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
<pevent_read_number>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x4,%edx
  je     L37
  jg     L2
  cmp    $0x1,%edx
  je     L68
  cmp    $0x2,%edx
  jne    L27
  movzwl (%rsi),%esi
  call   <__data2host2@plt>
  jmp    L27
L2:
  mov    $0x0,%eax
  cmp    $0x8,%edx
  jne    L27
  mov    %rsp,%rdi
  mov    $0x8,%edx
  call   <memcpy@plt>
  mov    (%rsp),%rsi
  mov    %rbx,%rdi
  call   <__data2host8@plt>
  cltq
  jmp    L27
L68:
  movzbl (%rsi),%eax
L27:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L82
  add    $0x10,%rsp
  pop    %rbx
  ret
L37:
  mov    (%rsi),%esi
  call   <__data2host4@plt>
  jmp    L27
L82:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long pevent_read_number(struct pevent *pevent, const void *ptr, int size) {
  switch (size) {
  case 1:
    return *(unsigned char *)ptr;
  case 2:
    return __data2host2(pevent, *(unsigned short *)(ptr));
  case 4:
    return __data2host4(pevent, *(unsigned int *)(ptr));
  case 8:
    return ({
      unsigned long long __val;
      memcpy(&__val, (ptr), sizeof(unsigned long long));
      __data2host8(pevent, __val);
    });
  default:
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pevent_read_number>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x4,%edx
  je     L25
  jg     L40
  cmp    $0x1,%edx
  je     L62
  cmp    $0x2,%edx
  jne    L89
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  movzwl (%rsi),%esi
  add    $0x10,%rsp
  pop    %rbp
  jmp    10d0 <__data2host2@plt>
  nopw   0x0(%rax,%rax,1)
L40:
  xor    %eax,%eax
  cmp    $0x8,%edx
  jne    L89
  mov    %rsp,%rdi
  mov    $0x8,%edx
  call   <memcpy@plt>
  mov    (%rsp),%rsi
  mov    %rbp,%rdi
  call   <__data2host8@plt>
  cltq
L89:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x10,%rsp
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L62:
  movzbl (%rsi),%eax
  jmp    L89
  nopl   (%rax)
L25:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  mov    (%rsi),%esi
  add    $0x10,%rsp
  pop    %rbp
  jmp    10b0 <__data2host4@plt>
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long pevent_read_number(struct pevent *pevent, const void *ptr, int size) {
  switch (size) {
  case 1:
    return *(unsigned char *)ptr;
  case 2:
    return __data2host2(pevent, *(unsigned short *)(ptr));
  case 4:
    return __data2host4(pevent, *(unsigned int *)(ptr));
  case 8:
    return ({
      unsigned long long __val;
      memcpy(&__val, (ptr), sizeof(unsigned long long));
      __data2host8(pevent, __val);
    });
  default:
    return 0;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<pevent_read_number>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x4,%edx
  je     L77
  jg     L17
  cmp    $0x1,%edx
  je     L10
  cmp    $0x2,%edx
  jne    L82
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L76
  movzwl (%rsi),%esi
  add    $0x10,%rsp
  pop    %rbp
  jmp    10d0 <__data2host2@plt>
  nopw   0x0(%rax,%rax,1)
L17:
  xor    %eax,%eax
  cmp    $0x8,%edx
  jne    L82
  mov    %rsp,%rdi
  mov    $0x8,%edx
  call   <memcpy@plt>
  mov    (%rsp),%rsi
  mov    %rbp,%rdi
  call   <__data2host8@plt>
  cltq
L82:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L76
  add    $0x10,%rsp
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  movzbl (%rsi),%eax
  jmp    L82
  nopl   (%rax)
L77:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L76
  mov    (%rsi),%esi
  add    $0x10,%rsp
  pop    %rbp
  jmp    10b0 <__data2host4@plt>
L76:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long pevent_read_number(struct pevent *pevent, const void *ptr, int size) {
