CODEWITHSUNDEEP
c++comma-operator
Saurabh Raje
Saurabh Raje

Reputation: 77

Comma operator used in assignment

Consider the code:

int i, j;
i = (i = 1, j = 3);
i++;
cout<<i<<" "<<j;

This printed 4 3 (c++14).

I read that the comma operator evaluates the expression on its left and returns the one on its right. Now if that is correct, I want to know what is the return value of j = 3? Is it the rvalue? Or a reference to the lvalue?

How does this actually work?

Upvotes: 0

Views: 177

Answers (3)

user31264
user31264

Reputation: 6737

To calculate (i=1, j=3), it calculates, from left to right, the expressions separated by comma, and returns the value of the last (rightmost) expression. So it calculates i=1 (i becomes 1), then it calculates j=3 (j becomes 3), then returns 3.

After calculating (i=1, j=3), which returned 3, it performs the assignment, which sets i to 3.

Then i++ is calculated, which sets i to 4.

Then i and j are printed.

Upvotes: 2

flogram_dev
flogram_dev

Reputation: 42888

I want to know what is the return value of j = 3?

Assignment operations* return (or "evaluate to") a reference to the left-hand side of the operation, in this case j.

So i = (i = 1, j = 3); is identical to:

i = 1;
j = 3;
i = j;

*For built-in types that is. Custom operator= overloads may return whatever they want, although it's recommended to return a reference to *this as it is the expected behavior among C++ programmers.

Upvotes: 2

Barry
Barry

Reputation: 304132

The assignment operator returns an lvalue referring to the left operand. It groups right-to-left. ([expr.ass]). Note that saying returning a reference to the lvalue doesn't make sense - it either returns an lvalue or not.

The comma operator performs the value computations and side effects from the left operand, discards them, and then does the same for the right operand. ([expr.comma])

So refactoring the comma operator would produce the following equivalent code:

i = 1;      // left operand, value discarded
i = j = 3;  // right operand, value preserved
i++;

and then refactoring the compound assignment would produce the following still equivalent code:

i = 1;
j = 3;     // rightmost assignment
i = j;     // leftmost assignment
i++;

Upvotes: 0

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