ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D23(%rip),%rsi
  lea    D34(%rip),%rdi
  call   <printf@plt>
  lea    D20(%rip),%rsi
  lea    D30(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D23
type_infered: string
name: unpack_data
arguments: data_label: D34
type_infered: string
name: unpack_data
arguments: data_label: D20
type_infered: string
name: unpack_data
arguments: data_label: D30
type_infered: string

tool: D23: zero

tool: D34: a is %s


tool: D20: positive

tool: D30: b is %s


assistant: ```c
int main(void) {
  int a = 0;
  printf("a is %s\n", a == 0 ? "zero" : a > 0 ? "positive" : "negative");
  int b = 1;
  printf("b is %s\n", b == 0 ? "zero" : b > 0 ? "positive" : "negative");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D25(%rip),%rsi
  lea    D33(%rip),%rdi
  call   <printf@plt>
  lea    D15(%rip),%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D25
type_guessed: string
name: unpack_data
arguments: tag: D33
type_guessed: string
name: unpack_data
arguments: tag: D15
type_guessed: string
name: unpack_data
arguments: tag: D1
type_guessed: string

tool: D25: zero

tool: D33: a is %s


tool: D15: positive

tool: D1: b is %s


assistant: ```c
int main(void) {
  int a = 0;
  printf("a is %s\n", a == 0 ? "zero" : a > 0 ? "positive" : "negative");
  int b = 1;
  printf("b is %s\n", b == 0 ? "zero" : b > 0 ? "positive" : "negative");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rsi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  lea    D16(%rip),%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D8
type_guessed: string
name: parse_data
arguments: data_label: D18
type_guessed: string
name: parse_data
arguments: data_label: D16
type_guessed: string
name: parse_data
arguments: data_label: D1
type_guessed: string

tool: D8: zero

tool: D18: a is %s


tool: D16: positive

tool: D1: b is %s


assistant: ```c
int main(void) {
  int a = 0;
  printf("a is %s\n", a == 0 ? "zero" : a > 0 ? "positive" : "negative");
  int b = 1;
  printf("b is %s\n", b == 0 ? "zero" : b > 0 ? "positive" : "negative");
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<output_buf_reversed>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %edx,-0x44(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x3736353433323130,%rax
  movabs $0x6665646362613938,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  subl   $0x1,-0x44(%rbp)
  jmp    L15
L5:
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  movzbl -0x20(%rbp,%rax,1),%eax
  movsbl %al,%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <output_char@plt>
  subl   $0x1,-0x44(%rbp)
L15:
  cmpl   $0x0,-0x44(%rbp)
  jns    L5
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

assistant: ```c
void output_buf_reversed(struct string_formatter *formatter, const char *buf, int nchars) {
  const char hex_val[] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'};
  for (nchars = nchars - 1; nchars >= 0; --nchars) {
    unsigned digit = buf[nchars];
    output_char(formatter, hex_val[digit]);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<output_buf_reversed>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movabs $0x3736353433323130,%rcx
  movabs $0x6665646362613938,%rbx
  mov    %rcx,(%rsp)
  mov    %rbx,0x8(%rsp)
  mov    %edx,%ebx
  sub    $0x1,%ebx
  js     L18
  mov    %rdi,%rbp
  movslq %ebx,%rbx
  add    %rsi,%rbx
  movslq %edx,%rax
  lea    -0x2(%rsi,%rax,1),%r12
  lea    -0x1(%rdx),%eax
  sub    %rax,%r12
L9:
  movsbl (%rbx),%eax
  movsbl (%rsp,%rax,1),%esi
  mov    %rbp,%rdi
  call   <output_char@plt>
  sub    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L9
L18:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void output_buf_reversed(struct string_formatter *formatter, const char *buf, int nchars) {
  const char hex_val[] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'};
  for (nchars = nchars - 1; nchars >= 0; --nchars) {
    unsigned digit = buf[nchars];
    output_char(formatter, hex_val[digit]);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<output_buf_reversed>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  movdqa D3(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %edx,%eax
  movaps %xmm0,(%rsp)
  sub    $0x1,%eax
  js     L2
  movslq %edx,%rdx
  movslq %eax,%rbx
  mov    %eax,%eax
  mov    %rdi,%rbp
  lea    -0x2(%rsi,%rdx,1),%r12
  add    %rsi,%rbx
  sub    %rax,%r12
  nopl   (%rax)
L28:
  movsbl (%rbx),%eax
  mov    %rbp,%rdi
  sub    $0x1,%rbx
  movsbl (%rsp,%rax,1),%esi
  call   <output_char@plt>
  cmp    %r12,%rbx
  jne    L28
L2:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D3
data_type: byte[16]

