Sizeof operator on string array is giving different output in C++

Question

I am trying to compile the following code:

#include <iostream>
using namespace std;
void show1(string text1[]) {

    cout << "Size of array text1 in show1: " << sizeof(text1) << endl;
}
int main() {
    string text1[] = {"apple","melon","pineapple"};
    cout << "Size of array text1: " << sizeof(text1) << endl;
    cout << "Size of string in the compiler: " << sizeof(string) << endl;
    show1(text1);
    return 0;
}

And the output is shown below:

Size of array text1: 96
Size of string in the compiler: 32
Size of array text1 in show1: 8

I am not able to understand, why is the sizeof operator working on the same array giving two different outputs at two different points? Please explain.

Does this answer your question? When a function has a specific-size array parameter, why is it replaced with a pointer? — Daniel Langr
– Daniel Langr, Commented Jun 6, 2020 at 16:39
When you declare arguments, arrays are really pointers. So the argument text1 for your show1 function is really declared as string* text1. And getting the sizeof of a pointer is the size of the pointer itself, not what it might point to. Use std::vector or std::array instead. — Some programmer dude
– Some programmer dude, Commented Jun 6, 2020 at 16:39
Reminder: sizeof(string) is the size of the std::string structure, not the size of the text in the string. See also std::string::length(). — Thomas Matthews
– Thomas Matthews, Commented Jun 6, 2020 at 16:43

TonySalimi · Accepted Answer · 2020-06-07 07:13:36Z

3

The sizeof() operator returns the compile time size of the objects. It means that if your type allocates a memory chunk at run time from heap, that memory is not taken into account by sizeof().

For your first case, i.e.

 string text1[] = {"apple","melon","pineapple"};

You have an array of 3 strings, so sizeof should return 3*sizeof(std::string). (3*32 = 96 in your case)

For your second case:

sizeof(string)

It should simply print the size of an string. (32 in your case).

Finally for your last case, do not forget that arrays are passed using a pointer in C/C++. So, your parameter is simply a pointer and sizeof() should print the size of a pointer on your machine.

Edit: As @ThomasMatthews has mentioned in the comments, if you are interested in getting the real size of an string (i.e. the number of characters inside it), you can use std::string::length() or std::string::size().

edited Jun 7, 2020 at 7:13

answered Jun 6, 2020 at 17:08

TonySalimi

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1 Comment

Ruslan Over a year ago

sizeof is not a function, it's an operator. And the parentheses are only necessary when its operand is a type, which makes it even less of a function.

GobeRadJem32 · Accepted Answer · 2020-06-07 12:18:31Z

0

Try with member function 'size'.

Write this code:

#include <iostream>
using namespace std;
void show1(string text1[]) 
{
    cout << "Size of array text1 in show1: " << text1->size() << endl;
}

int main() 
{
    string text1[] = {"apple","melon","pineapple"};
    cout << "Size of array text1: " << text1->size() << endl;
    cout << "Size of string in the compiler: " << sizeof(string) << endl;
    show1(text1);
    return 0;
}

Description:

std::vector has a member function size(). And std::string too. In std::vector return size of vector(all elements). In std::string returns all elements in array.

edited Jun 7, 2020 at 12:18

answered Jun 7, 2020 at 10:25

GobeRadJem32

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3 Comments

Blastfurnace Over a year ago

text1-size() is a syntax error. If you change it to text1->size() it will compile but it doesn't show the size of the array. Do you post wrong answers on purpose?

Blastfurnace Over a year ago

Once again, text1->size() doesn't show the size of the array. An array passed to a function decays to a pointer. You are printing the size of the first std::string in the array.

GobeRadJem32 Over a year ago

Sorry, I don't know. Tkanks for this noftification.

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