无符号整数溢出 [英] Overflowing of Unsigned Int

问题描述

溢出时， unsigned int 包含什么？具体来说，我想和两个 unsigned int 做一个乘法，想知道 unsigned int >

What will the unsigned int contain when I overflow it? To be specific, I want to do a multiplication with two unsigned ints and want to know what will be in the unsigned int after the multiplication is finished?

unsigned int someint = 253473829*13482018273;

推荐答案

无符号数字不能溢出，而是使用模的属性包围。

unsigned numbers can't overflow, but instead wrap around using the properties of modulo.

例如，当 unsigned int 为32位时，结果为： a * b）mod 2 ^ 32 。

For instance, when unsigned int is 32 bits, the result would be: (a * b) mod 2^32.

正如CharlesBailey指出的， 253473829 * 13482018273 可以在转换之前使用有符号乘法，因此您应该在乘法之前明确 unsigned ：

As CharlesBailey pointed out, 253473829*13482018273 may use signed multiplication before being converted, and so you should be explicit about unsigned before the multiplication:

unsigned int someint = 253473829U * 13482018273U;

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