ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D5
datatype: string

tool: D5: %d

assistant: ```c
int main() {
  int a;
  int b;
  int comma;
  comma = (a = 5, b = a, a);
  printf("%d", comma);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D6
data_type: string

tool: D6: %d

assistant: ```c
int main() {
  int a;
  int b;
  int comma;
  comma = (a = 5, b = a, a);
  printf("%d", comma);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <vfork@plt>
  endbr64
  mov    %rax,-0x28(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L23
  lea    D0(%rip),%rax
  mov    %rax,-0x20(%rbp)
  lea    D36(%rip),%rax
  mov    %rax,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <execv@plt>
  mov    $0x0,%eax
  jmp    L2
L23:
  cmpq   $0x0,-0x28(%rbp)
  jle    L43
  lea    -0x2c(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <waitpid@plt>
  mov    -0x2c(%rbp),%eax
  mov    %eax,%edi
  call   <WIFEXITED@plt>
  test   %rax,%rax
  je     L31
  mov    $0x0,%eax
  jmp    L2
L31:
  mov    $0x1,%eax
  jmp    L2
L43:
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0x2,%eax
L2:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L49
  call   <__stack_chk_fail@plt>
L49:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D0
type_guessed: string
name: retrieve_data
arguments: data_label: D36
type_guessed: string
name: retrieve_data
arguments: data_label: D12
type_guessed: string
name: retrieve_data
arguments: data_label: D39
type_guessed: string

tool: D0: echo

tool: D36: hello

tool: D12: /bin/echo

tool: D39: vfork

assistant: ```c
int main(void) {
  pid_t res = vfork();
  if (res == 0) {
    char *args[] = {"echo", "hello", NULL};
    execv("/bin/echo", args);
  } else if (res > 0) {
    int status;
    waitpid(res, &status, 0);
    if (WIFEXITED(status))
      return 0;
    else
      return 1;
  } else {
    perror("vfork");
    return 2;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <vfork@plt>
  endbr64
  test   %rax,%rax
  je     L42
  jle    L29
  lea    0xc(%rsp),%rsi
  mov    $0x0,%edx
  mov    %rax,%rdi
  call   <waitpid@plt>
  mov    0xc(%rsp),%edi
  call   <WIFEXITED@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
L16:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x38,%rsp
  ret
L42:
  lea    D37(%rip),%rax
  mov    %rax,0x10(%rsp)
  lea    D20(%rip),%rax
  mov    %rax,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  lea    0x10(%rsp),%rsi
  lea    D32(%rip),%rdi
  call   <execv@plt>
  mov    $0x0,%eax
  jmp    L16
L29:
  lea    D21(%rip),%rdi
  call   <perror@plt>
  mov    $0x2,%eax
  jmp    L16
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D37
type: string
name: disassemble_data
arguments: tag: D20
type: string
name: disassemble_data
arguments: tag: D32
type: string
name: disassemble_data
arguments: tag: D21
type: string

tool: D37: echo

tool: D20: hello

tool: D32: /bin/echo

tool: D21: vfork

assistant: ```c
int main(void) {
  pid_t res = vfork();
  if (res == 0) {
    char *args[] = {"echo", "hello", NULL};
    execv("/bin/echo", args);
  } else if (res > 0) {
    int status;
    waitpid(res, &status, 0);
    if (WIFEXITED(status))
      return 0;
    else
      return 1;
  } else {
    perror("vfork");
    return 2;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <vfork@plt>
  endbr64
  test   %rax,%rax
  je     L43
  jle    L6
  xor    %edx,%edx
  lea    0xc(%rsp),%rsi
  mov    %rax,%rdi
  call   <waitpid@plt>
  mov    0xc(%rsp),%edi
  call   <WIFEXITED@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
L30:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x38,%rsp
  ret
L6:
  lea    D38(%rip),%rdi
  call   <perror@plt>
  mov    $0x2,%eax
  jmp    L30
L43:
  lea    D28(%rip),%rax
  lea    0x10(%rsp),%rsi
  movq   $0x0,0x20(%rsp)
  mov    %rax,0x10(%rsp)
  lea    D4(%rip),%rdi
  lea    D11(%rip),%rax
  mov    %rax,0x18(%rsp)
  call   <execv@plt>
  xor    %eax,%eax
  jmp    L30
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D38
type_guessed: string
name: disassemble_data
arguments: tag: D28
type_guessed: string
name: disassemble_data
arguments: tag: D4
type_guessed: string
name: disassemble_data
arguments: tag: D11
type_guessed: string

