ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<ce_close>:
  endbr64
  push   %rbp
  mov    $0x1,%esi
  mov    %edi,%ebp
  call   <ce_poller_lookup@plt>
  test   %rax,%rax
  jne    L12
L16:
  xor    %eax,%eax
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  mov    $0x2,%esi
  mov    %ebp,%edi
  call   <ce_poller_lookup@plt>
  test   %rax,%rax
  je     L16
  mov    %ebp,%edi
  call   <close@plt>
  test   %rax,%rax
  je     L16
  lea    D0(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D0
type_guessed: string

tool: D0: ERROR: Failed to close fd


assistant: ```c
int ce_close(int fd) {
  if (ce_poller_lookup(fd, 1) != 0 && ce_poller_lookup(fd, 2) != 0) {
    if (close(fd) != 0) {
      printf("ERROR: Failed to close fd\n");
      return -1;
    }
  }
  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
<ce_close>:
  endbr64
  push   %rbp
  mov    $0x1,%esi
  mov    %edi,%ebp
  call   <ce_poller_lookup@plt>
  test   %rax,%rax
  jne    L13
L5:
  xor    %eax,%eax
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L13:
  mov    $0x2,%esi
  mov    %ebp,%edi
  call   <ce_poller_lookup@plt>
  test   %rax,%rax
  je     L5
  mov    %ebp,%edi
  call   <close@plt>
  test   %rax,%rax
  je     L5
  lea    D9(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D9
type: string

tool: D9: ERROR: Failed to close fd


assistant: ```c
int ce_close(int fd) {
  if (ce_poller_lookup(fd, 1) != 0 && ce_poller_lookup(fd, 2) != 0) {
    if (close(fd) != 0) {
      printf("ERROR: Failed to close fd\n");
      return -1;
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<basename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L10
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  jmp    L15
L10:
  mov    -0x18(%rbp),%rax
L15:
  leave
  ret
```

assistant: ```c
char *basename(const char *filename) {
  char *p = strrchr(filename, '/');
  return p ? p + 1 : (char *)filename;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<basename>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x2f,%esi
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  cmovne %rdx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *basename(const char *filename) {
  char *p = strrchr(filename, '/');
  return p ? p + 1 : (char *)filename;
}
```
user: What is the c source code of the assembly code below:

```asm
<basename>:
  endbr64
  push   %rbx
  mov    $0x2f,%esi
  mov    %rdi,%rbx
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  pop    %rbx
  cmovne %rdx,%rax
  ret
```

assistant: ```c
char *basename(const char *filename) {
  char *p = strrchr(filename, '/');
  return p ? p + 1 : (char *)filename;
}
```
user: Translate the below assembly code into C programming language:

```asm
<basename>:
  endbr64
  push   %rbx
  mov    $0x2f,%esi
  mov    %rdi,%rbx
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  pop    %rbx
  cmovne %rdx,%rax
  ret
```

assistant: ```c
char *basename(const char *filename) {
  char *p = strrchr(filename, '/');
  return p ? p + 1 : (char *)filename;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<copy_tm_to_TM64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L11
  mov    -0x10(%rbp),%rax
  mov    $0x4,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L1
L11:
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L1:
  nop
  leave
  ret
```

assistant: ```c
void copy_tm_to_TM64(const struct tm *src, struct tm *dest) {
  if (src == NULL) {
    memset(dest, 0, sizeof(*dest));
  } else {
    memcpy(dest, src, sizeof(*dest));
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<copy_tm_to_TM64>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rax
  mov    %rsi,%rdi
  test   %rax,%rax
  je     L2
  mov    $0x4,%edx
  mov    %rax,%rsi
  call   <memcpy@plt>
L19:
  add    $0x8,%rsp
  ret
L2:
  mov    $0x4,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  jmp    L19
```

assistant: ```c
void copy_tm_to_TM64(const struct tm *src, struct tm *dest) {
  if (src == NULL) {
    memset(dest, 0, sizeof(*dest));
  } else {
    memcpy(dest, src, sizeof(*dest));
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<copy_tm_to_TM64>:
  endbr64
  mov    %rdi,%r8
  mov    $0x4,%edx
  mov    %rsi,%rdi
  test   %r8,%r8
  je     L7
  mov    %r8,%rsi
  jmp    1070 <memcpy@plt>
  nopl   0x0(%rax)
L7:
  xor    %esi,%esi
  jmp    1060 <memset@plt>
```

