ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<countSort>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    %edx,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  mov    %rax,%rcx
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  test   %r12d,%r12d
  jle    L45
  mov    %rbp,%rsi
  lea    -0x1(%r12),%eax
  lea    0x4(%rbp,%rax,4),%r8
L15:
  mov    (%rsi),%eax
  cltd
  idiv   %ebx
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%edi
  sar    $0x1f,%edi
  sub    %edi,%edx
  lea    (%rdx,%rdx,4),%edi
  add    %edi,%edi
  sub    %edi,%eax
  movslq %eax,%rdx
  addl   $0x1,(%rsp,%rdx,4)
  add    $0x4,%rsi
  cmp    %r8,%rsi
  jne    L15
L45:
  lea    0x4(%rsp),%rax
  lea    0x28(%rsp),%rsi
L26:
  mov    -0x4(%rax),%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L26
  mov    %r12d,%esi
  sub    $0x1,%esi
  js     L60
  movslq %esi,%rsi
L11:
  mov    0x0(%rbp,%rsi,4),%edi
  mov    %edi,%eax
  cltd
  idiv   %ebx
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%r8d
  sar    $0x1f,%r8d
  sub    %r8d,%edx
  lea    (%rdx,%rdx,4),%r8d
  add    %r8d,%r8d
  sub    %r8d,%eax
  movslq %eax,%rdx
  mov    (%rsp,%rdx,4),%eax
  movslq %eax,%r8
  mov    %edi,-0x4(%rcx,%r8,4)
  sub    $0x1,%eax
  mov    %eax,(%rsp,%rdx,4)
  sub    $0x1,%rsi
  test   %esi,%esi
  jns    L11
L60:
  test   %r12d,%r12d
  jle    L42
  mov    %r12d,%r12d
  lea    (%rcx,%r12,4),%rsi
  mov    %rcx,%rax
  sub    %rcx,%rbp
L69:
  mov    (%rax),%edx
  mov    %edx,0x0(%rbp,%rax,1)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L69
L42:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void countSort(int *arr, int n, int exp) {
  int *output = (int *)malloc(sizeof(int) * n);
  int i, cont[10] = {0};
  for (i = 0; i < n; i++)
    cont[(arr[i] / exp) % 10]++;
  for (i = 1; i < 10; i++)
    cont[i] += cont[i - 1];
  for (i = n - 1; i >= 0; i--) {
    output[cont[(arr[i] / exp) % 10] - 1] = arr[i];
    cont[(arr[i] / exp) % 10]--;
  }
  for (i = 0; i < n; i++)
    arr[i] = output[i];
}
```
user: Translate the below assembly code into C programming language:

