How to use `sizeof` operator inside the condition of for-loop properly?

Do the conditions of the for-loop always need constant? How can I put sizeof function there to run an output showing all the elements of an array?

#include<iostream>
using namespace std;

void array_output(int a[])
{
    for (int i = 0; i < (sizeof(a)) / (sizeof(a[0])); i++)
    {
        cout << a[i] << endl;
    }
}

int main()
{
    int a[] = { 22, 53, 13, 65, 80, 31, 46 };
    array_output(a);
    return 0;
}

i<(sizeof(a) output shows first 4 elements
i<(sizeof(a))/(sizeof(a[0])) output shows only the first element
instead of sizeof when 7 is directly used as a condition, it gives
the right output, showing all the elements.

Solution

If you use the actual array in the sizeof operator you will get the size of the array in bytes, meaning you can calculate the number of elements like you expected it using sizeof(array) / sizeof(array_type).

int x[] = {1, 1, 1, 1, 1, 1};    
int sum = 0;

for (int i = 0; i < sizeof(x) / sizeof(int); i++)
    sum += x[i];

// sum == 6

However if you pass the array as a function parameter you will encounter pointer decay. This means that the array size information is lost and you get the pointer size instead, which is the behavior that you described.

int sum(int arr[])  // arr decays to int*
{
    int sum = 0;

    for (int i = 0; i < sizeof(arr) / sizeof(int); i++)
        sum += arr[i];

    return sum;
}

int main() 
{   
    int x[] = {1, 1, 1, 1, 1, 1};    
    return sum(x);  // will return 1 or 2, depending on architecture
}

You can still get the array size in the function if you use a template function for it.

#include <cstddef>

template <std::size_t N>
int sum(int (&arr)[N])
{
    int sum = 0;

    for (int i = 0; i < N; i++)
        sum += arr[i];

    return sum;
}

int main() 
{   
    int x[] = {1, 1, 1, 1, 1, 1};    
    return sum(x);  // will return 6
}