CODEWITHSUNDEEP
carrays
Jake
Jake

Reputation: 11

Pass by reference in c array

Hi i want to ask about why i need to use printf("\n%d",x); instead of printf("\n%d",*x);? Thank you very much

#include<stdio.h>
#include<stdlib.h>
#define totalnum 8
void display(int **);
int main()
{
    int marks[totalnum]={55,65,75,85,90,78,95,60};
    printf("The marks of student A are:");  
    display(marks);
    return 0;
}
void display(int *x)
{
    int i;
    for(i=0;i<totalnum;i++)
    printf("\n%d",x);
}

Upvotes: 0

Views: 92

Answers (2)

Sadique
Sadique

Reputation: 22823

There is no pass by reference in C. The array decays to a pointer in the display function which you declared wrongly as int ** instead of int * - Compiler should have given you a warning at least about this:

http://ideone.com/R3skNj

This is how your display function should be like:

void display(int *x) 
{
    int i;
    for(i = 0; i < totalnum; i++) {
        printf("\n%d",*(x+i)); // or printf("\n%d",x[i]);
    }
}

Upvotes: 2

Rizier123
Rizier123

Reputation: 59681

I think your looking for something like this:

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

#define totalnum 8

void display(int *);  //Typ is 'int *' NOT 'int **'
int main() {

    int marks[totalnum] = {55,65,75,85,90,78,95,60};

    printf("The marks of student A are:");
    display(marks);

    return 0;

}

void display(int *x) {
    int i;
    for(i = 0; i < totalnum; i++) {
        printf("\n%d",*x);  //Prints the value
        x++;  //increments the pointer
    }
}

Upvotes: 0

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