tool: {"D3": [0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x61,0x62,0x63,0x64,0x65,0x66]}

assistant: ```c
void output_buf_reversed(struct string_formatter *formatter, const char *buf, int nchars) {
  const char hex_val[] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'};
  for (nchars = nchars - 1; nchars >= 0; --nchars) {
    unsigned digit = buf[nchars];
    output_char(formatter, hex_val[digit]);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<output_buf_reversed>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  movdqa D6(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %edx,%eax
  movaps %xmm0,(%rsp)
  sub    $0x1,%eax
  js     L23
  movslq %edx,%rdx
  movslq %eax,%rbx
  mov    %eax,%eax
  mov    %rdi,%rbp
  lea    -0x2(%rsi,%rdx,1),%r12
  add    %rsi,%rbx
  sub    %rax,%r12
  nopl   (%rax)
L7:
  movsbl (%rbx),%eax
  mov    %rbp,%rdi
  sub    $0x1,%rbx
  movsbl (%rsp,%rax,1),%esi
  call   <output_char@plt>
  cmp    %r12,%rbx
  jne    L7
L23:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D6
type_infered: byte[16]

tool: {"D6": [0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x61,0x62,0x63,0x64,0x65,0x66]}

assistant: ```c
void output_buf_reversed(struct string_formatter *formatter, const char *buf, int nchars) {
  const char hex_val[] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'};
  for (nchars = nchars - 1; nchars >= 0; --nchars) {
    unsigned digit = buf[nchars];
    output_char(formatter, hex_val[digit]);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%edi
  call   <close@plt>
  mov    $0x1,%edi
  call   <close@plt>
  mov    $0x3,%edi
  call   <close@plt>
  mov    $0x4,%edi
  call   <close@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <pipe@plt>
  test   %rax,%rax
  jne    L20
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <pipe@plt>
  test   %rax,%rax
  je     L45
L20:
  mov    $0x4d,%eax
  jmp    L50
L45:
  mov    $0x0,%eax
  call   <test1@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L80
  mov    -0x14(%rbp),%eax
  jmp    L50
L80:
  mov    $0x0,%eax
  call   <test2@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L29
  mov    -0x14(%rbp),%eax
  jmp    L50
L29:
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x2a,%esi
  mov    $0x0,%edi
  call   <ppoll@plt>
  test   %rax,%rax
  jns    L88
  mov    $0x0,%eax
  jmp    L50
L88:
  mov    $0x4d,%eax
L50:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L56
  call   <__stack_chk_fail@plt>
L56:
  leave
  ret
```

assistant: ```c
int main(void) {
  int rc;
  int fds[2];
  (void)close(0);
  (void)close(1);
  (void)close(3);
  (void)close(4);
  if (pipe(fds) || pipe(fds))
    return 77;
  if ((rc = test1()))
    return rc;
  if ((rc = test2()))
    return rc;
  return ppoll(NULL, 42, NULL, NULL) < 0 ? 0 : 77;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%edi
  call   <close@plt>
  mov    $0x1,%edi
  call   <close@plt>
  mov    $0x3,%edi
  call   <close@plt>
  mov    $0x4,%edi
  call   <close@plt>
  mov    %rsp,%rdi
  call   <pipe@plt>
  mov    $0x4d,%ebx
  test   %rax,%rax
  jne    L18
  mov    %rsp,%rdi
  call   <pipe@plt>
  test   %rax,%rax
  jne    L26
  call   <test1@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L18
  mov    $0x0,%eax
  call   <test2@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L18
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x2a,%esi
  mov    $0x0,%edi
  call   <ppoll@plt>
  test   %rax,%rax
  mov    $0x4d,%eax
  cmovns %eax,%ebx
  jmp    L18
L26:
  mov    $0x4d,%ebx
L18:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    %ebx,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  int rc;
  int fds[2];
  (void)close(0);
  (void)close(1);
  (void)close(3);
  (void)close(4);
  if (pipe(fds) || pipe(fds))
    return 77;
  if ((rc = test1()))
    return rc;
  if ((rc = test2()))
    return rc;
  return ppoll(NULL, 42, NULL, NULL) < 0 ? 0 : 77;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  xor    %edi,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  call   <close@plt>
  mov    $0x1,%edi
  call   <close@plt>
  mov    $0x3,%edi
  call   <close@plt>
  mov    $0x4,%edi
  call   <close@plt>
  mov    %r12,%rdi
  call   <pipe@plt>
  test   %rax,%rax
  jne    L4
  mov    %r12,%rdi
  call   <pipe@plt>
  test   %rax,%rax
  jne    L4
  call   <test1@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L19
  xor    %eax,%eax
  call   <test2@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L19
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %edi,%edi
  mov    $0x2a,%esi
  call   <ppoll@plt>
  test   %rax,%rax
  js     L19
L4:
  mov    $0x4d,%r12d
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L51
