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Home/ Questions/Q 774025
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T19:03:45+00:00 2026-05-14T19:03:45+00:00

I have an selfmade Stringclass: //String.h String & operator = (const String &); String

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I have an selfmade Stringclass:

//String.h
String & operator = (const String &);
String & operator = (char*);
const String operator+ (String& s);
const String operator+ (char* sA);
.
.

//in main:
String s1("hi");
String s2("hello");

str2 = str1 + "ok";//this is ok to do 
str2 = "ok" + str1;//but not this way

//Shouldn't it automatically detect that one argument is a string and in both cases?
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  1. Editorial Team
    Editorial Team
    2026-05-14T19:03:45+00:00Added an answer on May 14, 2026 at 7:03 pm

    The + operator should not be a member function, but a free function, so that conversions can be performed on either of its operands. The easiest way to do this is to write operator += as a member and then use it to implement the free function for operator +. Something like:

    String operator +( const String & s1, const String & s2 ) {
        String result( s1 );
        return result += s2;
    }
    

    As others have suggested, you can overload for const char * for possible efficiency reasons, but the single function above is all you actually need.

    Please note that your code as it stands should give an error for:

    String s1("hi");
    String s2("hello");
    str2 = str1 + "ok";    // not OK!!! 
    

    something like:

    warning: deprecated conversion from string constant to 'char*'
    

    as the string literal (constant) “ok” is a const char *, not a char *. If your compiler does not give this warning, you should seriously think about upgrading it.

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