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

我正在尝试将值添加到两个寄存器中，并以8为模.因此，在C代码中，就像这样

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

不是所有的ARM处理器都有直接的除法或模运算指令，因此在大多数情况下，对模运算的调用最终会作为对例如的函数调用而结束. ___ 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，对8取模时，如果可以认为这些值是非负的，则可以将％8 部分作为&7 .在这种情况下，适合您的情况的程序集将是:

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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