user: 请从给定的汇编指令推导出C源代码：

```asm
<convert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    %esi,%ebx
  sub    $0x28,%rsp
  mov    %esi,-0x4c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,-0x50(%rbp)
  cmp    $0x1,%ebx
  jle    L57
  cmp    %ebx,%eax
  jle    L57
  movslq %ebx,%r15
  mov    %rsp,%rcx
  lea    0xf(,%r15,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L81
L28:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L28
L81:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L92
  orq    $0x0,-0x8(%rsp,%rdx,1)
L92:
  mov    -0x4c(%rbp),%eax
  mov    %rsp,%r14
  xor    %esi,%esi
  mov    %r14,%rdi
  lea    0x0(,%rax,4),%edx
  call   <memset@plt>
  mov    -0x50(%rbp),%eax
  mov    %rsp,%rdx
  mov    %eax,%r13d
  shr    $0x1f,%r13d
  add    %eax,%r13d
  sar    %r13d
  add    $0x1,%r13d
  movslq %r13d,%rax
  imul   %rax,%r15
  mov    %rax,-0x48(%rbp)
  add    $0xf,%r15
  mov    %r15,%rax
  and    $0xfffffffffffff000,%r15
  sub    %r15,%rdx
  and    $0xfffffffffffffff0,%rax
  cmp    %rdx,%rsp
  je     L88
L34:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rdx,%rsp
  jne    L34
L88:
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L51
  orq    $0x0,-0x8(%rsp,%rax,1)
L51:
  mov    -0x50(%rbp),%eax
  mov    %rsp,%r15
  test   %eax,%eax
  jle    L144
  mov    -0x4c(%rbp),%eax
  mov    $0x1,%esi
  xor    %edx,%edx
  mov    $0x1,%r9d
  lea    -0x1(%rax),%r8d
  mov    -0x50(%rbp),%eax
  sub    $0x1,%eax
  lea    0x1(%r12,%rax,1),%rdi
  jmp    L108
  nopw   0x0(%rax,%rax,1)
L147:
  test   %edx,%edx
  cmove  %r9d,%esi
  add    $0x1,%r12
  add    %esi,%edx
  cmp    %rdi,%r12
  je     L144
L108:
  movslq %edx,%r10
  movslq (%r14,%r10,4),%rax
  lea    0x1(%rax),%r11d
  add    %r15,%rax
  mov    %r11d,(%r14,%r10,4)
  imul   -0x48(%rbp),%r10
  movzbl (%r12),%r11d
  mov    %r11b,(%rax,%r10,1)
  cmp    %r8d,%edx
  jne    L147
  mov    $0xffffffff,%esi
  add    $0x1,%r12
  add    %esi,%edx
  cmp    %rdi,%r12
  jne    L108
L144:
  mov    -0x50(%rbp),%eax
  mov    %r15,%r13
  xor    %ebx,%ebx
  xor    %r15d,%r15d
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  nopl   0x0(%rax)
L18:
  mov    (%r14,%r15,4),%edx
  movslq %ebx,%rdi
  mov    %r13,%rsi
  add    %r12,%rdi
  call   <memcpy@plt>
  add    (%r14,%r15,4),%ebx
  add    -0x48(%rbp),%r13
  add    $0x1,%r15
  cmp    %r15d,-0x4c(%rbp)
  jg     L18
  movslq -0x50(%rbp),%r14
  movb   $0x0,(%r12,%r14,1)
L57:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  lea    -0x28(%rbp),%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L29:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *convert(char *s, int rows) {
  int n = strlen(s);
  if (rows <= 1 || rows >= n)
    return s;
  int count[rows];
  memset(count, 0, sizeof(count));
  char rows_result[rows][n / 2 + 1];
  int step = 0;
  int direct = 1;
  int i = 0, j = 0;
  for (i = 0; i < n; i++) {
    rows_result[step][count[step]++] = s[i];
    if (step == rows - 1)
      direct = -1;
    if (step == 0)
      direct = 1;
    step += direct;
  }
  char *result = malloc(n + 1);
  int cusor = 0;
  for (i = 0; i < rows; i++) {
    memcpy(result + cusor, rows_result[i], count[i]);
    cusor += count[i];
  }
  result[n] = '\0';
  return result;
}
```