  switch (size) {
  case 1:
    return *(unsigned char *)ptr;
  case 2:
    return __data2host2(pevent, *(unsigned short *)(ptr));
  case 4:
    return __data2host4(pevent, *(unsigned int *)(ptr));
  case 8:
    return ({
      unsigned long long __val;
      memcpy(&__val, (ptr), sizeof(unsigned long long));
      __data2host8(pevent, __val);
    });
  default:
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_http_content>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %ecx,-0x1c(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void parse_http_content(struct TCPRECORD *sess, struct NetFrame *frame, const unsigned char *px, unsigned length) {
  sess = (sess);
  frame = (frame);
  px = (px);
  length = (length);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<parse_http_content>:
  endbr64
  ret
```

assistant: ```c
void parse_http_content(struct TCPRECORD *sess, struct NetFrame *frame, const unsigned char *px, unsigned length) {
  sess = (sess);
  frame = (frame);
  px = (px);
  length = (length);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<parse_http_content>:
  endbr64
  ret
```

assistant: ```c
void parse_http_content(struct TCPRECORD *sess, struct NetFrame *frame, const unsigned char *px, unsigned length) {
  sess = (sess);
  frame = (frame);
  px = (px);
  length = (length);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<parse_http_content>:
  endbr64
  ret
```

assistant: ```c
void parse_http_content(struct TCPRECORD *sess, struct NetFrame *frame, const unsigned char *px, unsigned length) {
  sess = (sess);
  frame = (frame);
  px = (px);
  length = (length);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<Swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void Swap(int *a, int *b) {
  int t;
  t = *a;
  *a = *b;
  *b = t;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Swap>:
  endbr64
  mov    (%rdi),%eax
  mov    (%rsi),%edx
  mov    %edx,(%rdi)
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void Swap(int *a, int *b) {
  int t;
  t = *a;
  *a = *b;
  *b = t;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<Swap>:
  endbr64
  mov    (%rdi),%eax
  mov    (%rsi),%edx
  mov    %edx,(%rdi)
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void Swap(int *a, int *b) {
  int t;
  t = *a;
  *a = *b;
  *b = t;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Swap>:
  endbr64
  mov    (%rdi),%eax
  mov    (%rsi),%edx
  mov    %edx,(%rdi)
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void Swap(int *a, int *b) {
  int t;
  t = *a;
  *a = *b;
  *b = t;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<get_word_count>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x1d8,%rsp
  mov    %rdi,-0x1d8(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c4(%rbp)
  movq   $0x0,-0x1b8(%rbp)
  movq   $0x0,-0x1c0(%rbp)
  lea    -0x1b0(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x32,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movb   $0x0,-0x1c5(%rbp)
  lea    -0x1c0(%rbp),%rdx
  mov    -0x1d8(%rbp),%rax
  lea    D9(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtok_s@plt>
  mov    %rax,-0x1b8(%rbp)
  jmp    L8
L53:
  mov    -0x1b8(%rbp),%rdx
  lea    -0x1b0(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <dup_word@plt>
  cmp    $0x1,%eax
  je     L52
  mov    -0x1c4(%rbp),%ebx
  lea    0x1(%rbx),%eax
  mov    %eax,-0x1c4(%rbp)
  mov    -0x1b8(%rbp),%rax
  mov    %rax,%rdi
  call   <_strdup@plt>
  movslq %ebx,%rdx
  mov    %rax,-0x1b0(%rbp,%rdx,8)
L52:
  lea    -0x1c0(%rbp),%rax
  mov    %rax,%rdx
