ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ltrim>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  test   %rdi,%rdi
  je     L9
  mov    %rdi,%rbx
  jmp    L3
L20:
  add    $0x1,%rbx
L3:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L20
  cmp    %rbp,%rbx
  je     L9
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <memmove@plt>
L9:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *ltrim(char *s) {
  if (s == NULL)
    return NULL;
  char *ptr = s;
  while (isspace(*ptr))
    ptr++;
  if (s != ptr)
    memmove(s, ptr, strlen(ptr) + 1);
  return s;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ltrim>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L14
  mov    %rdi,%rbp
  jmp    L7
  nopl   0x0(%rax,%rax,1)
L2:
  add    $0x1,%rbp
L7:
  movsbl 0x0(%rbp),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L2
  cmp    %r12,%rbp
  je     L14
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  lea    0x1(%rax),%rdx
  call   <memmove@plt>
L14:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *ltrim(char *s) {
  if (s == NULL)
    return NULL;
  char *ptr = s;
  while (isspace(*ptr))
    ptr++;
  if (s != ptr)
    memmove(s, ptr, strlen(ptr) + 1);
  return s;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ltrim>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L25
  mov    %rdi,%rbp
  jmp    L38
  nopl   0x0(%rax,%rax,1)
L21:
  add    $0x1,%rbp
L38:
  movsbl 0x0(%rbp),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L21
  cmp    %r12,%rbp
  je     L25
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  lea    0x1(%rax),%rdx
  call   <memmove@plt>
L25:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *ltrim(char *s) {
  if (s == NULL)
    return NULL;
  char *ptr = s;
  while (isspace(*ptr))
    ptr++;
  if (s != ptr)
    memmove(s, ptr, strlen(ptr) + 1);
  return s;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<problem10>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  call   <sqrt@plt>
  mov    %rax,%rcx
  mov    %ebx,%eax
  lea    0xf(,%rax,4),%rax
  shr    $0x4,%rax
  shl    $0x4,%rax
  mov    %rax,%rsi
  and    $0xfffffffffffff000,%rsi
  mov    %rsp,%rdx
  sub    %rsi,%rdx
L82:
  cmp    %rdx,%rsp
  je     L40
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L82
L40:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L85
  orq    $0x0,-0x8(%rsp,%rdx,1)
L85:
  mov    %rsp,%rsi
  cmp    $0x2,%ebx
  jbe    L106
  lea    0x8(%rsi),%rax
  lea    -0x3(%rbx),%edx
  lea    0xc(%rsi,%rdx,4),%rdx
L109:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L109
L106:
  cmp    $0x1,%ecx
  jbe    L39
  lea    -0x2(%rcx),%r9d
  add    $0x3,%r9
  mov    $0x4,%edi
  mov    $0x2,%edx
  mov    $0x0,%r8d
  jmp    L96
L28:
  add    $0x1,%rdx
  add    $0x2,%edi
  cmp    %r9,%rdx
  je     L140
L96:
  cmpl   $0x0,(%rsi,%rdx,4)
  je     L28
  add    %edx,%r8d
  cmp    %edi,%ebx
  jbe    L28
  movslq %edi,%rax
L8:
  movl   $0x0,(%rsi,%rax,4)
  add    %rdx,%rax
  cmp    %eax,%ebx
  ja     L8
  jmp    L28
L39:
  mov    $0x0,%r8d
L140:
  lea    0x1(%rcx),%eax
  cmp    %ebx,%eax
  jae    L99
  cltq
  jmp    L98
L97:
  add    $0x1,%rax
  cmp    %eax,%ebx
  jbe    L99
L98:
  cmpl   $0x0,(%rsi,%rax,4)
  je     L97
  add    %eax,%r8d
  jmp    L97
L99:
  mov    %r8d,%edx
  mov    %ebx,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L168
  mov    -0x8(%rbp),%rbx
  leave
  ret
L168:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
type: string

