Dynamic array definition in C++

Hello-- could someone please explain to me why the following C++ code is printing 2 and not 5?

# include <iostream>
using namespace std;

int main(){

    int *foo;
    foo = new int[5];

    cout << sizeof(foo) / sizeof(foo[0]) << endl;

    delete[] foo;

    return 0;


Thank you!
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Because sizeof is not returning the length of the array but rather the size of the type:


In this case the size of a pointer (which is 8 on a 64-bit system) and the size of an integer (which is 4).

Just as addition: What you try only works with statically initialized arrays, i.e.:

    int foo[] = { 1, 2, 3, 4, 5 };
    cout << sizeof(foo) / sizeof(foo[0]) << endl;

This should output 5.
to add to above comments:

note, it is an inheritance of c that pointers to the first array element can be used like arrays but actually are bearing no information of the size of the array. things become worse if you pass an array to a function and the array turned to a pointer (silently).

int f(int array[10])   // the 10 was ignored and could (should) omitted
      return sizeof(array)/sizeof(array[0]));  // returns pointer size divided by int size

int arr[10] = 0´;
size_t num_elements = sizeof(arr)/sizeof(arr[0]);  // ok
size_t size = f(arr);   // size = 2

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though f explicitly used array syntax for the argument, the array was turned to pointer.

to come out of this, you have two choices:

1. use an array which is a member of a struct

struct Array10
     int arr[10];

int f(Array10 array)
       return sizeof(array.arr)/sizeof(array.arr[0]));  // returns 10

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2. use std::vector instead of c arrays
int f(std::vector<int>& array)
       return array.size();

std::vector<int> array(10);
int num = f(array); // returns 10

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note, the internal array of the std::vector is guaranteed to be a normal c array. so, you can use a std::vector<int> even if a function wants a c array:

void g(int arr[], int numitems)

std::vector<int> array(10);
g(&array[0], array.size());

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