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
L51:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  int rc;
  int fds[2];
  (void)close(0);
  (void)close(1);
  (void)close(3);
  (void)close(4);
  if (pipe(fds) || pipe(fds))
    return 77;
  if ((rc = test1()))
    return rc;
  if ((rc = test2()))
    return rc;
  return ppoll(NULL, 42, NULL, NULL) < 0 ? 0 : 77;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  xor    %edi,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  call   <close@plt>
  mov    $0x1,%edi
  call   <close@plt>
  mov    $0x3,%edi
  call   <close@plt>
  mov    $0x4,%edi
  call   <close@plt>
  mov    %r12,%rdi
  call   <pipe@plt>
  test   %rax,%rax
  jne    L37
  mov    %r12,%rdi
  call   <pipe@plt>
  test   %rax,%rax
  jne    L37
  call   <test1@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L32
  xor    %eax,%eax
  call   <test2@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L32
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %edi,%edi
  mov    $0x2a,%esi
  call   <ppoll@plt>
  test   %rax,%rax
  js     L32
L37:
  mov    $0x4d,%r12d
L32:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  int rc;
  int fds[2];
  (void)close(0);
  (void)close(1);
  (void)close(3);
  (void)close(4);
  if (pipe(fds) || pipe(fds))
    return 77;
  if ((rc = test1()))
    return rc;
  if ((rc = test2()))
    return rc;
  return ppoll(NULL, 42, NULL, NULL) < 0 ? 0 : 77;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  lea    D78(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  cmp    $0x3,%eax
  je     L55
  cmp    $0x3,%eax
  jg     L80
  cmp    $0x2,%eax
  je     L7
  cmp    $0x2,%eax
  jg     L80
  test   %eax,%eax
  je     L54
  cmp    $0x1,%eax
  je     L63
  jmp    L80
L54:
  lea    D92(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L4
L63:
  lea    D71(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L4
L7:
  lea    D32(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L4
L55:
  lea    D12(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L4
L80:
  lea    D97(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L4
L69:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L4:
  subl   $0x1,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L69
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  lea    D73(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D78
datatype: string
name: read_data
arguments: label: D0
datatype: string
name: read_data
arguments: label: D92
datatype: string
name: read_data
arguments: label: D71
datatype: string
name: read_data
arguments: label: D32
datatype: string
name: read_data
arguments: label: D12
datatype: string
name: read_data
arguments: label: D97
datatype: string
name: read_data
arguments: label: D57
datatype: string
name: read_data
arguments: label: D20
datatype: string
name: read_data
arguments: label: D73
datatype: string

tool: D78: -----------test2 with DEBUG macro = 0----------

tool: D0: argc:%d


tool: D92: zero

tool: D71: one

tool: D32: two

tool: D12: three

tool: D97: many

tool: D57: %s 

tool: D20: 
Argument count: %s


tool: D73: -----------------------------------------------

assistant: ```c
int main(int argc, char *argv[]) {
  printf("-----------test2 with DEBUG macro = 0----------\n");
  printf("argc:%d\n", argc);
  char *num;
  switch (argc - 1) {
  case 0:
    num = "zero";
    if (0) {
    case 1:
      num = "one";
      ;
    }
    if (0) {
    case 2:
      num = "two";
      ;
    }
    if (0) {
    case 3:
      num = "three";
      ;
    }
    if (0) {
    default:
      num = "many";
      ;
    }
    while (--argc)
      printf("%s ", argv[argc]);
    printf("\nArgument count: %s\n", num);
    break;
  }
  printf("-----------------------------------------------\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r12d
  mov    %rsi,%rbp
  lea    D105(%rip),%rdi
  call   <puts@plt>
  mov    %r12d,%esi
  lea    D90(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1(%r12),%ebx
  cmp    $0x3,%r12d
  je     L24
  cmp    $0x2,%ebx
  jg     L39
  lea    D60(%rip),%r13
  test   %ebx,%ebx
  je     L28
  cmp    $0x1,%ebx
  lea    D69(%rip),%r13
  lea    D7(%rip),%rax
  cmovne %rax,%r13
L54:
  movslq %ebx,%rbx
  lea    D85(%rip),%r12
L18:
  mov    0x0(%rbp,%rbx,8),%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  sub    $0x1,%rbx
  test   %ebx,%ebx
  jne    L18
L28:
  mov    %r13,%rsi
  lea    D102(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D50(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L39:
  cmp    $0x3,%ebx
  lea    D83(%rip),%r13
  lea    D7(%rip),%rax
  cmovne %rax,%r13
  jmp    L54
L24:
  lea    D27(%rip),%r13
  jmp    L54
```

