My program returns 4 instead of expected value 40

February 7, 2022

#include <stdio.h>
    
int value(int *a);
    
int main(){
    int num = 4;
    value(&num);
    printf("value of number is = %d", num);
    return 0;
}

int value(int *a){
    int c = (*a)*10;
    return c;
}

In this code, I transfer the address in function but it does not change, Why?

>Solution :

You have 2 ways of having a function change data outside it, you can pass a variable by reference and update it or you can return a value and use that. You are mixing half of each method (I reordered the code to make my commentary clearer)

  #include <stdio.h>
    
    int value(int *a);
    

// here you are passing by reference, good
int value(int *a){
    int c = (*a)*10; // but you dont change a
    return c;  // instead you return a new value
}

 int main(){
        int num = 4;
        value(&num); // but here you ignore the new value. 
        printf("value of number is = %d", num);
        return 0;
    }

to make reference way work you need

#include <stdio.h>
    
void value(int *a);
    
int main(){
    int num = 4;
    value(&num);
    printf("value of number is = %d", num);
    return 0;
}

void value(int *a){
    *a = (*a)*10; // chnage the value that a points at
    // no need to retunr anything
}

or to make the return way work

#include <stdio.h>
    // no need to pass by reference
int value(int a);
    
int main(){
    int num = 4;
    num = value(num); // use value return by function
    printf("value of number is = %d", num);
    return 0;
}

int value(int a){
    int c = a * 10;
    return c;
}