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

```asm
<byteorder_swapf32>:
  endbr64
  movd   %xmm0,%ecx
  xor    %eax,%eax
  movd   %xmm0,%edx
  shr    $0x18,%ecx
  mov    %cl,%al
  movd   %xmm0,%ecx
  shr    $0x10,%ecx
  mov    %cl,%ah
  movzbl %dh,%ecx
  shl    $0x18,%edx
  shl    $0x10,%ecx
  movzwl %ax,%eax
  or     %ecx,%eax
  or     %edx,%eax
  movd   %eax,%xmm0
  ret
```

assistant: ```c
float byteorder_swapf32(float f) {
  unsigned char c, *a;
  a = (unsigned char *)&f;
  c = a[3];
  a[3] = a[0];
  a[0] = c;
  c = a[2];
  a[2] = a[1];
  a[1] = c;
  return f;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<wcslen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  jmp    L8
L15:
  addq   $0x4,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L8:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L15
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
size_t wcslen(const wchar_t *s) {
  const wchar_t *p = s;
  size_t res = 0;
  assert(s != NULL);
  while (*p != L'\0') {
    p++;
    res++;
  }
  return res;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<wcslen>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmpl   $0x0,(%rbx)
  je     L11
  mov    $0x0,%eax
L13:
  add    $0x1,%rax
  cmpl   $0x0,(%rbx,%rax,4)
  jne    L13
L9:
  pop    %rbx
  ret
L11:
  mov    $0x0,%eax
  jmp    L9
```

assistant: ```c
size_t wcslen(const wchar_t *s) {
  const wchar_t *p = s;
  size_t res = 0;
  assert(s != NULL);
  while (*p != L'\0') {
    p++;
    res++;
  }
  return res;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<wcslen>:
  endbr64
  test   %rdi,%rdi
  push   %rbx
  mov    %rdi,%rbx
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    (%rbx),%ecx
  xor    %eax,%eax
  test   %ecx,%ecx
  je     L9
L5:
  add    $0x1,%rax
  mov    (%rbx,%rax,4),%edx
  test   %edx,%edx
  jne    L5
L9:
  pop    %rbx
  ret
```

assistant: ```c
size_t wcslen(const wchar_t *s) {
  const wchar_t *p = s;
  size_t res = 0;
  assert(s != NULL);
  while (*p != L'\0') {
    p++;
    res++;
  }
  return res;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<wcslen>:
  endbr64
  test   %rdi,%rdi
  push   %rbx
  mov    %rdi,%rbx
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    (%rbx),%ecx
  xor    %eax,%eax
  test   %ecx,%ecx
  je     L11
L2:
  add    $0x1,%rax
  mov    (%rbx,%rax,4),%edx
  test   %edx,%edx
  jne    L2
L11:
  pop    %rbx
  ret
```

assistant: ```c
size_t wcslen(const wchar_t *s) {
  const wchar_t *p = s;
  size_t res = 0;
  assert(s != NULL);
  while (*p != L'\0') {
    p++;
    res++;
  }
  return res;
}
```
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    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D10(%rip),%rax
  mov    %rax,-0x20(%rbp)
  lea    D12(%rip),%rax
  mov    %rax,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <execv@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D10
type: string
name: extract_symbolic_data
arguments: data_label: D12
type: string
name: extract_symbolic_data
arguments: data_label: D29
type: string

tool: D10: ps

tool: D12: -ef

tool: D29: /bin/ps

assistant: ```c
int main() {
  char *argvs[] = {"ps", "-ef", NULL};
  execv("/bin/ps", argvs);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D8(%rip),%rax
  mov    %rax,(%rsp)
  lea    D29(%rip),%rax
  mov    %rax,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  mov    %rsp,%rsi
  lea    D7(%rip),%rdi
  call   <execv@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D8
type_guessed: string
name: parse_data
arguments: data_label: D29
type_guessed: string
name: parse_data
arguments: data_label: D7
type_guessed: string

