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

```asm
<stat>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <smbw_path@plt>
  test   %rax,%rax
  jne    L11
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <real_stat@plt>
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L11:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <smbw_stat@plt>
  jmp    L17
```

assistant: ```c
int stat(const char *name, void *st) {
  if (smbw_path(name)) {
    return smbw_stat(name, st);
  }
  return real_stat(name, st);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<stat>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <smbw_path@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  test   %rax,%rax
  jne    L3
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  jmp    1090 <real_stat@plt>
  nopl   (%rax)
L3:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  jmp    1070 <smbw_stat@plt>
```

assistant: ```c
int stat(const char *name, void *st) {
  if (smbw_path(name)) {
    return smbw_stat(name, st);
  }
  return real_stat(name, st);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<stat>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <smbw_path@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  test   %rax,%rax
  jne    L3
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  jmp    1090 <real_stat@plt>
  nopl   (%rax)
L3:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  jmp    1070 <smbw_stat@plt>
```

assistant: ```c
int stat(const char *name, void *st) {
  if (smbw_path(name)) {
    return smbw_stat(name, st);
  }
  return real_stat(name, st);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<AESTagRead>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  nop
L8:
  mov    -0x4(%rbp),%eax
  add    $0x50,%eax
  cltq
  mov    (%rax),%eax
  and    $0x40000000,%eax
  test   %eax,%eax
  je     L8
  mov    -0x4(%rbp),%eax
  add    $0x70,%eax
  cltq
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x4(%rbp),%eax
  add    $0x74,%eax
  cltq
  mov    %rax,%rcx
  mov    -0x10(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
  mov    -0x4(%rbp),%eax
  add    $0x78,%eax
  cltq
  mov    %rax,%rcx
  mov    -0x10(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
  mov    -0x4(%rbp),%eax
  add    $0x7c,%eax
  cltq
  mov    %rax,%rcx
  mov    -0x10(%rbp),%rax
  lea    0xc(%rax),%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void AESTagRead(uint32_t ui32Base, uint32_t *pui32TagData) {
  ;
  while ((0x40000000 & ((*((volatile uint32_t *)(ui32Base + 0x00000050))))) == 0) {
  }
  pui32TagData[0] = (*((volatile uint32_t *)((ui32Base + 0x00000070))));
  pui32TagData[1] = (*((volatile uint32_t *)((ui32Base + 0x00000074))));
  pui32TagData[2] = (*((volatile uint32_t *)((ui32Base + 0x00000078))));
  pui32TagData[3] = (*((volatile uint32_t *)((ui32Base + 0x0000007C))));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<AESTagRead>:
  endbr64
  mov    %edi,%ecx
  lea    0x50(%rdi),%edx
  movslq %edx,%rdx
L1:
  mov    (%rdx),%eax
  test   $0x40000000,%eax
  je     L1
  lea    0x70(%rcx),%eax
  cltq
  mov    (%rax),%eax
  mov    %eax,(%rsi)
  lea    0x74(%rcx),%eax
  cltq
  mov    (%rax),%eax
  mov    %eax,0x4(%rsi)
  lea    0x78(%rcx),%eax
  cltq
  mov    (%rax),%eax
  mov    %eax,0x8(%rsi)
  add    $0x7c,%ecx
  movslq %ecx,%rcx
  mov    (%rcx),%eax
  mov    %eax,0xc(%rsi)
  ret
```

assistant: ```c
void AESTagRead(uint32_t ui32Base, uint32_t *pui32TagData) {
  ;
  while ((0x40000000 & ((*((volatile uint32_t *)(ui32Base + 0x00000050))))) == 0) {
  }
  pui32TagData[0] = (*((volatile uint32_t *)((ui32Base + 0x00000070))));
  pui32TagData[1] = (*((volatile uint32_t *)((ui32Base + 0x00000074))));
  pui32TagData[2] = (*((volatile uint32_t *)((ui32Base + 0x00000078))));
  pui32TagData[3] = (*((volatile uint32_t *)((ui32Base + 0x0000007C))));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<AESTagRead>:
  endbr64
  lea    0x50(%rdi),%edx
  mov    %edi,%ecx
  movslq %edx,%rdx
  nopl   0x0(%rax)
L7:
  mov    (%rdx),%eax
  test   $0x40000000,%eax
  je     L7
  lea    0x70(%rcx),%eax
  cltq
  mov    (%rax),%eax
  mov    %eax,(%rsi)
  lea    0x74(%rcx),%eax
  cltq
  mov    (%rax),%eax
  mov    %eax,0x4(%rsi)
  lea    0x78(%rcx),%eax
  add    $0x7c,%ecx
  cltq
  movslq %ecx,%rcx
  mov    (%rax),%eax
  mov    %eax,0x8(%rsi)
  mov    (%rcx),%eax
  mov    %eax,0xc(%rsi)
  ret
```

