difference between bracket operator [] and its constant version

[puzzlecracker]

In the following example,

String s;

std::cout<<s[3];
s[3]='x';


How does a compiler know which one to call? Does it try to call const
first, and if it fails it rolls back to non-const one. As is
cout<<s[3], How does it know which one to call?

Thanks

 

[Alan Woodland]

In the following example,

String s;

std::cout<<s[3];
s[3]='x';


How does a compiler know which one to call? Does it try to call const
first, and if it fails it rolls back to non-const one. As is
cout<<s[3], How does it know which one to call?

Both of those cases will call the non-const version of operator[] in
your 'String' class. (Assuming such a thing exits in this 'String' class).

to call the const one explicitly:

const String &cs = s;
std::cout << cs[3]; //can only match the const one.

Alan