（C）调用函数时，将多个值存储在数组中

Question

So, when calling a function I wish to put two values in, then those two values to be saved in an array in the function argument.

ie

function(1,2); (calling the function)

void function(int x[2]); (declaration of function)

with x[0] and x[1] being the two arguments when calling the function (ie the 1 and 2).

Any help would be appreciated.

Answer 1

You can use compound literals if you have two array members, but no array, and wish to call such a function:

void function(int x[2]);     // declaration of function

function((int [2]) {1, 2});  // calling the function

You can also use variables within the compound literal. Here is a simple example program:

#include <stdio.h>

void function(int x[2]);

int main(void)
{
    int arg1 = 3;
    int arg2 = 4;

    puts("Calling function with constants in compound literal:");
    function((int [2]) {1, 2});

    puts("Calling function with variables in compound literal:");
    function((int [2]) {arg1, arg2});

    return 0;
}

void function(int x[2])
{
    printf("x[0] = %d, x[1] = %d\n", x[0], x[1]);
}

Program output:

Calling function with constants in compound literal:
x[0] = 1, x[1] = 2
Calling function with variables in compound literal:
x[0] = 3, x[1] = 4

Answer 2

If you want to call a function that expects an array of two, passing 2 separate values is not the same thing.

This is as close as I can get:

void function(int x[2]) {....}

void caller(int a, int b)
{
    int args[2] = {a, b};  /* Convert two args into an array */
    function(args);        /* Call the function with an array */
}

int main(void)
{
    caller(1, 2);
}

Answer 3

Maybe Variable-length argument is what you want :

void function(int i, ...) { 
    int tmp; 
    va_list num_list; 

    va_start(num_list, i); 

    for(int j = 0; j < i; j++) 
        cout << va_arg(num_list, int) << endl; 

    va_end(num_list); 
}

the first 'i' is tell the function how much argument you input. using this like:

function(2,3,4); // pass 2 argument, 3 and 4

Answer 4

function(1,2); the function call will look for definition of type (int , int).

void function(int x[2]); this function argument is of an array type. so it is not possible to call the function by function(1,2);