size of array passed to function

Question

void quickSort(int a[],int s,int e){
    if(s>e){
        return;
    }

    int pivot = parti(a,s,e);
    cout<<"size "<<sizeof(a)<<endl;

When i am passing the array as argument to function, an array of size more than 4 is being treated as an array of size 4. why is it so?

@tkausl: That is not why. – Lightness Races in Orbit Jul 02 '16 at 13:50 — Lightness Races in Orbit, Jul 02 '16 at 13:50

score 0 · Answer 1 · answered Jul 02 '16 at 13:50

0

You cannot "pass the array to a function". It looks like you can, but you can't.

This:

void quickSort(int a[],int s, int e)

is actually:

void quickSort(int* a,int s, int e)

And, on your system, the size of a pointer is apparently four bytes.

This is why we shall not use sizeof to determine the size of a native array, because it's easy to accidentally take the size of something else instead.

answered Jul 02 '16 at 13:50

Lightness Races in Orbit

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Well, you can pass an array to a function, but the syntax is horrific and you have to know the size at compile time anyway. See http://en.cppreference.com/w/cpp/language/array#Array-to-pointer_decay – N. Shead Jul 02 '16 at 13:58
@N.Shead: No, that's passing a reference-to-an-array to a function. – Lightness Races in Orbit Jul 02 '16 at 13:58

songyuanyao · Answer 2 · 2016-07-02T13:59:29.487

0

Array-to-pointer decay happens here (passing an array as function argument). It means whatever array you passed to quickSort, sizeof(a) will always return the size of the pointer to int.

edited Jul 02 '16 at 13:59

answered Jul 02 '16 at 13:50

songyuanyao

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size of array passed to function

2 Answers2