool.h def.

My original def was:

typedef enum {false, true} bool;

Answer 1

This was a case of not understanding the compiler warning flags.

Use -Wconversion not -Wtraditional-conversion to get warnings that warn about implicit conversion. -Wtraditional-conversion is for warning about conversions in the absence of a prototype.

Was caught out because the typdef enum creates a default integer bool type (32bit normally), where as stdbool.h defines bool as an 8bit, which is compatible with the C++ bool.