assistant: ```c
void copy_tm_to_TM64(const struct tm *src, struct tm *dest) {
  if (src == NULL) {
    memset(dest, 0, sizeof(*dest));
  } else {
    memcpy(dest, src, sizeof(*dest));
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<copy_tm_to_TM64>:
  endbr64
  mov    %rdi,%r8
  mov    $0x4,%edx
  mov    %rsi,%rdi
  test   %r8,%r8
  je     L0
  mov    %r8,%rsi
  jmp    1070 <memcpy@plt>
  nopl   0x0(%rax)
L0:
  xor    %esi,%esi
  jmp    1060 <memset@plt>
```

assistant: ```c
void copy_tm_to_TM64(const struct tm *src, struct tm *dest) {
  if (src == NULL) {
    memset(dest, 0, sizeof(*dest));
  } else {
    memcpy(dest, src, sizeof(*dest));
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<mul>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <stack_ok@plt>
  test   %eax,%eax
  jne    L4
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <stack_dump@plt>
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <assert@plt>
L4:
  movsd  D27(%rip),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  D27(%rip),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <stack_pop@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <stack_pop@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  movsd  -0x10(%rbp),%xmm0
  mulsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rdx
  mov    -0x18(%rbp),%rax
  movq   %rdx,%xmm0
  mov    %rax,%rdi
  call   <stack_push@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <stack_ok@plt>
  test   %eax,%eax
  jne    L12
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <stack_dump@plt>
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <assert@plt>
L12:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
type: string
name: retrieve_data
arguments: data_tag: D27
type: byte[8]
name: retrieve_data
arguments: data_tag: D10
type: string

tool: D8: INVALID STACK IN THE BEGINNING OF mul!!!


tool: {"D27": [0x00,0x00,0x00,0x00,0x00,0xd0,0x84,0xc0]}

tool: D10: INVALID STACK IN THE END OF mul!!!


assistant: ```c
int mul(struct Stack *stk_p) {
  if (!stack_ok(stk_p)) {
    stack_dump(stk_p);
    printf("INVALID STACK IN THE BEGINNING OF mul!!!\n");
    assert(NULL);
  }
  double operand1 = -666;
  double operand2 = -666;
  operand1 = stack_pop(stk_p);
  operand2 = stack_pop(stk_p);
  stack_push(stk_p, operand1 * operand2);
  if (!stack_ok(stk_p)) {
    stack_dump(stk_p);
    printf("INVALID STACK IN THE END OF mul!!!\n");
    assert(NULL);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mul>:
  endbr64
  push   %r14
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <stack_ok@plt>
  test   %eax,%eax
  je     L9
L23:
  mov    %rbx,%rdi
  call   <stack_pop@plt>
  movq   %xmm0,%r14
  mov    %rbx,%rdi
  call   <stack_pop@plt>
  movq   %r14,%xmm1
  mulsd  %xmm1,%xmm0
  mov    %rbx,%rdi
  call   <stack_push@plt>
  mov    %rbx,%rdi
  call   <stack_ok@plt>
  test   %eax,%eax
  je     L30
L27:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %r14
  ret
L9:
  mov    %rbx,%rdi
  call   <stack_dump@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  jmp    L23
L30:
  mov    %rbx,%rdi
  call   <stack_dump@plt>
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  jmp    L27
```

tool_calls: name: retrieve_data
arguments: tag: D0
type: string
name: retrieve_data
arguments: tag: D15
type: string

tool: D0: INVALID STACK IN THE BEGINNING OF mul!!!


tool: D15: INVALID STACK IN THE END OF mul!!!