tool: D38: vfork

tool: D28: echo

tool: D4: /bin/echo

tool: D11: hello

assistant: ```c
int main(void) {
  pid_t res = vfork();
  if (res == 0) {
    char *args[] = {"echo", "hello", NULL};
    execv("/bin/echo", args);
  } else if (res > 0) {
    int status;
    waitpid(res, &status, 0);
    if (WIFEXITED(status))
      return 0;
    else
      return 1;
  } else {
    perror("vfork");
    return 2;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <vfork@plt>
  endbr64
  test   %rax,%rax
  je     L14
  jle    L22
  xor    %edx,%edx
  lea    0xc(%rsp),%rsi
  mov    %rax,%rdi
  call   <waitpid@plt>
  mov    0xc(%rsp),%edi
  call   <WIFEXITED@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
L8:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L48
  add    $0x38,%rsp
  ret
L22:
  lea    D35(%rip),%rdi
  call   <perror@plt>
  mov    $0x2,%eax
  jmp    L8
L14:
  lea    D32(%rip),%rax
  lea    D6(%rip),%rcx
  movq   $0x0,0x20(%rsp)
  movq   %rcx,%xmm0
  movq   %rax,%xmm1
  lea    0x10(%rsp),%rsi
  punpcklqdq %xmm1,%xmm0
  lea    D8(%rip),%rdi
  movaps %xmm0,0x10(%rsp)
  call   <execv@plt>
  xor    %eax,%eax
  jmp    L8
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D35
data_type: string
name: retrieve_data
arguments: label: D32
data_type: string
name: retrieve_data
arguments: label: D6
data_type: string
name: retrieve_data
arguments: label: D8
data_type: string

tool: D35: vfork

tool: D32: hello

tool: D6: echo

tool: D8: /bin/echo

assistant: ```c
int main(void) {
  pid_t res = vfork();
  if (res == 0) {
    char *args[] = {"echo", "hello", NULL};
    execv("/bin/echo", args);
  } else if (res > 0) {
    int status;
    waitpid(res, &status, 0);
    if (WIFEXITED(status))
      return 0;
    else
      return 1;
  } else {
    perror("vfork");
    return 2;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<packet_echoed_tcp_header>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <packet_echoed_ip_protocol@plt>
  mov    D8(%rip),%rdx
  mov    (%rdx),%rdx
  cmp    %rdx,%rax
  jne    L2
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <packet_echoed_layer4_header@plt>
  jmp    L14
L2:
  mov    $0x0,%eax
L14:
  leave
  ret
```

assistant: ```c
struct tcp *packet_echoed_tcp_header(struct packet *packet) {
  if (packet_echoed_ip_protocol(packet) == IPPROTO_TCP)
    return (struct tcp *)(packet_echoed_layer4_header(packet));
  return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<packet_echoed_tcp_header>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <packet_echoed_ip_protocol@plt>
  mov    %rax,%rcx
  mov    $0x0,%eax
  mov    D2(%rip),%rdx
  cmp    (%rdx),%rcx
  je     L13
L7:
  pop    %rbx
  ret
L13:
  mov    %rbx,%rdi
  call   <packet_echoed_layer4_header@plt>
  jmp    L7
```

assistant: ```c
struct tcp *packet_echoed_tcp_header(struct packet *packet) {
  if (packet_echoed_ip_protocol(packet) == IPPROTO_TCP)
    return (struct tcp *)(packet_echoed_layer4_header(packet));
  return NULL;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<packet_echoed_tcp_header>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <packet_echoed_ip_protocol@plt>
  mov    D5(%rip),%rdx
  cmp    (%rdx),%rax
  je     L3
  xor    %eax,%eax
  pop    %rbp
  ret
  nopl   (%rax)
L3:
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <packet_echoed_layer4_header@plt>
```

assistant: ```c
struct tcp *packet_echoed_tcp_header(struct packet *packet) {
  if (packet_echoed_ip_protocol(packet) == IPPROTO_TCP)
    return (struct tcp *)(packet_echoed_layer4_header(packet));
  return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<packet_echoed_tcp_header>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <packet_echoed_ip_protocol@plt>
  mov    D0(%rip),%rdx
  cmp    (%rdx),%rax
  je     L5
  xor    %eax,%eax
  pop    %rbp
  ret
  nopl   (%rax)
L5:
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <packet_echoed_layer4_header@plt>
```

assistant: ```c
struct tcp *packet_echoed_tcp_header(struct packet *packet) {
  if (packet_echoed_ip_protocol(packet) == IPPROTO_TCP)
    return (struct tcp *)(packet_echoed_layer4_header(packet));
  return NULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<power>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movsd  -0x18(%rbp),%xmm0
  mov    -0x20(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <pow@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  movsd  -0x18(%rbp),%xmm0
  mov    -0x20(%rbp),%rax
  movsd  -0x10(%rbp),%xmm1
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D44
type_guessed: string
name: extract_symbolic_data
arguments: tag: D13
type_guessed: string
name: extract_symbolic_data
arguments: tag: D5
type_guessed: string
name: extract_symbolic_data
arguments: tag: D28
type_guessed: string