  lea    D9(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok_s@plt>
  mov    %rax,-0x1b8(%rbp)
L8:
  cmpq   $0x0,-0x1b8(%rbp)
  jne    L53
  movb   $0x0,-0x1c5(%rbp)
  jmp    L45
L64:
  movsbl -0x1c5(%rbp),%eax
  cltq
  mov    -0x1b0(%rbp,%rax,8),%rax
  test   %rax,%rax
  je     L24
  movsbl -0x1c5(%rbp),%eax
  cltq
  mov    -0x1b0(%rbp,%rax,8),%rax
  mov    %rax,%rdi
  call   <free@plt>
  movsbl -0x1c5(%rbp),%eax
  cltq
  movq   $0x0,-0x1b0(%rbp,%rax,8)
L24:
  movzbl -0x1c5(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x1c5(%rbp)
L45:
  movsbl -0x1c5(%rbp),%eax
  cmp    %eax,-0x1c4(%rbp)
  jg     L64
  mov    -0x1c4(%rbp),%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D9
data_type: string

tool: D9:  


assistant: ```c
int get_word_count(char *p_str) {
  int cnt = 0;
  char *token = NULL, *nextContext = NULL;
  char *wordList[50] = {0}, i = 0;
  token = strtok_s(p_str, " \n\r", &nextContext);
  while (token != NULL) {
    if (dup_word(wordList, token) == 1) {
    } else {
      wordList[cnt++] = _strdup(token);
    }
    token = strtok_s(NULL, " \n\r", &nextContext);
  }
  for (i = 0; i < cnt; i++) {
    if (wordList[i] != NULL) {
      free(wordList[i]);
      wordList[i] = NULL;
    }
  }
  return cnt;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_word_count>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1b8,%rsp
  mov    %rdi,%r8
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x8(%rsp)
  lea    0x10(%rsp),%rdi
  mov    $0x32,%ecx
  rep stos %rax,%es:(%rdi)
  lea    0x8(%rsp),%rdx
  lea    D12(%rip),%rsi
  mov    %r8,%rdi
  call   <strtok_s@plt>
  test   %rax,%rax
  je     L85
  mov    %rax,%rbx
  mov    $0x0,%ebp
  lea    D12(%rip),%r12
  jmp    L91
L11:
  lea    0x8(%rsp),%rdx
  mov    %r12,%rsi
  mov    $0x0,%edi
  call   <strtok_s@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L90
L91:
  lea    0x10(%rsp),%rdi
  mov    %rbx,%rsi
  call   <dup_word@plt>
  cmp    $0x1,%eax
  je     L11
  lea    0x1(%rbp),%r13d
  mov    %rbx,%rdi
  call   <_strdup@plt>
  movslq %ebp,%rbp
  mov    %rax,0x10(%rsp,%rbp,8)
  mov    %r13d,%ebp
  jmp    L11
L90:
  test   %ebp,%ebp
  jle    L88
  mov    $0x0,%r12d
  jmp    L53
L42:
  add    $0x1,%r12d
  movsbl %r12b,%ebx
  cmp    %ebp,%ebx
  jge    L88
L53:
  movslq %ebx,%rax
  mov    0x10(%rsp,%rax,8),%rdi
  test   %rdi,%rdi
  je     L42
  call   <free@plt>
  movslq %ebx,%rbx
  movq   $0x0,0x10(%rsp,%rbx,8)
  jmp    L42
L85:
  mov    $0x0,%ebp
L88:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  mov    %ebp,%eax
  add    $0x1b8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D12
type_guessed: string

tool: D12:  


assistant: ```c
int get_word_count(char *p_str) {
  int cnt = 0;
  char *token = NULL, *nextContext = NULL;
  char *wordList[50] = {0}, i = 0;
  token = strtok_s(p_str, " \n\r", &nextContext);
  while (token != NULL) {
    if (dup_word(wordList, token) == 1) {
    } else {
      wordList[cnt++] = _strdup(token);
    }
    token = strtok_s(NULL, " \n\r", &nextContext);
  }
  for (i = 0; i < cnt; i++) {
    if (wordList[i] != NULL) {
      free(wordList[i]);
      wordList[i] = NULL;
    }
  }
  return cnt;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_word_count>:
  endbr64
  push   %r15
  mov    %rdi,%r8
  mov    $0x32,%ecx
  push   %r14
  push   %r13
  push   %r12
  lea    D9(%rip),%r12
  push   %rbp
  mov    %r12,%rsi
  push   %rbx
  sub    $0x1b8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  lea    0x8(%rsp),%r13
  movq   $0x0,0x8(%rsp)
  mov    %rbx,%rdi
  mov    %r13,%rdx
  rep stos %rax,%es:(%rdi)
  mov    %r8,%rdi
  call   <strtok_s@plt>