tool: D8: Problem 10: The sum of all primes below %u is %llu


assistant: ```c
void problem10(unsigned limit) {
  uint_fast64_t sum = 0;
  unsigned sqrt_limit = (int)sqrt(limit);
  bool sieve[limit];
  for (int i = 2; i < limit; ++i) {
    sieve[i] = true;
  }
  for (int i = 2; i <= sqrt_limit; ++i) {
    if (sieve[i]) {
      sum += i;
      for (int j = i * 2; j < limit; j += i) {
        sieve[j] = false;
      }
    }
  }
  for (int i = sqrt_limit + 1; i < limit; ++i) {
    if (sieve[i]) {
      sum += i;
    }
  }
  printf("Problem 10: The sum of all primes below %u is %llu\n", limit, sum);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<problem10>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  mov    %edi,%r12d
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  call   <sqrt@plt>
  mov    %rsp,%rdx
  mov    %rax,%rcx
  mov    %r12d,%eax
  lea    0xf(,%rax,4),%rax
  shr    $0x4,%rax
  shl    $0x4,%rax
  mov    %rax,%rsi
  and    $0xfffffffffffff000,%rsi
  sub    %rsi,%rdx
  cmp    %rdx,%rsp
  je     L82
L95:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rdx,%rsp
  jne    L95
L82:
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  jne    L8
L10:
  mov    %rsp,%rsi
  cmp    $0x2,%r12d
  jbe    L73
  lea    -0x3(%r12),%edx
  lea    0x8(%rsi),%rax
  lea    0xc(%rsi,%rdx,4),%rdx
  nopl   (%rax)
L55:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L55
L73:
  cmp    $0x1,%ecx
  jbe    L170
  lea    -0x2(%rcx),%r9d
  mov    $0x4,%edi
  mov    $0x2,%edx
  xor    %r8d,%r8d
  add    $0x3,%r9
  jmp    L83
  nop
L153:
  add    $0x1,%rdx
  add    $0x2,%edi
  cmp    %r9,%rdx
  je     L135
L83:
  mov    (%rsi,%rdx,4),%r10d
  test   %r10d,%r10d
  je     L153
  add    %edx,%r8d
  cmp    %edi,%r12d
  jbe    L153
  movslq %edi,%rax
L45:
  movl   $0x0,(%rsi,%rax,4)
  add    %rdx,%rax
  cmp    %eax,%r12d
  ja     L45
  add    $0x1,%rdx
  add    $0x2,%edi
  cmp    %r9,%rdx
  jne    L83
L135:
  lea    0x1(%rcx),%eax
  cmp    %r12d,%eax
  jae    L123
  nopl   0x0(%rax,%rax,1)
L127:
  mov    (%rsi,%rax,4),%edx
  test   %edx,%edx
  je     L90
  add    %eax,%r8d
L90:
  add    $0x1,%rax
  cmp    %eax,%r12d
  ja     L127
L123:
  mov    %r8d,%edx
  mov    %r12d,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L146
  mov    -0x8(%rbp),%r12
  leave
  ret
L8:
  orq    $0x0,-0x8(%rsp,%rax,1)
  jmp    L10
L170:
  xor    %r8d,%r8d
  jmp    L135
L146:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D6
type_infered: string

tool: D6: Problem 10: The sum of all primes below %u is %llu


assistant: ```c
void problem10(unsigned limit) {
  uint_fast64_t sum = 0;
  unsigned sqrt_limit = (int)sqrt(limit);
  bool sieve[limit];
  for (int i = 2; i < limit; ++i) {
    sieve[i] = true;
  }
  for (int i = 2; i <= sqrt_limit; ++i) {
    if (sieve[i]) {
      sum += i;
      for (int j = i * 2; j < limit; j += i) {
        sieve[j] = false;
      }
    }
  }
  for (int i = sqrt_limit + 1; i < limit; ++i) {
    if (sieve[i]) {
      sum += i;
    }
  }
  printf("Problem 10: The sum of all primes below %u is %llu\n", limit, sum);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<UsageFaultISR>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L4:
  jmp    L4
```

assistant: ```c
void UsageFaultISR(void) {
  while (1) {
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<UsageFaultISR>:
  endbr64
L8:
  jmp    L8
```

assistant: ```c
void UsageFaultISR(void) {
  while (1) {
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<UsageFaultISR>:
  endbr64
  nopl   0x0(%rax)
L4:
  jmp    L4
```

assistant: ```c
void UsageFaultISR(void) {
  while (1) {
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<UsageFaultISR>:
  endbr64
  nopl   0x0(%rax)
L4:
  jmp    L4
```

assistant: ```c
void UsageFaultISR(void) {
  while (1) {
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xd,-0xc(%rbp)
L17:
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jne    L17
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D11
type_infered: string
name: unpack_data
arguments: label: D2
type_infered: string
name: unpack_data
arguments: label: D37
type_infered: string
name: unpack_data
arguments: label: D1
type_infered: string

tool: D11: To enter the triskaidekaphobia therapy club,


tool: D2: please enter the secret code number: 

tool: D37: %d

tool: D1: Congratulations! You are cured!


