Function to remove a struct from an array doesn't work if I pass the array as a parameter

Question

I have a weird problem and I don't know the reason, so I can't think of a solution to fix it.

My problem:

I have a removeEntry function with a array of structs as parameter, but somehow this function doesn't work.

In an earlier version of my code, I declared my array of structs outside the main function (outside every function), and it worked then. But since I now create my array of struct in my main function, I have to give it as parameter, and now my removeEntry function doesn't work.

Code that isn't working:

void removeEntry(struct entry entries[])
{
    int entryNumber;
    int nrBytes = sizeof(entries);
    int arrayLength = nrBytes / sizeof(entries[0]);
    printf("\nEnter entry number to delete: ");
    scanf("%d",&entryNumber);
    while (scanfOnlyNumber(entryNumber) == false || entryNumber == 0 ||  entries[entryNumber].entry_number == 0)
    {
        printf("\nEnter a legit entry number to delete: ");
        scanf("%d", &entryNumber);
        // Tell the user that the entry was invalid
    }
    int i = 0;
    for(i = entryNumber; i < arrayLength - 1; i++)
    {
        entries[i] = entries[i + 1]; //removes element and moves every element after that one place back.
    }
    updateEntryNumber(entries);
    printf("\nEntry %d removed succesfully, and entry numbers updated!\n", entryNumber);
}

My teacher told me that my arraylength calculation doesn't work when I create my array of structs inside my main function (what I do now), but I can't tell why it doesn't work. If anybody can explain that, then I might be able to fix my removeEnty problem by myself.

If anyone wants the working code (where I don't give my array as parameter, because I create my array outside every function), then tell me and I will post it.

Answer 1

Problem

When you pass an array to a function, you don't pass the entire array, instead you pass a pointer to the array. Hence, when you do int nrBytes = sizeof(entries); you're actually getting the size of pointer variable rather than the size of array.

Solution

Pass your array length along with a pointer to the array to the function, something like this:

void removeEntry(struct entry entries[], int arrayLength){
    // your code
}

Answer 2

int nrBytes = sizeof(entries);     // basically size of struct node * 
int arrayLength = nrBytes / sizeof(entries[0]);  

In this sizeof(entries) will give size of struct pointer , and not the array , and also in second statement , it wont correctly .

This is because array decays into pointer after passing it to function and referring it .

What you need to do is calculate number of elements in struct array in main and then pass it to the function .

void removeEntry(struct entry entries[],int arrayLength); 

Calculate arrayLength in main as you do.

Answer 3

if you compile your example, you will probably see a warning like this one:

warning: sizeof on array function parameter will return size of 'entry *' instead of 'entry []' [-Wsizeof-array-argument]

As noted in the answer by V. Kravchenko, this suggests that nrBytes will contain merely the size of the pointer, while sizeof(entries[0]) returns the "true" size of the entire structure. Therefore, arrayLength will be most probably zero.

Answer 4

You should pass array's size to function, because int nrBytes = sizeof(entries); is always 4 or 8 (on x64 systems). It's just pointer's size.

Answer 5

int nrBytes = sizeof(entries);

Arrays of any type (even arrays of structs) decay into pointers when passed to a function. The lenght of this pointer is always fixed and gives no indication as to the lenght whatsoever. So pass the lenght with the function call.