ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r14
  xor    %edi,%edi
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  lea    0x50(%rsp),%r12
  call   <time@plt>
  mov    %r14,%rbx
  mov    %r14,%rbp
  mov    %eax,%edi
  call   <srand@plt>
  nopl   0x0(%rax,%rax,1)
L15:
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <rand@plt>
  mov    %r13,%rdi
  movslq %eax,%rsi
  cltd
  imul   $0x10624dd3,%rsi,%rsi
  sar    $0x26,%rsi
  sub    %edx,%esi
  imul   $0x3e8,%esi,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L15
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x13,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  call   <merge_sort@plt>
  nopl   0x0(%rax)
L5:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L5
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x60,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D0
data_type: string

tool: D0: %d 

assistant: ```c
int main(int argc, char **argv) {
  int a[20];
  srand(time(NULL));
  int i;
  for (i = 0; i < 20; i++) {
    a[i] = rand() % (1000 - 0) + 0;
    printf("%d ", a[i]);
  }
  printf("\n");
  merge_sort(a, 0, 20 - 1);
  for (i = 0; i < 20; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<g>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D9(%rip),%eax
  pop    %rbp
  ret
```

assistant: ```c
int g() {
  static int x = 2;
  return x;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<g>:
  endbr64
  mov    $0x2,%eax
  ret
```

assistant: ```c
int g() {
  static int x = 2;
  return x;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<g>:
  endbr64
  mov    $0x2,%eax
  ret
```

assistant: ```c
int g() {
  static int x = 2;
  return x;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<g>:
  endbr64
  mov    $0x2,%eax
  ret
```

assistant: ```c
int g() {
  static int x = 2;
  return x;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1c(%rbp),%rsi
  lea    -0x20(%rbp),%rcx
  lea    -0x24(%rbp),%rdx
  lea    -0x28(%rbp),%rax
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x28(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jg     L156
  mov    -0x18(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jg     L156
  mov    -0x18(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jg     L156
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L46
L156:
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jg     L56
  mov    -0x14(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jg     L56
  mov    -0x14(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jg     L56
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L46
L56:
  mov    -0x10(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jg     L121
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jg     L121
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jg     L121
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L46
L121:
  mov    -0xc(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jg     L46
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jg     L46
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jg     L46
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L46:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D56
data_type: string
name: unpack_data
arguments: tag: D11
data_type: string
name: unpack_data
arguments: tag: D29
data_type: string

tool: D56: Please enter 4 numbers separated by spaces > 

tool: D11: %d%d%d%d

tool: D29: %d is the smallest


assistant: ```c
int main() {
  int a, b, c, d, num_1, num_2, num_3, num_4;
  printf("Please enter 4 numbers separated by spaces > ");
  scanf("%d%d%d%d", &num_1, &num_2, &num_3, &num_4);
  a = (num_1);
  b = (num_2);
  c = (num_3);
  d = (num_4);
  if (a <= b && a <= c && a <= d)
    printf("%d is the smallest\n", a);
  else if (b <= a && b <= c && b <= d)
    printf("%d is the smallest\n", b);
  else if (c <= a && c <= b && c <= d)
    printf("%d is the smallest\n", c);
  else if (d <= a && d <= b && d <= c)
    printf("%d is the smallest\n", d);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D47(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%esi
  mov    0xc(%rsp),%ecx
  mov    0x10(%rsp),%eax
  mov    0x14(%rsp),%edx
  cmp    %eax,%ecx
  mov    %eax,%edi
  cmovle %ecx,%edi
  cmp    %edx,%edi
  cmovg  %edx,%edi
  cmp    %edi,%esi
  jle    L12
  cmp    %eax,%esi
  mov    %eax,%edi
  cmovle %esi,%edi
  cmp    %edx,%edi
  cmovg  %edx,%edi
  cmp    %edi,%ecx
  jle    L84
  cmp    %ecx,%esi
  cmovg  %ecx,%esi
  cmp    %esi,%edx
  mov    %esi,%ecx
  cmovle %edx,%ecx
  cmp    %ecx,%eax
  jle    L46
  cmp    %esi,%eax
  cmovg  %esi,%eax
  cmp    %eax,%edx
  jle    L7
L47:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L12:
  lea    D29(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L47
L84:
  mov    %ecx,%esi
  lea    D29(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L47
L46:
  mov    %eax,%esi
  lea    D29(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L47
L7:
  mov    %edx,%esi
  lea    D29(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L47
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D47
type_guessed: string
name: retrieve_data
arguments: data_label: D25
type_guessed: string
name: retrieve_data
arguments: data_label: D29
type_guessed: string

