C 中 function 的外部链接，不带“extern”可以吗？

Question

Hello I'm viewing an example of external linkage of function and variable in C.

This is an example that produces random variables.

1. random.c

#define SEED 20
#define MULT 3124
#define INC 2345
#define MOD 5436

unsigned int call_count = 0; 

static unsigned seed = SEED;  

unsigned random_i(void) {

    seed = (MULT * seed + INC) % MOD;

    call_count++;

    return seed;

}

double random_f(void) {

    seed = (MULT * seed + INC) % MOD;
    call_count++;

    return seed / (double)MOD;
}

2. and this is main.c

#include <stdio.h>
#include <stdlib.h> 

#pragma warning (disable:4996)

unsigned random_i(void); // I wonder why this prototype has no "extern" specifier
double random_f(void); // I wonder why this prototype has no "extern" specifier

extern unsigned call_count;

int main(void) {

    register int i;

    for (i = 0; i < 10; i++)
        printf("%d ", random_i());

    printf("\n");

    for (i = 0; i < 10; i++)
        printf("%lf ", random_f());

    printf("\nCall count : %d\n", call_count);
    
    return 0;
}

in this short program, I wonder why those of two function prototype has no " extern " specifier AND why this code is compiled without an error.

Because what I know is that when I use variables or functions that is in other source code, I have to do it with extern specifier for example extern int a=10; .

Please let me know.

Thanks.

Answer 1

Both functions and data definitions outside of a function block have implicit external linkage. However when it cones to a declaration , the extern keyword is required for data object otherwise it is a definition . For functions a prototype is unambiguously not a definition because it has no body.

Regarding:

I have to do it with extern specifier for example extern int a=10;

Not quite, in this case extern int a=10; is a definition not a declaration because you have included an initialiser. It is the same as int a=10; and would result in a duplicate definition error if also defined elsewhere.

Answer 2

Symbols at global scope are externally visible by default. This applies to both variables and functions. The extern keyword is only needed to prevent multiple definitions of a symbol. It states that the current declaration is defined elsewhere.

extern is required for call_count since it would otherwise be a definition. It is not required for the declaration of global functions since there is no definition.

Note that

extern int a=10;

generates a warning in gcc:

warning: ‘a’ initialized and declared ‘extern’ 

Answer 3

All functions are extern by default so prototypes do not need extern . But you can, of course, use it if you wish.

unsigned random_i(void); === extern unsigned random_i(void);