user: 请从给定的汇编指令推导出C源代码:

```asm
<get_common_chars>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x38,%rsp
  mov    %edx,-0x48(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rsp,%rcx
  add    $0x1,%eax
  mov    %eax,-0x58(%rbp)
  cltq
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L40
L160:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L160
L40:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L16
L36:
  mov    -0x50(%rbp),%rdi
  mov    %rsp,-0x60(%rbp)
  call   <strlen@plt>
  mov    %rsp,%rcx
  add    $0x1,%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L163
L4:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L4
L163:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L154
L35:
  mov    -0x50(%rbp),%rsi
  mov    %rsp,%rdi
  xor    %r15d,%r15d
  mov    %rsp,%r13
  call   <strcpy@plt>
  mov    %rbx,%rdi
  movl   $0x0,-0x54(%rbp)
  call   <strlen@plt>
  cmp    %r15d,%eax
  jle    L152
  nopl   (%rax)
L6:
  mov    -0x48(%rbp),%esi
  mov    -0x50(%rbp),%rdi
  mov    %r15d,%r14d
  mov    $0x0,%eax
  movzbl (%rbx,%r15,1),%r12d
  sub    %esi,%r14d
  lea    (%rsi,%r15,1),%ecx
  mov    %ecx,-0x44(%rbp)
  cmovs  %eax,%r14d
  call   <strlen@plt>
  mov    -0x44(%rbp),%ecx
  sub    $0x1,%eax
  cmp    %ecx,%eax
  jle    L38
L155:
  cmp    %r14d,%ecx
  jle    L12
  movslq %r14d,%rax
  jmp    L166
  nop
L65:
  add    $0x1,%rax
  cmp    %eax,%ecx
  jle    L12
L166:
  movslq %eax,%rdx
  cmp    %r12b,0x0(%r13,%rax,1)
  jne    L65
  movslq -0x54(%rbp),%rax
  mov    -0x60(%rbp),%rsi
  movb   $0x0,0x0(%r13,%rdx,1)
  mov    %rax,%rdi
  mov    %r12b,(%rsi,%rax,1)
  add    $0x1,%edi
  mov    %edi,-0x54(%rbp)
L12:
  mov    %rbx,%rdi
  add    $0x1,%r15
  call   <strlen@plt>
  cmp    %r15d,%eax
  jg     L6
L152:
  mov    -0x60(%rbp),%rbx
  movslq -0x54(%rbp),%r13
  movb   $0x0,(%rbx,%r13,1)
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    -0x58(%rbp),%edi
  imul   %eax,%edi
  add    $0x1,%edi
  call   <malloc@plt>
  mov    %rbx,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L168
  lea    -0x28(%rbp),%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L16:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L36
  nopl   0x0(%rax,%rax,1)
L38:
  mov    -0x50(%rbp),%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%ecx
  jmp    L155
L154:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L35
L168:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *get_common_chars(const char *str1, const char *str2, const int d_step) {
  char *ret_str;
  char ret[strlen(str1) + 1];
  char copy[strlen(str2) + 1];
  int i, j, k = 0;
  strcpy(copy, str2);
  for (i = 0; i < (int)strlen(str1); i++) {
    char tmp = str1[i];
    int max = (((0 > (i - d_step))) ? (0) : ((i - d_step)));
    int min = (((i + d_step < ((int)strlen(str2) - 1))) ? (i + d_step) : (((int)strlen(str2) - 1)));
    for (j = max; j < min; j++) {
      if (copy[j] == tmp) {
        ret[k] = tmp;
        copy[j] = (char)0;
        k++;
        break;
      }
    }
  }
  ret[k++] = '\0';
  ret_str = malloc((sizeof(ret) * strlen(ret)) + 1);
  strcpy(ret_str, ret);
  return (ret_str);
}
```