```asm
<countSort>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  lea    0x0(,%r12,4),%edi
  push   %rbx
  mov    %edx,%ebx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  pxor   %xmm0,%xmm0
  movq   $0x0,0x20(%rsp)
  mov    %rax,%rcx
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  test   %r12d,%r12d
  jle    L29
  lea    -0x1(%r12),%eax
  mov    %rbp,%rdi
  mov    %rax,%rsi
  lea    0x4(%rbp,%rax,4),%r8
  nopl   0x0(%rax,%rax,1)
L69:
  mov    (%rdi),%eax
  add    $0x4,%rdi
  cltd
  idiv   %ebx
  movslq %eax,%rdx
  mov    %eax,%r9d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%r9d
  sar    $0x22,%rdx
  sub    %r9d,%edx
  lea    (%rdx,%rdx,4),%r9d
  add    %r9d,%r9d
  sub    %r9d,%eax
  movslq %eax,%rdx
  addl   $0x1,(%rsp,%rdx,4)
  cmp    %rdi,%r8
  jne    L69
  mov    0x4(%rsp),%edx
  mov    (%rsp),%edi
L52:
  lea    0x4(%rsp),%rax
  lea    0x28(%rsp),%r8
  jmp    L83
  nopw   0x0(%rax,%rax,1)
L90:
  mov    (%rax),%edx
  mov    -0x4(%rax),%edi
L83:
  add    %edi,%edx
  add    $0x4,%rax
  mov    %edx,-0x4(%rax)
  cmp    %rax,%r8
  jne    L90
  movslq %esi,%rdi
  test   %esi,%esi
  js     L93
  nopw   0x0(%rax,%rax,1)
L70:
  mov    0x0(%rbp,%rdi,4),%esi
  sub    $0x1,%rdi
  mov    %esi,%eax
  cltd
  idiv   %ebx
  movslq %eax,%rdx
  mov    %eax,%r8d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%r8d
  sar    $0x22,%rdx
  sub    %r8d,%edx
  lea    (%rdx,%rdx,4),%r8d
  add    %r8d,%r8d
  sub    %r8d,%eax
  movslq %eax,%rdx
  movslq (%rsp,%rdx,4),%r8
  mov    %r8,%rax
  mov    %esi,-0x4(%rcx,%r8,4)
  sub    $0x1,%eax
  mov    %eax,(%rsp,%rdx,4)
  test   %edi,%edi
  jns    L70
L93:
  lea    (%rcx,%r12,4),%rdx
  sub    %rcx,%rbp
  test   %r12d,%r12d
  jle    L15
  nopl   0x0(%rax)
L22:
  mov    (%rcx),%eax
  mov    %eax,0x0(%rbp,%rcx,1)
  add    $0x4,%rcx
  cmp    %rcx,%rdx
  jne    L22
L15:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L29:
  xor    %edi,%edi
  xor    %edx,%edx
  lea    -0x1(%r12),%esi
  jmp    L52
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void countSort(int *arr, int n, int exp) {
  int *output = (int *)malloc(sizeof(int) * n);
  int i, cont[10] = {0};
  for (i = 0; i < n; i++)
    cont[(arr[i] / exp) % 10]++;
  for (i = 1; i < 10; i++)
    cont[i] += cont[i - 1];
  for (i = n - 1; i >= 0; i--) {
    output[cont[(arr[i] / exp) % 10] - 1] = arr[i];
    cont[(arr[i] / exp) % 10]--;
  }
  for (i = 0; i < n; i++)
    arr[i] = output[i];
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<check_authentication>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x24(%rbp)
  mov    -0x38(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    -0x20(%rbp),%rax
  lea    D8(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L6
  movl   $0x1,-0x24(%rbp)
L6:
  mov    -0x24(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D8
type: string

tool: D8: p@ssw0rd

assistant: ```c
int check_authentication(char *password) {
  int auth_flag = 0;
  char password_buffer[16];
  strcpy(password_buffer, password);
  if (strcmp(password_buffer, "p@ssw0rd") == 0) {
    auth_flag = 1;
  }
  return auth_flag;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<check_authentication>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <strcpy@plt>
  lea    D9(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %al
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  movzbl %al,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D9
type_infered: string

tool: D9: p@ssw0rd

assistant: ```c
int check_authentication(char *password) {
  int auth_flag = 0;
  char password_buffer[16];
  strcpy(password_buffer, password);
  if (strcmp(password_buffer, "p@ssw0rd") == 0) {
    auth_flag = 1;
  }
  return auth_flag;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<check_authentication>:
  endbr64
  push   %rbp
  mov    %rdi,%rsi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <strcpy@plt>
  lea    D9(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %al
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x20,%rsp
  movzbl %al,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D9
datatype: string

tool: D9: p@ssw0rd

assistant: ```c
int check_authentication(char *password) {
  int auth_flag = 0;
  char password_buffer[16];
  strcpy(password_buffer, password);
  if (strcmp(password_buffer, "p@ssw0rd") == 0) {
    auth_flag = 1;
  }
  return auth_flag;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<check_authentication>:
  endbr64
  push   %rbp
  mov    %rdi,%rsi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <strcpy@plt>
  lea    D1(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %al
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x20,%rsp
  movzbl %al,%eax
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D1
type_guessed: string