assistant: ```c
void AESTagRead(uint32_t ui32Base, uint32_t *pui32TagData) {
  ;
  while ((0x40000000 & ((*((volatile uint32_t *)(ui32Base + 0x00000050))))) == 0) {
  }
  pui32TagData[0] = (*((volatile uint32_t *)((ui32Base + 0x00000070))));
  pui32TagData[1] = (*((volatile uint32_t *)((ui32Base + 0x00000074))));
  pui32TagData[2] = (*((volatile uint32_t *)((ui32Base + 0x00000078))));
  pui32TagData[3] = (*((volatile uint32_t *)((ui32Base + 0x0000007C))));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<AESTagRead>:
  endbr64
  lea    0x50(%rdi),%edx
  mov    %edi,%ecx
  movslq %edx,%rdx
  nopl   0x0(%rax)
L7:
  mov    (%rdx),%eax
  test   $0x40000000,%eax
  je     L7
  lea    0x70(%rcx),%eax
  cltq
  mov    (%rax),%eax
  mov    %eax,(%rsi)
  lea    0x74(%rcx),%eax
  cltq
  mov    (%rax),%eax
  mov    %eax,0x4(%rsi)
  lea    0x78(%rcx),%eax
  add    $0x7c,%ecx
  cltq
  movslq %ecx,%rcx
  mov    (%rax),%eax
  mov    %eax,0x8(%rsi)
  mov    (%rcx),%eax
  mov    %eax,0xc(%rsi)
  ret
```

assistant: ```c
void AESTagRead(uint32_t ui32Base, uint32_t *pui32TagData) {
  ;
  while ((0x40000000 & ((*((volatile uint32_t *)(ui32Base + 0x00000050))))) == 0) {
  }
  pui32TagData[0] = (*((volatile uint32_t *)((ui32Base + 0x00000070))));
  pui32TagData[1] = (*((volatile uint32_t *)((ui32Base + 0x00000074))));
  pui32TagData[2] = (*((volatile uint32_t *)((ui32Base + 0x00000078))));
  pui32TagData[3] = (*((volatile uint32_t *)((ui32Base + 0x0000007C))));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x7b,-0xc(%rbp)
  movss  D73(%rip),%xmm0
  movss  %xmm0,-0x8(%rbp)
  movb   $0x7a,-0x10(%rbp)
  lea    D109(%rip),%rax
  mov    %al,-0xf(%rbp)
  lea    D57(%rip),%rax
  mov    %al,-0xe(%rbp)
  lea    D28(%rip),%rax
  mov    %al,-0xd(%rbp)
  movl   $0x516,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D94(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movss  -0x8(%rbp),%xmm0
  cvttss2si %xmm0,%eax
  mov    %eax,%esi
  lea    D103(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movsbl -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movsbl -0xf(%rbp),%eax
  mov    %eax,%esi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movsbl -0xe(%rbp),%eax
  mov    %eax,%esi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movsbl -0xd(%rbp),%eax
  mov    %eax,%esi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D94(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D73
type_infered: f32
name: retrieve_data
arguments: label: D109
type_infered: string
name: retrieve_data
arguments: label: D57
type_infered: string
name: retrieve_data
arguments: label: D28
type_infered: string
name: retrieve_data
arguments: label: D94
type_infered: string
name: retrieve_data
arguments: label: D103
type_infered: string
name: retrieve_data
arguments: label: D54
type_infered: string

tool: D73: 999.999

tool: D109: E

tool: D57: G

tool: D28: A

tool: D94: %d

tool: D103: %f

tool: D54: %c

assistant: ```c
int main() {
  int MyInt = 123;
  float MyFloat = 999.999;
  char MyChar = 'z';
  char E = "E";
  char G = "G";
  char A = "A";
  int CourseNumber = 1302;
  printf("%d", MyInt);
  printf("%f", MyFloat);
  printf("%c", MyChar);
  printf("%c", E);
  printf("%c", G);
  printf("%c", A);
  printf("%d", CourseNumber);
  return 0;
}
```
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
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x7b,%esi
  lea    D42(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x3e7,%esi
  lea    D59(%rip),%rdi
  call   <printf@plt>
  mov    $0x7a,%esi
  lea    D6(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    D1(%rip),%rsi
  movsbl %sil,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    D52(%rip),%rsi
  movsbl %sil,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    D34(%rip),%rsi
  movsbl %sil,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x516,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D42
type_guessed: string
name: disassemble_data
arguments: data_tag: D59
type_guessed: string
name: disassemble_data
arguments: data_tag: D6
type_guessed: string
name: disassemble_data
arguments: data_tag: D1
type_guessed: string
name: disassemble_data
arguments: data_tag: D52
type_guessed: string
name: disassemble_data
arguments: data_tag: D34
type_guessed: string