tool: D44: +-------*--------*--------*--------*-------+

tool: D13: Power!
 please input two numbers:

tool: D5: %lf %lf

tool: D28: pow(%lf, %lf) = %lf


assistant: ```c
void power() {
  double result, num1, num2;
  printf("+-------*--------*--------*--------*-------+\n");
  printf("Power!\n please input two numbers:");
  scanf("%lf %lf", &num1, &num2);
  result = pow(num1, num2);
  printf("pow(%lf, %lf) = %lf\n", num1, num2, result);
  printf("+-------*--------*--------*--------*-------+\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<power>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D3(%rip),%rbx
  mov    %rbx,%rdi
  call   <puts@plt>
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%r12
  mov    0x8(%rsp),%rbp
  movq   %r12,%xmm1
  movq   %rbp,%xmm0
  call   <pow@plt>
  movapd %xmm0,%xmm2
  movq   %r12,%xmm1
  movq   %rbp,%xmm0
  lea    D30(%rip),%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <puts@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D3
type: string
name: unpack_data
arguments: label: D18
type: string
name: unpack_data
arguments: label: D24
type: string
name: unpack_data
arguments: label: D30
type: string

tool: D3: +-------*--------*--------*--------*-------+

tool: D18: Power!
 please input two numbers:

tool: D24: %lf %lf

tool: D30: pow(%lf, %lf) = %lf


assistant: ```c
void power() {
  double result, num1, num2;
  printf("+-------*--------*--------*--------*-------+\n");
  printf("Power!\n please input two numbers:");
  scanf("%lf %lf", &num1, &num2);
  result = pow(num1, num2);
  printf("pow(%lf, %lf) = %lf\n", num1, num2, result);
  printf("+-------*--------*--------*--------*-------+\n");
}
```
user: What is the c source code of the assembly code below?

```asm
<power>:
  endbr64
  push   %rbp
  lea    D39(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x20(%rsp),%rdx
  lea    0x18(%rsp),%rsi
  lea    D38(%rip),%rdi
  call   <scanf@plt>
  movsd  0x18(%rsp),%xmm3
  movsd  0x20(%rsp),%xmm1
  movapd %xmm3,%xmm0
  movsd  %xmm1,0x8(%rsp)
  movsd  %xmm3,(%rsp)
  call   <pow@plt>
  movsd  (%rsp),%xmm3
  movsd  0x8(%rsp),%xmm1
  lea    D28(%rip),%rdi
  movapd %xmm0,%xmm2
  mov    $0x3,%eax
  movapd %xmm3,%xmm0
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x30,%rsp
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D39
type_guessed: string
name: disassemble_data
arguments: tag: D16
type_guessed: string
name: disassemble_data
arguments: tag: D38
type_guessed: string
name: disassemble_data
arguments: tag: D28
type_guessed: string

tool: D39: +-------*--------*--------*--------*-------+

tool: D16: Power!
 please input two numbers:

tool: D38: %lf %lf

tool: D28: pow(%lf, %lf) = %lf


assistant: ```c
void power() {
  double result, num1, num2;
  printf("+-------*--------*--------*--------*-------+\n");
  printf("Power!\n please input two numbers:");
  scanf("%lf %lf", &num1, &num2);
  result = pow(num1, num2);
  printf("pow(%lf, %lf) = %lf\n", num1, num2, result);
  printf("+-------*--------*--------*--------*-------+\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<power>:
  endbr64
  push   %rbp
  lea    D22(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x20(%rsp),%rdx
  lea    0x18(%rsp),%rsi
  lea    D10(%rip),%rdi
  call   <scanf@plt>
  movsd  0x18(%rsp),%xmm3
  movsd  0x20(%rsp),%xmm1
  movapd %xmm3,%xmm0
  movsd  %xmm1,0x8(%rsp)
  movsd  %xmm3,(%rsp)
  call   <pow@plt>
  movsd  (%rsp),%xmm3
  movsd  0x8(%rsp),%xmm1
  lea    D23(%rip),%rdi
  movapd %xmm0,%xmm2
  mov    $0x3,%eax
  movapd %xmm3,%xmm0
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x30,%rsp
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D22
type_infered: string
name: disassemble_data
arguments: data_label: D0
type_infered: string
name: disassemble_data
arguments: data_label: D10
type_infered: string
name: disassemble_data
arguments: data_label: D23
type_infered: string

tool: D22: +-------*--------*--------*--------*-------+

tool: D0: Power!
 please input two numbers:

tool: D10: %lf %lf

tool: D23: pow(%lf, %lf) = %lf


assistant: ```c
void power() {
  double result, num1, num2;
  printf("+-------*--------*--------*--------*-------+\n");
  printf("Power!\n please input two numbers:");
  scanf("%lf %lf", &num1, &num2);
  result = pow(num1, num2);
  printf("pow(%lf, %lf) = %lf\n", num1, num2, result);
  printf("+-------*--------*--------*--------*-------+\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L35
L19:
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  mov    $0xf,%ecx
  mov    %eax,%edx
  mov    $0x0,%esi
  mov    $0x1,%edi
  call   <gen@plt>
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L35:
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L32
  mov    -0xc(%rbp),%eax
  test   %eax,%eax
  jg     L19
L32:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D11
data_type: string
name: unpack_data
arguments: data_tag: D13
data_type: string
name: unpack_data
arguments: data_tag: D8
data_type: string

tool: D11: %d:

tool: 

tool: D8: %d

assistant: ```c
int main() {
  int n;
  while (scanf("%d", &n) == 1 && n > 0) {
    printf("%d:", n);
    gen(1, 0, n, 15);
    puts("");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  lea    D2(%rip),%rbx
  lea    D11(%rip),%r13
  lea    D20(%rip),%r12
  jmp    L23
L33:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    $0xf,%ecx
  mov    0x4(%rsp),%edx
  mov    $0x0,%esi
  mov    $0x1,%edi
  call   <gen@plt>
  mov    %r12,%rdi
  call   <puts@plt>
L23:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L31
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jg     L33
L31:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
type_guessed: string
name: disassemble_data
arguments: data_tag: D11
type_guessed: string
name: disassemble_data
arguments: data_tag: D20
type_guessed: string

tool: D2: %d

tool: D11: %d:

tool: 

assistant: ```c
int main() {
  int n;
  while (scanf("%d", &n) == 1 && n > 0) {
    printf("%d:", n);
    gen(1, 0, n, 15);
    puts("");
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  lea    D27(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D10(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  jmp    L38
  nopw   0x0(%rax,%rax,1)
L39:
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L21
  mov    %r13,%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%edx
  mov    $0x1,%edi
  xor    %esi,%esi
  mov    $0xf,%ecx
  call   <gen@plt>
  mov    %r12,%rdi
  call   <puts@plt>
L38:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L39
L21:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D0
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D27
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D10
type_guessed: string

tool: D0: %d:

tool: 

tool: D10: %d

assistant: ```c
int main() {
  int n;
  while (scanf("%d", &n) == 1 && n > 0) {
    printf("%d:", n);
    gen(1, 0, n, 15);
    puts("");
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r13
  lea    D6(%rip),%r13
  push   %r12
  lea    D13(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D18(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  jmp    L6
  nopw   0x0(%rax,%rax,1)
L27:
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L39
  mov    %r13,%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%edx
  mov    $0x1,%edi
  xor    %esi,%esi
  mov    $0xf,%ecx
  call   <gen@plt>
  mov    %r12,%rdi
  call   <puts@plt>
L6:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L27
L39:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D6
type_infered: string
name: unpack_data
arguments: data_label: D13
type_infered: string
name: unpack_data
arguments: data_label: D18
type_infered: string

tool: D6: %d:

tool: 

tool: D18: %d

assistant: ```c
int main() {
  int n;
  while (scanf("%d", &n) == 1 && n > 0) {
    printf("%d:", n);
    gen(1, 0, n, 15);
    puts("");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_export_address>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <get_export@plt>
  cmp    $0xffffffff,%eax
  je     L18
  mov    -0x10(%rbp),%rax
  jmp    L14
L18:
  mov    $0x0,%eax
L14:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  leave
  ret
```

assistant: ```c
void *get_export_address(const char *name) {
  void *address;
  if (get_export(name, &address) != -1)
    return address;
  return NULL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<get_export_address>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <get_export@plt>
  cmp    $0xffffffff,%eax
  mov    $0x0,%eax
  cmovne (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *get_export_address(const char *name) {
  void *address;
  if (get_export(name, &address) != -1)
    return address;
  return NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_export_address>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <get_export@plt>
  cmp    $0xffffffff,%eax
  mov    $0x0,%eax
  cmovne (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *get_export_address(const char *name) {
  void *address;
  if (get_export(name, &address) != -1)
    return address;
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<get_export_address>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <get_export@plt>
  cmp    $0xffffffff,%eax
  mov    $0x0,%eax
  cmovne (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *get_export_address(const char *name) {
  void *address;
  if (get_export(name, &address) != -1)
    return address;
  return NULL;
}
```