tool: D47: Please enter 4 numbers separated by spaces > 

tool: D25: %d%d%d%d

tool: D29: %d is the smallest


assistant: ```c
int main() {
  int a, b, c, d, num_1, num_2, num_3, num_4;
  printf("Please enter 4 numbers separated by spaces > ");
  scanf("%d%d%d%d", &num_1, &num_2, &num_3, &num_4);
  a = (num_1);
  b = (num_2);
  c = (num_3);
  d = (num_4);
  if (a <= b && a <= c && a <= d)
    printf("%d is the smallest\n", a);
  else if (b <= a && b <= c && b <= d)
    printf("%d is the smallest\n", b);
  else if (c <= a && c <= b && c <= d)
    printf("%d is the smallest\n", c);
  else if (d <= a && d <= b && d <= c)
    printf("%d is the smallest\n", d);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D46(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%r10d
  mov    0x10(%rsp),%r8d
  mov    0x14(%rsp),%r9d
  mov    0x8(%rsp),%esi
  cmp    %r8d,%r10d
  mov    %r8d,%eax
  cmovle %r10d,%eax
  cmp    %r9d,%eax
  cmovg  %r9d,%eax
  cmp    %eax,%esi
  jle    L71
  cmp    %r8d,%esi
  mov    %r8d,%eax
  cmovle %esi,%eax
  cmp    %r9d,%eax
  cmovg  %r9d,%eax
  cmp    %eax,%r10d
  jle    L27
  cmp    %r10d,%esi
  cmovg  %r10d,%esi
  cmp    %esi,%r9d
  mov    %esi,%eax
  cmovle %r9d,%eax
  cmp    %eax,%r8d
  jle    L24
  cmp    %esi,%r8d
  cmovg  %esi,%r8d
  cmp    %r8d,%r9d
  jle    L47
L31:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L47:
  mov    %r9d,%esi
L71:
  lea    D23(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L31
L27:
  mov    %r10d,%esi
  lea    D23(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L31
L24:
  mov    %r8d,%esi
  lea    D23(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L31
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D46
type: string
name: parse_data
arguments: tag: D3
type: string
name: parse_data
arguments: tag: D23
type: string

tool: D46: Please enter 4 numbers separated by spaces > 

tool: D3: %d%d%d%d

tool: D23: %d is the smallest


assistant: ```c
int main() {
  int a, b, c, d, num_1, num_2, num_3, num_4;
  printf("Please enter 4 numbers separated by spaces > ");
  scanf("%d%d%d%d", &num_1, &num_2, &num_3, &num_4);
  a = (num_1);
  b = (num_2);
  c = (num_3);
  d = (num_4);
  if (a <= b && a <= c && a <= d)
    printf("%d is the smallest\n", a);
  else if (b <= a && b <= c && b <= d)
    printf("%d is the smallest\n", b);
  else if (c <= a && c <= b && c <= d)
    printf("%d is the smallest\n", c);
  else if (d <= a && d <= b && d <= c)
    printf("%d is the smallest\n", d);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D47(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%r10d
  mov    0x10(%rsp),%r8d
  mov    0x14(%rsp),%r9d
  mov    0x8(%rsp),%esi
  cmp    %r8d,%r10d
  mov    %r8d,%eax
  cmovle %r10d,%eax
  cmp    %r9d,%eax
  cmovg  %r9d,%eax
  cmp    %eax,%esi
  jle    L32
  cmp    %r8d,%esi
  mov    %r8d,%eax
  cmovle %esi,%eax
  cmp    %r9d,%eax
  cmovg  %r9d,%eax
  cmp    %eax,%r10d
  jle    L30
  cmp    %r10d,%esi
  cmovg  %r10d,%esi
  cmp    %esi,%r9d
  mov    %esi,%eax
  cmovle %r9d,%eax
  cmp    %eax,%r8d
  jle    L37
  cmp    %esi,%r8d
  cmovg  %esi,%r8d
  cmp    %r8d,%r9d
  jle    L72
L61:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L72:
  mov    %r9d,%esi
L32:
  lea    D39(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L61
L30:
  mov    %r10d,%esi
  lea    D39(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L61
L37:
  mov    %r8d,%esi
  lea    D39(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L61
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D47
type_infered: string
name: read_data
arguments: tag: D25
type_infered: string
name: read_data
arguments: tag: D39
type_infered: string

