Why copy constructor not invoked here

Question

stri(){}

stri(char *s);//constructor used to initilize object with constant string

stri(stri &s1);//copy constructor performs memberwise copy

friend stri operator+(stri &s1,stri &s2);//conccats two string objects

void operator=(stri &s);//performs memberwise copy

//In main
//s1 and s2 are initilized with constant strings
stri s3=s1+s2; //Gives error? However when copy constructor is removed works fine

Answer 1

You declared the copy constructor like this:

stri(stri &s1);

This line, specifically the expression on the right hand side of =, produces a temporary:

stri s3 = s1+s2;
       // ^^^^^ the result of this expression is a temporary

As this is copy initialization, it needs to call the copy constructor. But as temporaries cannot bind to references to non-const objects, you get an error.

When you comment out the copy constructor, the compiler generates one for you. Its signature is then

stri(stri const&);

Now it takes a reference to const and a temporary can bind to it. The fix should be obvious now.

Note that even though a well formed copy initialization requires an accesible copy constructor, the compiler can choose to elide the call to it during optimization, even when that elision changes the observable behavior of your program.

Answer 2

Let me dust off my crystal ball and guess that the error you're getting is somewhere along the lines of "temporary cannot bind to reference." That's because your copy constructor takes its parameter as stri & instead of const stri &; in other words, a reference to non-const. Such references cannot bind to temporaries. s1 + s2 returns a temporary, so the copy ctor invocation fails.

Unless you're doing really whacky stuff in your copy ctor (modifying the object copied from), change it to take its parameter as const stri &.