  test   %rax,%rax
  je     L68
  mov    %rax,%rbp
  xor    %r14d,%r14d
  nopl   (%rax)
L12:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <dup_word@plt>
  cmp    $0x1,%eax
  je     L1
  mov    %rbp,%rdi
  lea    0x1(%r14),%r15d
  call   <_strdup@plt>
  mov    %rax,0x10(%rsp,%r14,8)
  movslq %r15d,%r14
L1:
  mov    %r13,%rdx
  mov    %r12,%rsi
  xor    %edi,%edi
  call   <strtok_s@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L12
  test   %r14d,%r14d
  je     L39
  xor    %eax,%eax
L4:
  movslq %eax,%rbx
  mov    0x10(%rsp,%rbx,8),%rdi
  test   %rdi,%rdi
  je     L24
  call   <free@plt>
  movq   $0x0,0x10(%rsp,%rbx,8)
L24:
  add    $0x1,%ebp
  movsbl %bpl,%eax
  cmp    %r14d,%eax
  jl     L4
L39:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L70
  add    $0x1b8,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L68:
  xor    %r14d,%r14d
  jmp    L39
L70:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D9
type_infered: string

tool: D9:  


assistant: ```c
int get_word_count(char *p_str) {
  int cnt = 0;
  char *token = NULL, *nextContext = NULL;
  char *wordList[50] = {0}, i = 0;
  token = strtok_s(p_str, " \n\r", &nextContext);
  while (token != NULL) {
    if (dup_word(wordList, token) == 1) {
    } else {
      wordList[cnt++] = _strdup(token);
    }
    token = strtok_s(NULL, " \n\r", &nextContext);
  }
  for (i = 0; i < cnt; i++) {
    if (wordList[i] != NULL) {
      free(wordList[i]);
      wordList[i] = NULL;
    }
  }
  return cnt;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_word_count>:
  endbr64
  push   %r15
  mov    %rdi,%r8
  mov    $0x32,%ecx
  push   %r14
  push   %r13
  lea    D4(%rip),%r13
  push   %r12
  mov    %r13,%rsi
  push   %rbp
  push   %rbx
  sub    $0x1b8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  lea    0x8(%rsp),%r14
  movq   $0x0,0x8(%rsp)
  mov    %r12,%rdi
  mov    %r14,%rdx
  rep stos %rax,%es:(%rdi)
  mov    %r8,%rdi
  call   <strtok_s@plt>
  test   %rax,%rax
  je     L21
  mov    %rax,%rbp
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L11:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <dup_word@plt>
  cmp    $0x1,%eax
  je     L13
L22:
  mov    %rbp,%rdi
  lea    0x1(%rbx),%r15d
  call   <_strdup@plt>
  mov    %r14,%rdx
  mov    %r13,%rsi
  xor    %edi,%edi
  mov    %rax,0x10(%rsp,%rbx,8)
  call   <strtok_s@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L58
  mov    %rbp,%rsi
  mov    %r12,%rdi
  movslq %r15d,%rbx
  call   <dup_word@plt>
  cmp    $0x1,%eax
  jne    L22
L13:
  mov    %r14,%rdx
  mov    %r13,%rsi
  xor    %edi,%edi
  call   <strtok_s@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L11
  test   %ebx,%ebx
  je     L21
  mov    %ebx,%r15d
L58:
  xor    %ebp,%ebp
  xor    %eax,%eax
  nopl   0x0(%rax)
L32:
  movslq %eax,%rbx
  mov    0x10(%rsp,%rbx,8),%rdi
  test   %rdi,%rdi
  je     L33
  call   <free@plt>
  movq   $0x0,0x10(%rsp,%rbx,8)
L33:
  add    $0x1,%ebp
  movsbl %bpl,%eax
  cmp    %r15d,%eax
  jl     L32
L81:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x1b8,%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L21:
  xor    %r15d,%r15d
  jmp    L81
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D4
datatype: string

tool: D4:  


assistant: ```c
int get_word_count(char *p_str) {
  int cnt = 0;
  char *token = NULL, *nextContext = NULL;
  char *wordList[50] = {0}, i = 0;
  token = strtok_s(p_str, " \n\r", &nextContext);
  while (token != NULL) {
    if (dup_word(wordList, token) == 1) {
    } else {
      wordList[cnt++] = _strdup(token);
    }
    token = strtok_s(NULL, " \n\r", &nextContext);
  }
  for (i = 0; i < cnt; i++) {
