CODEWITHSUNDEEP
c++if-statementcomma-operator
Boris Month
Boris Month

Reputation: 463

Using the comma operator in if statements

I tried the following:

if(int i=6+4==10)
    cout << "works!" << i;

if(int i=6+4,i==10)
    cout << "doesn't even compile" << i;

The first works fine while the second doesn't compile. Why is this?

EDIT: Now I know that the first one may not work as I intend it to. The value of i inside the if scope will be 1, not 10. (as pointed out by one of the comments on this question).

So is there a way to initialize and use a variable inside of an if statement at the same time similar to for(int i=0;i<10;i++)? So that you could produce something like if((int i=6+4)==10) (which will not compile) where the value of I inside the if scope would be 10? I know you could declare and initialize I before the if statement but is there a way to do this within the statement itself?

To give you an idea why I think this would be usefull.

 if(int v1=someObject1.getValue(), int v2=someObject2.getValue(), v1!=v2)
    {
        //v1 and v2 are visible in this scope 
        //and can be used for further calculation without the need to call
        //someObject1.getValue() und someObject2.getValue() again.
    }
    //if v1==v2 there is nothing to be done which is why v1 und v2
    //only need to be visible in the scope of the if.

Upvotes: 1

Views: 2883

Answers (4)

fwyzard
fwyzard

Reputation: 2687

As of C++17 what you were trying to do is finally possible:

if (int i=6+4; i==10)
    cout << "works, and i is " << i << endl;

Note the use of ; of instead of , to separate the declaration and the actual condition.

Upvotes: 0

CB Bailey
CB Bailey

Reputation: 792229

The expression used as an initializer expression must be an assignment-expression so if you want to use a comma operator you must parenthesize the initializer.

E.g. (not that what you are attempting makes much sense as 6 + 4 has no side effects and the value is discarded and i == 10 uses the uninitialized value of i in its own initializer.)

if (int i = (6 + 4, i == 10)) // behaviour is undefined

Did you really mean something like this?

int i = 6 + 4;
if (i == 10)

When using the form of if that declares a new variable the condition checked is always the value of the initialized variable converted to bool. If you want the condition to be an expression involving the new variable you must declare the variable before the if statement and use the expression that you want to test as the condition.

E.g.

int i;
if ((i = 6 + 4) == 10)

Upvotes: 5

David Rodr&#237;guez - dribeas
David Rodr&#237;guez - dribeas

Reputation: 208373

There are different alternatives, because what you want cannot be done (you cannot mix the comma operator with declarations). You could, for example, declare the variable outside of the if condition:

int i = 6+4;
if ( i == 10 ) ...

Or you can change the value of i to be 0 instead of 10 and recalculate i inside the else block:

if ( int i = (6+4)-10 ) ; else {
   i += 10;
   // ...
}

Much simpler, don't declare the variable at all, since you know the value inside the loop:

if ( (6+4)==10 ) {
   int i = 10;
   // ...
}

Unless of course you need the value of i in the case where it is not 10, in which case the second option is the most appropriate.

Upvotes: 0

Charlie Martin
Charlie Martin

Reputation: 112376

I doubt seriously either example works to do anything useful. All that it does is evaluate to "true" in a complicated fashions.

But the reason the second one doesn't compile is that it's interpreted as two declarations: int i = 6+4; int i==10 and int i==10 isn't valid because that's an equality operator, not an assignment.

Upvotes: 3

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