user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<convert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %esi,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rsp,%rcx
  movslq %eax,%rdx
  movslq %r14d,%rax
  imul   %rdx,%rax
  mov    %rdx,-0x50(%rbp)
  mov    %rdx,%rbx
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L145
L91:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L91
L145:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L12
L115:
  mov    %r12,%rdi
  mov    %rsp,-0x58(%rbp)
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r14
  mov    -0x48(%rbp),%eax
  cmp    $0x1,%eax
  je     L45
  mov    %ebx,%edx
  mov    $0x20,%esi
  mov    %rsp,%rdi
  xor    %ebx,%ebx
  imul   %eax,%edx
  xor    %r13d,%r13d
  xor    %r15d,%r15d
  call   <memset@plt>
  mov    -0x48(%rbp),%eax
  movb   $0x0,(%r14)
  movl   $0x0,-0x44(%rbp)
  sub    $0x1,%eax
  mov    %eax,-0x5c(%rbp)
  jmp    L43
  nopl   0x0(%rax)
L70:
  add    $0x1,%r15d
L85:
  cmp    %r15d,-0x5c(%rbp)
  mov    $0x1,%eax
  cmove  %eax,%r13d
  xor    %eax,%eax
  test   %r15d,%r15d
  cmove  %eax,%r13d
  add    $0x1,%rbx
L43:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L109
  movslq %r15d,%rdi
  imul   -0x50(%rbp),%rdi
  movzbl (%r12,%rbx,1),%r8d
  movslq -0x44(%rbp),%rax
  add    -0x58(%rbp),%rax
  mov    %r8b,(%rax,%rdi,1)
  test   %r13d,%r13d
  je     L70
  addl   $0x1,-0x44(%rbp)
  sub    $0x1,%r15d
  jmp    L85
L45:
  mov    %r12,%r14
L93:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L103
  lea    -0x28(%rbp),%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L109:
  mov    -0x48(%rbp),%eax
  test   %eax,%eax
  jle    L93
  movl   $0x0,-0x44(%rbp)
  mov    -0x58(%rbp),%r13
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L30:
  xor    %r15d,%r15d
  jmp    L138
  nopl   (%rax)
L0:
  movzbl 0x0(%r13,%r15,1),%eax
  cmp    $0x20,%al
  je     L124
  movslq %ebx,%rcx
  add    $0x1,%ebx
  mov    %al,(%r14,%rcx,1)
  movslq %ebx,%rax
  movb   $0x0,(%r14,%rax,1)
L124:
  add    $0x1,%r15
L138:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %r15d,%eax
  jg     L0
  addl   $0x1,-0x44(%rbp)
  add    -0x50(%rbp),%r13
  mov    -0x44(%rbp),%eax
  cmp    %eax,-0x48(%rbp)
  jne    L30
  jmp    L93
L12:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L115
L103:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *convert(char *s, int numRows) {
  char map[numRows][strlen(s)];
  int stx = 0;
  int sty = 0;
  int dir = 0;
  int i, j, k = 0;
  char *ans = malloc(sizeof(char) * (strlen(s) + 1));
  if (numRows == 1)
    return s;
  memset(map, ' ', sizeof(map));
  ans[0] = '\0';
  for (i = 0; i < strlen(s); i++) {
    map[stx][sty] = s[i];
    if (dir == 0) {
      stx++;
    } else {
      sty++;
      stx--;
    }
    if (stx == numRows - 1)
      dir = 1;
    if (stx == 0)
      dir = 0;
  }
  for (i = 0; i < numRows; i++) {
    for (j = 0; j < strlen(s); j++) {
      if (map[i][j] != ' ') {
        ans[k] = map[i][j];
        ans[++k] = '\0';
      }
    }
  }
  return ans;
}
```