user: 这段汇编代码用C语言表示是什么样的？

```asm
<GrabGETVar>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    $0x3f,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x4a8,%rsp
  mov    %rdx,0x8(%rsp)
  lea    0x90(%rsp),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x498(%rsp)
  xor    %eax,%eax
  jmp    L6
  nopl   0x0(%rax)
L101:
  movzbl (%r15,%rbx,1),%eax
  cmp    $0x26,%al
  cmove  %ebp,%eax
  mov    %al,(%r12,%rbx,1)
  movb   $0x0,0x91(%rsp,%rbx,1)
  add    $0x1,%rbx
L6:
  mov    %r15,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L101
  xor    %ebx,%ebx
  xor    %ebp,%ebp
  lea    0x10(%rsp),%r15
  xor    %r14d,%r14d
  jmp    L84
  nopl   0x0(%rax)
L53:
  cmp    $0x3f,%cl
  je     L140
L160:
  add    $0x1,%r14d
L84:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %r14d,%eax
  jle    L128
  movslq %r14d,%rax
  movzbl 0x90(%rsp,%rax,1),%ecx
  test   %ebp,%ebp
  je     L53
  cmp    $0x1,%ebp
  jne    L73
L127:
  movzbl 0x90(%rsp,%rax,1),%eax
  cmp    $0x3d,%al
  je     L125
  movslq %ebx,%rcx
  add    $0x1,%ebx
  mov    $0x1,%ebp
  mov    %al,0x10(%rsp,%rcx,1)
  movslq %ebx,%rax
  movb   $0x0,0x10(%rsp,%rax,1)
  jmp    L160
  nopw   0x0(%rax,%rax,1)
L125:
  add    $0x1,%r14d
  xor    %ebx,%ebx
  movslq %r14d,%rax
L73:
  movzbl 0x90(%rsp,%rax,1),%eax
  cmp    $0x3f,%al
  je     L90
  movslq %ebx,%rcx
  add    $0x1,%ebx
  mov    $0x2,%ebp
  mov    %al,0x50(%rsp,%rcx,1)
  movslq %ebx,%rax
  movb   $0x0,0x50(%rsp,%rax,1)
  jmp    L160
  nopl   0x0(%rax)
L140:
  add    $0x1,%r14d
  xor    %ebx,%ebx
  movslq %r14d,%rax
  jmp    L127
  nopl   0x0(%rax,%rax,1)
L90:
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <IsEqual@plt>
  test   %rax,%rax
  jne    L22
  xor    %ebx,%ebx
  mov    $0x1,%ebp
  jmp    L160
  nopl   0x0(%rax)
L128:
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <IsEqual@plt>
  test   %rax,%rax
  jne    L22
  mov    $0xffffffff,%eax
L62:
  mov    0x498(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L86
  add    $0x4a8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L22:
  mov    0x8(%rsp),%rsi
  lea    0x50(%rsp),%rdi
  xor    %edx,%edx
  call   <CopyString@plt>
  mov    $0x1,%eax
  jmp    L62
L86:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int GrabGETVar(char *getData, char *var, char *ret) {
  int i, v, n, c, nr;
  char AllVars[1024];
  char gVar[50];
  char gVal[50];
  c = 0;
  n = 0;
  for (i = 0; i < strlen(getData); i++) {
    if (getData[i] == '?') {
      c++;
      v = 1;
    }
    if (v == 1) {
      AllVars[n] = getData[i];
      if (getData[i] == '&') {
        c++;
        AllVars[n] = '?';
      }
      AllVars[n + 1] = 0x00;
      n++;
    }
  }
  n = 0;
  v = 0;
  for (i = 0; i < strlen(AllVars); i++) {
    if (v == 0) {
      if (AllVars[i] == '?') {
        n = 0;
        v = 1;
        i++;
      }
    }
    if (v == 1) {
      if (AllVars[i] == '=') {
        n = 0;
        v = 2;
        i++;
      } else {
        gVar[n] = AllVars[i];
        gVar[n + 1] = 0x00;
        n++;
      }
    }
    if (v == 2) {
      if (AllVars[i] == '?') {
        if (IsEqual(gVar, var)) {
          CopyString(gVal, ret, 0);
          return 1;
        }
        n = 0;
        v = 1;
      } else {
        gVal[n] = AllVars[i];
        gVal[n + 1] = 0x00;
        n++;
      }
    }
  }
  if (IsEqual(gVar, var)) {
    CopyString(gVal, ret, 0);
    return 1;
  }
  return -1;
}
```