array of characters: it doesn't print the exact size!

[soul111]

in this code, the output is 6 instead of 5, i didn't get it!
the condition is to not include '\0', is the compiler counting it ?

#include <iostream>
using namespace std;
int main(){
char a[5]={'1','2','3','4','5'};
int c=0;
while(a[c]!='\0')
    c++;
cout<<c<<endl;
}

[Peter87]

How do you expect the loop to find a '\0' when the array doesn't contain such a character.

[soul111]

How do you expect the loop to find a '\0' when the array doesn't contain such a character.

\0 is exist in the array, it's the termination character.

if i was wrong, can you explain please ?

[jlb]

\0 is exist in the array, it's the termination character.

No it doesn't exist in the array. There is no automatic termination character for character arrays.

[Ganado]

...and to clarify further, you're thinking of string literals. They will put a null terminator on the end. But you have to give it enough room (the actual length of the string + 1).

char a[6] = "hello";
// a --> { 'h', 'e', 'l', 'l', 'o', '\0' } 

You're also trying to implement strlen, it seems.
http://www.cplusplus.com/reference/cstring/strlen/

Oh yeah, and... since you're using C++, you should use std::string objects instead.

http://www.cplusplus.com/reference/string/string/length/

// string::length
#include <iostream>
#include <string>

int main ()
{
  std::string str ("Test string");
  std::cout << "The size of str is " << str.length() << " bytes.\n";
  return 0;
}