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

Using the really excellent example from Greggo, I got this to work as a bubble sort with passing an array as a pointer and doing a simple -1 manipulation.

#include<stdio.h>

void sub_one(int (*arr)[7])
{
     int i; 
     for(i=0;i<7;i++)
    {
        (*arr)[i] -= 1 ; // subtract 1 from each point
        printf("%i\n", (*arr)[i]);

    }

}   

int main()
{
    int a[]= { 180, 185, 190, 175, 200, 180, 181};
    int pos, j, i;
    int n=7;
    int temp;
    for (pos =0; pos < 7; pos ++){
        printf("\nPosition=%i Value=%i", pos, a[pos]);
    }
    for(i=1;i<=n-1;i++){
        temp=a[i];
        j=i-1;
        while((temp<a[j])&&(j>=0)) // while selected # less than a[j] and not j isn't 0
        {
            a[j+1]=a[j];    //moves element forward
            j=j-1;
        }
         a[j+1]=temp;    //insert element in proper place
    }

    printf("\nSorted list is as follows:\n");
    for(i=0;i<n;i++)
    {
        printf("%d\n",a[i]);
    }
    printf("\nmedian = %d\n", a[3]);
    sub_one(&a);

    return 0;
}

I need to read up on how to encapsulate pointers because that threw me off.

Answer 2

In new code assignment should be,

B[0] = 5

In func(B), you are just passing address of the pointer which is pointing to array B. You can do change in func() as B[i] or *(B + i). Where i is the index of the array.

In the first code the declaration says,

int *B[10]

says that B is an array of 10 elements, each element of which is a pointer to a int. That is, B[i] is a int pointer and *B[i] is the integer it points to the first integer of the i-th saved text line.

Answer 3

The argument of func is accepting double-pointer variable. Hope this helps...

#include <stdio.h>

int func(int **B){

}

int main(void){

    int *B[10];

    func(B);

    return 0;
}

Answer 4

Make use of *(B) instead of *B[0]. Here, *(B+i) implies B[i] and *(B) implies B[0], that is
*(B+0)=*(B)=B[0].

#include <stdio.h>

int func(int *B){
    *B = 5;     
    // if you want to modify ith index element in the array just do *(B+i)=<value>
}

int main(void){

    int B[10] = {};
    printf("b[0] = %d\n\n", B[0]);
    func(B);
    printf("b[0] = %d\n\n", B[0]);
    return 0;
}

Answer 5

In your new code,

int func(int *B){
    *B[0] = 5;
}

B is a pointer to int, thus B[0] is an int, and you can't dereference an int. Just remove the *,

int func(int *B){
    B[0] = 5;
}

and it works.

In the initialisation

int B[10] = {NULL};

you are initialising anint with a void* (NULL). Since there is a valid conversion from void* to int, that works, but it is not quite kosher, because the conversion is implementation defined, and usually indicates a mistake by the programmer, hence the compiler warns about it.

int B[10] = {0};

is the proper way to 0-initialise an int[10].

Answer 6

In the function declaration you have to type as

VOID FUN(INT *a[]);
/*HERE YOU CAN TAKE ANY FUNCTION RETURN TYPE HERE I CAN TAKE VOID AS THE FUNCTION RETURN  TYPE FOR THE FUNCTION FUN*/
//IN THE FUNCTION HEADER WE CAN WRITE AS FOLLOWS
void fun(int *a[])
//in the function body we can use as
a[i]=var