我可以指出,ARM指令是:
strd.w r0, r1, [r2]
我知道,这是一份储存指示,储存在<编码>*r2上,但我并不完全确定什么。 为什么有两个来源登记册(r0和r1)和d.wsuffix意味着什么?
这项职能将2个32个轨道登记册的64个字面内容储存在记忆中。 8至4层金刚开始储存在<代码>r2的地址。 头4个缩略语来自r0,第二4个则来自r1。
粗略当量 C是:
int32 *ptr=(int32 *) r2;
*(ptr) = r0;
*(ptr+1) = r1; // ptr+1 adds four bytes to the memory position
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