Arrays getting passed through a function-CodePudding

I am learning how variables are passed to functions by value, while arrays are passed by reference.

I ran my script to verify that, however the pointer addresses are different. Why is that?

void arrayFunction(int array[4]);

int main(void){
    int foo[4] = {0, 1, 2, 3};
    printf("This is the address to foo in main: %p\n",&foo);
    arrayFunction(foo);
}

void arrayFunction(int array[4]){
    printf("This is the address to array in method: %p\n", &array);
    array[0] = 22;
}

CodePudding user response：

The is the address of foo aka &foo[0] is conceptually copied to a new variable. Or said differently the pointer is still passed by value.

CodePudding user response：

&array is the address of the array variable in the stack of the called function. However that's implemented as a pointer that points to the same array elements.

Here is a modified version of your program showing that:

#include <stdio.h>
void arrayFunction(int array[4]);

int main(void){
    int foo[4] = {0, 1, 2, 3};
    printf("This is the address to foo in main:     \t%p\n",&foo[0]);
    arrayFunction(foo);
}

void arrayFunction(int array[4]){
    printf("This is the address to array in method: \t%p\n", &array[0]);
    array[0] = 22;
}

CodePudding user response：

There is a subtle difference between the array and the normal element. If arr[10] is declared, then only (arr) gives the address of the first emement. In your code, you have print address of the pointer pointing to that first element '&arr'. Just remove the '&' and both will show the same address.

#include<stdio.h>

void arrayFunction(int array[4]);

int main(void){
    int foo[4] = {0, 1, 2, 3};
    printf("This is the address to foo in main: %p\n",foo);
    arrayFunction(foo);
}

void arrayFunction(int array[4]){
    printf("This is the address to array in method: %p\n",array);
    array[0] = 22;
}