所以我有一个程序可以识别一个词（s1）是否是另一个词（s2）的变位词。 但我不断收到错误 A2070：第 15、34 和 36 行的指令操作数无效。我已经查找了错误，但似乎找不到解决方案或无法理解它。
我所尝试的只是将指定行的 movzx 更改为 mov，但这会使程序抛出异常。如果程序正常运行，应该将 eax 中的值从 0 更改为 1。
.386
.model flat,stdcall
.stack 4096
ExitProcess proto,dwExitCode:dword
.data
s1 byte "GARDEN"
s2 byte "DANGER"
c1 byte 26 dup(0) ;counter for each letter in s1
c2 byte 26 dup(0) ;counter for each letter in s2
.code
main proc
mov eax, 0 ;we will assume that we do not have an anagram
movzx ebx, lengthof s1 ;(1) iterate lengthof s1 times
mov esi, 0 ;start at the first byte of s1 and s2
CounterLoop: ;this will increment the proper 'elements' of c1 and c2
movzx edi, s1[esi] ;move the value from s1 into edi
inc c1[edi - 65] ;(2) increment the counter at the value - 65.
;subtract 65 because the ASCII value of A is 65, B is 66, C is 67...
;when you subtract 65 then the sum of all the As will be stored in 'index' 0
;Bs in 'index' 1, Cs in 'index' 2...
movzx edi, s2[esi] ;(3) Do the same procedure for s2
inc c2[edi - 65] ;(4) increment the second counter at the value - 65.
inc esi ;increment esi
loop CounterLoop ;after this loop terminates our couter arrays will have the proper values
movzx esi, 0 ;(5) start checking the counter arrays at 'index' 0
movzx ebx, lengthof c1 ;(6)iterate lengthof c1 times
VerifyLoop:
mov bl, c1 ;(7) move value of c1 into bl
cmp bl, c2 ;(8) check bl vs the value of c2
jmp NoAna ;(9) if they are not equal then we do not have an anagram. jump to NoAna
inc esi ;increment esi
loop VerifyLoop
mov eax, 1 ;if the loop terminates and we have not jumped then we know we have an anagram
NoAna:
invoke ExitProcess, 0
main endp
end main