assistant: ```c
int mul(struct Stack *stk_p) {
  if (!stack_ok(stk_p)) {
    stack_dump(stk_p);
    printf("INVALID STACK IN THE BEGINNING OF mul!!!\n");
    assert(NULL);
  }
  double operand1 = -666;
  double operand2 = -666;
  operand1 = stack_pop(stk_p);
  operand2 = stack_pop(stk_p);
  stack_push(stk_p, operand1 * operand2);
  if (!stack_ok(stk_p)) {
    stack_dump(stk_p);
    printf("INVALID STACK IN THE END OF mul!!!\n");
    assert(NULL);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mul>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  call   <stack_ok@plt>
  test   %eax,%eax
  je     L22
L29:
  mov    %rbp,%rdi
  call   <stack_pop@plt>
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <stack_pop@plt>
  mulsd  0x8(%rsp),%xmm0
  mov    %rbp,%rdi
  call   <stack_push@plt>
  mov    %rbp,%rdi
  call   <stack_ok@plt>
  test   %eax,%eax
  je     L18
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
  nopl   (%rax)
L22:
  mov    %rbp,%rdi
  call   <stack_dump@plt>
  lea    D18(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <assert@plt>
  jmp    L29
  nopl   (%rax)
L18:
  mov    %rbp,%rdi
  call   <stack_dump@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <assert@plt>
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D18
datatype: string
name: retrieve_data
arguments: label: D0
datatype: string

tool: D18: INVALID STACK IN THE BEGINNING OF mul!!!


tool: D0: INVALID STACK IN THE END OF mul!!!


assistant: ```c
int mul(struct Stack *stk_p) {
  if (!stack_ok(stk_p)) {
    stack_dump(stk_p);
    printf("INVALID STACK IN THE BEGINNING OF mul!!!\n");
    assert(NULL);
  }
  double operand1 = -666;
  double operand2 = -666;
  operand1 = stack_pop(stk_p);
  operand2 = stack_pop(stk_p);
  stack_push(stk_p, operand1 * operand2);
  if (!stack_ok(stk_p)) {
    stack_dump(stk_p);
    printf("INVALID STACK IN THE END OF mul!!!\n");
    assert(NULL);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<mul>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  call   <stack_ok@plt>
  test   %eax,%eax
  je     L5
L29:
  mov    %rbp,%rdi
  call   <stack_pop@plt>
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <stack_pop@plt>
  mulsd  0x8(%rsp),%xmm0
  mov    %rbp,%rdi
  call   <stack_push@plt>
  mov    %rbp,%rdi
  call   <stack_ok@plt>
  test   %eax,%eax
  je     L11
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
  nopl   (%rax)
L5:
  mov    %rbp,%rdi
  call   <stack_dump@plt>
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <assert@plt>
  jmp    L29
  nopl   (%rax)
L11:
  mov    %rbp,%rdi
  call   <stack_dump@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <assert@plt>
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
data_type: string
name: extract_symbolic_data
arguments: data_label: D0
data_type: string

tool: D5: INVALID STACK IN THE BEGINNING OF mul!!!


tool: D0: INVALID STACK IN THE END OF mul!!!