    if (wordList[i] != NULL) {
      free(wordList[i]);
      wordList[i] = NULL;
    }
  }
  return cnt;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<kill>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jne    L0
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <_exit@plt>
L0:
  mov    D7(%rip),%rax
  mov    (%rax),%edx
  mov    D1(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  leave
  ret
```

assistant: ```c
int kill(int pid, int sig) {
  if (pid == 1) {
    _exit(sig);
  }
  errno = ESRCH;
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<kill>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  je     L9
L1:
  mov    D19(%rip),%rax
  mov    (%rax),%edx
  mov    D13(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
L9:
  mov    %esi,%edi
  call   <_exit@plt>
  jmp    L1
```

assistant: ```c
int kill(int pid, int sig) {
  if (pid == 1) {
    _exit(sig);
  }
  errno = ESRCH;
  return -1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<kill>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  je     L7
L1:
  mov    D9(%rip),%rax
  mov    (%rax),%edx
  mov    D1(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax)
L7:
  mov    %esi,%edi
  call   <_exit@plt>
  jmp    L1
```

assistant: ```c
int kill(int pid, int sig) {
  if (pid == 1) {
    _exit(sig);
  }
  errno = ESRCH;
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<kill>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  je     L8
L5:
  mov    D15(%rip),%rax
  mov    (%rax),%edx
  mov    D7(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax)
L8:
  mov    %esi,%edi
  call   <_exit@plt>
  jmp    L5
```

assistant: ```c
int kill(int pid, int sig) {
  if (pid == 1) {
    _exit(sig);
  }
  errno = ESRCH;
  return -1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<xdr_u_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x82,%eax
  je     L43
  cmp    $0x82,%eax
  jg     L84
  cmp    $0x80,%eax
  je     L74
  cmp    $0x81,%eax
  jne    L84
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%eax
  mov    %rax,-0x10(%rbp)
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <XDR_PUTLONG@plt>
  jmp    L35
L43:
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <XDR_GETLONG@plt>
  test   %eax,%eax
  jne    L29
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  jmp    L35
L29:
  mov    -0x10(%rbp),%rax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
L74:
  mov    D12(%rip),%rax
  mov    (%rax),%eax
  jmp    L35
L84:
  mov    D6(%rip),%rax
  mov    (%rax),%eax
L35:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
bool_t xdr_u_int(XDR *xdrs, unsigned int *up) {
  if (sizeof(unsigned int) == sizeof(unsigned long)) {
    return (xdr_u_long(xdrs, (unsigned long *)up));
  } else if (sizeof(unsigned int) < sizeof(unsigned long)) {
    unsigned long l;
    switch (xdrs->x_op) {
    case XDR_ENCODE:
      l = (unsigned long)*up;
      return XDR_PUTLONG(xdrs, (long *)&l);
    case XDR_DECODE:
      if (!XDR_GETLONG(xdrs, (long *)&l))
        return FALSE;
      *up = (unsigned int)l;
    case XDR_FREE:
      return TRUE;
    }
    return FALSE;
  } else {
    return (xdr_short(xdrs, (short *)up));
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<xdr_u_int>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    (%rdi),%eax
  cmp    $0x81,%eax
  je     L71
  cmp    $0x82,%eax
  je     L1
  cmp    $0x80,%eax
  je     L19
  mov    D17(%rip),%rax
  mov    (%rax),%eax
  jmp    L3
L71:
  mov    (%rsi),%eax
  mov    %rax,(%rsp)
  mov    %rsp,%rsi
  call   <XDR_PUTLONG@plt>
  jmp    L3
L1:
  mov    %rsp,%rsi
