Assign value in ternary operator

Question

When using std::weak_ptr, it is best practice to access the corresponding std::shared_ptr with the lock() method, as so:

std::weak_ptr<std::string> w; std::shared_ptr<std::string> s = std::make_shared<std::string>("test"); w = s; if (auto p = w.lock()) std::cout << *p << "\n"; else std::cout << "Empty";

If I wanted to use the ternary operator to short hand this, it would seem that this:

std::cout << (auto p = w.lock()) ? *p : "Empty";

would be valid code, but this does not compile.

Is it possible to use this approach with the ternary operator?

All parts of a conditional expression are in turn expressions. You can't have definition of variables in expressions. — Some programmer dude
– Some programmer dude, Commented Oct 1, 2021 at 22:44
Somewhat related: what's an expression and expression statement in c++? and Expression Versus Statement. Declarations are not expressions in C++, so it won't be possible to declare p inside of the first expression of ?:. However, assignments are expressions, so something like (p = w.lock()) ? *p : "Empty" would be valid, assuming p was already declared. — Brian61354270
– Brian61354270, Commented Oct 1, 2021 at 22:49
You could declare p outside the ternary and then assign it inside, but then you can't use the placeholder auto. It'd be std::shared_ptr<std::string> p; std::cout << (p = w.lock()) ? *p : "Empty"; — Nate Eldredge
– Nate Eldredge, Commented Oct 1, 2021 at 23:34
@NateEldredge you might not be able to use auto in that case, but you can still deduce the shared_ptr type from the weak_ptr, by using std::shared_ptr<decltype(w)::element_type> p;, or even simpler decltype(w.lock()) p; — Remy Lebeau
– Remy Lebeau, Commented Oct 2, 2021 at 15:21

eerorika · Accepted Answer · 2021-10-01 23:36:29Z

auto p = w.lock() is not an assignment. It's a declaration of a variable. You can declare a variable in the condition of an if statement, but you cannot declare variables within a conditional expression.

You can write:

auto p = w.lock(); std::cout << ( p ? *p : "Empty" );

Davis Herring · Accepted Answer · 2021-10-02 16:38:45Z

If you want to introduce a variable for an expression (like let in Lisp or Haskell), you can use a lambda:

std::cout << [p = w.lock()] {return p ? *p : "Empty";}();

This also confines the ?:, which interprets std::cout as part of the condition in your question.

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