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And the error says it's an int, but it's declared long. Am I missing something about long in C++ not being 64 bits?



That's the whole point. =] This is a surprising aspect of C++: the shift expression doesn't have the type of the declared variable.

The integer literals we are shifting are of 'int' type, and the shift occurs at that type (based on the usual arithmetic conversions). There is stack overflow question with explanations and a good blog post here about it:

http://stackoverflow.com/questions/836544/usual-arithmetic-c...

http://blogs.msdn.com/b/oldnewthing/archive/2004/03/10/87247...

Also, you can look through the C++98 standard to understand all the details. Relevant sections are [expr]p9 and [expr.shift].


The left operand is an int, so the result of the shift expression is an int. The fact that the shift expression is used to initialize a long is not relevant.


long is 32-bit on x86_32 Linux, 64-bit on x86_64 Linux, and 32-bit on x86_64 Windows.




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