  call   <XDR_GETLONG@plt>
  test   %eax,%eax
  jne    L25
  mov    D17(%rip),%rax
  mov    (%rax),%eax
  jmp    L3
L25:
  mov    (%rsp),%rax
  mov    %eax,(%rbx)
L19:
  mov    D22(%rip),%rax
  mov    (%rax),%eax
L3:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x10,%rsp
  pop    %rbx
  ret
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
bool_t xdr_u_int(XDR *xdrs, unsigned int *up) {
  if (sizeof(unsigned int) == sizeof(unsigned long)) {
    return (xdr_u_long(xdrs, (unsigned long *)up));
  } else if (sizeof(unsigned int) < sizeof(unsigned long)) {
    unsigned long l;
    switch (xdrs->x_op) {
    case XDR_ENCODE:
      l = (unsigned long)*up;
      return XDR_PUTLONG(xdrs, (long *)&l);
    case XDR_DECODE:
      if (!XDR_GETLONG(xdrs, (long *)&l))
        return FALSE;
      *up = (unsigned int)l;
    case XDR_FREE:
      return TRUE;
    }
    return FALSE;
  } else {
    return (xdr_short(xdrs, (short *)up));
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<xdr_u_int>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    (%rdi),%eax
  cmp    $0x81,%eax
  je     L17
  cmp    $0x82,%eax
  je     L69
  cmp    $0x80,%eax
  je     L40
L32:
  mov    D18(%rip),%rax
  mov    (%rax),%eax
L37:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L57
  add    $0x10,%rsp
  pop    %rbx
  ret
L69:
  mov    %rsp,%rsi
  call   <XDR_GETLONG@plt>
  test   %eax,%eax
  je     L32
  mov    (%rsp),%rax
  mov    %eax,(%rbx)
L40:
  mov    D19(%rip),%rax
  mov    (%rax),%eax
  jmp    L37
  nopl   (%rax)
L17:
  mov    (%rsi),%eax
  mov    %rsp,%rsi
  mov    %rax,(%rsp)
  call   <XDR_PUTLONG@plt>
  jmp    L37
L57:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
bool_t xdr_u_int(XDR *xdrs, unsigned int *up) {
  if (sizeof(unsigned int) == sizeof(unsigned long)) {
    return (xdr_u_long(xdrs, (unsigned long *)up));
  } else if (sizeof(unsigned int) < sizeof(unsigned long)) {
    unsigned long l;
    switch (xdrs->x_op) {
    case XDR_ENCODE:
      l = (unsigned long)*up;
      return XDR_PUTLONG(xdrs, (long *)&l);
    case XDR_DECODE:
      if (!XDR_GETLONG(xdrs, (long *)&l))
        return FALSE;
      *up = (unsigned int)l;
    case XDR_FREE:
      return TRUE;
    }
    return FALSE;
  } else {
    return (xdr_short(xdrs, (short *)up));
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xdr_u_int>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    (%rdi),%eax
  cmp    $0x81,%eax
  je     L62
  cmp    $0x82,%eax
  je     L26
  cmp    $0x80,%eax
  je     L21
L32:
  mov    D16(%rip),%rax
  mov    (%rax),%eax
L1:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x10,%rsp
  pop    %rbx
  ret
L26:
  mov    %rsp,%rsi
  call   <XDR_GETLONG@plt>
  test   %eax,%eax
  je     L32
  mov    (%rsp),%rax
  mov    %eax,(%rbx)
L21:
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  jmp    L1
  nopl   (%rax)
L62:
  mov    (%rsi),%eax
  mov    %rsp,%rsi
  mov    %rax,(%rsp)
  call   <XDR_PUTLONG@plt>
  jmp    L1
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
bool_t xdr_u_int(XDR *xdrs, unsigned int *up) {
  if (sizeof(unsigned int) == sizeof(unsigned long)) {
    return (xdr_u_long(xdrs, (unsigned long *)up));
  } else if (sizeof(unsigned int) < sizeof(unsigned long)) {
    unsigned long l;
    switch (xdrs->x_op) {
    case XDR_ENCODE:
      l = (unsigned long)*up;
      return XDR_PUTLONG(xdrs, (long *)&l);
    case XDR_DECODE:
      if (!XDR_GETLONG(xdrs, (long *)&l))
        return FALSE;
      *up = (unsigned int)l;
    case XDR_FREE:
      return TRUE;
    }
    return FALSE;
  } else {
    return (xdr_short(xdrs, (short *)up));
  }
}
```