assistant: ```c
int mul(struct Stack *stk_p) {
  if (!stack_ok(stk_p)) {
    stack_dump(stk_p);
    printf("INVALID STACK IN THE BEGINNING OF mul!!!\n");
    assert(NULL);
  }
  double operand1 = -666;
  double operand2 = -666;
  operand1 = stack_pop(stk_p);
  operand2 = stack_pop(stk_p);
  stack_push(stk_p, operand1 * operand2);
  if (!stack_ok(stk_p)) {
    stack_dump(stk_p);
    printf("INVALID STACK IN THE END OF mul!!!\n");
    assert(NULL);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<Hex2Byte>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpq   $0x0,-0x28(%rbp)
  je     L51
  cmpq   $0x0,-0x30(%rbp)
  je     L51
  cmpq   $0x0,-0x38(%rbp)
  je     L51
  movq   $0x0,-0x18(%rbp)
  jmp    L1
L64:
  mov    -0x18(%rbp),%rax
  lea    (%rax,%rax,1),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xb(%rbp)
  mov    -0x18(%rbp),%rax
  add    %rax,%rax
  lea    0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xa(%rbp)
  movb   $0x0,-0x9(%rbp)
  lea    -0x20(%rbp),%rcx
  lea    -0xb(%rbp),%rax
  mov    $0x10,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x18(%rbp)
L1:
  mov    -0x18(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jb     L64
  jmp    L21
L51:
  nop
L21:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L43
  call   <__stack_chk_fail@plt>
L43:
  leave
  ret
```

assistant: ```c
void Hex2Byte(const char *hexString, unsigned char *byteArray, size_t byteArraySize) {
  if (NULL == hexString || NULL == byteArray || 0 == byteArraySize) {
    return;
  }
  char temp[3];
  char *end;
  for (size_t i = 0; i < byteArraySize; i++) {
    temp[0] = hexString[i * 2];
    temp[1] = hexString[i * 2 + 1];
    temp[2] = 0;
    byteArray[i] = (unsigned char)strtoul(temp, &end, 16);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<Hex2Byte>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdx,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  sete   %al
  test   %rdx,%rdx
  sete   %dl
  or     %dl,%al
  jne    L31
  mov    %rdi,%rbp
  mov    %rsi,%r12
  test   %rdi,%rdi
  je     L31
  mov    $0x0,%ebx
  lea    0x8(%rsp),%r14
L20:
  movzbl 0x0(%rbp,%rbx,2),%eax
  mov    %al,0x15(%rsp)
  movzbl 0x1(%rbp,%rbx,2),%eax
  mov    %al,0x16(%rsp)
  movb   $0x0,0x17(%rsp)
  lea    0x15(%rsp),%rdi
  mov    $0x10,%edx
  mov    %r14,%rsi
  call   <strtoul@plt>
  mov    %al,(%r12,%rbx,1)
  add    $0x1,%rbx
  cmp    %rbx,%r13
  ja     L20
L31:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void Hex2Byte(const char *hexString, unsigned char *byteArray, size_t byteArraySize) {
  if (NULL == hexString || NULL == byteArray || 0 == byteArraySize) {
    return;
  }
  char temp[3];
  char *end;
  for (size_t i = 0; i < byteArraySize; i++) {
    temp[0] = hexString[i * 2];
    temp[1] = hexString[i * 2 + 1];
    temp[2] = 0;
    byteArray[i] = (unsigned char)strtoul(temp, &end, 16);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<Hex2Byte>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  sete   %al
  test   %rdx,%rdx
  sete   %dl
  or     %dl,%al
  jne    L39
  mov    %rdi,%rbp
  test   %rdi,%rdi
  je     L39
  mov    %rsi,%r12
  lea    0x8(%rsp),%r15
  lea    0x15(%rsp),%r14
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L25:
  movzbl 0x0(%rbp,%rbx,2),%eax
  mov    $0x10,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  movb   $0x0,0x17(%rsp)
  mov    %al,0x15(%rsp)
  movzbl 0x1(%rbp,%rbx,2),%eax
  mov    %al,0x16(%rsp)
  call   <strtoul@plt>
  mov    %al,(%r12,%rbx,1)
  add    $0x1,%rbx
  cmp    %rbx,%r13
  ja     L25
L39:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L30:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void Hex2Byte(const char *hexString, unsigned char *byteArray, size_t byteArraySize) {
  if (NULL == hexString || NULL == byteArray || 0 == byteArraySize) {
    return;
  }
  char temp[3];
  char *end;
  for (size_t i = 0; i < byteArraySize; i++) {
    temp[0] = hexString[i * 2];
    temp[1] = hexString[i * 2 + 1];
    temp[2] = 0;
    byteArray[i] = (unsigned char)strtoul(temp, &end, 16);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<Hex2Byte>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  sete   %al
  test   %rdx,%rdx
  sete   %dl
  or     %dl,%al
  jne    L21
  mov    %rdi,%r15
  test   %rdi,%rdi
  je     L21
  mov    %rsi,%rbp
  lea    0x8(%rsp),%r14
  lea    0x14(%rsp),%r13
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L39:
  movzbl 0x1(%r15,%rbx,2),%ecx
  xor    %eax,%eax
  mov    (%r15,%rbx,2),%al
  mov    %r14,%rsi
  mov    $0x10,%edx
  mov    %r13,%rdi
  movb   $0x0,0x16(%rsp)
  mov    %cl,%ah
  mov    %ax,0x14(%rsp)
  call   <strtoul@plt>
  mov    %al,0x0(%rbp,%rbx,1)
  add    $0x1,%rbx
  cmp    %rbx,%r12
  ja     L39
L21:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void Hex2Byte(const char *hexString, unsigned char *byteArray, size_t byteArraySize) {
  if (NULL == hexString || NULL == byteArray || 0 == byteArraySize) {
    return;
  }
  char temp[3];
  char *end;
  for (size_t i = 0; i < byteArraySize; i++) {
    temp[0] = hexString[i * 2];
    temp[1] = hexString[i * 2 + 1];
    temp[2] = 0;
    byteArray[i] = (unsigned char)strtoul(temp, &end, 16);
  }
}
```
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    $0x660,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x658(%rbp)
  jmp    L5
L44:
  movl   $0x0,-0x654(%rbp)
  jmp    L21
L8:
  lea    -0x650(%rbp),%rcx
  mov    -0x654(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x658(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rsi,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x654(%rbp)
L21:
  cmpl   $0x13,-0x654(%rbp)
  jle    L8
  addl   $0x1,-0x658(%rbp)
L5:
  cmpl   $0x13,-0x658(%rbp)
  jle    L44
  lea    -0x650(%rbp),%rax
  mov    %rax,%rdi
  call   <find@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D9
data_type: string

