user: 请从给定的汇编指令推导出C源代码:

```asm
<editDistance>:
  endbr64
  push   %rbp
  mov    %edx,%r10d
  mov    %rsi,%r11
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    0x1(%rdx),%eax
  mov    %rsp,%rcx
  cltq
  lea    0x0(,%rax,4),%r13
  imul   %rax,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L103
L116:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L116
L103:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L54
L45:
  mov    %r13,%rax
  mov    %rsp,%r12
  movslq %r10d,%r14
  shr    $0x2,%rax
  mov    %rax,-0x50(%rbp)
  xor    %eax,%eax
  mov    -0x50(%rbp),%rcx
  test   %r10d,%r10d
  js     L106
  nopl   (%rax)
L124:
  movslq %eax,%rdx
  imul   %rcx,%rdx
  mov    %eax,(%r12,%rdx,4)
  mov    %rax,%rdx
  mov    %eax,(%r12,%rax,4)
  add    $0x1,%rax
  cmp    %r14,%rdx
  jne    L124
  test   %r10d,%r10d
  jle    L106
  lea    -0x1(%r10),%eax
  mov    %r12,-0x48(%rbp)
  lea    (%r12,%r13,1),%r8
  mov    %r12,%rdi
  lea    0x1(%rbx,%rax,1),%rax
  mov    %r10d,%r10d
  mov    %rax,%r12
  cs nopw 0x0(%rax,%rax,1)
L14:
  movzbl (%rbx),%r9d
  xor    %eax,%eax
  jmp    L24
  nopl   0x0(%rax,%rax,1)
L117:
  lea    0x1(%rdx),%r15d
  cmp    %edx,%esi
  lea    0x1(%rsi),%ecx
  cmovge %r15d,%ecx
  mov    %ecx,%edx
L76:
  mov    %edx,0x4(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r10
  je     L107
L24:
  mov    (%rdi,%rax,4),%edx
  cmp    (%r11,%rax,1),%r9b
  je     L76
  mov    (%r8,%rax,4),%ecx
  mov    0x4(%rdi,%rax,4),%esi
  cmp    %esi,%ecx
  jg     L117
  cmp    %edx,%ecx
  lea    0x1(%rdx),%r15d
  lea    0x1(%rcx),%esi
  cmovge %r15d,%esi
  mov    %esi,%edx
  mov    %edx,0x4(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r10
  jne    L24
L107:
  add    $0x1,%rbx
  add    %r13,%r8
  add    %r13,%rdi
  cmp    %r12,%rbx
  jne    L14
  mov    -0x48(%rbp),%r12
L106:
  mov    -0x50(%rbp),%rax
  add    $0x1,%rax
  imul   %r14,%rax
  mov    (%r12,%rax,4),%eax
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L30
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L54:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L45
L30:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int editDistance(const char *S, const char *T, int kmerLength) {
  int matrix[kmerLength + 1][kmerLength + 1];
  for (int i = 0; i <= kmerLength; i++) {
    matrix[i][0] = i;
    matrix[0][i] = i;
  }
  for (int j = 1; j <= kmerLength; j++) {
    for (int k = 1; k <= kmerLength; k++) {
      if (S[j - 1] == T[k - 1])
        matrix[j][k] = matrix[j - 1][k - 1];
      else
        matrix[j][k] = (matrix[j][k - 1] + 1) > (matrix[j - 1][k] + 1) ? ((matrix[j - 1][k - 1] + 1) > (matrix[j - 1][k] + 1) ? (matrix[j - 1][k] + 1) : (matrix[j - 1][k - 1] + 1)) : ((matrix[j - 1][k - 1] + 1) > (matrix[j][k - 1] + 1) ? (matrix[j][k - 1] + 1) : (matrix[j - 1][k - 1] + 1));
    }
  }
  return matrix[kmerLength][kmerLength];
}
```