tool: D42: %d

tool: D59: %f

tool: D6: %c

tool: D1: E

tool: D52: G

tool: D34: A

assistant: ```c
int main() {
  int MyInt = 123;
  float MyFloat = 999.999;
  char MyChar = 'z';
  char E = "E";
  char G = "G";
  char A = "A";
  int CourseNumber = 1302;
  printf("%d", MyInt);
  printf("%f", MyFloat);
  printf("%c", MyChar);
  printf("%c", E);
  printf("%c", G);
  printf("%c", A);
  printf("%d", CourseNumber);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  lea    D20(%rip),%r12
  mov    $0x7b,%esi
  push   %rbp
  mov    %r12,%rdi
  lea    D18(%rip),%rbp
  sub    $0x8,%rsp
  call   <printf@plt>
  mov    $0x3e7,%esi
  lea    D59(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x7a,%esi
  call   <printf@plt>
  lea    D9(%rip),%rsi
  mov    %rbp,%rdi
  movsbl %sil,%esi
  call   <printf@plt>
  lea    D5(%rip),%rsi
  mov    %rbp,%rdi
  movsbl %sil,%esi
  call   <printf@plt>
  lea    D35(%rip),%rsi
  mov    %rbp,%rdi
  movsbl %sil,%esi
  call   <printf@plt>
  mov    %r12,%rdi
  mov    $0x516,%esi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D20
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D18
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D59
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D9
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D5
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D35
type_infered: string

tool: D20: %d

tool: D18: %c

tool: D59: %f

tool: D9: E

tool: D5: G

tool: D35: A

assistant: ```c
int main() {
  int MyInt = 123;
  float MyFloat = 999.999;
  char MyChar = 'z';
  char E = "E";
  char G = "G";
  char A = "A";
  int CourseNumber = 1302;
  printf("%d", MyInt);
  printf("%f", MyFloat);
  printf("%c", MyChar);
  printf("%c", E);
  printf("%c", G);
  printf("%c", A);
  printf("%d", CourseNumber);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  lea    D54(%rip),%r12
  mov    $0x7b,%esi
  push   %rbp
  mov    %r12,%rdi
  lea    D42(%rip),%rbp
  sub    $0x8,%rsp
  call   <printf@plt>
  mov    $0x3e7,%esi
  lea    D52(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x7a,%esi
  call   <printf@plt>
  lea    D21(%rip),%rsi
  mov    %rbp,%rdi
  movsbl %sil,%esi
  call   <printf@plt>
  lea    D16(%rip),%rsi
  mov    %rbp,%rdi
  movsbl %sil,%esi
  call   <printf@plt>
  lea    D11(%rip),%rsi
  mov    %rbp,%rdi
  movsbl %sil,%esi
  call   <printf@plt>
  mov    %r12,%rdi
  mov    $0x516,%esi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D54
type: string
name: read_data
arguments: label: D42
type: string
name: read_data
arguments: label: D52
type: string
name: read_data
arguments: label: D21
type: string
name: read_data
arguments: label: D16
type: string
name: read_data
arguments: label: D11
type: string

