linux - 什么时候CPSR GE[3 :0] bits be modified

Question

我在 ARM 文档中读到:

GE[3:0], bits[19:16] The instructions described in Parallel addition and subtraction instructions on page A4-171 update these flags to indicate the results from individual bytes or halfwords of the operation. These flags can control a later SEL instruction.

显然 GE[3:0] 代表“eq/lt/gt”？

我遇到了几个奇怪的问题，我还没有任何线索，但它们都有 CPSR 值 xxxf0030，所以 GE 位是 0b1111？那代表什么？这些GE位正常吗？

提前致谢!

Answer 1

在 ARMv7 ARM(匹配该文本)中，如何设置 GE 标志的细节仅在并行指令本身的操作伪代码中。可悲的是，他们似乎已经删除了 ARMv6 ARM 中的这个漂亮的散文描述:

Instructions that operate on halfwords:

• set or clear GE[3:2] together, based on the result of the top halfword calculation
• set or clear GE[1:0] together, based on the result of the bottom halfword calculation.

Instructions that operate on bytes:

• set or clear GE[3] according to the result of the top byte calculation
• set or clear GE[2] according to the result of the second byte calculation
• set or clear GE[1] according to the result of the third byte calculation
• set or clear GE[0] according to the result of the bottom byte calculation.

Each bit is set (otherwise cleared) if the results of the corresponding calculation are as follows:

• for unsigned byte addition, if the result is greater than or equal to 2^8
• for unsigned halfword addition, if the result is greater than or equal to 2^16
• for unsigned subtraction, if the result is greater than or equal to zero
• for signed arithmetic, if the result is greater than or equal to zero.

作为算术标志，它们可以有任何旧值(在重置时未定义，并且可以通过 APSR 自由写入)，所以除非您专门使用其中一条设置它们的指令，否则它们几乎没有意义并且可以忽略。