user: 这段汇编代码用C语言表示是什么样的：

```asm
<radf2>:
  endbr64
  push   %r15
  mov    %edi,%r9d
  mov    %esi,%r10d
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rcx,-0x10(%rsp)
  test   %esi,%esi
  jle    L76
  movslq %edi,%r11
  lea    (%rdi,%rdi,1),%esi
  mov    %r10d,%edi
  mov    %rdx,%rax
  imul   %r9d,%edi
  movslq %esi,%rsi
  mov    -0x10(%rsp),%rdx
  shl    $0x3,%r11
  shl    $0x3,%rsi
  xor    %ecx,%ecx
  movslq %edi,%rdi
  nopl   (%rax)
L54:
  movsd  (%rax),%xmm0
  addsd  (%rax,%rdi,8),%xmm0
  add    $0x1,%ecx
  movsd  %xmm0,(%rdx)
  movsd  (%rax),%xmm0
  subsd  (%rax,%rdi,8),%xmm0
  add    %r11,%rax
  movsd  %xmm0,-0x8(%rdx,%rsi,1)
  add    %rsi,%rdx
  cmp    %ecx,%r10d
  jne    L54
L76:
  cmp    $0x1,%r9d
  jle    L83
  cmp    $0x2,%r9d
  je     L67
  mov    %r9d,%eax
  and    $0x1,%eax
  mov    %eax,-0x4(%rsp)
  test   %r10d,%r10d
  jle    L83
  mov    %r10d,%r12d
  lea    (%r9,%r9,1),%ebp
  movslq %r9d,%r11
  mov    -0x10(%rsp),%rsi
  imul   %r9d,%r12d
  movslq %ebp,%rbp
  lea    0x0(,%r11,8),%r15
  xor    %ebx,%ebx
  shl    $0x3,%rbp
  lea    -0x18(%rbp),%r14
  movslq %r12d,%r12
  shl    $0x3,%r12
  lea    0x0(%r13,%r12,1),%rdi
  nopl   (%rax)
L49:
  mov    %rdi,%rdx
  lea    (%r14,%rsi,1),%rcx
  mov    $0x2,%eax
  sub    %r12,%rdx
  nop
L22:
  movsd  -0x10(%r8,%rax,8),%xmm0
  movsd  -0x8(%rdi,%rax,8),%xmm2
  add    $0x10,%rdx
  sub    $0x10,%rcx
  movsd  -0x8(%r8,%rax,8),%xmm3
  movsd  (%rdi,%rax,8),%xmm4
  movapd %xmm0,%xmm1
  mulsd  %xmm2,%xmm1
  movapd %xmm3,%xmm5
  mulsd  %xmm3,%xmm2
  mulsd  %xmm4,%xmm0
  mulsd  %xmm4,%xmm5
  subsd  %xmm2,%xmm0
  movsd  (%rdx),%xmm2
  addsd  %xmm5,%xmm1
  addsd  %xmm0,%xmm2
  movsd  %xmm2,(%rsi,%rax,8)
  subsd  (%rdx),%xmm0
  movsd  %xmm0,0x18(%rcx)
  movsd  -0x8(%rdx),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rsi,%rax,8)
  movsd  -0x8(%rdx),%xmm0
  add    $0x2,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,0x10(%rcx)
  cmp    %eax,%r9d
  jg     L22
  add    $0x1,%ebx
  add    %r15,%rdi
  add    %rbp,%rsi
  cmp    %ebx,%r10d
  jne    L49
  mov    -0x4(%rsp),%eax
  test   %eax,%eax
  jne    L83
L7:
  lea    (%r9,%r9,1),%esi
  mov    -0x10(%rsp),%rdx
  shl    $0x3,%r11
  xor    %ecx,%ecx
  imul   %r10d,%r9d
  movslq %esi,%rsi
  lea    -0x8(%r13,%r11,1),%rax
  movq   D1(%rip),%xmm1
  shl    $0x3,%rsi
  add    %r11,%rdx
  movslq %r9d,%rdi
  nopl   0x0(%rax)
L106:
  movsd  (%rax,%rdi,8),%xmm0
  add    $0x1,%ecx
  xorpd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx)
  movsd  (%rax),%xmm0
  add    %r11,%rax
  movsd  %xmm0,-0x8(%rdx)
  add    %rsi,%rdx
  cmp    %ecx,%r10d
  jg     L106
L83:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L67:
  test   %r10d,%r10d
  jle    L83
  mov    $0x2,%r11d
  jmp    L7
```

tool_calls: name: read_data
arguments: label: D1
type: byte[8]

tool: {"D1": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
void radf2(int ido, int l1, const double cc[], double ch[], const double wa1[]) {
  int i, k, ic;
  double ti2, tr2;
  for (k = 0; k < l1; k++) {
    ch[2 * k * ido] = cc[k * ido] + cc[(k + l1) * ido];
    ch[(2 * k + 1) * ido + ido - 1] = cc[k * ido] - cc[(k + l1) * ido];
  }
  if (ido < 2)
    return;
  if (ido != 2) {
    for (k = 0; k < l1; k++) {
      for (i = 2; i < ido; i += 2) {
        ic = ido - i;
        tr2 = wa1[i - 2] * cc[i - 1 + (k + l1) * ido] + wa1[i - 1] * cc[i + (k + l1) * ido];
        ti2 = wa1[i - 2] * cc[i + (k + l1) * ido] - wa1[i - 1] * cc[i - 1 + (k + l1) * ido];
        ch[i + 2 * k * ido] = cc[i + k * ido] + ti2;
        ch[ic + (2 * k + 1) * ido] = ti2 - cc[i + k * ido];
        ch[i - 1 + 2 * k * ido] = cc[i - 1 + k * ido] + tr2;
        ch[ic - 1 + (2 * k + 1) * ido] = cc[i - 1 + k * ido] - tr2;
      }
    }
    if (ido % 2 == 1)
      return;
  }
  for (k = 0; k < l1; k++) {
    ch[(2 * k + 1) * ido] = -cc[ido - 1 + (k + l1) * ido];
    ch[ido - 1 + 2 * k * ido] = cc[ido - 1 + k * ido];
  }
}
```