user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<imprime_matriz_lu>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  lea    D38(%rip),%rdi
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  lea    D40(%rip),%rbx
  sub    $0x28,%rsp
  call   <puts@plt>
  lea    -0x1(%r13),%rax
  mov    %rax,0x18(%rsp)
  test   %r13,%r13
  jle    L67
  nop
L152:
  lea    0x0(,%rbp,8),%rax
  mov    %rax,0x8(%rsp)
  mov    0x18(%rsp),%rax
  cmp    %rbp,%rax
  mov    %rax,%r14
  cmovg  %rbp,%r14
  xor    %r15d,%r15d
  jmp    L35
  nopl   0x0(%rax)
L94:
  mov    (%r12,%r15,8),%rax
  mov    0x8(%rsp),%rdx
  mov    %rbx,%rdi
  add    $0x1,%r15
  movsd  (%rax,%rdx,1),%xmm0
  mov    $0x1,%eax
  call   <printf@plt>
  cmp    %r14,%r15
  jg     L17
L35:
  cmp    %r15,%rbp
  jg     L94
  movsd  D11(%rip),%xmm0
  mov    %rbx,%rdi
  mov    $0x1,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r14,%r15
  jle    L35
L17:
  cmp    %r15,%r13
  jle    L80
  nopl   0x0(%rax)
L20:
  pxor   %xmm0,%xmm0
  mov    %rbx,%rdi
  mov    $0x1,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%r13
  jg     L20
L80:
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    0x1(%rbp),%rax
  mov    %rax,0x10(%rsp)
  cmp    %rax,%r13
  je     L100
  mov    %rax,%rbp
  jmp    L152
  nopw   0x0(%rax,%rax,1)
L100:
  lea    D26(%rip),%rdi
  call   <puts@plt>
  movq   $0x0,0x8(%rsp)
  mov    0x8(%rsp),%rax
  lea    0x0(,%rax,8),%r13
  test   %rax,%rax
  jne    L118
  nop
L134:
  xor    %r15d,%r15d
  nopl   0x0(%rax,%rax,1)
L81:
  mov    (%r12,%r15,8),%rax
  mov    %rbx,%rdi
  movsd  (%rax,%r13,1),%xmm0
  mov    $0x1,%eax
  call   <printf@plt>
  mov    %r15,%rax
  add    $0x1,%r15
  cmp    %rax,%rbp
  jg     L81
L40:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%rcx
  lea    0x1(%rcx),%rax
  cmp    %rcx,%rbp
  je     L114
  mov    %rax,0x8(%rsp)
  lea    0x0(,%rax,8),%r13
  test   %rax,%rax
  je     L134
L118:
  mov    %rax,%r14
  mov    0x10(%rsp),%rax
  cmp    %r14,%rax
  cmovle %rax,%r14
  xor    %r15d,%r15d
  nopw   0x0(%rax,%rax,1)
L91:
  pxor   %xmm0,%xmm0
  mov    %rbx,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    %r15,%rax
  add    $0x1,%r15
  cmp    %r14,%r15
  jl     L91
  cmp    %rbp,%rax
  jl     L81
  jmp    L40
  nopl   0x0(%rax,%rax,1)
L114:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L67:
  add    $0x28,%rsp
  lea    D26(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1080 <puts@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D38
data_type: string
name: parse_data
arguments: data_tag: D40
data_type: string
name: parse_data
arguments: data_tag: D11
data_type: f64
name: parse_data
arguments: data_tag: D26
data_type: string

tool: D38: 
 *************** MATRIZ L:****************

tool: D40: %f 

tool: D11: 1

tool: D26: 
 *************** MATRIZ U:****************

assistant: ```c
void imprime_matriz_lu(double **A, long n) {
  long i, j;
  float u = 1, c = 0;
  printf("\n *************** MATRIZ L:****************\n");
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      if (i < j)
        printf("%f ", c);
      else if (i > j)
        printf("%f ", A[j][i]);
      else
        printf("%f ", u);
    }
    printf("\n");
  }
  printf("\n *************** MATRIZ U:****************\n");
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      if (i <= j)
        printf("%f ", A[j][i]);
      else
        printf("%f ", c);
    }
    printf("\n");
  }
}
```