user: 这段汇编代码用C语言表示是什么样的：

```asm
<rfftf1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    0x8(%r8),%rax
  mov    %rdi,0x8(%rsp)
  mov    %rsi,0x20(%rsp)
  mov    %rdx,0x28(%rsp)
  mov    %rcx,0x18(%rsp)
  mov    %rax,0x10(%rsp)
  test   %rax,%rax
  je     L19
  lea    0x8(%r8,%rax,8),%r15
  mov    %rdi,%r12
  lea    -0x1(%rdi),%rbx
  mov    %rdx,%r14
  mov    %r15,%rax
  mov    $0x1,%ebp
  mov    %rsi,%r15
  mov    %rax,%r13
  jmp    L7
  nopw   0x0(%rax,%rax,1)
L32:
  cmp    $0x2,%rsi
  je     L65
  cmp    $0x3,%rsi
  je     L138
  cmp    $0x5,%rsi
  je     L16
  cmp    $0x1,%rdi
  jne    L33
  mov    %r15,%rdx
  mov    %r14,%r15
  mov    %rdx,%r14
L33:
  sub    $0x8,%rsp
  mov    %rdi,%rcx
  mov    %rax,%rdx
  mov    %r14,%r9
  push   %r8
  imul   %rax,%rcx
  mov    %r15,%r8
  call   <radfg@plt>
  pop    %rax
  pop    %rdx
L111:
  add    $0x1,%rbp
  sub    $0x8,%r13
  cmp    %rbp,0x10(%rsp)
  jb     L62
L7:
  mov    0x8(%rsp),%rax
  xor    %edx,%edx
  mov    0x0(%r13),%rsi
  mov    0x18(%rsp),%rcx
  div    %r12
  xor    %edx,%edx
  mov    %rax,%rdi
  mov    %r12,%rax
  div    %rsi
  lea    -0x1(%rsi),%rdx
  imul   %rdi,%rdx
  sub    %rdx,%rbx
  lea    (%rcx,%rbx,8),%r8
  mov    %rax,%r12
  cmp    $0x4,%rsi
  jne    L32
  mov    %r14,%rcx
  mov    %r15,%rdx
  mov    %rax,%rsi
  add    $0x1,%rbp
  call   <radf4@plt>
  mov    %r14,%rax
  sub    $0x8,%r13
  mov    %r15,%r14
  mov    %rax,%r15
  cmp    %rbp,0x10(%rsp)
  jae    L7
L62:
  mov    %r14,%r13
L136:
  cmp    %r13,0x20(%rsp)
  je     L48
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L65:
  mov    %r14,%rcx
  mov    %r15,%rdx
  mov    %rax,%rsi
  call   <radf2@plt>
  mov    %r14,%rax
  mov    %r15,%r14
  mov    %rax,%r15
  jmp    L111
  nopl   0x0(%rax)
L138:
  mov    %r14,%rcx
  mov    %r15,%rdx
  mov    %rax,%rsi
  call   <radf3@plt>
  mov    %r14,%rax
  mov    %r15,%r14
  mov    %rax,%r15
  jmp    L111
  nopl   0x0(%rax)
L16:
  mov    %r14,%rcx
  mov    %r15,%rdx
  mov    %rax,%rsi
  call   <radf5@plt>
  mov    %r14,%rax
  mov    %r15,%r14
  mov    %rax,%r15
  jmp    L111
  nopl   0x0(%rax)
L48:
  mov    0x8(%rsp),%rdx
  mov    0x28(%rsp),%rsi
  mov    0x20(%rsp),%rdi
  add    $0x38,%rsp
  pop    %rbx
  shl    $0x3,%rdx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    10d0 <memcpy@plt>
  xchg   %ax,%ax
L19:
  mov    %rdx,%r13
  jmp    L136
```

assistant: ```c
void rfftf1(size_t n, double c[], double ch[], const double wa[], const size_t ifac[]) {
  size_t k1, l1 = n, nf = ifac[1], iw = n - 1;
  double *p1 = ch, *p2 = c;
  for (k1 = 1; k1 <= nf; ++k1) {
    size_t ip = ifac[nf - k1 + 2];
    size_t ido = n / l1;
    l1 /= ip;
    iw -= (ip - 1) * ido;
    do {
      double *tmp_ = (p1);
      (p1) = (p2);
      (p2) = tmp_;
    } while (0);
    if (ip == 4)
      radf4(ido, l1, p1, p2, wa + iw);
    else if (ip == 2)
      radf2(ido, l1, p1, p2, wa + iw);
    else if (ip == 3)
      radf3(ido, l1, p1, p2, wa + iw);
    else if (ip == 5)
      radf5(ido, l1, p1, p2, wa + iw);
    else {
      if (ido == 1)
        do {
          double *tmp_ = (p1);
          (p1) = (p2);
          (p2) = tmp_;
        } while (0);
      radfg(ido, ip, l1, ido * l1, p1, p2, wa + iw);
      do {
        double *tmp_ = (p1);
        (p1) = (p2);
        (p2) = tmp_;
      } while (0);
    }
  }
  if (p1 == c)
    memcpy(c, ch, n * sizeof(double));
}
```