Thread: Fairly Simple Char Array Question

I'm not understanding why the string provided for initialization is too long?

Code:

int main()
{
   char array[1] = "A";


   return 0;
}

So when I have a string in quotes, there is an implied null byte at the end?

Yes, at least for an initialization of a char array, since that is the definition of a C-string (char array terminated with a null character). However, I'm not a C++ person, so I'm not qualified to speak about C++ strings in general.

See for yourself:

Code:

#include <iostream>

int main()
{
   char array[2] = "A";

   if(array[1] == '\0')
        std::cout << "null char\n";

   return 0;
}

Originally Posted by Matticus

Yes, at least for an initialization of a char array, since that is the definition of a C-string (char array terminated with a null character). However, I'm not a C++ person, so I'm not qualified to speak about C++ strings in general.

String literals in C++ (like "A") are specified in the same way as string literals in C.

Well, there is a slight difference in that in C, the string literal would not be too long, but rather array would be initialised such that array[0] would have the value of 'A', since the null character implied in the string literal is only used for initialisation if there is sufficient space, unlike in C++.

In other words, this is legal in C:

char array[1] = "A";

but not in C++.
But beware that if it was legal, you no longer have a C-style string, so attempts to use this as a string (using string functions such as strcpy, stcat, etc, or even initializing a std::string with it, e.g. std::string s(array)) will result in undefined behaviour.
There are ways around it, but the point is be careful. If at all possible, don't do it at all.