user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<multiply>:
  endbr64
  lea    D2(%rip),%r8
  cmpb   $0x30,(%rdi)
  je     L45
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbx
  cmpb   $0x30,(%rsi)
  je     L93
  call   <strlen@plt>
  mov    %eax,%r15d
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%r14d
  mov    %r15d,0x4(%rsp)
  lea    (%r15,%rax,1),%ebp
  movslq %ebp,%rdi
  mov    %rdi,0x8(%rsp)
  lea    0x0(,%rbp,4),%r15d
  mov    %r15d,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  mov    %r15d,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    0x4(%rsp),%eax
  mov    %eax,%edi
  sub    $0x1,%edi
  js     L104
  lea    -0x1(%r14),%edx
  cltq
  movslq %edx,%rcx
  add    %rcx,%rax
  lea    0x0(%r13,%rax,4),%r8
  movslq %edi,%rsi
  add    %r12,%rsi
  mov    %edx,%r9d
  mov    %rcx,%r10
  movslq %r14d,%r14
  lea    -0x2(%rbx,%r14,1),%r11
  mov    %edx,%edx
  sub    %rdx,%r11
L59:
  test   %r9d,%r9d
  js     L30
  lea    (%rbx,%r10,1),%rcx
  mov    %r8,%rdx
L51:
  movsbl (%rsi),%eax
  sub    $0x30,%eax
  movsbl (%rcx),%r12d
  sub    $0x30,%r12d
  imul   %r12d,%eax
  add    %eax,(%rdx)
  sub    $0x4,%rdx
  sub    $0x1,%rcx
  cmp    %r11,%rcx
  jne    L51
L30:
  sub    $0x1,%edi
  sub    $0x4,%r8
  sub    $0x1,%rsi
  cmp    $0xffffffff,%edi
  jne    L59
L104:
  lea    -0x1(%rbp),%eax
  test   %eax,%eax
  jle    L24
  cltq
  lea    -0x4(%r13,%rax,4),%rdx
  mov    0x8(%rsp),%rax
  lea    -0xc(%r13,%rax,4),%rsi
  lea    -0x2(%rbp),%eax
  shl    $0x2,%rax
  sub    %rax,%rsi
L20:
  mov    0x4(%rdx),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  add    %eax,(%rdx)
  mov    0x4(%rdx),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,0x4(%rdx)
  sub    $0x4,%rdx
  cmp    %rsi,%rdx
  jne    L20
L24:
  lea    0x1(%rbp),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,%r8
  cmpl   $0x0,0x0(%r13)
  sete   %dl
  sete   %al
  movzbl %al,%eax
  cmp    %eax,%ebp
  jle    L73
  movzbl %dl,%eax
  lea    0x0(%r13,%rax,4),%rcx
  mov    %rdi,%rax
  movzbl %dl,%r9d
  lea    -0x1(%rbp),%edx
  sub    %r9d,%edx
  lea    0x1(%rdi,%rdx,1),%rsi
L50:
  movzbl (%rcx),%ebx
  lea    0x30(%rbx),%edx
  mov    %dl,(%rax)
  add    $0x4,%rcx
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L50
  sub    %r9d,%ebp
L29:
  movslq %ebp,%rax
  movb   $0x0,(%rdi,%rax,1)
L93:
  mov    %r8,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L73:
  mov    $0x0,%ebp
  jmp    L29
L45:
  mov    %r8,%rax
  ret
```

tool_calls: name: read_data
arguments: tag: D2
data_type: string

tool: D2: 0

assistant: ```c
char *multiply(char *num1, char *num2) {
  if (*num1 == '0' || *num2 == '0')
    return "0";
  int len1 = strlen(num1);
  int len2 = strlen(num2);
  int len = len1 + len2;
  int *arr = (int *)malloc(sizeof(int) * len);
  memset(arr, 0, sizeof(int) * len);
  for (int i = len1 - 1; i > -1; i--)
    for (int j = len2 - 1; j > -1; j--)
      arr[i + j + 1] += (num1[i] - '0') * (num2[j] - '0');
  for (int i = len - 1; i > 0; i--) {
    arr[i - 1] += arr[i] / 10;
    arr[i] %= 10;
  }
  char *s = (char *)malloc(sizeof(char) * (len + 1));
  int index = 0;
  int i = 0;
  if (arr[i] == 0)
    i++;
  while (i < len)
    s[index++] = arr[i++] + '0';
  s[index] = '\0';
  return s;
}
```