tool: D54: %d

tool: D42: %c

tool: D52: %f

tool: D21: E

tool: D16: G

tool: D11: A

assistant: ```c
int main() {
  int MyInt = 123;
  float MyFloat = 999.999;
  char MyChar = 'z';
  char E = "E";
  char G = "G";
  char A = "A";
  int CourseNumber = 1302;
  printf("%d", MyInt);
  printf("%f", MyFloat);
  printf("%c", MyChar);
  printf("%c", E);
  printf("%c", G);
  printf("%c", A);
  printf("%d", CourseNumber);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L16
L3:
  mov    -0x20(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x20(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,(%rbx)
  addl   $0x1,-0x20(%rbp)
L16:
  mov    -0x20(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L3
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <longestConsecutive@plt>
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D7
type_infered: string

tool: D7: %d


assistant: ```c
int main(int argc, char **argv) {
  int i, size = argc - 1;
  int *nums = malloc(size * sizeof(int));
  for (i = 0; i < size; i++) {
    nums[i] = atoi(argv[i + 1]);
  }
  printf("%d\n", longestConsecutive(nums, size));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %rsi,%r12
  lea    -0x1(%rdi),%r14d
  lea    0x0(,%r14,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %r14d,%r14d
  jle    L16
  lea    -0x2(%rbx),%r13d
  add    $0x2,%r13
  mov    $0x1,%ebx
L8:
  mov    (%r12,%rbx,8),%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp,%rbx,4)
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L8
L16:
  mov    %r14d,%esi
  mov    %rbp,%rdi
  call   <longestConsecutive@plt>
  mov    %eax,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D9
type_guessed: string

tool: D9: %d


assistant: ```c
int main(int argc, char **argv) {
  int i, size = argc - 1;
  int *nums = malloc(size * sizeof(int));
  for (i = 0; i < size; i++) {
    nums[i] = atoi(argv[i + 1]);
  }
  printf("%d\n", longestConsecutive(nums, size));
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r14
  lea    -0x1(%rdi),%r14d
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  lea    0x0(,%r14,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %r14d,%r14d
  jle    L18
  lea    -0x2(%rbx),%r13d
  mov    $0x1,%ebx
  add    $0x2,%r13
  nopw   0x0(%rax,%rax,1)
L16:
  mov    (%r12,%rbx,8),%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp,%rbx,4)
  add    $0x1,%rbx
  cmp    %rbx,%r13
  jne    L16
L18:
  mov    %r14d,%esi
  mov    %rbp,%rdi
  call   <longestConsecutive@plt>
  lea    D3(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: tag: D3
type_guessed: string

tool: D3: %d


assistant: ```c
int main(int argc, char **argv) {
  int i, size = argc - 1;
  int *nums = malloc(size * sizeof(int));
  for (i = 0; i < size; i++) {
    nums[i] = atoi(argv[i + 1]);
  }
  printf("%d\n", longestConsecutive(nums, size));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r14
  lea    -0x1(%rdi),%r14d
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  lea    0x0(,%r14,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %r14d,%r14d
  jle    L16
  lea    -0x2(%rbx),%r13d
  mov    $0x1,%ebx
  add    $0x2,%r13
  nopw   0x0(%rax,%rax,1)
L5:
  mov    (%r12,%rbx,8),%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp,%rbx,4)
  add    $0x1,%rbx
  cmp    %rbx,%r13
  jne    L5
L16:
  mov    %r14d,%esi
  mov    %rbp,%rdi
  call   <longestConsecutive@plt>
  lea    D3(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: tag: D3
data_type: string

