Explicit (long long) conversion of arithmetic expression returns 0

Question

When I multiply a long long value with a double value, I get a double value.

To convert it back to long long, I use (long long).

But the output I get is 0, when it should be 10.

This is my code:

#include <bits/stdc++.h>
using namespace std;
int main()
{
    long long n = 100;
    printf("%lld",(long long)0.1*n);

    return 0;
}

Why is this? How can I fix this?

Two things: First please read [Why should I not #include ?](https://stackoverflow.com/questions/31816095/why-should-i-not-include-bits-stdc-h) Secondly, `(long long)0.1*n` is equal to `((long long)0.1)*n` — Some programmer dude, Jun 25 '18 at 09:53

score 1 · Answer 1 · answered Jun 25 '18 at 09:53

1

You are casting 0.1 to long long. Which is 0. This will work:

(long long) (0.1 * ...)

answered Jun 25 '18 at 09:53

Kaveh Vahedipour

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moooeeeep · Answer 2 · 2018-06-25T10:14:23.560

C-style casts have a higher operator precedence than arithmetic operators, like the multiplication operator.

So, in order to receive a non-zero result here, you'd have to put parentheses around the arithmetic expression to have it evaluated before the type conversion takes place.

For reference:

https://en.cppreference.com/w/cpp/language/operator_precedence

Quote from there (emphasis mine):

When parsing an expression, an operator which is listed on some row of the table above with a precedence will be bound tighter (as if by parentheses) to its arguments than any operator that is listed on a row further below it with a lower precedence.

Explicit (long long) conversion of arithmetic expression returns 0

2 Answers2