Passing modified array values back to main function in C

Question

Sorry if the title is still ambiguous.

I'm doing this assignment for school and below are my defined function prototypes, the main function and the change_array function.

The overall objective of this program is to allow users to input 5 different numbers and be stored into an array. Then what the change_array function does is to double (multiply by 2) any numbers that are below 10, however, it is currently not doing what it is intended to do. I'm really stuck, so I was wondering if anyone can point out my mistakes. I'm not asking for an exact answer, I just need some pointers and guidance.

What is going wrong is that the change_array function is not changing any of the values given by the users. So for example, if the user inputs, "3, 5, 6, 12, 32", the output of my program is still "3, 5, 6, 12, 32". But what I really want is, "6, 10, 12, 12, 32" after the arrays are passed back from the change_array function.

EDITED with complete program:

#include <stdio.h>
#define SIZE 5
void fill_array(double x[], int s);
void change_array(double x[], int s);
void print_array(double x[], int s);

main()
{
    double x[SIZE];

    fill_array(x, SIZE);
    printf("The array is as: \n");
    print_array(x, SIZE);
    change_array(x, SIZE);
    printf("After change, the array is: \n");
    print_array(x, SIZE);
}

void fill_array(double x[], int s)
{
    int i=0;

    printf("Please input 5 Non-Negative double numbers\n");
    for (i=0; i<s; i++) {
        printf("Number %d: ", i+1);
        scanf("%d", &x[i]);
        printf("\n");
    }
}

void change_array(double x[], int s)
{
    int i=0;

    for (i=0; i<s; i++) {
        if (x[i] < 10) {
              (x[i] = (x[i] * 2));
        }
    }
}

void print_array(double x[], int s)
{
    int i=0;

    for (i=0; i<s; i++) {
        printf("%ld \t", x[i]);
    }
    printf("\n");
}

My code is written in C.

change_array indeed modifies the array so that the modification are visible to the caller. You need to post the real code, preferably a complete program. — David Heffernan
– David Heffernan, Commented Mar 9, 2012 at 20:21
OK, the new code makes it clear that the problem is simply the printf format string. — David Heffernan
– David Heffernan, Commented Mar 9, 2012 at 20:49

Taylor Southwick · Accepted Answer · 2012-03-09 20:19:34Z

2

Arrays are always passed by reference, so that is not the issue. It works as expected when I add a print function and an extra end curly brace. Can you provide a minimal working example (MWE) so that we can just run the code? (I know this should probably be a comment, but I'm 3 points shy of being able to comment)

answered Mar 9, 2012 at 20:19

Taylor Southwick

1,49613 silver badges14 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Steven Chen Over a year ago

Since both you and Als approved that it works, then I think it might be my other functions that are causing the unwanted output. I will try to look at my work closely again.

David Heffernan · Accepted Answer · 2012-03-09 20:43:37Z

1

There is nothing wrong with change_array. It does indeed modify the values in the caller's array.

The problem lies the printing function, print_array. You are using the wrong format string for your printf. You need to use %f rather than %ld since these are floating point values and not integers.

answered Mar 9, 2012 at 20:43

David Heffernan

616k46 gold badges1.1k silver badges1.5k bronze badges

Comments

Alok Save · Accepted Answer · 2012-03-09 20:26:32Z

1

I modified your code to make in to a minimal sample and it works fine for me:

  #include <stdio.h>
  #define SIZE 5
  void change_array(double x[], int s);


   int main()
   {
        double x[SIZE] = {3, 5, 6, 12, 32};
        printf("The array is as: \n");
        for(int i = 0;i<SIZE;i++)
            printf("\n%f",x[i]);

        //fill_array(x, SIZE);

        //print_array(x, SIZE);
        change_array(x, SIZE);
        printf("After change, the array is: \n");
        for(int i = 0;i<SIZE;i++)
            printf("\n%f",x[i]);
        // print_array(x, SIZE);
        return 0;
    }


    void change_array(double x[], int s)
    {
        int i=0;

        for (i=0; i<s; i++) 
        {
            if (x[i] < 10) 
            {
                 (x[i] = (x[i] * 2));
            }
        }
    }

Here is the output:

The array is as:

After change, the array is:

edited Mar 9, 2012 at 20:26

answered Mar 9, 2012 at 20:21

Alok Save

208k56 gold badges448 silver badges539 bronze badges

2 Comments

Steven Chen Over a year ago

Thank you, as I said to TSouthwick, I think it might be my other functions that are causing the unwanted output. I will try to look at my work closely again.

Steven Chen Over a year ago

Just posted my entire program.

Collectives™ on Stack Overflow

Passing modified array values back to main function in C

3 Answers 3

1 Comment

Comments

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related