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

```asm
<convert>:
  endbr64
  test   %rdi,%rdi
  je     L19
  push   %r15
  push   %r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <strlen@plt>
  mov    %eax,%ebx
  cmp    $0x2,%eax
  jle    L130
  cmp    $0x1,%r13d
  je     L130
  lea    0x1(%rax),%edi
  lea    -0x1(%r13),%r14d
  call   <malloc@plt>
  cmp    %r13d,%ebx
  lea    (%r14,%r14,1),%r15d
  cmovle %ebx,%r13d
  test   %r13d,%r13d
  jle    L132
  cmp    %r13d,%ebx
  mov    %r13d,%edi
  mov    %r13d,0xc(%rsp)
  movslq %r15d,%r12
  cmovle %ebx,%edi
  xor    %r10d,%r10d
  xor    %ecx,%ecx
  xor    %r9d,%r9d
  mov    %edi,%r13d
  nopw   0x0(%rax,%rax,1)
L153:
  test   %r9d,%r9d
  je     L4
  movslq %r14d,%rdx
  cmp    %r9d,%r14d
  je     L104
  mov    %r15d,%r11d
  sub    %r10d,%r11d
  cmp    %r9d,%ebx
  jle    L226
  lea    0x1(%rcx),%edx
  mov    %r9d,%esi
  mov    $0x1,%edi
  movslq %edx,%rdx
  jmp    L180
L110:
  add    %r11d,%esi
  xor    %edi,%edi
  add    $0x1,%rdx
  cmp    %esi,%ebx
  jle    L226
L180:
  movslq %esi,%r8
  movslq %edx,%rcx
  movzbl 0x0(%rbp,%r8,1),%r8d
  mov    %r8b,-0x1(%rax,%rdx,1)
  test   %edi,%edi
  jne    L110
  add    %r10d,%esi
  mov    $0x1,%edi
  add    $0x1,%rdx
  cmp    %esi,%ebx
  jg     L180
L226:
  add    $0x1,%r9d
  add    $0x2,%r10d
  cmp    %r13d,%r9d
  jl     L153
L238:
  mov    0xc(%rsp),%r13d
  lea    (%r9,%r9,1),%r10d
  cmp    %r13d,%r9d
  jge    L217
  nopl   (%rax)
L214:
  cmp    %r9d,%r14d
  je     L234
  mov    %r15d,%r11d
  sub    %r10d,%r11d
  cmp    %r9d,%ebx
  jle    L234
  add    $0x1,%ecx
  mov    %r9d,%edx
  mov    $0x1,%edi
  movslq %ecx,%rsi
  jmp    L193
L3:
  add    %r11d,%edx
  xor    %edi,%edi
  add    $0x1,%rsi
  cmp    %edx,%ebx
  jle    L234
L193:
  movslq %edx,%r8
  movslq %esi,%rcx
  movzbl 0x0(%rbp,%r8,1),%r8d
  mov    %r8b,-0x1(%rax,%rsi,1)
  test   %edi,%edi
  jne    L3
  add    %r10d,%edx
  mov    $0x1,%edi
  add    $0x1,%rsi
  cmp    %edx,%ebx
  jg     L193
L234:
  add    $0x1,%r9d
  add    $0x2,%r10d
  cmp    %r9d,%r13d
  jg     L214
L217:
  add    %rax,%rcx
L155:
  movb   $0x0,(%rcx)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L4:
  xor    %edx,%edx
L104:
  add    $0x1,%ecx
  movslq %ecx,%rsi
L103:
  movzbl 0x0(%rbp,%rdx,1),%edi
  add    %r12,%rdx
  movslq %esi,%rcx
  mov    %dil,-0x1(%rax,%rsi,1)
  add    $0x1,%rsi
  cmp    %edx,%ebx
  jg     L103
  add    $0x1,%r9d
  add    $0x2,%r10d
  cmp    %r13d,%r9d
  jl     L153
  jmp    L238
  xchg   %ax,%ax
L130:
  add    $0x18,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1090 <strdup@plt>
L132:
  mov    %rax,%rcx
  jmp    L155
L19:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *convert(char *s, int numRows) {
  if (!s)
    return (0);
  char *res;
  int i, j, l, step1, step2, flag, p, k;
  l = strlen(s);
  if (l <= 2 || numRows == 1)
    return (strdup(s));
  res = malloc(l + 1);
  k = 2 * numRows - 2;
  p = 0;
  for (i = 0; i < numRows && i < l; i++) {
    if (i == 0 || i == (numRows - 1)) {
      for (j = i; j < l; j += k)
        res[p++] = s[j];
      continue;
    }
    step1 = k - i * 2;
    step2 = k - step1;
    flag = 1;
    for (j = i; j < l;) {
      res[p++] = s[j];
      if (flag) {
        flag = 0;
        j += step1;
      } else {
        flag = 1;
        j += step2;
      }
    }
  }
  res[p] = 0;
  return (res);
}
```