Im学校正在撰写ARM模拟器。 In the ARM ARM ( 计算该分行的第二和第三最高分界线如下:I1 = NOT(J1 EORS);和I2 = NOT(J2 EORS);(ARM ARM pg 239)。 是否有任何人知道为什么这样呢? 出于某种原因,这似乎给我造成错误。
它属于一种迷惑人的范围,但似乎是一种将I1和I2的比喻描绘成有效的指示编码的技术。
umb分行指示被编码为16轨次构件。 每一次16轨次构造都需要自己的单独指示编码。
缩略语是个标语,因此,我们看不出有什么办法区分编码T3和编码T4与第一个16个轨道次构造。
In the second sub-instruction bit 12 distinguishes Encoding T3 and T4. However, using I1 and I2 directly would clash with existing instructions so they re munged into one of four encodings, each determining the range of the branch.
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