ARM汇编中的当即数

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在 mov r2,#0xffb 这句中，不是MOV的用法犯错，而是当即数用法犯错。
当即数的用法界说在Arm Architechture Reference Manual(简称ARMARM)的A5-4页开端

很重要的一段：

An immdediate operand value is formed by rotating an 8-bit constant (in a 32-bit word) by an even number of bits (0,2,4,8,26,28,30). Therefore, each instruction contains an 8-bit constant and a 4-bit rotate to be applied to that constant.

Some valid constants are:
0xFF, 0x104, 0xFF0, 0xFF00, 0xFF000, 0xFF000000, 0xF000000F

Some invalid constants are:
0x101, 0x102, 0xFF1, 0xFF04, 0xFF003, 0xFFFFFFFF, 0xF000001F

而在下面的A5-6页中说到
Specifies the immediate constant wanted. It is encoded in the instruction as an 8-bit immediate (immed_8) and a 4-bit immediate (rotate_imm), so that is equal to the result of rotating immed_8 right by (2*rotate_imm) bits.

shifter_operand = immed_8 Rotate_Right (rotate_imm * 2)

以及
Some values of have more than one possible encoding. When more than one encoding is available, an assembler needs to choose the correct one to use. For more precise control of the encoding, the instruction fields can be specified directly byusing the syntax: #,

所以，综上所述，首要解说清楚了 mov r2,#0xFF,28 一句。28并不是第三个操作数，而是和0xFF并在一同作为当即数运用，将0xFF循环右移28位。而这儿有必要着重右移XX位有必要是个偶数，由于它将等于 rotate_imm*2，那么在该指令的机器码中rotate_imm = 14, 也就是在32-BIT的机器码中第11到第8 bit = 1110B

然后再来看 mov r2,#0xFFB 的犯错状况
0xFFB = 111111111011B，很显然依照 shifter_operand = immed_8 Rotate_Right (rotate_imm * 2) 的公式， shifter_operand = 0xFFB时无法得到有用的 immed_8 和 rotate_imm, 所以编译呈现过错 error A0092: no immediate rotate operand can be created: 4091 也能够理解了，它应该是说无法生成rotate_imm，实际上immed_8也是无法生成的。

关于当即数怎么分解成immed_8和rotate_imm，能够参阅上面给出的valid constants和invalid constants，简而言之，假如该当即数能够分解成一个8-bit的二进制数循环右移偶数位，那么这个当即数是有用的，反之无效。

在上面的比如中，想要得到 r2 = 0xFFB 那么在汇编里就有必要走两步了，一步是无论怎么无法抵达的。