user: What would the C source look like for the assembly code provided here?

```asm
<get_common_chars>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %edx,-0x54(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,-0x64(%rbp)
  cltq
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L88:
  cmp    %rdx,%rsp
  je     L98
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L88
L98:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L61
  orq    $0x0,-0x8(%rsp,%rax,1)
L61:
  mov    %rsp,-0x60(%rbp)
  mov    -0x50(%rbp),%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffffff0,%rdx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rcx
  sub    %rax,%rcx
L80:
  cmp    %rcx,%rsp
  je     L9
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L80
L9:
  mov    %rdx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L1
  orq    $0x0,-0x8(%rsp,%rax,1)
L1:
  mov    %rsp,%r14
  mov    -0x50(%rbp),%rsi
  mov    %rsp,%rdi
  call   <strcpy@plt>
  mov    $0x0,%r15d
  movl   $0x0,-0x58(%rbp)
  jmp    L140
L144:
  mov    -0x50(%rbp),%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%r12d
  jmp    L100
L136:
  mov    -0x58(%rbp),%esi
  movslq %esi,%rdx
  mov    -0x60(%rbp),%rdi
  mov    %r13b,(%rdi,%rdx,1)
  cltq
  movb   $0x0,(%r14,%rax,1)
  add    $0x1,%esi
  mov    %esi,-0x58(%rbp)
L10:
  add    $0x1,%r15
L140:
  mov    -0x48(%rbp),%rdi
  call   <strlen@plt>
  cmp    %r15d,%eax
  jle    L40
  mov    -0x48(%rbp),%rax
  movzbl (%rax,%r15,1),%r13d
  mov    %r15d,%ebx
  mov    -0x54(%rbp),%ecx
  sub    %ecx,%ebx
  mov    $0x0,%eax
  cmovs  %eax,%ebx
  lea    (%rcx,%r15,1),%r12d
  mov    -0x50(%rbp),%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  cmp    %r12d,%eax
  jle    L144
L100:
  cmp    %ebx,%r12d
  jle    L10
  movslq %ebx,%rbx
L84:
  mov    %ebx,%eax
  cmp    %r13b,(%r14,%rbx,1)
  je     L136
  add    $0x1,%rbx
  cmp    %ebx,%r12d
  jg     L84
  jmp    L10
L40:
  movslq -0x58(%rbp),%rax
  mov    -0x60(%rbp),%r15
  movb   $0x0,(%r15,%rax,1)
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    -0x64(%rbp),%edi
  imul   %eax,%edi
  add    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %r15,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  mov    %rbx,%rax
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L30:
  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);
}
```