How does sizeof() behaves in C++11?

Question

The sizeof() function in c++ is behaving very weird. I am unable to make any sense out of it. I was writing code to implement the binary search algorithm to find an element in an array. To have a lesser number of arguments in the function, I decided to get the length of the array using sizeof() function. Here is the code:

#include <bits/stdc++.h>

using namespace std;

int binarySearch(int arr[], int target) {
    int low = 0;
    int high = sizeof(arr) / sizeof(arr[0]) - 1;

    cout << "high is: " << high << endl;

    while (low <= high) {
        int mid = low + (high - low) / 2;

        if (arr[mid] == target) return mid;

        else if (arr[mid] > target) high = mid - 1;

        else low = mid + 1;
    }
    return -1;
}

int main() {
    int array[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
    cout << binarySearch(array, 10) << endl;
    cout << "size of behaviour in main: " << sizeof(array);
    return 0;
}

The output is:

high is: 0
-1
size of behaviour in main: 40

On debugging, I realised that inside binarySearch function, the sizeof(arr) is giving me the size of int type i.e. 4. so sizeof(arr)/sizeof(arr[0])-1 = 4/4-1 = 0.

However, in int main, the sizeof(array) is giving me the size of int times length of the array i.e. 4*10 = 40.

What explains this change in behaviour of the sizeof() function with respect to its usage in int main and in function definition?

`int arr[]` is not an array in this context as you might think. Therefore `int high = sizeof(arr) / sizeof(arr[0]) - 1;` makes no sense. — Niclas Larsson, May 12 '20 at 09:04
@NiclasLarsson I understand it now that arr is only a pointer, not an array. However, I still do not want to explicitly give int size in function argument. Is there any way around? — user9194161, May 12 '20 at 09:09
change the function to `template int binarySearch(int (&arr)[N], int target)`, then you have the size of the array as `N` in the function. The call site is the same. https://ideone.com/bt0Du1 — mch, May 12 '20 at 09:11
[Why should I not #include ?](https://stackoverflow.com/questions/31816095/why-should-i-not-include-bits-stdc-h) — Evg, May 12 '20 at 09:13

Niclas Larsson · Accepted Answer · 2020-05-12T09:22:50.770

That doesn't work as you're getting the sizeof of a pointer. Refactoring the code to use a template is one solution:

#include <iostream>

using namespace std;

template<std::size_t N>
int binarySearch(int (&arr)[N], int target) {
    int low = 0;
    int high = sizeof(arr) / sizeof(arr[0]) - 1;

    cout << "high is: " << high << endl;

    while (low <= high) {
        int mid = low + (high - low) / 2;

        if (arr[mid] == target) return mid;

        else if (arr[mid] > target) high = mid - 1;

        else low = mid + 1;
    }
    return -1;
}

int main() {
    int array[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
    cout << binarySearch(array, 10) << endl;
    cout << "size of behaviour in main: " << sizeof(array);
    return 0;
}

Using std::array is probably a better option. Both templates and std::array have sizes known at compile time. If you want a runtime approach, take a look at std::vector.

[Why should I not #include ?](https://stackoverflow.com/questions/31816095/why-should-i-not-include-bits-stdc-h) — Evg, May 12 '20 at 09:13

score 0 · Answer 2 · answered May 12 '20 at 09:04

0

When u pass an array as an argument to a function.. it is passed as pointer ( int *arr).. thats why it is 4 ìn the binary search function.

answered May 12 '20 at 09:04

suresh

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How does sizeof() behaves in C++11?

2 Answers2