x86汇编转C代码后无法生成有效英文单词的问题排查求助

我理解下面这段x86汇编代码转换成C代码应该是这样的：

#include <stdio.h>

void f(const int *arr, int length) {
    char buffer[32] = {0};
    const char *table = "DCBAHGFELKJIPONMTSRQXWVUZY";
    for (int i = 0; i < length; i++) {
        buffer[i] = table[arr[i]];
    }
    puts(buffer);
}

int main() {
    int arr[] = {21, 7, 0, 16, 10, 12, 18};
    int length = 7;
    f(arr, length);
    return 0;
}

对应的汇编代码如下：

.text:00000627 ; ||||||||||||||| S U B R O U T I N E ||||||||||||||||||||||||||||||||
.text:00000627
.text:00000627 ; Attributes: bp-based frame
.text:00000627
.text:00000627 public _Z1fPhj
.text:00000627 _Z1fPhj proc near ; CODE XREF: main+53 p
.text:00000627
.text:00000627 var_5C = dword ptr -5Ch
.text:00000627 var_50 = dword ptr -50h
.text:00000627 var_4C = dword ptr -47h
.text:00000627 var_43 = dword ptr -43h
.text:00000627 var_3F = dword ptr -3Fh
.text:00000627 var_3B = dword ptr -3Bh
.text:00000627 var_37 = dword ptr -37h
.text:00000627 var_33 = dword ptr -33h
.text:00000627 var_2F = word ptr -2Fh
.text:00000627 var_2D = byte ptr -2Dh
.text:00000627 var_2C = byte ptr -2Ch
.text:00000627 var_C = dword ptr -0Ch
.text:00000627 var_4 = dword ptr -4
.text:00000627 arg_0 = dword ptr 8
.text:00000627 arg_4 = dword ptr 0Ch
.text:00000627
.text:00000627 push ebp
.text:00000628 mov ebp, esp
.text:0000062A push ebx
.text:0000062B sub esp, 64h
.text:0000062E call __x86_get_pc_thunk_bx
.text:00000633 add ebx, 199Dh
.text:00000639 mov eax, [ebp+arg_0]
.text:0000063C mov [ebp+var_5C], eax
.text:0000063F mov eax, large gs:14h
.text:00000645 mov [ebp+var_C], eax
.text:00000648 xor eax, eax
.text:0000064A mov [ebp+var_47], 'DCBA'
.text:00000651 mov [ebp+var_43], 'HGFE'
.text:00000658 mov [ebp+var_3F], 'LKJI'
.text:0000065F mov [ebp+var_3B], 'PONM'
.text:00000666 mov [ebp+var_37], 'TSRQ'
.text:0000066D mov [ebp+var_33], 'XWVU'
.text:00000674 mov [ebp+var_2F], 'ZY'
.text:0000067A mov [ebp+var_2D], 0
.text:0000067E sub esp, 4
.text:00000681 push 20h ; size_t
.text:00000683 push 0 ; int
.text:00000685 lea eax, [ebp+var_2C]
.text:00000688 push eax ; void *
.text:00000689 call _memset
.text:0000068E add esp, 10h
.text:00000691 mov [ebp+var_50], 0
.text:00000698
.text:00000698 loc_698: ; CODE XREF: _Z1fPhj+A5¯j
.text:00000698 mov eax, [ebp+var_50]
.text:0000069B cmp [ebp+arg_4], eax
.text:0000069E jbe short loc_6CE
.text:000006A0 mov edx, [ebp+var_50]
.text:000006A3 mov eax, [ebp+var_5C]
.text:000006A6 add eax, edx
.text:000006A8 movzx eax, byte ptr [eax]
.text:000006AB movzx eax, al
.text:000006AE mov [ebp+var_4C], eax
.text:000006B1 mov edx, [ebp+var_4C]
.text:000006B4 mov eax, [ebp+var_50]
.text:000006B7 add eax, edx
.text:000006B9 movzx eax, byte ptr [ebp+eax+var_47]
.text:000006BE lea ecx, [ebp+var_2C]
.text:000006C1 mov edx, [ebp+var_50]
.text:000006C4 add edx, ecx
.text:000006C6 mov [edx], al
.text:000006C8 add [ebp+var_50], 1
.text:000006CC jmp short loc_698
.text:000006CE ; ---------------------------------------------------------------------------
.text:000006CE
.text:000006CE loc_6CE: ; CODE XREF: _Z1fPhj+77 j
.text:000006CE sub esp, 0Ch
.text:000006D1 lea eax, [ebp+var_2C]
.text:000006D4 push eax ; char *
.text:000006D5 call _puts
.text:000006DA add esp, 10h
.text:000006DD mov eax, 0
.text:000006E2 mov ecx, [ebp+var_C]
.text:000006E5 xor ecx, large gs:14h
.text:000006EC jz short loc_6F3
.text:000006EE call __stack_chk_fail_local
.text:000006F3
.text:000006F3 loc_6F3: ; CODE XREF: _Z1fPhj+C5 j
.text:000006F3 mov ebx, [ebp+var_4]
.text:000006F6 leave
.text:000006F7 retn
.text:000006F7 _Z1fPhj endp

这段C代码本应该输出一个真实的英文单词，但实际运行结果是WEDTJPR，完全不是有效单词。我试过各种调整，都没法得到接近真实单词的结果，到底哪里理解错了？有没有大佬能帮忙指点一下？非常感谢！

编辑补充：
终于找到问题所在并解决了！修正的关键点有两个：

• 字母表使用标准顺序：*table = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
• 索引计算要加上循环变量i：buffer[i] = table[i + arr[i]];

备注：内容来源于stack exchange，提问作者bob