user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x100,%edi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <malloc@plt>
  mov    $0x100,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    $0x80,%edi
  mov    %rbp,%r15
  mov    %rbp,%rbx
  mov    %rax,%r13
  mov    %rax,0x8(%rsp)
  lea    0x100(%rbp),%r14
  call   <malloc@plt>
  sub    %rbp,%r13
  mov    %rax,%r12
  xchg   %ax,%ax
L46:
  mov    $0x80,%edi
  add    $0x8,%rbx
  call   <malloc@plt>
  mov    $0x80,%edi
  mov    %rax,-0x8(%rbx)
  call   <malloc@plt>
  mov    %rax,-0x8(%rbx,%r13,1)
  cmp    %r14,%rbx
  jne    L46
  movss  D4(%rip),%xmm1
  mov    %r12,%rbx
  mov    %r12,%rsi
  movss  D29(%rip),%xmm0
  lea    0x80(%r12),%r14
  nopl   0x0(%rax)
L123:
  mov    (%r15),%rcx
  mov    (%r15,%r13,1),%rdx
  xor    %eax,%eax
  nopl   0x0(%rax)
L37:
  movss  %xmm1,(%rcx,%rax,1)
  movss  %xmm0,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    $0x80,%rax
  jne    L37
  movl   $0x0,(%rsi)
  add    $0x4,%rsi
  add    $0x8,%r15
  cmp    %rsi,%r14
  jne    L123
  mov    0x8(%rsp),%rsi
  mov    $0x20,%ecx
  mov    %r12,%rdx
  mov    %rbp,%rdi
  mov    $0x20,%r9d
  mov    $0x20,%r8d
  mov    %r12,%r15
  call   <foo@plt>
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L99:
  add    $0x4,%r15
  cmp    %r14,%r15
  je     L56
L15:
  movss  (%r15),%xmm0
  ucomiss D21(%rip),%xmm0
  jp     L12
  je     L99
L12:
  xor    %eax,%eax
  add    $0x4,%r15
  call   <abort@plt>
  cmp    %r14,%r15
  jne    L15
L56:
  lea    0x8(%r12),%rdi
  mov    %r12,%rcx
  xor    %eax,%eax
  mov    $0x20,%r9d
  and    $0xfffffffffffffff8,%rdi
  movq   $0x0,(%r12)
  xor    %r8d,%r8d
  mov    %r12,%rdx
  movq   $0x0,0x78(%r12)
  sub    %rdi,%rcx
  sub    $0xffffff80,%ecx
  shr    $0x3,%ecx
  rep stos %rax,%es:(%rdi)
  mov    $0x20,%ecx
  mov    %rbp,%rdi
  mov    0x8(%rsp),%rsi
  call   <foo@plt>
  pxor   %xmm0,%xmm0
  jmp    L75
  nopl   0x0(%rax,%rax,1)
L87:
  add    $0x4,%rbx
  cmp    %rbx,%r14
  je     L128
L75:
  ucomiss (%rbx),%xmm0
  jp     L110
  je     L87
L110:
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <abort@plt>
  pxor   %xmm0,%xmm0
  cmp    %rbx,%r14
  jne    L75
L128:
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_label: D4
type_guessed: byte[4]
name: parse_data
arguments: data_label: D29
type_guessed: byte[4]
name: parse_data
arguments: data_label: D21
type_guessed: byte[4]

tool: {"D4": [0x00,0x00,0x00,0x40]}

tool: {"D29": [0x00,0x00,0xc0,0x3f]}

tool: {"D21": [0x00,0x00,0x40,0x45]}

assistant: ```c
int main() {
  const int N = 32;
  float **ar1, **ar2;
  float *res;
  int i, j;
  ar1 = (float **)malloc(N * sizeof(float *));
  ar2 = (float **)malloc(N * sizeof(float *));
  res = (float *)malloc(N * sizeof(float));
  for (i = 0; i < N; i++) {
    ar1[i] = (float *)malloc(N * sizeof(float));
    ar2[i] = (float *)malloc(N * sizeof(float));
  }
  for (i = 0; i < N; i++) {
    for (j = 0; j < N; j++) {
      ar1[i][j] = 2.0f;
      ar2[i][j] = 1.5f;
    }
    res[i] = 0.0f;
  }
  foo(ar1, ar2, res, N, N, N);
  for (i = 0; i < N; i++)
    if (res[i] != 3072.0f)
      abort();
  for (i = 0; i < N; i++)
    res[i] = 0.0f;
  foo(ar1, ar2, res, N, 0, N);
  for (i = 0; i < N; i++)
    if (res[i] != 0.0f)
      abort();
  return 0;
}
```