RV32无MUL移位乘法结果异常求助：0xffffffff×0x19计算错误

为无硬件乘法器的CH32V003芯片实现无MUL指令的32位无符号乘法，将AVR平台的移位加法代码移植到RISC-V汇编，输出64位结果。多数测试用例（含0xffffffff×0xffffffff）计算正确，但0xffffffff×0x19得到结果0x12ffffffe7，正确结果应为0x18ffffffe7。尝试过ChatGPT、Gemini、DeepSeek等生成的多种代码变体，问题仍存在。目前仅通过重复加法实现的乘法能得到正确结果，怀疑移位加法方案的进位检测存在问题。在RARS中模拟代码，恳请提供解决方案。

移植的RISC-V汇编代码 #

.data 
result_lo: .word 0
result_hi: .word 0
modulo:    .word 0

.text
 li a1,0xffffffff   # multiplicant
 li a2,0x19 # multiplier
 li a3,0x00000000   # result_lo
 li a4,0x00000000   # result_hi
 li a5,0        # working register 

 start:
 call ROR       # rotate right multiplier to test lsb is 0 or 1
 bnez x3,multiply   # if lsb =1 branch to repeated adding of multiplicant to result register
 finishmul:
 call RLL2      # shift multiplicand left or multiply by 2
 beqz a2,exit_proc
 J start        # repeat loop
 exit_proc:
    j exit_proc
    #ret
 multiply:
 add a5,a3,a1       # add multiplicant to low result register and store final result in a5 for processing
 sltu a0,a5,a3      # set a0 to 1 if result of addition a3:a1 i a5 is greater than a3
 sltu x3,a5,a1      # set x3 to 1 if result of addition a3:a1 in a5 is greater than a1
 or a0,a0,x3        # or a0 and x3 , if 1 carry if a0 = 0 no carry
 bnez a0,carryset   # if a0 = 1 carry set, branch to label carry set
 mv a3,a5       # result in working register copied to a3 low result register
 J finishmul        # jump to label finishmul
 carryset:      # reach here only if carryset
 mv a3,a5       # copy a5 to low result a3
 addi a4,a4,1       # add carry to a4 high register result
 J finishmul        # jump to label finishmul

 ROR:
 li x3,0        # clear carry
 mv t0,a2       # copy number in a2 to t0
 andi t0,t0,1       # extract lsb is 0 or 1
 beqz t0,zzz        # if lab is 0 branch to zzz
 li x3,1        # if lsb is 1 carry occured , load 1 in carry register x3
 srli a2,a2,1       # shift right a2 by 1 postion 
 ret            # return to caller
 zzz:           # reach here if lsb =0
 li x3,0        # load x3 0 indicating carry bit is 0
 srli a2,a2,1       # right shift multiplier once. divide multiplier by 2
 ret            # return to caller

 ROL:
 li x3,0        # 
 mv t0,a2
 li x3,0x80000000
 and t0,t0,x3
 beqz t0,zzz1
 li x3,1        # carry
 slli a2,a2,1
  ret
 zzz1:
 li x3,0
 slli a2,a2,1
 ret

 RLL2:          # rotate left 2 registers a3:a5
 mv a5,a4       # copy contents of a4 to a5
 li x3,0        # clear x3
 mv t0,a1       # copy multiplicant to t0
 li x3 ,0x80000000  # load x3 MSB bitmask
 and t0,t0,x3       # and with 0x800000000 to extract the MSB
 bnez t0,OR1        # if MSB = 1 branch to OR1 label
 slli a1,a1,1       # shift left 1 position a1 register ( multiplicant)
 slli a5,a5,1       # shift left 1 position working register with value of a4 register ( multiplicant)
 beqz a2,exit       # if multiplier register is 0 exit
 mv a4,a5       # copy back the shifter multiplicant to a4
 ret
 OR1:
 mv a5,a4
 slli a1,a1,1
 slli a5,a5,1
 li x3,1
 or a5,a5,x3
 beqz a2,exit
 mv a4,a5
 ret
exit:
    ret

经测试，上述代码在RARS中计算0xffffffff×0x19得到0x12ffffffe7，而正确结果应为0x18ffffffe7；仅通过重复加法实现的乘法能得到正确结果。

内容的提问来源于stack exchange，提问作者sajeev sankaran