strange behaviour using sizeof in C++

Question

I'm very new to C++. From examples, I find this use of sizeof in order to retrieve the length of an array

int main()
{
int newdens[10];
// the first line returns 10 which is correct
std::cout << "\nLength of array = " << (sizeof(v1)/sizeof(*v1)) << std::endl;
std::cout << "\nLength of array = " << (sizeof(v1) << std::endl; //returns 40
std::cout << "\nLength of array = " << (sizeof(*v1)) << std::endl; //returns 4
    }

But if i wrote a function like this

#include <iostream>

void myCounter(int v1[])
    {
    int L, L2, L3;
    L = (sizeof(v1)/sizeof(*v1)); 
    L2 = (sizeof(v1)); 
    L3 = (sizeof(*v1)); 
    std::cout << "\nLength of array = " << L << std::endl;
    std::cout << "\nLength of array = " << L2 << std::endl;
    std::cout << "\nLength of array = " << L3 << std::endl;

    }

int main()
{
int v1[10];

std::cout << "\nLength of array = " << (sizeof(v1)/sizeof(*v1)) << std::endl;
std::cout << "\nLength of array = " << (sizeof(v1)) << std::endl;
std::cout << "\nLength of array = " << (sizeof(*v1)) << std::endl;

myCounter(v1);
    }

the outputs are L=2, L2 = 8, L3 = 4. I can't understand where the problem is.

How to retrieve the correct lenght of v1 inside the function?

in `void myCounter(int v1[])`, `v1` has decayed to a pointer. All of its array-ness has been stripped away, so it no longer knows its length. — user4581301, Mar 20 '18 at 16:33
and sizeof(pointer) == 8 (generally), sizeof(int) == 4, so do the math... — papagaga, Mar 20 '18 at 16:33
you should take a look at `std::vector` which has a `size` method — 463035818_is_not_an_ai, Mar 20 '18 at 16:34
Other handy reading: [What is array decaying?](https://stackoverflow.com/questions/1461432/what-is-array-decaying) First answer explains quite well and gives a few avoidance techniques (which you may or may not be able to use). — user4581301, Mar 20 '18 at 16:36

Benjamin Barrois · Answer 1 · 2018-03-20T16:44:57.883

2

Your problem here is that sizeof() is resolved at compile time. As it has no information about how large is your array, it cannot tell its size. It interprets it as a pointer to an int, which is 64-bit on your machine.

The best way for you is to use std::vector instead of C-style array and use its method size().

edited Mar 20 '18 at 16:44

answered Mar 20 '18 at 16:35

Benjamin Barrois

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`sizeof` most definitely gives you the size of an array. The issues is in the function you no longer have an array. – NathanOliver Mar 20 '18 at 16:36
Yes you are right, I didn't know that actually! – Benjamin Barrois Mar 20 '18 at 16:38
Recommend an update to the answer to take that into account. No point leaving it mostly correct when it can be all correct. In addition, an answer should give a solution to the problem where possible. – user4581301 Mar 20 '18 at 16:40
The answer has been edited along your advice. – Benjamin Barrois Mar 20 '18 at 16:45

strange behaviour using sizeof in C++

1 Answers1