如何在ARM汇编中进行模运算? [英] How to do the modulo operation in ARM assembly?

问题描述

我正在尝试将两个寄存器中的值相加，然后对它们取模 8.所以，在 C 代码中，它会是这样的

I'm trying to add the values in two registers, and modulo them by 8. So, in C code, it would be like this

a = a + b;
c = a % 8;

如何在 ARM 汇编中执行上述操作.

how to do this above operation in ARM assembly.

推荐答案

并非所有 ARM 处理器都有直接的除法或取模指令，因此在大多数情况下，对取模运算的调用最终会作为对例如___modsi3.

Not all ARM processors have a direct instruction for division or modulo, so in most cases, a call to the modulo operation would end up as a function call to e.g. ___modsi3.

在这种特殊情况下，当对 8 取模时，如果可以假设这些值是非负的，则可以将 % 8 部分作为 &7.在这种情况下，您的案例的程序集将是:

In this particular case, when doing modulo for 8, if the values can be assumed to be nonnegative, you can do the % 8 part as & 7. In that case, the assembly for your case would be:

add rA, rA, rB
and rC, rA, #7

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