Pass by reference in c array

Question

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);
}

Hint: check the data types. but your code and usage doesn't look proper. — Sourav Ghosh
– Sourav Ghosh, Commented Dec 16, 2014 at 5:49
There are several (obscure) differences between the two versions of display() However, using int [] in the prototype is not a clear method . better to use int * as that is clear. Then both versions should use: printf("\n%d",x[i]); as it is (almost) always better to not change the passed by value parameter(s). — user3629249
– user3629249, Commented Dec 16, 2014 at 11:56
What's up with the Stealth edit, taking someone's code in their answer and putting it in your question? Don't do that. Do DO a rollback, or I will. That is not considered fair; it's called "dirty pool". Edit: Never mind about you doing a rollback, I did it. — Funk Forty Niner
– Funk Forty Niner, Commented Dec 16, 2014 at 13:32

Sadique · Accepted Answer · 2014-12-16 06:33:08Z

2

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]);
    }
}

edited Dec 16, 2014 at 6:33

answered Dec 16, 2014 at 5:50

Sadique

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Comments

Rizier123 · Accepted Answer · 2014-12-16 05:51:37Z

0

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
    }
}

answered Dec 16, 2014 at 5:51

Rizier123

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5 Comments

Jake Over a year ago

i want to ask what is the difference between your code and this code

Jake Over a year ago

'/* lab10_3.c pass by reference*/ #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[i]); }'

Rizier123 Over a year ago

@Jake In your Code your just printing the address of the first element of the array. In my Code i print all values of the array.

Jake Over a year ago

Sorry i have another code now updated plz check again plz THANK YOU i updated it in the question thanks

Rizier123 Over a year ago

@Jake In the first one you pass the address of the first element of your array and print the address x times(x elements of the array). In the second example you pass the full array and you go through the values with the loop and you're printing all values of the array.

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