user: 这段汇编代码用C语言表示是什么样的：

```asm
<booth>:
  endbr64
  push   %r15
  mov    %edi,%r15d
  push   %r14
  mov    $0x1,%r14d
  push   %r13
  push   %r12
  lea    D55(%rip),%r12
  push   %rbp
  lea    D32(%rip),%rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  lea    D29(%rip),%rdi
  call   <puts@plt>
  lea    D47(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  lea    (%r15,%r15,1),%ecx
  mov    %ebx,%edx
  xor    %edi,%edi
  and    $0x1fe,%ecx
  lea    D86(%rip),%rsi
  mov    %r14d,%r15d
  mov    %ecx,%r13d
  call   <println@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    %ebx,%eax
  shl    $0x9,%eax
  and    $0x1fe00,%eax
  mov    %eax,0x8(%rsp)
  mov    %ebx,%eax
  neg    %eax
  shl    $0x9,%eax
  and    $0x1fe00,%eax
  mov    %eax,0x4(%rsp)
  jmp    L37
  nopw   0x0(%rax,%rax,1)
L59:
  mov    %r13d,%ecx
  mov    %ebx,%edx
  lea    D46(%rip),%rsi
  mov    %r15d,%edi
  call   <println@plt>
  mov    %r13d,%r14d
L28:
  and    $0x10000,%r13d
  mov    %r15d,%edi
  mov    %ebx,%edx
  mov    %r12,%rsi
  lea    0x20000(%r14),%eax
  cmovne %eax,%r14d
  add    $0x1,%r15d
  mov    %r14d,%r13d
  shr    %r13d
  mov    %r13d,%ecx
  call   <println@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  cmp    $0x9,%r15d
  je     L19
L37:
  mov    %r13d,%eax
  and    $0x3,%eax
  cmp    $0x1,%eax
  je     L6
  cmp    $0x2,%eax
  jne    L59
  add    0x4(%rsp),%r13d
  mov    %ebx,%edx
  lea    D87(%rip),%rsi
  mov    %r15d,%edi
  mov    %r13d,%ecx
  mov    %r13d,%r14d
  call   <println@plt>
  and    $0x1ffff,%r14d
  jmp    L28
  cs nopw 0x0(%rax,%rax,1)
L6:
  add    0x8(%rsp),%r13d
  mov    %ebx,%edx
  lea    D28(%rip),%rsi
  mov    %r15d,%edi
  mov    %r13d,%ecx
  mov    %r13d,%r14d
  call   <println@plt>
  and    $0x1ffff,%r14d
  jmp    L28
  nopl   0x0(%rax,%rax,1)
L19:
  mov    0xc(%rsp),%esi
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  imul   %ebx,%esi
  call   <printf@plt>
  add    $0x18,%rsp
  mov    %r14d,%edi
  xor    %esi,%esi
  pop    %rbx
  shr    $0x2,%edi
  pop    %rbp
  mov    $0x20,%edx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    10b0 <printbin@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D55
datatype: string
name: unpack_data
arguments: data_label: D32
datatype: string
name: unpack_data
arguments: data_label: D29
datatype: string
name: unpack_data
arguments: data_label: D47
datatype: string
name: unpack_data
arguments: data_label: D86
datatype: string
name: unpack_data
arguments: data_label: D46
datatype: string
name: unpack_data
arguments: data_label: D87
datatype: string
name: unpack_data
arguments: data_label: D28
datatype: string
name: unpack_data
arguments: data_label: D4
datatype: string

tool: D55: Shift Right

tool: D32: |____________|__________________|________________________________________|_________________________________________|

tool: D29: 
____________________________________________________________________________________________________________________

tool: D47: | Iteration  |     Operation    | 		Multiplicand		 |    		 Product 		   |

tool: D86: Initialization

tool: D46: No Operation

tool: D87: prod=prod-mcand

tool: D28: prod=prod+mcand

tool: D4: 

The answer is: %d

Binary: 

assistant: ```c
void booth(unsigned int multiplier, int multiplicand) {
  unsigned int product = multiplier;
  product &= 0xFF;
  printf("\n____________________________________________________________________________________________________________________\n");
  printf("| Iteration  |     Operation    | \t\tMultiplicand\t\t |    \t\t Product \t\t   |\n");
  printf("|____________|__________________|________________________________________|_________________________________________|\n");
  product <<= 1;
  println(0, "Initialization", multiplicand, product);
  printf("|____________|__________________|________________________________________|_________________________________________|\n");
  unsigned int add = (multiplicand << (8 + 1)) & 130560;
  unsigned int sub = (-multiplicand << (8 + 1)) & 130560;
  for (int i = 1; i <= 8; i++) {
    if ((product & 3) == 1) {
      product += add;
      println(i, "prod=prod+mcand", multiplicand, product);
    } else if ((product & 3) == 2) {
      product += sub;
      println(i, "prod=prod-mcand", multiplicand, product);
    } else
      println(i, "No Operation", multiplicand, product);
    product &= 0x1FFFF;
    if ((product & 65536) == 65536)
      product += 131072;
    product >>= 1;
    println(i, "Shift Right", multiplicand, product);
    printf("|____________|__________________|________________________________________|_________________________________________|\n");
  }
  product = product >> 1;
  int val = product;
  printf("\n\nThe answer is: %d\n\nBinary: ", multiplicand * multiplier);
  printbin(product, 0, 32);
}
```