Array of type int doesn't get modified from the function call

Question

For some reason that I'm not aware of, array from main() doesn't get to be modified when it should. Isn't "a" passed by reference? Can someone please guide me a little bit?

Here is the code:

#include "stdio.h"

void sort(int list[])
{
    //int list[] = { 4, 3, 2, 1, 10 };
    int n = sizeof(list) / sizeof(int);
    int min = 0;
    int temp = 0;
    for (int i = 0; i < n; i++)
    {
        min = i;
        //profiler.countOperation("compSort", n, 1);
        for (int j = i + 1; j < n; j++)
        {
            if (list[j] < list[min])
            {
                min = j;
                //profiler.countOperation("compSort", n, 1);
                    }
        }
        if (min != i)
        {
            //profiler.countOperation("compSort", n, 1);
            temp = list[min];
            list[min] = list[i];
            list[i] = temp;
        }
    }
}

int main()
{
    int a[5] = {4, 3, 2, 1, 10};
    sort(a);
    printf("%d\n", a[0]);
    for (int i = 0; i < 5; i++)
    {
        printf("%d", a[i]);
    }
    return 0;
    /*int arr[MAX_SIZE];
    for(int t = 0; t < 1; t++)
    {
        for (int n = 100; n < 3000; n = n + 300)
        {
            FillRandomArray(arr, n);
            sort();
            printf("done %d \n", n);

            if (!IsSorted(arr, n)
            {
                printf("error sort \n");
            }
        }
    }
    profiler.addSeries("TotalSort", "cmpSel", "atribSel");
    profiler.createGroup("SortMediu", "TotalSort", "cmpSel", "atribSel");
    profiler.showReport();*/
}

What is the result of `int n = sizeof(list) / sizeof(int);`....? — LPs, Mar 31 '17 at 08:31

score 4 · Accepted Answer · answered Mar 31 '17 at 08:31

The problem is with

 int n = sizeof(list) / sizeof(int);

arrays, once passed as function arguments decays to the pointer to the first element. They don't have array properties anymore. So, the list here is adjusted as a pointer to int.

Quoting C11, chapter §6.7.6.3

A declaration of a parameter as ‘‘array of type’’ shall be adjusted to ‘‘qualified pointer to type’’, where the type qualifiers (if any) are those specified within the [ and ] of the array type derivation.

Solution: You need to calculate the size at the caller (before the array decay happens) and pass that as a different argument to the called function.

LPs · Answer 2 · 2017-03-31T08:42:48.357

Array passed to function in that way decays to pointer so

int n = sizeof(list) / sizeof(int);

Result in sizeof(int *)/sizeof(int)

You can use VLAs to sole the problem

#include "stdio.h"

void sort(size_t n, int list[n])
{
    size_t min = 0;
    int temp = 0;

    for (size_t i = 0; i < n; i++)
    {
        min = i;
        //profiler.countOperation("compSort", n, 1);
        for (size_t j = i + 1; j < n; j++)
        {
            if (list[j] < list[min])
            {
                min = j;
                //profiler.countOperation("compSort", n, 1);
            }
        }
        if (min != i)
        {
            //profiler.countOperation("compSort", n, 1);
            temp = list[min];
            list[min] = list[i];
            list[i] = temp;
        }
    }
}

int main(void)
{
    int a[5] = { 4, 3, 2, 1, 10 };
    sort(sizeof(a) / sizeof(a[0]), a);
    printf("%d\n", a[0]);
    for (int i = 0; i < 5; i++)
    {
        printf("%d", a[i]);
    }
    printf("\n");

    return 0;
}

Array of type int doesn't get modified from the function call

2 Answers2