assistant: ```c
int main() {
  const int secret_code = 13;
  int code_entered;
  do {
    printf("To enter the triskaidekaphobia therapy club,\n");
    printf("please enter the secret code number: ");
    scanf("%d", &code_entered);
  } while (code_entered != secret_code);
  printf("Congratulations! You are cured!\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```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    D15(%rip),%r13
  lea    D16(%rip),%r12
  lea    0x4(%rsp),%rbp
  lea    D19(%rip),%rbx
L5:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmpl   $0xd,0x4(%rsp)
  jne    L5
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D15
type_infered: string
name: extract_symbolic_data
arguments: tag: D16
type_infered: string
name: extract_symbolic_data
arguments: tag: D19
type_infered: string
name: extract_symbolic_data
arguments: tag: D11
type_infered: string

tool: D15: To enter the triskaidekaphobia therapy club,


tool: D16: please enter the secret code number: 

tool: D19: %d

tool: D11: Congratulations! You are cured!


assistant: ```c
int main() {
  const int secret_code = 13;
  int code_entered;
  do {
    printf("To enter the triskaidekaphobia therapy club,\n");
    printf("please enter the secret code number: ");
    scanf("%d", &code_entered);
  } while (code_entered != secret_code);
  printf("Congratulations! You are cured!\n");
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  lea    D39(%rip),%r13
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D14(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  nopl   0x0(%rax,%rax,1)
L14:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmpl   $0xd,0x4(%rsp)
  jne    L14
  lea    D37(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D39
type_guessed: string
name: disassemble_data
arguments: data_label: D0
type_guessed: string
name: disassemble_data
arguments: data_label: D14
type_guessed: string
name: disassemble_data
arguments: data_label: D37
type_guessed: string

tool: D39: To enter the triskaidekaphobia therapy club,


tool: D0: please enter the secret code number: 

tool: D14: %d

tool: D37: Congratulations! You are cured!


assistant: ```c
int main() {
  const int secret_code = 13;
  int code_entered;
  do {
    printf("To enter the triskaidekaphobia therapy club,\n");
    printf("please enter the secret code number: ");
    scanf("%d", &code_entered);
  } while (code_entered != secret_code);
  printf("Congratulations! You are cured!\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D39(%rip),%r13
  push   %r12
  lea    D26(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D12(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  nopl   0x0(%rax,%rax,1)
L12:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmpl   $0xd,0x4(%rsp)
  jne    L12
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D39
datatype: string
name: extract_symbolic_data
arguments: tag: D26
datatype: string
name: extract_symbolic_data
arguments: tag: D12
datatype: string
name: extract_symbolic_data
arguments: tag: D15
datatype: string

tool: D39: To enter the triskaidekaphobia therapy club,


tool: D26: please enter the secret code number: 

tool: D12: %d

tool: D15: Congratulations! You are cured!


assistant: ```c
int main() {
  const int secret_code = 13;
  int code_entered;
  do {
    printf("To enter the triskaidekaphobia therapy club,\n");
    printf("please enter the secret code number: ");
    scanf("%d", &code_entered);
  } while (code_entered != secret_code);
  printf("Congratulations! You are cured!\n");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<print_icmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  mov    0x4(%rax),%eax
  mov    %eax,%edi
  call   <inet_ntoa@plt>
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  lea    D15(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <inet_ntoa@plt>
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  lea    D8(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%ecx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  lea    D18(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L22
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%eax
  test   %eax,%eax
  jne    L22
  mov    -0x18(%rbp),%rax
  mov    0x8(%rax),%edx
  mov    -0x8(%rbp),%rax
  lea    D0(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
L22:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x8,%eax
  jne    L18
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%eax
  test   %eax,%eax
  jne    L18
  mov    -0x18(%rbp),%rax
  mov    0x8(%rax),%edx
  mov    -0x8(%rbp),%rax
  lea    D13(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
L18:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xa,%edi
  call   <fputc@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D15
datatype: string
name: retrieve_data
arguments: data_tag: D8
datatype: string
name: retrieve_data
arguments: data_tag: D18
datatype: string
name: retrieve_data
arguments: data_tag: D0
datatype: string
name: retrieve_data
arguments: data_tag: D13
datatype: string

