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
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%r13
  mov    %esi,%r15d
  mov    %esi,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,%ebx
  movslq %eax,%rcx
  mov    %rcx,-0x50(%rbp)
  movslq %r15d,%rax
  imul   %rcx,%rax
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L56:
  cmp    %rdx,%rsp
  je     L13
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L56
L13:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L110
  orq    $0x0,-0x8(%rsp,%rax,1)
L110:
  mov    %rsp,%r14
  mov    %r14,-0x58(%rbp)
  mov    %r13,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r15
  mov    -0x48(%rbp),%eax
  cmp    $0x1,%eax
  je     L148
  mov    %ebx,%edx
  imul   %eax,%edx
  mov    $0x20,%esi
  mov    %r14,%rdi
  call   <memset@plt>
  movb   $0x0,(%r15)
  mov    $0x0,%r14d
  mov    $0x0,%r12d
  movl   $0x0,-0x44(%rbp)
  mov    $0x0,%ebx
  mov    -0x48(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x5c(%rbp)
  jmp    L70
L84:
  addl   $0x1,-0x44(%rbp)
  sub    $0x1,%ebx
L36:
  cmp    %ebx,-0x5c(%rbp)
  mov    $0x1,%eax
  cmove  %eax,%r12d
  test   %ebx,%ebx
  cmove  %ebx,%r12d
  add    $0x1,%r14
L70:
  mov    %r13,%rdi
  call   <strlen@plt>
  cmp    %r14d,%eax
  jle    L100
  movslq -0x44(%rbp),%rax
  movslq %ebx,%rdx
  imul   -0x50(%rbp),%rdx
  add    -0x58(%rbp),%rax
  movzbl 0x0(%r13,%r14,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  test   %r12d,%r12d
  jne    L84
  add    $0x1,%ebx
  jmp    L36
L100:
  cmpl   $0x0,-0x48(%rbp)
  jle    L43
  mov    -0x58(%rbp),%r14
  mov    $0x0,%r12d
  movl   $0x0,-0x44(%rbp)
  jmp    L25
L81:
  add    $0x1,%rbx
L122:
  mov    %r13,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L39
  movzbl (%r14,%rbx,1),%eax
  cmp    $0x20,%al
  je     L81
  movslq %r12d,%rdx
  mov    %al,(%r15,%rdx,1)
  add    $0x1,%r12d
  movslq %r12d,%rax
  movb   $0x0,(%r15,%rax,1)
  jmp    L81
L39:
  addl   $0x1,-0x44(%rbp)
  mov    -0x44(%rbp),%eax
  add    -0x50(%rbp),%r14
  cmp    %eax,-0x48(%rbp)
  je     L43
L25:
  mov    $0x0,%ebx
  jmp    L122
L148:
  mov    %r13,%r15
L43:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L99
  mov    %r15,%rax
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L99:
  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;
}
```