c++: Multiple assignment inside if block syntax for short circuiting

Question

I wanted to be able to do multiple assignments in an if block and short-circuit if the first one fails. However, this does not compile and says, expected primary-expression before ‘auto’

#include <iostream>
#include <optional>

std::optional<int> foo()
{
    return 0;
}

int main() {
    if (auto a = foo() && auto b = foo())
    {
        std::cout << "a = " << *a << ", b = " << *b << std::endl;
    }
}

The following works though and does what I want.

    if (auto a = foo())
    {
      if (auto b = foo()) {
        std::cout << "a = " << *a << ", b = " << *b << std::endl;
      }
    }

But is there a way for me to use the syntax in the first one? Using parenthesis to surround the expressions does not work.

According to http://jatinganhotra.com/blog/2013/02/09/c-plus-plus-variable-declaration-in-if-expression/ the standard says we can only do one assignment in an expression — Collin, Jun 13 '19 at 16:21
Just saw that there is an excellent [answer](https://stackoverflow.com/a/7837092/9199167) in that other question, which addresses your issue as well. — Max Vollmer, Jun 13 '19 at 16:22
Thank you. I didn't come across the fact that only a single variable declaration is allowed. Thank you. — Mochan, Jun 13 '19 at 16:23
@Mochan You can actually declare more than one variable now with C++17. Those answers are outdated. — NathanOliver, Jun 13 '19 at 16:24

Bob__ · Accepted Answer · 2019-06-13T16:59:11.767

6

Since C++17, you could write

if (decltype(foo()) a, b; (a = foo()) && (b = foo()))
{
    std::cout << "a = " << *a << ", b = " << *b << std::endl;
}

But if you want to see the short circuit in action, you should change the condition:

#include <iostream>
#include <optional>

std::optional<int> foo()
{
    std::cout << "foo called.\n";
    return 1;
}

int main()
{
    if (decltype(foo()) a, b; (a = foo()) or (b = foo()))
    {
        // For exposition only            ^^
        std::cout << "a = " << *a << ", b = " << *b << std::endl;
        // Note that, now, this is UB            ^^
    }
}

Compare the output of this

foo called.
a = 1, b = 0

Versus the former.

edited Jun 13 '19 at 16:59

answered Jun 13 '19 at 16:19

Bob__

12,361
3
28
42

Clever idea to use `decltype()` instead of `auto` . – πάντα ῥεῖ Jun 13 '19 at 16:45
1

Do note that `*b` is UB since `b` doesn't have a value. You'll need to guard against uninitialized variables in the body, or assign them a default value to begin with. – NathanOliver Jun 13 '19 at 16:53
@NathanOliver Yes, the OP correctly used `&&`. The `||` version is only to trigger the short circuit. – Bob__ Jun 13 '19 at 16:55

Max Vollmer · Answer 2 · 2019-06-13T16:29:39.790

2

It has nothing to do with auto. Declarations cannot be part of an expression. You can either use an expression or a single declaration inside an if statement.

So even this will not compile:

#include <iostream>

int foo()
{
    return 0;
}

int main()
{
    if (int a = foo() && int b = foo())
    {
        std::cout << "a = " << a << ", b = " << b << std::endl;
    }
    return 0;
}

Edit: In C++17 you can have an initializer statement before the condition statement, like so: if (auto a = foo(), b = foo(); a && b). However that will not give you short circuiting.

edited Jun 13 '19 at 16:29

answered Jun 13 '19 at 16:19

Max Vollmer

8,412
9
28
43

2

You can have more than a single declaration now. `if (auto a = foo(), b = foo(); a && b)` is legal in C++17, but it doesn't stop `b` from being declared so `foo` is called twice. Might be OK for the OP though. – NathanOliver Jun 13 '19 at 16:26
@NathanOliver That works for me in gcc 8. Thank you. – Mochan Jun 13 '19 at 16:29
1

@NathanOliver Thanks, I edited that in. Doesn't give OP the desired short circuiting, though. – Max Vollmer Jun 13 '19 at 16:30
What if a and b do not have the same type? – Mochan Jun 13 '19 at 16:30
@MaxVollmer I didn't realize I don't get short-circuiting. – Mochan Jun 13 '19 at 16:31
1

@Mochan They have to have the same type, or the same type but with reference or pointer qualifications. It is the same thing as the init part of a for loop. – NathanOliver Jun 13 '19 at 16:32

πάντα ῥεῖ · Answer 3 · 2019-06-13T16:40:27.973

The only way I see is that you reformulate (without declaring the variables inside the if condition):

#include <iostream>
#include <optional>

std::optional<int> foo()
{
    return 0;
}

int main() {
    std::optional<int> a , b;
    if ((a = foo()) && (b = foo()))
    {
        std::cout << "a = " << *a << ", b = " << *b << std::endl;
    }
}

See a working example here please.

Since c++17 you can even do the variable declarations inside the if condition like this:

if (std::optional<int> a , b; (a = foo()) && (b = foo()))

Though usage of auto isn't possible in that case.

Another live demo.

c++: Multiple assignment inside if block syntax for short circuiting

3 Answers3