Why can't the compiler conclude that a large number is long?

Question

It compiles

var fourGb = (long)4*1024*1024*1024;

But it fails

var fourGb = 4*1024*1024*1024;

C "Overflow of operation during compilation in check mode."

So, if the compiler knows that this will be an overflow, why not conclude that the type of the variable should be long?

Answer 1

Imagine the noise that can cause. "But the compiler can understand that an expression has to be evaluated for so long, why can't it?"

And that won't happen, too expensive.

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Answer 2

. http://msdn.microsoft.com/en-us/library/ctetwysk%28VS.80%29.aspx

int, int #. L, .

var fourGb = 4L * 1024 * 1024 * 1024;

Answer 3

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Answer 4

:

• "var"

Perhaps you have:

• recalculated "why am I getting zero" for this: double varname = 1/2
• duplicated his Instead of an error, why do both operands get boosted or doubled?