tool: D8: ps

tool: D29: -ef

tool: D7: /bin/ps

assistant: ```c
int main() {
  char *argvs[] = {"ps", "-ef", NULL};
  execv("/bin/ps", argvs);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D14(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  lea    D28(%rip),%rax
  mov    %rsp,%rsi
  movq   $0x0,0x10(%rsp)
  mov    %rax,(%rsp)
  lea    D7(%rip),%rax
  mov    %rax,0x8(%rsp)
  call   <execv@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D14
type_infered: string
name: read_data
arguments: tag: D28
type_infered: string
name: read_data
arguments: tag: D7
type_infered: string

tool: D14: /bin/ps

tool: D28: ps

tool: D7: -ef

assistant: ```c
int main() {
  char *argvs[] = {"ps", "-ef", NULL};
  execv("/bin/ps", argvs);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D13(%rip),%rdx
  lea    D15(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  lea    D12(%rip),%rax
  movq   %rdx,%xmm0
  mov    %rsp,%rsi
  movq   $0x0,0x10(%rsp)
  movq   %rax,%xmm1
  punpcklqdq %xmm1,%xmm0
  movaps %xmm0,(%rsp)
  call   <execv@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D13
type: string
name: extract_symbolic_data
arguments: tag: D15
type: string
name: extract_symbolic_data
arguments: tag: D12
type: string

tool: D13: ps

tool: D15: /bin/ps

tool: D12: -ef

assistant: ```c
int main() {
  char *argvs[] = {"ps", "-ef", NULL};
  execv("/bin/ps", argvs);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parse_array_end>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <node_stack_at@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <node_stack_at@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L24
  cmpl   $0x4,-0x8(%rbp)
  je     L36
  cmpl   $0x5,-0x8(%rbp)
  jne    L24
L36:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  cmpl   $0x5,-0x8(%rbp)
  jne    L100
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <pop_node_stack@plt>
  jmp    L99
L100:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L99
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <pop_node_stack@plt>
  jmp    L99
L24:
  cmpl   $0x4,-0x4(%rbp)
  je     L55
  cmpl   $0x5,-0x4(%rbp)
  jne    L20
L55:
  cmpl   $0x8,-0x8(%rbp)
  jne    L20
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <pop_node_stack@plt>
  jmp    L20
L99:
  nop
L20:
  nop
  leave
  ret
```

assistant: ```c
void parse_array_end(unsigned int **stack, unsigned int *stack_size, unsigned int *array_depth) {
  int stack_top = node_stack_at(0, *stack, *stack_size);
  int parent = node_stack_at(1, *stack, *stack_size);
  if (parent == 0x00000000 && (stack_top == 0x00000004 || stack_top == 0x00000005)) {
    (*array_depth)--;
    if (stack_top == 0x00000005) {
      pop_node_stack(stack, stack_size);
    } else {
      if ((*array_depth) == 0) {
        pop_node_stack(stack, stack_size);
      }
    }
  } else if ((parent == 0x00000004 || parent == 0x00000005) && stack_top == 0x00000008) {
    (*array_depth)--;
    pop_node_stack(stack, stack_size);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parse_array_end>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rdx,%r13
  mov    (%rsi),%edx
  mov    (%rdi),%rsi
  mov    $0x0,%edi
  call   <node_stack_at@plt>
  mov    %eax,%r12d
  mov    0x0(%rbp),%edx
  mov    (%rbx),%rsi
  mov    $0x1,%edi
  call   <node_stack_at@plt>
  test   %eax,%eax
  jne    L32
  lea    -0x4(%r12),%eax
  cmp    $0x1,%eax
  jbe    L58
L76:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L58:
  mov    0x0(%r13),%eax
  sub    $0x1,%eax
  mov    %eax,0x0(%r13)
  cmp    $0x5,%r12d
  je     L52
  test   %eax,%eax
  jne    L76
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <pop_node_stack@plt>
  jmp    L76
L52:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <pop_node_stack@plt>
  jmp    L76
L32:
  sub    $0x4,%eax
  cmp    $0x1,%eax
  ja     L76
  cmp    $0x8,%r12d
  jne    L76
  subl   $0x1,0x0(%r13)
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <pop_node_stack@plt>
  jmp    L76
```

