data type conversion in the expressions of C++

Asked Jan 04 '18 at 15:36

Active Jan 04 '18 at 15:55

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Let's consider the scenario. I declared

int n;
long int sum;

and consider expression sum = (n*(n-1))/2;

Is n converted to long int and then the multiplication is done, or does the multiplication happen first and then the truncated value (if it exceeds int) is converted to long int and stored in sum?

edited Jan 04 '18 at 15:55

melpomene

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asked Jan 04 '18 at 15:36

Sainathreddy

3

If `n` is signed and `(n*(n-1))/2 > INT_MAX` this is undefined behaviour. – George Jan 04 '18 at 15:37
2

No, conversion does only happen at the assignment, not for the calculation – UnholySheep Jan 04 '18 at 15:38
http://en.cppreference.com/w/c/language/conversion – Jesper Juhl Jan 04 '18 at 15:42
[Similiar question](https://stackoverflow.com/questions/31662792/multiplication-of-two-integers-in-c) – kishore Jan 04 '18 at 16:31

data type conversion in the expressions of C++

0 Answers0