tool: D15: ICMP:	 %s 

tool: D8:  --> %s 

tool: D18: Type %d , code %d

tool: D0:  echo reply: SEQNR = %d 

tool: D13:  echo reqest: SEQNR = %d 

assistant: ```c
void print_icmp(FILE *dst, const struct ip *ip, const struct icmphdr *icmp) {
  fprintf(dst, "ICMP:\t %s ", inet_ntoa(ip->ip_src));
  fprintf(dst, " --> %s ", inet_ntoa(ip->ip_dst));
  fprintf(dst, "Type %d , code %d", icmp->type, icmp->code);
  if (icmp->type == 0 && icmp->code == 0) {
    fprintf(dst, " echo reply: SEQNR = %d ", icmp->un.echo.sequence);
  }
  if (icmp->type == 8 && icmp->code == 0) {
    fprintf(dst, " echo reqest: SEQNR = %d ", icmp->un.echo.sequence);
  }
  fprintf(dst, "\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<print_icmp>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%rbx
  mov    0x4(%rsi),%edi
  call   <inet_ntoa@plt>
  mov    %rax,%rdx
  lea    D5(%rip),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  mov    (%r12),%edi
  call   <inet_ntoa@plt>
  mov    %rax,%rdx
  lea    D20(%rip),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  mov    0x4(%rbx),%ecx
  mov    (%rbx),%edx
  lea    D34(%rip),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  cmpl   $0x0,(%rbx)
  jne    L41
  cmpl   $0x0,0x4(%rbx)
  je     L11
L49:
  mov    %rbp,%rsi
  mov    $0xa,%edi
  call   <fputc@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  mov    0x8(%rbx),%edx
  lea    D29(%rip),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
L41:
  cmpl   $0x8,(%rbx)
  jne    L49
  cmpl   $0x0,0x4(%rbx)
  jne    L49
  mov    0x8(%rbx),%edx
  lea    D48(%rip),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  jmp    L49
```

tool_calls: name: read_data
arguments: tag: D5
data_type: string
name: read_data
arguments: tag: D20
data_type: string
name: read_data
arguments: tag: D34
data_type: string
name: read_data
arguments: tag: D29
data_type: string
name: read_data
arguments: tag: D48
data_type: string

tool: D5: ICMP:	 %s 

tool: D20:  --> %s 

tool: D34: Type %d , code %d

tool: D29:  echo reply: SEQNR = %d 

tool: D48:  echo reqest: SEQNR = %d 

assistant: ```c
void print_icmp(FILE *dst, const struct ip *ip, const struct icmphdr *icmp) {
  fprintf(dst, "ICMP:\t %s ", inet_ntoa(ip->ip_src));
  fprintf(dst, " --> %s ", inet_ntoa(ip->ip_dst));
  fprintf(dst, "Type %d , code %d", icmp->type, icmp->code);
  if (icmp->type == 0 && icmp->code == 0) {
    fprintf(dst, " echo reply: SEQNR = %d ", icmp->un.echo.sequence);
  }
  if (icmp->type == 8 && icmp->code == 0) {
    fprintf(dst, " echo reqest: SEQNR = %d ", icmp->un.echo.sequence);
  }
  fprintf(dst, "\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<print_icmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    0x4(%rsi),%edi
  mov    %rdx,%rbx
  call   <inet_ntoa@plt>
  lea    D1(%rip),%rsi
  mov    %rbp,%rdi
  mov    %rax,%rdx
  xor    %eax,%eax
  call   <fprintf@plt>
  mov    (%r12),%edi
  call   <inet_ntoa@plt>
  lea    D5(%rip),%rsi
  mov    %rbp,%rdi
  mov    %rax,%rdx
  xor    %eax,%eax
  call   <fprintf@plt>
  mov    0x4(%rbx),%ecx
  mov    (%rbx),%edx
  mov    %rbp,%rdi
  lea    D39(%rip),%rsi
  xor    %eax,%eax
  call   <fprintf@plt>
  mov    (%rbx),%eax
  test   %eax,%eax
  jne    L18
  mov    0x4(%rbx),%edx
  test   %edx,%edx
  je     L27
L42:
  pop    %rbx
  mov    %rbp,%rsi
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1080 <fputc@plt>
  xchg   %ax,%ax
L27:
  mov    0x8(%rbx),%edx
  lea    D35(%rip),%rsi
  mov    %rbp,%rdi
  call   <fprintf@plt>
  mov    (%rbx),%eax
L18:
  cmp    $0x8,%eax
  jne    L42
  mov    0x4(%rbx),%eax
  test   %eax,%eax
  jne    L42
  mov    0x8(%rbx),%edx
  lea    D28(%rip),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <fprintf@plt>
  jmp    L42
```