tool: D1: p@ssw0rd

assistant: ```c
int check_authentication(char *password) {
  int auth_flag = 0;
  char password_buffer[16];
  strcpy(password_buffer, password);
  if (strcmp(password_buffer, "p@ssw0rd") == 0) {
    auth_flag = 1;
  }
  return auth_flag;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x3c(%rbp)
  jmp    L10
L0:
  mov    -0x3c(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x3c(%rbp)
L10:
  cmpl   $0x9,-0x3c(%rbp)
  jle    L0
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D5(%rip),%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x38(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <bufwrite@plt>
  mov    -0x38(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <bufwrite_a@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D6
datatype: string
name: unpack_data
arguments: label: D1
datatype: string
name: unpack_data
arguments: label: D5
datatype: string

tool: D6: %d 

tool: D1: 
%d


tool: D5: I don't know how to do,possibly fllow my heart!

assistant: ```c
int main() {
  int i, a[10];
  for (i = 1; i <= 9; i++) {
    printf("%d ", i);
  }
  printf("\n%d\n", *(a + 10));
  void bufwrite(char *, int);
  void bufwrite_a(char *, int);
  char *p = "I don't know how to do,possibly fllow my heart!";
  bufwrite(p, 10);
  bufwrite_a(p, 10);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    $0x1,%ebx
  lea    D24(%rip),%rbp
L16:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  jne    L16
  mov    0x28(%rsp),%esi
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%esi
  lea    D6(%rip),%rbx
  mov    %rbx,%rdi
  call   <bufwrite@plt>
  mov    $0xa,%esi
  mov    %rbx,%rdi
  call   <bufwrite_a@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D24
type_infered: string
name: disassemble_data
arguments: label: D28
type_infered: string
name: disassemble_data
arguments: label: D6
type_infered: string

tool: D24: %d 

tool: D28: 
%d


tool: D6: I don't know how to do,possibly fllow my heart!

assistant: ```c
int main() {
  int i, a[10];
  for (i = 1; i <= 9; i++) {
    printf("%d ", i);
  }
  printf("\n%d\n", *(a + 10));
  void bufwrite(char *, int);
  void bufwrite_a(char *, int);
  char *p = "I don't know how to do,possibly fllow my heart!";
  bufwrite(p, 10);
  bufwrite_a(p, 10);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D11(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L3:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0xa,%ebx
  jne    L3
  mov    0x28(%rsp),%esi
  lea    D18(%rip),%rdi
  lea    D29(%rip),%rbp
  call   <printf@plt>
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <bufwrite@plt>
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <bufwrite_a@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D11
type: string
name: parse_data
arguments: label: D18
type: string
name: parse_data
arguments: label: D29
type: string

tool: D11: %d 

tool: D18: 
%d


tool: D29: I don't know how to do,possibly fllow my heart!

assistant: ```c
int main() {
  int i, a[10];
  for (i = 1; i <= 9; i++) {
    printf("%d ", i);
  }
  printf("\n%d\n", *(a + 10));
  void bufwrite(char *, int);
  void bufwrite_a(char *, int);
  char *p = "I don't know how to do,possibly fllow my heart!";
  bufwrite(p, 10);
  bufwrite_a(p, 10);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D11(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L13:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0xa,%ebx
  jne    L13
  mov    0x28(%rsp),%esi
  lea    D25(%rip),%rdi
  lea    D3(%rip),%rbp
  call   <printf@plt>
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <bufwrite@plt>
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <bufwrite_a@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D11
type_guessed: string
name: disassemble_data
arguments: data_label: D25
type_guessed: string
name: disassemble_data
arguments: data_label: D3
type_guessed: string

tool: D11: %d 

tool: D25: 
%d


tool: D3: I don't know how to do,possibly fllow my heart!