tool: D47: Please enter 4 numbers separated by spaces > 

tool: D25: %d%d%d%d

tool: D39: %d is the smallest


assistant: ```c
int main() {
  int a, b, c, d, num_1, num_2, num_3, num_4;
  printf("Please enter 4 numbers separated by spaces > ");
  scanf("%d%d%d%d", &num_1, &num_2, &num_3, &num_4);
  a = (num_1);
  b = (num_2);
  c = (num_3);
  d = (num_4);
  if (a <= b && a <= c && a <= d)
    printf("%d is the smallest\n", a);
  else if (b <= a && b <= c && b <= d)
    printf("%d is the smallest\n", b);
  else if (c <= a && c <= b && c <= d)
    printf("%d is the smallest\n", c);
  else if (d <= a && d <= b && d <= c)
    printf("%d is the smallest\n", d);
  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
<positive_or_not>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L32
  mov    $0x5a,%edi
  call   <print_char@plt>
  jmp    L39
L32:
  cmpl   $0x0,-0x4(%rbp)
  jle    L8
  mov    $0x50,%edi
  call   <print_char@plt>
  jmp    L39
L8:
  mov    $0x4e,%edi
  call   <print_char@plt>
L39:
  nop
  leave
  ret
```

assistant: ```c
void positive_or_not(int i) {
  if (i == 0) {
    print_char('Z');
  } else if (i > 0) {
    print_char('P');
  } else {
    print_char('N');
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<positive_or_not>:
  endbr64
  sub    $0x8,%rsp
  test   %edi,%edi
  je     L36
  jle    L25
  mov    $0x50,%edi
  call   <print_char@plt>
L20:
  add    $0x8,%rsp
  ret
L36:
  mov    $0x5a,%edi
  call   <print_char@plt>
  jmp    L20
L25:
  mov    $0x4e,%edi
  call   <print_char@plt>
  jmp    L20
```

assistant: ```c
void positive_or_not(int i) {
  if (i == 0) {
    print_char('Z');
  } else if (i > 0) {
    print_char('P');
  } else {
    print_char('N');
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<positive_or_not>:
  endbr64
  test   %edi,%edi
  je     L12
  jle    L8
  mov    $0x50,%edi
  jmp    1050 <print_char@plt>
  nopl   0x0(%rax)
L8:
  mov    $0x4e,%edi
  jmp    1050 <print_char@plt>
  nopw   0x0(%rax,%rax,1)
L12:
  mov    $0x5a,%edi
  jmp    1050 <print_char@plt>
```

assistant: ```c
void positive_or_not(int i) {
  if (i == 0) {
    print_char('Z');
  } else if (i > 0) {
    print_char('P');
  } else {
    print_char('N');
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<positive_or_not>:
  endbr64
  test   %edi,%edi
  je     L9
  jle    L18
  mov    $0x50,%edi
  jmp    1050 <print_char@plt>
  nopl   0x0(%rax)
L18:
  mov    $0x4e,%edi
  jmp    1050 <print_char@plt>
  nopw   0x0(%rax,%rax,1)
L9:
  mov    $0x5a,%edi
  jmp    1050 <print_char@plt>
```

