算术溢出与算术进位 [英] Arithmetic Overflow vs. Arithmetic Carry

问题描述

我的演讲幻灯片中的一个给出了算术溢出的例子，并在ARM芯片中引入了条件分支标志的主题，引用如下：

V（溢出）
- 7FFFFFFF + 1
（进位）
- FFFFFFFF + 1

假设为了举例，地址只能容纳8个字节。所以对我来说，似乎喜欢加1到7FFFFFFF给出80000000.我认为80000000仍然适合8字节的地址。

为什么这是一个算术溢出？这张幻灯片是否是错误的？或者是我的理解存在缺陷？

感谢您的回应

当寄存器无法正确地将结果表示为有符号值（您溢出到符号位中）时，溢出标志被置位。

如果寄存器无法正确表示结果为无符号值（不需要符号位），则会设置进位标志。

One of my lecture slides gives an example of arithmetic overflow and carry in a topic for conditional branching flags on an ARM chip, quoted below:

• V (overflow) -  7FFFFFFF+1
• C (carry) -  FFFFFFFF+1

Presumably for the sake of the example, the address can only hold 8 bytes. So to me, it seems likes adding 1 to 7FFFFFFF gives 80000000. I thought 80000000 would still fit into an 8-byte address.

Why is this an arithmetic overflow? Is it the wrong way around on the slide? Or is my understanding flawed?

Thanks for any responses

解决方案

• Overflow flags get set when the register cannot properly represent the result as a signed value (you overflowed into the sign bit).
• Carry flags are set when the register cannot properly represent the result as an unsigned value (no sign bit required).

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