assistant: ```c
int main() {
  int i, a[10];
  for (i = 1; i <= 9; i++) {
    printf("%d ", i);
  }
  printf("\n%d\n", *(a + 10));
  void bufwrite(char *, int);
  void bufwrite_a(char *, int);
  char *p = "I don't know how to do,possibly fllow my heart!";
  bufwrite(p, 10);
  bufwrite_a(p, 10);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x1505,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L9
L7:
  mov    -0x14(%rbp),%eax
  shl    $0x5,%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  addq   $0x1,-0x8(%rbp)
L9:
  mov    -0x8(%rbp),%rax
  cmp    -0x30(%rbp),%rax
  jb     L7
  mov    -0x14(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int hash(void *data, size_t size) {
  const unsigned char *ptr;
  size_t i;
  unsigned int hashValue = 5381;
  ptr = data;
  for (i = 0; i < size; i++)
    hashValue = (hashValue << 5) + hashValue + (unsigned int)*ptr++;
  return hashValue;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<hash>:
  endbr64
  test   %rsi,%rsi
  je     L18
  add    %rdi,%rsi
  mov    $0x1505,%eax
L9:
  mov    %eax,%edx
  shl    $0x5,%edx
  add    %edx,%eax
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%edx
  add    %edx,%eax
  cmp    %rdi,%rsi
  jne    L9
  ret
L18:
  mov    $0x1505,%eax
  ret
```

assistant: ```c
unsigned int hash(void *data, size_t size) {
  const unsigned char *ptr;
  size_t i;
  unsigned int hashValue = 5381;
  ptr = data;
  for (i = 0; i < size; i++)
    hashValue = (hashValue << 5) + hashValue + (unsigned int)*ptr++;
  return hashValue;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<hash>:
  endbr64
  test   %rsi,%rsi
  je     L9
  add    %rdi,%rsi
  mov    $0x1505,%eax
  nopl   0x0(%rax)
L12:
  mov    %eax,%edx
  add    $0x1,%rdi
  shl    $0x5,%edx
  add    %edx,%eax
  movzbl -0x1(%rdi),%edx
  add    %edx,%eax
  cmp    %rdi,%rsi
  jne    L12
  ret
  nop
L9:
  mov    $0x1505,%eax
  ret
```