assistant: ```c
void positive_or_not(int i) {
  if (i == 0) {
    print_char('Z');
  } else if (i > 0) {
    print_char('P');
  } else {
    print_char('N');
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<parity>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    %eax
  xor    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x2,%eax
  xor    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x11111111,%eax
  imul   $0x11111111,%eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x1c,%eax
  and    $0x1,%eax
  pop    %rbp
  ret
```

assistant: ```c
int parity(unsigned int x) {
  x ^= x >> 1;
  x ^= x >> 2;
  x = (x & 0x11111111U) * 0x11111111U;
  return (x >> 28) & 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<parity>:
  endbr64
  mov    %edi,%edx
  shr    %edx
  xor    %edi,%edx
  mov    %edx,%eax
  shr    $0x2,%eax
  xor    %edx,%eax
  and    $0x11111111,%eax
  imul   $0x11111111,%eax,%eax
  shr    $0x1c,%eax
  and    $0x1,%eax
  ret
```

assistant: ```c
int parity(unsigned int x) {
  x ^= x >> 1;
  x ^= x >> 2;
  x = (x & 0x11111111U) * 0x11111111U;
  return (x >> 28) & 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<parity>:
  endbr64
  mov    %edi,%edx
  shr    %edx
  xor    %edi,%edx
  mov    %edx,%eax
  shr    $0x2,%eax
  xor    %edx,%eax
  and    $0x11111111,%eax
  imul   $0x11111111,%eax,%eax
  shr    $0x1c,%eax
  and    $0x1,%eax
  ret
```

assistant: ```c
int parity(unsigned int x) {
  x ^= x >> 1;
  x ^= x >> 2;
  x = (x & 0x11111111U) * 0x11111111U;
  return (x >> 28) & 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<parity>:
  endbr64
  mov    %edi,%edx
  shr    %edx
  xor    %edi,%edx
  mov    %edx,%eax
  shr    $0x2,%eax
  xor    %edx,%eax
  and    $0x11111111,%eax
  imul   $0x11111111,%eax,%eax
  shr    $0x1c,%eax
  and    $0x1,%eax
  ret
```

assistant: ```c
int parity(unsigned int x) {
  x ^= x >> 1;
  x ^= x >> 2;
  x = (x & 0x11111111U) * 0x11111111U;
  return (x >> 28) & 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ftok>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stat@plt>
  test   %rax,%rax
  jns    L19
  mov    $0xffffffff,%eax
  jmp    L18
L19:
  mov    -0x1c(%rbp),%eax
  shl    $0x18,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  shl    $0x10,%eax
  and    $0xff0000,%eax
  or     %eax,%edx
  mov    -0xc(%rbp),%eax
  movzwl %ax,%eax
  or     %edx,%eax
L18:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

