Why size of c++ dynamic array is not changed?

Question

Hi guys I tested c++ dynamic array and I changed 2 inputs every time. but array size is not changed. Why?

#include <iostream>

using namespace std;

int main()
{
int r = 0;
int c = 0;
int cnt = 1;

cin >> r;
cin >> c;

int** arr1 = new int* [r];


for (int i = 0; i < r; i++)
{
    arr1[i] = new int[c];
}

cout << sizeof(arr1) << endl;
cout << sizeof(arr1[0]);
}

I knew that If I entered two value 3 and 4 then results are 3 and 4

but the results are 4 and 4

`sizeof` doesn't do what you think it does. – Sam Varshavchik Jun 28 '20 at 20:31 — Sam Varshavchik, Jun 28 '20 at 20:31
Unrelated: You should probably use a `std::vector` instead. – Ted Lyngmo Jun 28 '20 at 20:31 — Ted Lyngmo, Jun 28 '20 at 20:31
Note: variable length array is not C++ standard. – Louis Go Jun 29 '20 at 00:58 — Louis Go, Jun 29 '20 at 00:58
Oh .. That is interesting. Thank you guys – Hamp Jun 29 '20 at 02:44 — Hamp, Jun 29 '20 at 02:44

score 2 · Accepted Answer · answered Jun 28 '20 at 20:29

2

You are testing the size of a pointer which is always 4 on 32 bit. That's because sizeof is a compile time thing; it cannot determine an array size.

Generally arrays do not carry size information, that is why functions like strlen need a null terminator.

Suggestion: use std::vector.

answered Jun 28 '20 at 20:29

Michael Chourdakis

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To add to this answer, you should debug your program put a break point after your for loop. Modern IDEs will let you see each variable being used and its values/e,element. While vector is a great option, because you have the sizes input, just keep those handy whenever you need the size of the array – arc-menace Jun 28 '20 at 20:33
@Michael Chourdakis Thank you Thank you so much – Hamp Jun 29 '20 at 02:47

Why size of c++ dynamic array is not changed?

1 Answers1