sizeof char* array in C/C++

Question

There are plenty of similar inquires, but in my case I don't understand what isn't working:

int mysize = 0;
mysize = sizeof(samplestring) / sizeof(*samplestring);
std::cout << mysize << '\n' << samplestring;

This outputs:

4

Press 'q' to quit.

How is it possible? 4 definitely isn't the size of this string. I even tried the following, with the same result:

mysize = sizeof(samplestring) / sizeof(samplestring[0]);

EDIT: Ok, this is the declaration:

char *samplestring = "Start."; 

I'm on C++, but I need to use functions that only accept char *. Later in the code I assign new strings to that variable, like:

samplestring = "Press 'r' for red text.";

Yes, the compiler gives me warnings, but I have no idea how can I use different strings if I can't overwrite them...

Answer 1

4 isn't the size of the string, because samplestring isn't a string. It's a char*, whose size is (on your platform) 4, divided by 1 (size of char) is, correctly, 4.

In C++, you'd use std::string and the length() method.

In C, you'd use strlen which takes as parameter a NULL-terminated char pointer.

Answer 2

First of all, sizeof(samplestring[0]) is the same as sizeof(*samplestring), they're both returning the size of the first element of the samplestring array. And, assuming samplestring is an array of chars, sizeof(char) is defined to be 1.

You haven't shown how samplestring is declared. It could be one of the following:

char const *samplestring = "Hello, World!";

or

char *samplestring = malloc( ... );

or

char samplestring[10];

In the first 2 cases the type of samplestring is char *, so sizeof(samplestring) returns sizeof(char *), which, on your platform is 4.

In the third case, the type of samplestring is char[10] (array of 10 chars), but if you call a function that takes a char * as its parameter, the char array will decay to a pointer pointing to the first element of the array. In this case, trying to print sizeof within the function will still result in the size of a pointer being printed.

If you want the size of the original array to be printed from within the function, then the function parameter needs to be a pointer to the original array type (and the type includes size of the original array).

#include <stdio.h>

void foo( char (*arr)[42] )
{
  printf( "%u", (unsigned)sizeof(*arr) );
}

int main()
{
  char arr[42];
  foo( &arr );

  return 0;
}  

Output:

42

This fixes the size of the array that can be passed to the function and is not desirable in a lot of cases. The only other solution is to keep track of the array yourself (or use strlen if you have a NULL terminated string).