Why do sizeof statsements produce different results?

Question

Possible Duplicate:
Calculating size of an array
Sizeof an array in the C programming language?

Why does sizeof(arr)/(double)sizeof(arr[0]) work and sizeof(arr+0)/(double)sizeof(arr[0]) not work?

And why does sizeof(arr)/(double)sizeof(arr[0]) also not work when arr is passed to a function? Here is the full code example:

#include <iostream>

using namespace std;

int givesize(int arr[])
{
   cout<<sizeof(arr)/(double)sizeof(arr[0])<<"\n";
}

int main()
{
   int arr[] = {1,2,3,4,5};
   cout<<sizeof(arr)/(double)sizeof(arr[0])<<"\n";
   cout<<sizeof(arr+2)/(double)sizeof(arr[0])<<"\n";
   givesize(arr);
   return 0;
}

Output 5 1 1

See http://stackoverflow.com/questions/720077/calculating-size-of-an-array, and hundreds of others — Mat, Jul 07 '12 at 15:36

score 4 · Answer 1 · edited May 23 '17 at 10:09

4

This happens because arrays decay to pointers when they are passed to functions that take pointers as arguments.

Inside your main, the

int arr[] = {1,2,3,4,5};

declaration is equivalent to

int arr[5] = {1,2,3,4,5};

because the compiler has enough information to calculate the 5 from the aggregate initializer. In the function header, however, int arr[] does not mean the same thing: there is no context around it to tell the compiler that it's anything but a pointer to an integer.

edited May 23 '17 at 10:09

Community

1
1

answered Jul 07 '12 at 15:35

Sergey Kalinichenko

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1

Also the same thing happens with `arr+2`. – casablanca Jul 07 '12 at 15:36
1

Not just any functions, only functions that take pointers. – Benjamin Lindley Jul 07 '12 at 15:36
@BenjaminLindley That's an excellent point, thank you very much! – Sergey Kalinichenko Jul 07 '12 at 15:39
To the OP: And then just in case you get confused by why the same doesn't apply to sizeof (like I was earlier this week) this one might be useful later: http://stackoverflow.com/questions/11283361/why-does-decay-to-pointer-for-array-argument-appear-not-to-apply-to-sizeof – mathematician1975 Jul 07 '12 at 15:58

score 2 · Answer 2 · answered Jul 07 '12 at 15:41

int arr[] in a function argument list is the same as int* arr. Which means that in:

int givesize(int arr[])
{
   cout<<sizeof(arr)/(double)sizeof(arr[0])<<"\n";
}

sizeof(arr)/(double)sizeof(arr[0]) is actually sizeof(int*) / (double)sizeof(int). As it outputs 1 that implies a 32-bit platform.

int arr[] as a local variable is not the same as int* arr, hence sizeof(arr)/(double)sizeof(arr[0]) is the same as sizeof(int[5]) / (double)sizeof(int) which evaluates to 5.

Next, sizeof(arr+2)/(double)sizeof(arr[0]) is again the same as sizeof(int*) / (double)sizeof(int), because the type of expression arr+2 is int*.

score 0 · Answer 3 · answered Jul 07 '12 at 15:39

0

Inside main, arr has type "array of five ints", but arr+0 has type "pointer to int". When you add zero to arr, you are treating it like a pointer, and as soon as you use an array like a pointer, it gets converted to a pointer.

answered Jul 07 '12 at 15:39

Vaughn Cato

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score 0 · Answer 4 · answered Jul 07 '12 at 15:45

Why does sizeof(arr)/(double)sizeof(arr[0]) work

Because sizeof(arr) returns the size of the entire array in bytes (let's say 20 assuming int is 32 bits), and sizeof(arr[0]) returns the size of the first int (which is 4 bytes). This will give us 5.

And why does sizeof(arr)/(double)sizeof(arr[0]) also not work when arr is passed to a function?

Because arrays decay into pointers. This means they lose an important piece of information: their size. We cannot directly calculate the size of an array just by a pointer. How would the function even know it is an array to begin with?

C++? Use a template function:

template <typename T, int N>
int getsize(const T (&arr)[N])
{
    return N;
}

Why do sizeof statsements produce different results?

4 Answers4