tool_calls: name: parse_data
arguments: tag: D105
type_guessed: string
name: parse_data
arguments: tag: D90
type_guessed: string
name: parse_data
arguments: tag: D60
type_guessed: string
name: parse_data
arguments: tag: D69
type_guessed: string
name: parse_data
arguments: tag: D7
type_guessed: string
name: parse_data
arguments: tag: D85
type_guessed: string
name: parse_data
arguments: tag: D102
type_guessed: string
name: parse_data
arguments: tag: D50
type_guessed: string
name: parse_data
arguments: tag: D83
type_guessed: string
name: parse_data
arguments: tag: D27
type_guessed: string

tool: D105: -----------test2 with DEBUG macro = 0----------

tool: D90: argc:%d


tool: D60: zero

tool: D69: one

tool: D7: many

tool: D85: %s 

tool: D102: 
Argument count: %s


tool: D50: -----------------------------------------------

tool: D83: three

tool: D27: two

assistant: ```c
int main(int argc, char *argv[]) {
  printf("-----------test2 with DEBUG macro = 0----------\n");
  printf("argc:%d\n", argc);
  char *num;
  switch (argc - 1) {
  case 0:
    num = "zero";
    if (0) {
    case 1:
      num = "one";
      ;
    }
    if (0) {
    case 2:
      num = "two";
      ;
    }
    if (0) {
    case 3:
      num = "three";
      ;
    }
    if (0) {
    default:
      num = "many";
      ;
    }
    while (--argc)
      printf("%s ", argv[argc]);
    printf("\nArgument count: %s\n", num);
    break;
  }
  printf("-----------------------------------------------\n");
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edi,%ebx
  lea    D27(%rip),%rdi
  sub    $0x8,%rsp
  call   <puts@plt>
  mov    %ebx,%esi
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  sub    $0x1,%ebx
  cmp    $0x2,%ebx
  je     L30
  jg     L37
  lea    D103(%rip),%r13
  test   %ebx,%ebx
  je     L21
  cmp    $0x1,%ebx
  lea    D22(%rip),%r13
  lea    D68(%rip),%rax
  cmovne %rax,%r13
L9:
  movslq %ebx,%rbx
  lea    D14(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L18:
  mov    0x0(%rbp,%rbx,8),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  sub    $0x1,%rbx
  call   <printf@plt>
  test   %ebx,%ebx
  jne    L18
L21:
  mov    %r13,%rsi
  lea    D93(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D72(%rip),%rdi
  call   <puts@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L37:
  cmp    $0x3,%ebx
  lea    D91(%rip),%r13
  lea    D68(%rip),%rax
  cmovne %rax,%r13
  jmp    L9
L30:
  lea    D75(%rip),%r13
  jmp    L9
```

tool_calls: name: read_data
arguments: data_tag: D27
data_type: string
name: read_data
arguments: data_tag: D29
data_type: string
name: read_data
arguments: data_tag: D103
data_type: string
name: read_data
arguments: data_tag: D22
data_type: string
name: read_data
arguments: data_tag: D68
data_type: string
name: read_data
arguments: data_tag: D14
data_type: string
name: read_data
arguments: data_tag: D93
data_type: string
name: read_data
arguments: data_tag: D72
data_type: string
name: read_data
arguments: data_tag: D91
data_type: string
name: read_data
arguments: data_tag: D75
data_type: string

tool: D27: -----------test2 with DEBUG macro = 0----------

tool: D29: argc:%d


tool: D103: zero

tool: D22: one

tool: D68: many

tool: D14: %s 

tool: D93: 
