 John Voss -4 years ago 125
C++ Question

# 3 * 1000000000 overflows as an int, but the variable is long long. Why?

I have a simple c++ app that performs the following calculations

``````long long calcOne = 3 * 100000000;     // 3e8, essentially
long long calcTwo = 3 * 1000000000;    // 3e9, essentially
long long calcThree = 3 * 10000000000; // 3e10, essentially
``````

If I write the result of each calculation I get the following output:

``````calcOne = 300000000
calcTwo = -1294967296
calcThree = 30000000000
``````

So why does the second calculation fail? As far as I can tell it is within the limits of a long long type (calcThree was larger...).

I am using Visual Studio 2015 on Windows 10. Thanks in advance. Sam Varshavchik

Integer constants are, by default `int`s.

``````1000000000
``````

That can fit into an `int`. So, this constant gets parsed as an `int`. But multiplying it by 3 overflows int.

``````10000000000
``````

This is too big to an int, so this constant is a `long long`, so the resulting multiplication does not overflow.

Solution: explicitly use `long long` constants:

``````long long calcOne = 3 * 100000000LL;     // 3e8, essentially
long long calcTwo = 3 * 1000000000LL;    // 3e9, essentially
long long calcThree = 3 * 10000000000LL; // 3e10, essentially
``````
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