assistant: ```c
key_t ftok(const char *path, int id) {
  struct stat st;
  if (stat(path, &st) < 0)
    return (key_t)-1;
  return (key_t)((id & 0xff) << 24 | (st.st_dev & 0xff) << 16 | (st.st_ino & 0xffff));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ftok>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %esi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  test   %rax,%rax
  js     L19
  mov    (%rsp),%eax
  shl    $0x10,%eax
  and    $0xff0000,%eax
  movzwl 0x4(%rsp),%edx
  or     %edx,%eax
  shl    $0x18,%ebx
  or     %ebx,%eax
L11:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x10,%rsp
  pop    %rbx
  ret
L19:
  mov    $0xffffffff,%eax
  jmp    L11
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
key_t ftok(const char *path, int id) {
  struct stat st;
  if (stat(path, &st) < 0)
    return (key_t)-1;
  return (key_t)((id & 0xff) << 24 | (st.st_dev & 0xff) << 16 | (st.st_ino & 0xffff));
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ftok>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  test   %rax,%rax
  js     L10
  mov    (%rsp),%eax
  movzwl 0x4(%rsp),%edx
  shl    $0x18,%ebx
  shl    $0x10,%eax
  and    $0xff0000,%eax
  or     %edx,%eax
  or     %ebx,%eax
L9:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x10,%rsp
  pop    %rbx
  ret
L10:
  mov    $0xffffffff,%eax
  jmp    L9
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
key_t ftok(const char *path, int id) {
  struct stat st;
  if (stat(path, &st) < 0)
    return (key_t)-1;
  return (key_t)((id & 0xff) << 24 | (st.st_dev & 0xff) << 16 | (st.st_ino & 0xffff));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ftok>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <stat@plt>
  test   %rax,%rax
  js     L20
  mov    (%rsp),%eax
  movzwl 0x4(%rsp),%edx
  shl    $0x18,%ebx
  shl    $0x10,%eax
  and    $0xff0000,%eax
  or     %edx,%eax
  or     %ebx,%eax
L9:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L25
  add    $0x10,%rsp
  pop    %rbx
  ret
L20:
  mov    $0xffffffff,%eax
  jmp    L9
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
key_t ftok(const char *path, int id) {
  struct stat st;
  if (stat(path, &st) < 0)
    return (key_t)-1;
  return (key_t)((id & 0xff) << 24 | (st.st_dev & 0xff) << 16 | (st.st_ino & 0xffff));
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<findLargestBST>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x18(%rbp)
  mov    D4(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  lea    -0x14(%rbp),%rcx
  lea    -0x18(%rbp),%rdx
  lea    -0x18(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <isBSTfind@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
int findLargestBST(struct Node *node) {
  int min = INT_MIN;
  int max = INT_MAX;
  int result = 0;
  int tmp = isBSTfind(node, &min, &min, &result);
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<findLargestBST>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,(%rsp)
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%rcx
  mov    %rsp,%rsi
  mov    %rsi,%rdx
  call   <isBSTfind@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int findLargestBST(struct Node *node) {
  int min = INT_MIN;
  int max = INT_MAX;
  int result = 0;
  int tmp = isBSTfind(node, &min, &min, &result);
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<findLargestBST>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    D8(%rip),%rax
  mov    %rsp,%rsi
  lea    0x4(%rsp),%rcx
  movl   $0x0,0x4(%rsp)
  mov    %rsi,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rsp)
  call   <isBSTfind@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int findLargestBST(struct Node *node) {
  int min = INT_MIN;
  int max = INT_MAX;
  int result = 0;
  int tmp = isBSTfind(node, &min, &min, &result);
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<findLargestBST>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    D7(%rip),%rax
  mov    %rsp,%rsi
  lea    0x4(%rsp),%rcx
  movl   $0x0,0x4(%rsp)
  mov    %rsi,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rsp)
  call   <isBSTfind@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int findLargestBST(struct Node *node) {
  int min = INT_MIN;
  int max = INT_MAX;
  int result = 0;
  int tmp = isBSTfind(node, &min, &min, &result);
  return result;
}
```
user: What is the c source code of the assembly code below:

```asm
<vp8_get_mb_ss_c>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
L6:
  mov    -0x8(%rbp),%eax
  lea    (%rax,%rax,1),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  movswl %ax,%edx
  mov    -0x8(%rbp),%eax
  lea    (%rax,%rax,1),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzwl (%rax),%eax
  cwtl
  imul   %edx,%eax
  add    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
  cmpl   $0xff,-0x8(%rbp)
  jbe    L6
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int vp8_get_mb_ss_c(short *src_ptr) {
  unsigned int i = 0, sum = 0;
  do {
    sum += (src_ptr[i] * src_ptr[i]);
    i++;
  } while (i < 256);
  return sum;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<vp8_get_mb_ss_c>:
  endbr64
  lea    0x200(%rdi),%rcx
  mov    $0x0,%edx
L8:
  movswl (%rdi),%eax
  imul   %eax,%eax
  add    %eax,%edx
  add    $0x2,%rdi
  cmp    %rcx,%rdi
  jne    L8
  mov    %edx,%eax
  ret
```

assistant: ```c
unsigned int vp8_get_mb_ss_c(short *src_ptr) {
  unsigned int i = 0, sum = 0;
  do {
    sum += (src_ptr[i] * src_ptr[i]);
    i++;
  } while (i < 256);
  return sum;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<vp8_get_mb_ss_c>:
  endbr64
  lea    0x200(%rdi),%rdx
  xor    %r8d,%r8d
  xchg   %ax,%ax
L4:
  movswl (%rdi),%eax
  add    $0x2,%rdi
  imul   %eax,%eax
  add    %eax,%r8d
  cmp    %rdi,%rdx
  jne    L4
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int vp8_get_mb_ss_c(short *src_ptr) {
  unsigned int i = 0, sum = 0;
  do {
    sum += (src_ptr[i] * src_ptr[i]);
    i++;
  } while (i < 256);
  return sum;
}
```