assistant: ```c
unsigned int hash(void *data, size_t size) {
  const unsigned char *ptr;
  size_t i;
  unsigned int hashValue = 5381;
  ptr = data;
  for (i = 0; i < size; i++)
    hashValue = (hashValue << 5) + hashValue + (unsigned int)*ptr++;
  return hashValue;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<hash>:
  endbr64
  test   %rsi,%rsi
  je     L2
  add    %rdi,%rsi
  mov    $0x1505,%eax
  nopl   0x0(%rax)
L1:
  mov    %eax,%edx
  add    $0x1,%rdi
  shl    $0x5,%edx
  add    %edx,%eax
  movzbl -0x1(%rdi),%edx
  add    %edx,%eax
  cmp    %rsi,%rdi
  jne    L1
  ret
  nop
L2:
  mov    $0x1505,%eax
  ret
```

assistant: ```c
unsigned int hash(void *data, size_t size) {
  const unsigned char *ptr;
  size_t i;
  unsigned int hashValue = 5381;
  ptr = data;
  for (i = 0; i < size; i++)
    hashValue = (hashValue << 5) + hashValue + (unsigned int)*ptr++;
  return hashValue;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x24(%rbp)
  lea    D23(%rip),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <ftell@plt>
  mov    %eax,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x28(%rbp)
  jmp    L83
L72:
  lea    -0x14(%rbp),%rdx
  lea    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <leer_numero@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <feof@plt>
  test   %rax,%rax
  jne    L38
  lea    -0x10(%rbp),%rdx
  lea    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <leer_numero@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <feof@plt>
  test   %rax,%rax
  jne    L49
  lea    -0xc(%rbp),%rax
  mov    $0x4,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rcx
  lea    -0x14(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <restar@plt>
  lea    -0xc(%rbp),%rdx
  lea    -0x2c(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <escribir_numero@plt>
  addl   $0x4,-0x24(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <feof@plt>
  test   %rax,%rax
  jne    L7
L83:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <feof@plt>
  test   %rax,%rax
  je     L72
  jmp    L89
L38:
  nop
  jmp    L89
L49:
  nop
  jmp    L89
L7:
  nop
L89:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    -0x24(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <truncate@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L80
  call   <__stack_chk_fail@plt>
L80:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D23
type_infered: string
name: retrieve_data
arguments: tag: D2
type_infered: string

tool: D23: rb+

tool: D2: enteros.dat

assistant: ```c
int main() {
  char n1[4];
  char n2[4];
  char res[4];
  int largo_archivo = 0;
  FILE *fp = fopen("enteros.dat", "rb+");
  int wr = ftell(fp);
  int rd = wr;
  while (feof(fp) == 0) {
    leer_numero(fp, &rd, n1);
    if (feof(fp) != 0)
      break;
    leer_numero(fp, &rd, n2);
    if (feof(fp) != 0)
      break;
    memset(&res, 0, sizeof(char) * 4);
    restar(n1, n2, res);
    escribir_numero(fp, &wr, res);
    largo_archivo += 4;
    if (feof(fp) != 0)
      break;
  }
  fclose(fp);
  truncate("enteros.dat", largo_archivo);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D20(%rip),%rsi
  lea    D26(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <ftell@plt>
  mov    %eax,0x4(%rsp)
  mov    %eax,0x8(%rsp)
  mov    $0x0,%r12d
  lea    0xc(%rsp),%r13
L41:
  mov    %rbx,%rdi
  call   <feof@plt>
  test   %rax,%rax
  jne    L22
  lea    0x8(%rsp),%rsi
  mov    %r13,%rdx
  mov    %rbx,%rdi
  call   <leer_numero@plt>
  mov    %rbx,%rdi
  call   <feof@plt>
  test   %rax,%rax
  jne    L22
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  mov    %rbx,%rdi
  call   <leer_numero@plt>
  mov    %rbx,%rdi
  call   <feof@plt>
  test   %rax,%rax
  jne    L22
  lea    0x14(%rsp),%rbp
  mov    $0x4,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdx
  mov    %r13,%rdi
  call   <restar@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdx
  mov    %rbx,%rdi
  call   <escribir_numero@plt>
  add    $0x4,%r12d
  mov    %rbx,%rdi
  call   <feof@plt>
  test   %rax,%rax
  je     L41
L22:
  mov    %rbx,%rdi
  call   <fclose@plt>
  mov    %r12d,%esi
  lea    D26(%rip),%rdi
  call   <truncate@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L45:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D20
datatype: string
name: unpack_data
arguments: label: D26
datatype: string

tool: D20: rb+

tool: D26: enteros.dat

assistant: ```c
int main() {
  char n1[4];
  char n2[4];
  char res[4];
  int largo_archivo = 0;
  FILE *fp = fopen("enteros.dat", "rb+");
  int wr = ftell(fp);
  int rd = wr;
  while (feof(fp) == 0) {
    leer_numero(fp, &rd, n1);
    if (feof(fp) != 0)
      break;
    leer_numero(fp, &rd, n2);
    if (feof(fp) != 0)
      break;
    memset(&res, 0, sizeof(char) * 4);
    restar(n1, n2, res);
    escribir_numero(fp, &wr, res);
    largo_archivo += 4;
    if (feof(fp) != 0)
      break;
  }
  fclose(fp);
  truncate("enteros.dat", largo_archivo);
  return 0;
}
```