C - gcc: no compiler warning with different functi

问题:

I try to figure out why my c-compiler gives me no warning/error with following (simplified) code.

The function-declaration have no parameters while the function-implementation have parameters:

some.h:

void foo();

some.c:

static uint32_t count = 0; 

void foo(uint32_t num) {
    count += num;
    print("Count: %u");
}

main.c:

foo(100);
foo();

Output:

Count: 100
Count: 100

Compiler for target build:

gcc-arm-none-eabi-4_9-2015q1-20150306-win32

Linker for target build:

gcc-arm-none-eabi-4_9-2015q1-20150306-win32

Compiler-Flags:

-Wall -Werror -DuECC_CURVE=uECC_secp256r1 -DMEMORY_CHECK -DDEBUG -Os -g3 -DBACKTRACE

回答1:

Because of backward compatibility, a declaration like

void foo();

doesn't declare a function that takes no argument, it declares a function which takes an unknown number of arguments of unknown type.

That means both your calls are correct, and the compiler can't really warn you about it.

The other problematic thing is that the declaration in the source file actually matches the declaration in the header file, it just makes it more precise. Therefore you will not get a warning or error there either.

回答2:

In C this function declaration

void foo();

means that there is nothing known about the function parameters in the point of the declaration.

The types and number of the parameters are deduced from a function call.

As for your program then this call

foo();

has undefined behavior because the number of parameters and the number of arguments mismatch.