tool: D3: %d


assistant: ```c
int main(int argc, char **argv) {
  int i, size = argc - 1;
  int *nums = malloc(size * sizeof(int));
  for (i = 0; i < size; i++) {
    nums[i] = atoi(argv[i + 1]);
  }
  printf("%d\n", longestConsecutive(nums, size));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<RandomFunc>:
  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
  movzwl (%rax),%eax
  add    $0x7296,%ax
  mov    %ax,-0xa(%rbp)
  movw   $0x0,-0xc(%rbp)
  nop
  movzwl -0xa(%rbp),%eax
  movzwl %ax,%eax
  add    %eax,%eax
  mov    %eax,%edx
  movzwl -0xa(%rbp),%eax
  shr    $0xf,%ax
  or     %edx,%eax
  xor    $0xdf00,%ax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %dx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  {
    state[0UL] = (input[0UL] - 51238316UL) - (unsigned short)47038;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      local1++;
    }
    output[0UL] = ((state[0UL] << (unsigned short)1) | (state[0UL] >> (unsigned short)15)) ^ (unsigned short)57088;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  add    $0x7296,%ax
  rol    %ax
  xor    $0xdf00,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  {
    state[0UL] = (input[0UL] - 51238316UL) - (unsigned short)47038;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      local1++;
    }
    output[0UL] = ((state[0UL] << (unsigned short)1) | (state[0UL] >> (unsigned short)15)) ^ (unsigned short)57088;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  add    $0x7296,%ax
  rol    %ax
  xor    $0xdf00,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  {
    state[0UL] = (input[0UL] - 51238316UL) - (unsigned short)47038;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      local1++;
    }
    output[0UL] = ((state[0UL] << (unsigned short)1) | (state[0UL] >> (unsigned short)15)) ^ (unsigned short)57088;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  add    $0x7296,%ax
  rol    %ax
  xor    $0xdf00,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  {
    state[0UL] = (input[0UL] - 51238316UL) - (unsigned short)47038;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      local1++;
    }
    output[0UL] = ((state[0UL] << (unsigned short)1) | (state[0UL] >> (unsigned short)15)) ^ (unsigned short)57088;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<encrypt_for_client>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x130,%rsp
  mov    %rdi,-0x128(%rbp)
  mov    %rsi,-0x130(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x130(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L43
  mov    -0x130(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <createRSA@plt>
  mov    %rax,-0x118(%rbp)
  mov    D3(%rip),%rax
  mov    (%rax),%esi
  mov    -0x118(%rbp),%rcx
  lea    -0x110(%rbp),%rdx
  mov    -0x128(%rbp),%rax
  mov    %esi,%r8d
  mov    %rax,%rsi
  mov    $0x100,%edi
  call   <RSA_public_encrypt@plt>
  movl   $0x0,-0x11c(%rbp)
  jmp    L29
L11:
  mov    -0x11c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x128(%rbp),%rax
  add    %rax,%rdx
  mov    -0x11c(%rbp),%eax
  cltq
  movzbl -0x110(%rbp,%rax,1),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x11c(%rbp)
L29:
  cmpl   $0xff,-0x11c(%rbp)
  jle    L11
  jmp    L36
L43:
  nop
L36:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L39
  call   <__stack_chk_fail@plt>
L39:
  leave
  ret
```

assistant: ```c
void encrypt_for_client(char *buffer, char *pubkey) {
  if (pubkey[0] == '\0')
    return;
  RSA *rsa = createRSA(pubkey, 1);
  char encrypted[256];
  RSA_public_encrypt(256, buffer, encrypted, rsa, RSA_PKCS1_PADDING);
  for (int i = 0; i < 256; ++i)
    buffer[i] = encrypted[i];
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<encrypt_for_client>:
  endbr64
  push   %rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  cmpb   $0x0,(%rsi)
  jne    L14
L8:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x110,%rsp
  pop    %rbx
  ret
L14:
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  mov    $0x1,%esi
  call   <createRSA@plt>
  mov    %rax,%rcx
  mov    %rsp,%rdx
  mov    D9(%rip),%rax
  mov    (%rax),%r8d
  mov    %rbx,%rsi
  mov    $0x100,%edi
  call   <RSA_public_encrypt@plt>
  mov    $0x0,%eax
L17:
  movzbl (%rsp,%rax,1),%edx
  mov    %dl,(%rbx,%rax,1)
  add    $0x1,%rax
  cmp    $0x100,%rax
  jne    L17
  jmp    L8
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void encrypt_for_client(char *buffer, char *pubkey) {
  if (pubkey[0] == '\0')
    return;
  RSA *rsa = createRSA(pubkey, 1);
  char encrypted[256];
  RSA_public_encrypt(256, buffer, encrypted, rsa, RSA_PKCS1_PADDING);
  for (int i = 0; i < 256; ++i)
    buffer[i] = encrypted[i];
}
```