converting char array to string type

std:: string  str;
char  array[]  =   " Hello World " ;
for ( int  i  =   0 ; array[i]  !=   0 ; i ++ )
   str  +=  array[i];
// -----------------------------------
std:: string  str;
char  array[]  =   " Hello World " ;

str  =  array;

Use of NULL is discouraged in C++ because it can be redefined to be anything one wants -- c++ standards do not dictate what NULL should be.

The '\0' and 0 are one in the same thing. The compiler will translate '\0' to 0 during compilation.

All C-style strings are said to be NULL-terminated -- that definition is carry-over from C language. It really means that the end of the string is indicated by the byte which contains a 0.

you cannot assign C-strings that have enbedded 0s to std::string as I posted earlier. As you found out the assignment stops at the first 0. You could do it one character at a time, but then std::string is no longer an ascii string but a binary string, and most of the std::string methods cannot be used, again because of embedded 0s.

In this example, the output of the first cout is jest "Hello" because of the embedded 0.

#include  < string >
#include  < iostream >
using   namespace  std;

int  main()
{
     int  i;
     char  str[]  =   " Hello \0World " ;
     string  s  =  str;
    cout  <<  s  <<  endl;  <<  output  =   " Hello "
     int  sz  =   sizeof (str);
    s  =   "" ;
     for (i  =   0 ; i  <  sz; i ++ )
        s  +=  str[i];

    cout  <<  s  <<  endl;
  //  now assign characters one at a time
    sz  =  s.length();
     for (i  =   0 ; i  <  sz; i ++ )
        cout  <<  s[i];
    cout  <<  endl; output  =   " Hello World "  
     return   0 ;

} 

int  main()
{
     int  i;
     string  s;
     char  str[]  =   " Hello \0World " ;
     int  sz  =   sizeof (str);
    s.assign(str,sz);
    sz  =  s.length();
     for (i  =   0 ; i  <  sz; i ++ )
        cout  <<  s[i];
    cout  <<  endl;
     return   0 ;

}