user: 这段汇编代码用C语言表示是什么样的：

```asm
<printProgress>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %esi,-0x58(%rbp)
  mov    %edx,-0x5c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x54(%rbp),%xmm1
  movsd  D17(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ssl -0x58(%rbp),%xmm1
  cvtss2sd %xmm1,%xmm1
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0x40(%rbp)
  pxor   %xmm2,%xmm2
  cvtsi2ssl -0x54(%rbp),%xmm2
  movd   %xmm2,%edx
  lea    -0x27(%rbp),%rax
  movd   %edx,%xmm0
  lea    D10(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    -0x40(%rbp),%edx
  lea    -0x32(%rbp),%rax
  movd   %edx,%xmm0
  lea    D37(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    -0x27(%rbp),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x32(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x5c(%rbp),%eax
  sub    %edx,%eax
  mov    %eax,%ebx
  lea    -0x27(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  mov    %eax,-0x3c(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0x3c(%rbp),%xmm0
  mulss  -0x40(%rbp),%xmm0
  movss  D54(%rip),%xmm1
  divss  %xmm1,%xmm0
  movd   %xmm0,%eax
  movd   %eax,%xmm0
  call   <floor@plt>
  mov    %eax,-0x38(%rbp)
  movl   $0x0,-0x44(%rbp)
  jmp    L47
L8:
  mov    $0x3d,%edi
  call   <putchar@plt>
  addl   $0x1,-0x44(%rbp)
L47:
  mov    -0x38(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x44(%rbp)
  jl     L8
  mov    -0x3c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x44(%rbp)
  jge    L14
  mov    $0x3e,%edi
  call   <putchar@plt>
L14:
  mov    -0x38(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x44(%rbp)
  jmp    L59
L44:
  mov    $0x20,%edi
  call   <putchar@plt>
  addl   $0x1,-0x44(%rbp)
L59:
  mov    -0x3c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x44(%rbp)
  jl     L44
  lea    -0x32(%rbp),%rax
  mov    %rax,%rsi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L30
  call   <__stack_chk_fail@plt>
L30:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D17
type: f64
name: unpack_data
arguments: data_tag: D10
type: string
name: unpack_data
arguments: data_tag: D37
type: string
name: unpack_data
arguments: data_tag: D21
type: string
name: unpack_data
arguments: data_tag: D54
type: byte[4]
name: unpack_data
arguments: data_tag: D38
type: string

tool: D17: 100

tool: D10: It = %i [

tool: D37: ] %.3f%%

tool: D21: %s

tool: {"D54": [0x00,0x00,0xc8,0x42]}

tool: D38:  %s


assistant: ```c
void printProgress(int it, int itmax, int nx) {
  char start[15], end[11];
  int indent, newNX, i;
  float percPrg = it * 100.0 / (float)itmax;
  sprintf(start, "It = %i [", it);
  sprintf(end, "] %.3f%%", percPrg);
  printf("%s", start);
  newNX = nx - strlen(end) - strlen(start);
  indent = floor(percPrg * newNX / 100.0);
  for (i = 0; i < indent - 1; i++)
    printf("=");
  if (i < newNX - 1)
    printf(">");
  for (i = indent + 1; i < newNX - 1; i++)
    printf(" ");
  printf(" %s\n", end);
}
```