Question

Currently I’m trying to rewrite the += operator for a class I wrote called mystring:

MyString& operator+=(MyString& s1, const MyString& s2)
{

    int newStringLength = s1.length + s2.length;
    char* newStorage = new char[newStringLength +  1];

    strcpy(newStorage, s1.data);

    strcpy(newStorage + s1.length, s2.data);
    delete[] s1.data;

    s1.length = newStringLength;
    s1.data = newStorage;

    return s1;

}

MyString operator+(const MyString& s1, const MyString& s2)
{

    MyString temp;
    delete[] temp.data;

    temp.length = s1.length;
    temp.data = new char[temp.length+1];

    strcpy(temp.data, s1.data);
    temp+=s2;

    return temp;

}

Where length is the length of the string and data is a string stored in char * format.

The program works fine when I try to do something like:

MyString test1 = "hi";
MyString test2 = "to"; 

test1 += test2;

But does not work when I try something like:

   MyString test;
    MyString test1 = "hi";
    MyString test2 = "to"; 

    test += test2 + test1
          += "you";

Basically when I start mixing += and + in an alternating way it doesn’t work. Here is the error at compilation:

testoutput.cpp:26: error: no match for ‘operator+=’ in ‘operator+(const MyString&, const MyString&)(((const MyString&)((const MyString*)(& test1)))) += "you"’
mystring.h:45: note: candidates are: MyString& operator+=(MyString&, const MyString&)

Does anyone have any idea how I can change my code in order to achieve this functionality?

Answer 1

It doesn’t make sense to mix + and += in such a way. I’m not really sure what your intended behavior is, but if you want the nested += to apply to test1, you’ll have to use parenthesis:

test += test2 + (test1 += "you");

This is not a problem with your assignment operator, but with operator precedence in the language. If you would replace MyString with int you would run into the same problems.

The precedence and associativity of the + and += operators causes the expression without parenthesis to be interpreted like this:

test += ((test2 + test1) += "you");

This tries to assign to test2 + test1, but that’s not possible (you can only assign to variables). This operator precedence cannot be changed and without parenthesis the expression will always be interpreted this way.