Arrays,Pointers and sizeof() operator

[ZephyrTR]

Hi,

I'm trying to calculate length of different types of objects with sizeof(), particularly arrays and pointers.But I'm confused with the result of str1. I'm also not very sure if I can calculate the lengths of all types of array with the:

size_t size = sizeof(str1) / sizeof(*str1)

Any help will be appreciated.

#include <iostream>
using namespace std;

int main(){

        int arr1[]= {1, 2, 3, 4, 5};
        char * str1 = "This is a test for sizeof()";
        char str2[] = "This is a test for sizeof()";
        char str3[] = { 'C', '+' , '+' };

        cout << "Sizeof arr1 " << sizeof(arr1) << endl;
        cout << "Sizeof str1 " << sizeof(str1) << endl;
        cout << "Sizeof str2 " << sizeof(str2) << endl;
        cout << "Sizeof str3 " << sizeof(str3) << endl;

        cout << "Length of arr1 by formula " <<  sizeof(arr1) / sizeof(*arr1) << endl;
        cout << "Length of str1 by formula " <<  sizeof(str1) / sizeof(*str1) << endl;
        cout << "Length of str2 by formula " <<  sizeof(str2) / sizeof(*str2) << endl;
        cout << "Length of str3 by formula " <<  sizeof(str3) / sizeof(*str3) << endl;

Prints:

Sizeof arr1 20 // 4*5=20 OK
Sizeof str1 8 // ??
Sizeof str2 28 // 1*28=28 OK
Sizeof str3 3 // 1*3=3 OK
Length of arr1 by formula 5
Length of str1 by formula 8
Length of str2 by formula 28
Length of str3 by formula 3

[ZephyrTR]

Oh my god, I was messing around and the following code shocked me into death.

#include <iostream>
using namespace std;

void foe(int[]);

int main(){

        int myarray[] = {0,1,2,3,4,5}

        cout << sizeof(myarray) / sizeof(*myarray) << endl;
        foe(myarray);

        return 0;
}


void foe(int a[]){

        cout << sizeof(a) / sizeof(*a) << endl;

}

Result:

6
2

[RisteMK]

in you first post you missed '*' in this 'cout << "Sizeof str1 " << sizeof(str1) << endl;'
the right is: 'cout << "Sizeof str1 " << sizeof(*str1) << endl;'

[coder777]

ZephyrTR wrote:

Oh my god, I was messing around and the following code shocked me into death.

So why? Your pointers are obviously 64 bit -> 8 bytes. While int is 32 bit -> 4 bytes.

So sizeof(a) / sizeof(*a) -> 8 / 4 -> 2 while myarray has 6 elements. I'd say everything is ok.

[ZephyrTR]

@RisteMK

The following gives 1.

cout << "Sizeof str1 " << sizeof(*str1) << endl;

@coder777

My confusion is about the behaviour of sizeof().

When we pass an array name(which is also a pointer) to sizeof() then it gives the total size of array in bytes. On the other hand, when we pass a plain pointer it gives the address length. So I can conclude that it can understand whether we passed an array or a plain pointer.

When we pass an array to a function the the function cannot understand whether we passed an array. This is what I understand from the second example. Is that correct?

[amnuwan]

As coder777 said everything is very clear here ( ref to first post ).

Sizeof arr1 20 // 4*5=20 OK
Sizeof str1 8 // ?? -- This is OK, sizeof(char*) = 8 , in your machine
Sizeof str2 28 // 1*28=28 OK
Sizeof str3 3 // 1*3=3 OK
Length of arr1 by formula 5
Length of str1 by formula 8 -- This is OK too, this means 8/1 = 8
Length of str2 by formula 28
Length of str3 by formula 3

Answer for the 'Oh my god' post,

void foe(int a[])

Here variable 'a' is considered as type int*. Yes this is a big harm for the uniformness of C++ language. My personal opinion in creator could have forbidden using array types as function arguments so that users could use do it only in following form

void foe(int* a)

[CharlieWoods]

Just in case you're (still) wondering how to get the length of str1 to show 28, you should use strlen(str1)