Question

I have this code say:

 std::string str("ashish");  
 str.append("\0\0");  
 printf("%d", str.length());  

It is printing 6 but if I have this code

 std::string str("ashish");  
 str.append("\0\0",2);  
 printf("%d", str.length());  

it is printing 8 ! Why?

Answer 1

It’s because str.append("\0\0") uses the null character to determine the end of the string. So “\0\0” is length zero. The other overload, str.append("\0\0",2), just takes the length you give it, so it appends two characters.

From the standard:

  basic_string&
   append(const charT* s, size_type n);

7         Requires: s points to an array of at least n elements of charT.

8         Throws: length_error if size() + n > max_size().

9         Effects: The function replaces the string controlled by *this with a string of length size() + n whose first size() elements are a copy of the original string controlled by *this and whose remaining elements are a copy of the initial n elements of s.

10         Returns: *this.

  basic_string& append(const charT* s);

11         Requires: s points to an array of at least traits::length(s) + 1 elements of charT.

12         Effects: Calls append(s, traits::length(s)).

13         Returns: *this.

                                                               — [string::append] 21.4.6.2 p7-13