tool_calls: name: parse_data
arguments: tag: D1
type_guessed: string
name: parse_data
arguments: tag: D5
type_guessed: string
name: parse_data
arguments: tag: D39
type_guessed: string
name: parse_data
arguments: tag: D35
type_guessed: string
name: parse_data
arguments: tag: D28
type_guessed: string

tool: D1: ICMP:	 %s 

tool: D5:  --> %s 

tool: D39: Type %d , code %d

tool: D35:  echo reply: SEQNR = %d 

tool: D28:  echo reqest: SEQNR = %d 

assistant: ```c
void print_icmp(FILE *dst, const struct ip *ip, const struct icmphdr *icmp) {
  fprintf(dst, "ICMP:\t %s ", inet_ntoa(ip->ip_src));
  fprintf(dst, " --> %s ", inet_ntoa(ip->ip_dst));
  fprintf(dst, "Type %d , code %d", icmp->type, icmp->code);
  if (icmp->type == 0 && icmp->code == 0) {
    fprintf(dst, " echo reply: SEQNR = %d ", icmp->un.echo.sequence);
  }
  if (icmp->type == 8 && icmp->code == 0) {
    fprintf(dst, " echo reqest: SEQNR = %d ", icmp->un.echo.sequence);
  }
  fprintf(dst, "\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<print_icmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    0x4(%rsi),%edi
  mov    %rdx,%rbx
  call   <inet_ntoa@plt>
  lea    D25(%rip),%rsi
  mov    %rbp,%rdi
  mov    %rax,%rdx
  xor    %eax,%eax
  call   <fprintf@plt>
  mov    (%r12),%edi
  call   <inet_ntoa@plt>
  lea    D5(%rip),%rsi
  mov    %rbp,%rdi
  mov    %rax,%rdx
  xor    %eax,%eax
  call   <fprintf@plt>
  mov    0x4(%rbx),%ecx
  mov    (%rbx),%edx
  mov    %rbp,%rdi
  lea    D14(%rip),%rsi
  xor    %eax,%eax
  call   <fprintf@plt>
  mov    (%rbx),%eax
  test   %eax,%eax
  jne    L6
  mov    0x4(%rbx),%edx
  test   %edx,%edx
  je     L28
L49:
  pop    %rbx
  mov    %rbp,%rsi
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1080 <fputc@plt>
  xchg   %ax,%ax
L28:
  mov    0x8(%rbx),%edx
  lea    D24(%rip),%rsi
  mov    %rbp,%rdi
  call   <fprintf@plt>
  mov    (%rbx),%eax
L6:
  cmp    $0x8,%eax
  jne    L49
  mov    0x4(%rbx),%eax
  test   %eax,%eax
  jne    L49
  mov    0x8(%rbx),%edx
  lea    D22(%rip),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <fprintf@plt>
  jmp    L49
```

tool_calls: name: retrieve_data
arguments: tag: D25
type_guessed: string
name: retrieve_data
arguments: tag: D5
type_guessed: string
name: retrieve_data
arguments: tag: D14
type_guessed: string
name: retrieve_data
arguments: tag: D24
type_guessed: string
name: retrieve_data
arguments: tag: D22
type_guessed: string

tool: D25: ICMP:	 %s 

tool: D5:  --> %s 

tool: D14: Type %d , code %d

tool: D24:  echo reply: SEQNR = %d 

tool: D22:  echo reqest: SEQNR = %d 

assistant: ```c
void print_icmp(FILE *dst, const struct ip *ip, const struct icmphdr *icmp) {
  fprintf(dst, "ICMP:\t %s ", inet_ntoa(ip->ip_src));
  fprintf(dst, " --> %s ", inet_ntoa(ip->ip_dst));
  fprintf(dst, "Type %d , code %d", icmp->type, icmp->code);
  if (icmp->type == 0 && icmp->code == 0) {
    fprintf(dst, " echo reply: SEQNR = %d ", icmp->un.echo.sequence);
  }
  if (icmp->type == 8 && icmp->code == 0) {
    fprintf(dst, " echo reqest: SEQNR = %d ", icmp->un.echo.sequence);
  }
  fprintf(dst, "\n");
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<inter_unit_ck_int_eq>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    D16(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jne    L2
  mov    D4(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
L2:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int inter_unit_ck_int_eq(int first_test_number, int second_test_number) {
  int truth = EXIT_FAILURE;
  if (first_test_number == second_test_number) {
    truth = EXIT_SUCCESS;
  }
  return truth;
}
```