Argument count: %s


tool: D72: -----------------------------------------------

tool: D91: three

tool: D75: two

assistant: ```c
int main(int argc, char *argv[]) {
  printf("-----------test2 with DEBUG macro = 0----------\n");
  printf("argc:%d\n", argc);
  char *num;
  switch (argc - 1) {
  case 0:
    num = "zero";
    if (0) {
    case 1:
      num = "one";
      ;
    }
    if (0) {
    case 2:
      num = "two";
      ;
    }
    if (0) {
    case 3:
      num = "three";
      ;
    }
    if (0) {
    default:
      num = "many";
      ;
    }
    while (--argc)
      printf("%s ", argv[argc]);
    printf("\nArgument count: %s\n", num);
    break;
  }
  printf("-----------------------------------------------\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edi,%ebx
  lea    D4(%rip),%rdi
  sub    $0x8,%rsp
  call   <puts@plt>
  mov    %ebx,%esi
  lea    D55(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  sub    $0x1,%ebx
  cmp    $0x2,%ebx
  je     L28
  jg     L15
  lea    D102(%rip),%r13
  test   %ebx,%ebx
  je     L16
  cmp    $0x1,%ebx
  lea    D75(%rip),%r13
  lea    D34(%rip),%rax
  cmovne %rax,%r13
L29:
  movslq %ebx,%rbx
  lea    D56(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L48:
  mov    0x0(%rbp,%rbx,8),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  sub    $0x1,%rbx
  call   <printf@plt>
  test   %ebx,%ebx
  jne    L48
L16:
  mov    %r13,%rsi
  lea    D51(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D47(%rip),%rdi
  call   <puts@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L15:
  cmp    $0x3,%ebx
  lea    D69(%rip),%r13
  lea    D34(%rip),%rax
  cmovne %rax,%r13
  jmp    L29
L28:
  lea    D19(%rip),%r13
  jmp    L29
```

tool_calls: name: retrieve_data
arguments: tag: D4
data_type: string
name: retrieve_data
arguments: tag: D55
data_type: string
name: retrieve_data
arguments: tag: D102
data_type: string
name: retrieve_data
arguments: tag: D75
data_type: string
name: retrieve_data
arguments: tag: D34
data_type: string
name: retrieve_data
arguments: tag: D56
data_type: string
name: retrieve_data
arguments: tag: D51
data_type: string
name: retrieve_data
arguments: tag: D47
data_type: string
name: retrieve_data
arguments: tag: D69
data_type: string
name: retrieve_data
arguments: tag: D19
data_type: string

tool: D4: -----------test2 with DEBUG macro = 0----------

tool: D55: argc:%d


tool: D102: zero

tool: D75: one

tool: D34: many

tool: D56: %s 

tool: D51: 
Argument count: %s


tool: D47: -----------------------------------------------

tool: D69: three

tool: D19: two

assistant: ```c
int main(int argc, char *argv[]) {
  printf("-----------test2 with DEBUG macro = 0----------\n");
  printf("argc:%d\n", argc);
  char *num;
  switch (argc - 1) {
  case 0:
    num = "zero";
    if (0) {
    case 1:
      num = "one";
      ;
    }
    if (0) {
    case 2:
      num = "two";
      ;
    }
    if (0) {
    case 3:
      num = "three";
      ;
    }
    if (0) {
    default:
      num = "many";
      ;
    }
    while (--argc)
      printf("%s ", argv[argc]);
    printf("\nArgument count: %s\n", num);
    break;
  }
  printf("-----------------------------------------------\n");
  return 0;
}
```