Purpose of returning by const value?

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What are the use cases for having a function return by const value for non-builtin type? 4 answers

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In the hypothetical situation where you could perform a potentially expensive non-const operation on an object, returning by const-value prevents you from accidentally calling this operation on a temporary. Imagine that + returned a non-const value, and you could write:

(a + b).expensive();

In the age of C++11, however, it is strongly advised to return values as non-const so that you can take full advantage of rvalue references, which only make sense on non-constant rvalues.

In summary, there is a rationale for this practice, but it is essentially obsolete.

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It's pretty pointless to return a const value from a function.

It's difficult to get it to have any effect on your code:

const int foo() {
   return 3;
}

int main() {
   int x = foo();  // copies happily
   x = 4;
}

and:

const int foo() {
   return 3;
}

int main() {
   foo() = 4;  // not valid anyway for built-in types
}

// error: lvalue required as left operand of assignment

Though you can notice if the return type is a user-defined type:

struct T {};

const T foo() {
   return T();
}

int main() {
   foo() = T();
}

// error: passing ‘const T’ as ‘this’ argument of ‘T& T::operator=(const T&)’ discards qualifiers

it's questionable whether this is of any benefit to anyone.

Returning a reference is different, but unless Object is some template parameter, you're not doing that.

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It makes sure that the returned object (which is an RValue at that point) can't be modified. This makes sure the user can't do thinks like this:

myFunc() = Object(...);

That would work nicely if myFunc returned by reference, but is almost certainly a bug when returned by value (and probably won't be caught by the compiler). Of course in C++11 with its rvalues this convention doesn't make as much sense as it did earlier, since a const object can't be moved from, so this can have pretty heavy effects on performance.

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