[英]armv8 RRX register usage
In one of my homework exercises, I'm asked to write an armv8 program that counts number of 1-bits in a register.在我的一个家庭作业练习中,我被要求编写一个 armv8 程序来计算寄存器中 1 位的数量。 Here's my implementation:这是我的实现:
.arch armv8-a // specifies the ARMv8 Architecture
.text
.align 2 // align to a multiple of 4 (1<<2)
.global start // arm64_emu.sh starts execution at start
.type start, %function
start:
movz x0, #8
movz x10, #0
movz x1, #0
loop:
rrx x0, x0 //rotate x0 and put the last bit into carry
bcs skip
add x10, x10, 1
skip:
cmp x1, #3
bne loop
svc 0 // dump registers
svc 999 // stop the emulation
.size start, .-start
A nice flow chart from this site gives a good overview of my program: http://www.8085projects.info/Program21.html这个站点的一个不错的流程图很好地概述了我的程序: http : //www.8085projects.info/Program21.html
However, it gives me this error:但是,它给了我这个错误:
zl5022@enterprise:~$ arm64_emu.sh c.s
c.s: Assembler messages:
c.s:11: Error: unknown mnemonic `rrx' -- `rrx x0,x0'
According to http://infocenter.arm.com/help/index.jsp?topic=/com.arm.doc.dui0204j/Cjacbgca.html , rrx takes two registers as input, so am I missing something here?根据http://infocenter.arm.com/help/index.jsp?topic=/com.arm.doc.dui0204j/Cjacbgca.html , rrx 将两个寄存器作为输入,所以我在这里遗漏了什么?
The ARMv8 rotate instruction is ROR. ARMv8 循环指令是 ROR。
ROR Xd, Xm, #uimm
Rotate Right (immediate, 64-bit)
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