tool: D9: %d

assistant: ```c
int main() {
  int arr[20][20];
  int i, j;
  for (i = 0; i < 20; i++) {
    for (j = 0; j < 20; j++) {
      scanf("%d", &arr[i][j]);
    }
  }
  find(arr);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x658,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x648(%rsp)
  xor    %eax,%eax
  lea    0x50(%rsp),%rbp
  lea    0x690(%rsp),%r13
  lea    D7(%rip),%r12
L26:
  lea    -0x50(%rbp),%rbx
L0:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L0
  add    $0x50,%rbp
  cmp    %r13,%rbp
  jne    L26
  mov    %rsp,%rdi
  call   <find@plt>
  mov    0x648(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  mov    $0x0,%eax
  add    $0x658,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D7
type: string

tool: D7: %d

assistant: ```c
int main() {
  int arr[20][20];
  int i, j;
  for (i = 0; i < 20; i++) {
    for (j = 0; j < 20; j++) {
      scanf("%d", &arr[i][j]);
    }
  }
  find(arr);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D8(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x658,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x648(%rsp)
  xor    %eax,%eax
  lea    0x50(%rsp),%rbp
  lea    0x690(%rsp),%r13
  nopl   0x0(%rax,%rax,1)
L8:
  lea    -0x50(%rbp),%rbx
  nopl   0x0(%rax)
L12:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %rbp,%rbx
  jne    L12
  lea    0x50(%rbx),%rbp
  cmp    %r13,%rbp
  jne    L8
  mov    %rsp,%rdi
  call   <find@plt>
  mov    0x648(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x658,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D8
type_infered: string

tool: D8: %d

assistant: ```c
int main() {
  int arr[20][20];
  int i, j;
  for (i = 0; i < 20; i++) {
    for (j = 0; j < 20; j++) {
      scanf("%d", &arr[i][j]);
    }
  }
  find(arr);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D3(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x658,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x648(%rsp)
  xor    %eax,%eax
  lea    0x50(%rsp),%rbp
  lea    0x690(%rsp),%r13
  nopl   0x0(%rax,%rax,1)
L27:
  lea    -0x50(%rbp),%rbx
  nopl   0x0(%rax)
L23:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %rbp,%rbx
  jne    L23
  lea    0x50(%rbx),%rbp
  cmp    %r13,%rbp
  jne    L27
  mov    %rsp,%rdi
  call   <find@plt>
  mov    0x648(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x658,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D3
datatype: string

tool: D3: %d

assistant: ```c
int main() {
  int arr[20][20];
  int i, j;
  for (i = 0; i < 20; i++) {
    for (j = 0; j < 20; j++) {
      scanf("%d", &arr[i][j]);
    }
  }
  find(arr);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parse_buffer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  movq   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x23,%al
  jne    L62
  mov    $0x0,%eax
  jmp    L32
L62:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L4
L11:
  addq   $0x1,-0x8(%rbp)
L4:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L11
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jne    L85
  mov    $0x0,%eax
  jmp    L32
L85:
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    $0x3d,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L76
  mov    $0xffffffff,%eax
  jmp    L32
L76:
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
L27:
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L27
  mov    -0x20(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    $0x1,%eax
L32:
  leave
  ret
```

assistant: ```c
int parse_buffer(char *buf, char **val) {
  char *ptr;
  *val = NULL;
  if (buf[0] == '#') {
    return 0;
  }
  ptr = buf;
  while (*ptr != '\n') {
    ptr++;
  }
  if (ptr == buf) {
    return 0;
  }
  *ptr = '\0';
  ptr = strstr(buf, "=");
  if (ptr == NULL) {
    return -1;
  }
  *ptr = '\0';
  do {
    ptr++;
  } while (*ptr == ' ');
  *val = ptr;
  return 1;
}
```