I was going through the go tutorial on golang.org and I came across an example that i partially understand...
MaxInt uint64 = 1<<64 - 1
Now I understand this to be shifting the bit 64 places to the left which would make it a 1 followed by 64 0's.
My question is why is this the max integer that can be achieved in a 64 bit number. Wouldn't the max integer be 111111111....(until the 64th 1) instead of 100000...(until the 64th one)?
分享 What happens here, step by step:
Take 1.
Shift it to the left 64 bits. This is tricky. The result actually needs 65 bits for representation - namely 1 followed by 64 zeroes. Since we are calculating a 64 bit value here why does this even compile instead of overflowing to 0 or 1 or producing a compile error?
It works because the arithmetic used to calculate constants in Go is a bit magic (https://blog.golang.org/constants) in that it has nothing to do whatsoever with the type of the named constant being calculated. You can say foo uint8 = 1<<415 / 1<<414 and foo is now 2.
Subtract 1. This brings us back into 64 bits numbers, as it's actually 11....1 (64 times), which is indeed the maximum value of uint64. Without this step, the compiler would complain about us trying to cram 65 bit value into uint64.
Name the constant MaxInt and give it type uint64. Success!
The magic arithmetic used to calculate constants still has limitations (obviously). Shifts greater than 500 or so produce funny named stupid shift errors.
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