C pass int array pointer as parameter into a function

Question

I want to pass the B int array pointer into func function and be able to change it from there and then view the changes in main function

#include 

        

Answer 1

main()
{
    int *arr[5];
    int i=31, j=5, k=19, l=71, m;

    arr[0]=&i;
    arr[1]=&j;
    arr[2]=&k;
    arr[3]=&l;
    arr[4]=&m;

    for(m=0; m<=4; m++)
    {
        printf("%d",*(arr[m]));
    }
    return 0;
}

Answer 2

If you actually want to pass an array pointer, it's

#include <stdio.h>

void func(int (*B)[10]){   // ptr to array of 10 ints.
        (*B)[0] = 5;   // note, *B[0] means *(B[0])
         //B[0][0] = 5;  // same, but could be misleading here; see below.
}

int main(void){

        int B[10] = {0};   // not NULL, which is for pointers.
        printf("b[0] = %d\n\n", B[0]);
        func(&B);            // &B is ptr to arry of 10 ints.
        printf("b[0] = %d\n\n", B[0]);

        return 0;
}

But as mentioned in other answers, it's not that common to do this. Usually a pointer-to-array is passed only when you want to pass a 2d array, where it suddenly looks a lot clearer, as below. A 2D array is actually passed as a pointer to its first row.

void func( int B[5][10] )  // this func is actually the same as the one above! 
{
         B[0][0] = 5;
}

int main(void){
    int Ar2D[5][10];
    func(Ar2D);   // same as func( &Ar2D[0] )
}

The parameter of func may be declared as int B[5][10], int B[][10], int (*B)[10], all are equivalent as parameter types.

Addendum: you can return a pointer-to-array from a function, but the syntax to declare the function is very awkward, the [10] part of the type has to go after the parameter list:

int MyArr[5][10];
int MyRow[10];

int (*select_myarr_row( int i ))[10] { // yes, really
   return (i>=0 && i<5)? &MyArr[i] : &MyRow;
}

This is usually done as below, to avoid eyestrain:

typedef int (*pa10int)[10];

pa10int select_myarr_row( int i ) {
   return (i>=0 && i<5)? &MyArr[i] : &MyRow;
}

Answer 3

Maybe you were trying to do this?

#include <stdio.h>

int func(int * B){

    /* B + OFFSET = 5 () You are pointing to the same region as B[OFFSET] */
    *(B + 2) = 5;
}

int main(void) {

    int B[10];

    func(B);

    /* Let's say you edited only 2 and you want to show it. */
    printf("b[0] = %d\n\n", B[2]);

    return 0;
}