有意溢出的python 32位和64位整数数学
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python 32-bit and 64-bit integer math with intentional overflow
提问by Jason S
What's the best way to do integer math in 32- and 64-bit, so that overflow happens like it does in C?
在 32 位和 64 位中进行整数数学运算的最佳方法是什么，以便像在 C 中那样发生溢出？
e.g. (65536*65536+1)*(65536*65536+1) should be 0x0000000200000001 in 64-bit math, and not its exact value (non-overflowing) 0x10000000200000001.
例如 (65536*65536+1)*(65536*65536+1) 在 64 位数学中应该是 0x0000000200000001，而不是它的确切值（非溢出）0x10000000200000001。
采纳答案by martineau
Just &the result with the appropriate 32- or 64-bit mask (0xffffffffor 0xffffffffffffffff).
只是&带有适当的 32 位或 64 位掩码（0xffffffff或0xffffffffffffffff）的结果。
回答by dawg
Use NumPywith the appropriate integer size and the overflow is more C like:
使用具有适当整数大小的NumPy，溢出更像 C：
32 bit:
32 位：
>>> np.uint32(2**32-3) + np.uint32(5)
__main__:1: RuntimeWarning: overflow encountered in uint_scalars
2
64 bit:
64 位：
>>> i64=np.uint64(65536*65536+1)
>>> hex(i64*i64)
'0x200000001L'
Compare with Python's native int:
与 Python 的原生 int 比较：
>>> hex((65536*65536+1)*(65536*65536+1))
'0x10000000200000001L'
You can see that NumPy is doing as you desire.
您可以看到 NumPy 正在按您的意愿行事。
