I am implementing a program to divide all value in a array by 100 then store them in b array using malloc. The problem is I got segmentation fault when printing value of b in main.

This is my code

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

void divide(int *a, int n, double *b){
    b=malloc(n*sizeof(double));
    
    for(int i=0; i<n; i++){
        b[i]=(double)a[i]/100.0;
    }

    //check: values still remain in b
    for (size_t i = 0; i < 5; i++)
    {
        printf("%.2f ", b[i]);
    }
}

int main(){
    int a[]={1,2,3,4,5};
    double *b;

    divide(a,5,b);
    
    //check: lost value and cause segmentation fault
    for (size_t i = 0; i < 5; i++)
    {
        printf("%.2f ", b[i]);
    }
    free(b);
    return 0;
}

So what cause this problem and how to fix it?

Thanks in advance.

>Solution :

You are passing the pointer b by value to the function divide

divide(a,5,b);

That is the function deals with a copy of the original pointer. Changing the copy does not influence on the original pointer.

You need either to pass the pointer by reference through a pointer to it or redesign the function such a way that it will return a pointer to the dynamically allocated memory within the function.

For example the function could be declared and defined the following way

double * divide( const int *a, size_t n )
{
    double *b = malloc( n * sizeof( double ) );

    if ( b != NULL )
    {
        for ( size_t i = 0; i < n; i++ )
        {
            b[i] = a[i] / 100.0;
        }

        //check: values still remain in b
        for ( size_t i = 0; i < n; i++ )
        {
            printf("%.2f ", b[i]);
        }
    }

    return b;
}

And in main you can write

double *b = divide( a, sizeof( a ) / sizeof( *a ) );

Otherwise the function can look like

void divide( const int *a, size_t n, double **b )
{
    *b = malloc( n * sizeof( double ) );

    if ( *b != NULL )
    {
        for ( size_t i = 0; i < n; i++ )
        {
            ( *b )[i] = a[i] / 100.0;
        }

        //check: values still remain in b
        for ( size_t i = 0; i < n; i++ )
        {
            printf("%.2f ", ( *b )[i]);
        }
    }
}

And called like

divide( a, sizeof( a ) / sizeof( *a ), &b );