Re: Implicit type conversion between strings and char *

From:

"Victor Bazarov" <v.Abazarov@comAcast.net>

Newsgroups:

comp.lang.c++

Date:

Fri, 22 Jun 2007 17:47:02 -0400

Message-ID:

<f5hg0n$dab$1@news.datemas.de>

utab wrote:

I was trying to write a more complex program, and while searching for
sth in my reference C++ primer, by Lippman. I had a question in the
mind, see the code below

#include <string>
#include <iostream>
#include <cstring>
int main(){
const char *lit="hello";
std::string str("hello"); // implicit conversion from const char*
to string,right?
if(str=="hello") // implicit conversion from const
char* to string, right?
std::cout << "the same";
else
std::cout<< "different";
return 0;
}

To be able to compare, one type should be converted to the other. I
could not find the answer in the book. Maybe I missed :-)

It depends on the type. Here are two examples:

    struct HasConversion {
        HasConversion(int) {}
        bool operator ==(HasConvesion const&) { return false; }
    };

    struct HasComparisonInstead {
        bool operator ==(int) { return false; }
    };

    int main() {
        HasConversion hc(42);
        hc == 666;
        HasComparisonInstead hci;
        hci == 42;
    }

In fact, there is an operator == defined for comparing 'std::string'
with a char const*. So, no implicit conversion should happen in the
latter case (from your posting). As for the creation of the string
in the former case, the conversion is not "implicit". It's just
parameterised construction.

V
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