itte在臂总成 [英] itte in arm assembly

问题描述

什么是以下行臂组件做的：

What does the following line do in arm assembly:

000031e6        2916    cmp r1, #22
000031e8        bf1a    itte    ne

我得到的第一行（R1相比至22日）但对于第二行（之前我从来没有见过的itte命令和谷歌搜索返回任何内容）

I get the first line (comparing r1 to 22) but what about the second line (I've never seen the itte command before and googling returned nothing)

推荐答案

这是ARM的IF-THEN-ELSE指令，这是在Thumb-2指令集介绍。 （根据您上述具体的例子，那将是有益的，如果你已经表明，未来3说明随后的 ITTE 指令时，你就会明白为什么当你完成看完这个答案。）

It is the ARM's IF-THEN-ELSE instruction, which was introduced in the Thumb-2 instruction set. (Based on your specific example above, it would have been helpful if you had shown the next 3 instructions that followed the ITTE instruction, you'll understand why when you're done reading this answer.)

此指令用于处理条件code小序列，多达4条指令。把它看成是实现ARM的条件执行不同的方式（例如BNE - 只执行分支指令，如果零标志没有被设置）

This instruction is used for handling small sequences of conditional code, up to 4 instructions. Think of it as a different way of implementing the ARM's conditional execution (e.g. BNE - the branch instruction is only executed if the zero flag is not set).

它的好处是，它避免了服用一个分支处罚（您已经了解管线等presumably。）

The benefit of it is that it avoids the penalty of taking a branch (presumably you've learned about pipelines etc.)

的指令是有点麻烦，但一旦你环绕它你的头，这是pretty优雅。

The instruction is a bit involved but once you wrap your head around it, it's pretty elegant.

形式是这样的：

IT＆LT; X＆GT;＆LT; Y＆GT;＆LT; Z＆GT;＆LT;＆COND GT;

其中， X ，是和以Z 是可选的，而且必须是 T （对于然后）或电子（对于其他）。 ＆LT; COND＆GT; 是任何的条件，如 NE 或 EQ 或 GT 等，这反映在APSR标记。

where x, y, and z are optional, and must be either T (for "then") or E (for "else"). <cond> is any of the conditions such as NE or EQ or GT, etc. that are reflected in the APSR flags.

所以，你总是有一个 T 继 I （指令是 IT 毕竟！），然后0-3 电子的或 T 的。对于每个 T ，每个电子，您必须在匹配了相同的顺序后续指令。每个匹配后续指令必须有匹配的 IT 指令的条件。

So you always have one T following the I (the instruction is IT after all!), and then 0-3 E's or T's. For each T and each E, you must have a subsequent instruction in the same order that matches up. Each matching subsequent instruction must have conditions that match up with the IT instruction.

熊我，我知道这是令人困惑的。我在这里给了几个例子来说明。

Bear with me, I know this is confusing. I'll give a couple examples here to illustrate.

指令的最小形式会是这样的：

The minimal form of the instruction would be something like:

IT LT
SUBLT.W  R2, R1

在此情况下，如果 LT 为真（每APSR标记），减法会发生。注意 LT 在 SUB 匹配 LT 在 IT 指令。

In this case, if LT is true (per the APSR flags), the subtraction will take place. Notice the LT in the SUB matches the LT in the IT instruction.

一个成熟的例子是这样的：

A full-blown example would be something like:

ITETT NE
ADDNE R0, R0, R1
ADDEQ R0, R0, R3
ADDNE R2, R4, #1
MOVNE R5, R3

因此​​，我们有THEN ELSE THEN THEN（邰蒂），其中 NE 状态。在随后的4个条件指令声明（邰蒂 4的指令，各1个），在然后指令具有 NE 的条件，与ELSE指令（在 IT 指令后的第二个指令 - 记E组的4 E和T的第2）具有相反的状态。它不可能是别的，即这将是一个错误，如果它是像 LT 而不是 EQ 。 EQ 是 NE 。

So we have THEN ELSE THEN THEN (TETT), with NE condition. Notice in the 4 conditional instructions that follow (4 instructions, 1 each for TETT), the "THEN" instructions have the NE condition, and the "ELSE" instruction (the 2nd instruction after the IT instruction - remember the E was the 2nd of 4 E's and T's) has the opposite condition. It cannot be anything else, i.e. it would be an error if it was something like LT instead of EQ. EQ is the opposite of NE.

所以，如果 NE 为真，则说明1，3和4会被处决。否则（ EQ ），只有2指令（ ADDEQ ）将被执行。

So if NE is true, then instructions 1, 3 and 4 would be executed. Otherwise (EQ), only instruction 2 (ADDEQ) would be executed.

我给了1个和4条指令的例子，但你也可以有2 （IT {T，E} ）和3个指令（ IT {T，E} {T，E} ）的形式为好。

I've given examples of 1 and 4 instructions, but you can also have 2 (IT{T,E})and 3 instruction (IT{T,E}{T,E}) forms as well.

最后，要带回家的地步，我给怎么下面的C code可以使用此指令来实现的例子：

Finally, to bring home the point, I'll give an example of how the following C code can be implemented using this instruction:

if (R4 == R5)
{
  R7 = R8 + R9;
  R7 /= 2;
}
else
{
  R7 = R10 + R11;
  R7 *= 2;
}

转换成

CMP R4, R5
ITTEE EQ
ADDEQ R7, R8, R9    ; if R4 = R5, R7 = R8 + R9
ASREQ R7, R7, #1    ; if R4 = R5, R7 /= 2
ADDNE R7, R10, R11  ; if R4 != R5, R7 = R10 + R11
LSLNE R7, R7, #1    ; if R4 != R5, R7 *=2

这应该给你足够的咀嚼了一会儿。

That should give you enough to chew on for a while.

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