Why does it receive only 1 integer?

Question

Why does this receive only one integer?
Here is the code:

#include <iostream>

int main () {
    int num1,num2,num3;
    std::cin>>num1,num2,num3;

    return 0;
}

You should read about comma operator. You want `std::cin >> num1 >> num2 >> num3`. — HolyBlackCat, Jan 18 '18 at 02:00
Because that's what you asked for.`std::cin>>num1>>num2>>num3;` — Retired Ninja, Jan 18 '18 at 02:00
By the way, compiling with `-Wall` would warn about things like this. — Silvio Mayolo, Jan 18 '18 at 02:01
What this comma means so? and what does it do with num2,num3? — Mohamed Magdy, Jan 18 '18 at 02:01
@HolyBlackCat Can you recommend me a good book as a beginner? — Mohamed Magdy, Jan 18 '18 at 02:05
A lot of books don't cover the comma operator. It's seldom used and often considered a design mistake of the language. — Silvio Mayolo, Jan 18 '18 at 02:06
@SilvioMayolo I don't need a book which covers commas, I need a book which builds me as a beginner. — Mohamed Magdy, Jan 18 '18 at 02:09
@MohamedMagdy [The Definitive C++ Book Guide and List](https://stackoverflow.com/q/388242/3309790) — songyuanyao, Jan 18 '18 at 02:13

songyuanyao · Accepted Answer · 2018-01-18T02:09:25.840

According to the Operator Precedence, comma operator has lower precedence than operator>>, so std::cin>>num1,num2,num3; is same as (std::cin>>num1), num2, num3;; the following num2, num3 does nothing in fact. (More precisely, std::cin>>num1 is evaluated firstly and its result is discarded; then num2 is evaluated, num3 is evaluated at last and its value is the result of the whole comma expression.)

What you want should be std::cin >> num1 >> num2 >> num3;.

score 0 · Answer 2 · answered Jan 18 '18 at 02:01

0

That is not the correct syntax. That's an application of the comma operator. You want

std::cin >> num1 >> num2 >> num3;

answered Jan 18 '18 at 02:01

Silvio Mayolo

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Why does it receive only 1 integer?

2 Answers2