Implicit Declaration

/* main.c */
#include <stdint.h>
#include <stdio.h>
/* int64_t foo(); */
int main(int argc, char *argv[]) {
    int64_t x = foo();
    printf("after return: %ld\n", x);
    return 0;
}

/* ----------------------------------------------- */

/* foo.c */
#include <stdint.h>
#include <stdio.h
int64_t foo() {
    int64_t ret = 1l << 31;
    printf("before return: %ld\n", ret);
    return ret;
}

$> gcc main.c foo.c
main.c: In function ‘main’:
main.c:6:17: warning: implicit declaration of function ‘foo’ [-Wimplicit-function-declaration]
     int64_t x = foo();

$> ./a.out
before return: 2147483648
after return: -2147483648

but if we add the declaration `int64_t foo();` in main.c

$> gcc main.c foo.c

$> ./a.out 
before return: 2147483648
after return: 2147483648

Did you notice the warning `implicit description of function ‘foo’ [-Wimplicit-function-declaration]`

When C doesn't find a declaration, it assumes this implicit declaration: `int f();`, which means the function can receive whatever you give it, and returns an int

What happends in the problematic example:

1. when foo returns, it will place 1l<<31, or 0x0000000080000000 in x0 (suppose under aarch64), which is a 64bit register.
2. now control returns back to main, the implicit declaration makes `main` believe that `foo` will return an int, so it will fetch the `int` value from w0 (the bottom half of x0), which is 0x80000000.
3. then the int value 0x80000000 is cast to int64_t, which will be 0xffffffff80000000 (guess you know why 0xffffffff80000000 instead of 0x0000000080000000)

C99 6.5.2.2/6 "Function calls":

If the expression that denotes the called function has a type that does not include a prototype, the integer promotions are performed on each argument, and arguments that have type float are promoted to double.

对于函数参数:

1. 整数类型会转换为 int
2. 浮点类型会转换为 double