Comma-separated lines of code inside braceless loop [duplicate]

General Tech Bugs & Fixes . 2 years ago

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Posted on 16 Aug 2022, this text provides information on Bugs & Fixes related to General Tech. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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manpreet Tuteehub forum best answer Best Answer 2 years ago

 

This question already has an answer here:

I came across this block of code when studying for my class. It was in my professor's presentation and I thought it was an error, so out of pure curiosity, I tried to compile it. To my surprise, it's actually a valid code. I tried looking it up online but couldn't find a clear answer as to why exactly does it work.

int n, num;
scanf("%d", &n);
vector<int> arr;
arr.reserve(n);

while (n--)
        scanf("%d", &num),
        arr.push_back(num);

for (int i : arr) printf("%d ", i);
printf("\n");

I expected the while-loop to cause a compilation error, but it runs smoothly. Is it standard or was that introduced recently?

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manpreet 2 years ago

This is the comma operator at work. It evaluates the left hand side of the operator (scanf("%d", &num)), then the right hand side (arr.push_back(num)) and returns the result of this expression.

I don't think the code snippet is a good usage scenario for the comma operator. It seems to be employed for not typing {} to create a body for the while loop, which is lazy and unnecessarily "smart". But it works.

If you're interested in legitimate use cases for the comma operator, have a look at this article.


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