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Home/ Questions/Q 6969761
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T16:37:20+00:00 2026-05-27T16:37:20+00:00

int a = 1; int b = (1,2,3); cout << a+b << endl; //

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int a = 1;
int b = (1,2,3);
cout << a+b << endl; // this prints 4
  1. Is (1,2,3) some sort of structure in c++ (some primitive type of list, maybe?)
  2. Why is b assigned the value 3? Does the compiler simply take the last value from the list?
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  1. Editorial Team
    Editorial Team
    2026-05-27T16:37:21+00:00Added an answer on May 27, 2026 at 4:37 pm

    Yes, that’s exactly it: the compiler takes the last value. That’s the comma operator, and it evaluates its operands left-to-right and returns the rightmost one. It also resolves left-to-right. Why anyone would write code like that, I have no idea 🙂

    So int b = (1, 2, 3) is equivalent to int b = 3. It’s not a primitive list of any kind, and the comma operator , is mostly used to evaluate multiple commands in the context of one expression, like a += 5, b += 4, c += 3, d += 2, e += 1, f for example.

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