assistant: ```c
void parse_array_end(unsigned int **stack, unsigned int *stack_size, unsigned int *array_depth) {
  int stack_top = node_stack_at(0, *stack, *stack_size);
  int parent = node_stack_at(1, *stack, *stack_size);
  if (parent == 0x00000000 && (stack_top == 0x00000004 || stack_top == 0x00000005)) {
    (*array_depth)--;
    if (stack_top == 0x00000005) {
      pop_node_stack(stack, stack_size);
    } else {
      if ((*array_depth) == 0) {
        pop_node_stack(stack, stack_size);
      }
    }
  } else if ((parent == 0x00000004 || parent == 0x00000005) && stack_top == 0x00000008) {
    (*array_depth)--;
    pop_node_stack(stack, stack_size);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_array_end>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rsi),%edx
  mov    (%rdi),%rsi
  xor    %edi,%edi
  call   <node_stack_at@plt>
  mov    (%r12),%edx
  mov    0x0(%rbp),%rsi
  mov    $0x1,%edi
  mov    %eax,%ebx
  call   <node_stack_at@plt>
  test   %eax,%eax
  jne    L36
  lea    -0x4(%rbx),%eax
  cmp    $0x1,%eax
  jbe    L50
L51:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L36:
  sub    $0x4,%eax
  cmp    $0x1,%eax
  ja     L51
  cmp    $0x8,%ebx
  jne    L51
  subl   $0x1,0x0(%r13)
  jmp    L12
  nopl   0x0(%rax)
L50:
  mov    0x0(%r13),%eax
  sub    $0x1,%eax
  mov    %eax,0x0(%r13)
  cmp    $0x5,%ebx
  je     L12
  test   %eax,%eax
  jne    L51
L12:
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <pop_node_stack@plt>
```

assistant: ```c
void parse_array_end(unsigned int **stack, unsigned int *stack_size, unsigned int *array_depth) {
  int stack_top = node_stack_at(0, *stack, *stack_size);
  int parent = node_stack_at(1, *stack, *stack_size);
  if (parent == 0x00000000 && (stack_top == 0x00000004 || stack_top == 0x00000005)) {
    (*array_depth)--;
    if (stack_top == 0x00000005) {
      pop_node_stack(stack, stack_size);
    } else {
      if ((*array_depth) == 0) {
        pop_node_stack(stack, stack_size);
      }
    }
  } else if ((parent == 0x00000004 || parent == 0x00000005) && stack_top == 0x00000008) {
    (*array_depth)--;
    pop_node_stack(stack, stack_size);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parse_array_end>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rsi),%edx
  mov    (%rdi),%rsi
  xor    %edi,%edi
  call   <node_stack_at@plt>
  mov    (%r12),%edx
  mov    0x0(%rbp),%rsi
  mov    $0x1,%edi
  mov    %eax,%ebx
  call   <node_stack_at@plt>
  test   %eax,%eax
  jne    L35
  lea    -0x4(%rbx),%eax
  cmp    $0x1,%eax
  jbe    L33
L29:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L35:
  sub    $0x4,%eax
  cmp    $0x1,%eax
  ja     L29
  cmp    $0x8,%ebx
  jne    L29
  subl   $0x1,0x0(%r13)
  jmp    L42
  nopl   0x0(%rax)
L33:
  mov    0x0(%r13),%eax
  sub    $0x1,%eax
  mov    %eax,0x0(%r13)
  cmp    $0x5,%ebx
  je     L42
  test   %eax,%eax
  jne    L29
L42:
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <pop_node_stack@plt>
```

assistant: ```c
void parse_array_end(unsigned int **stack, unsigned int *stack_size, unsigned int *array_depth) {
  int stack_top = node_stack_at(0, *stack, *stack_size);
  int parent = node_stack_at(1, *stack, *stack_size);
  if (parent == 0x00000000 && (stack_top == 0x00000004 || stack_top == 0x00000005)) {
    (*array_depth)--;
    if (stack_top == 0x00000005) {
      pop_node_stack(stack, stack_size);
    } else {
      if ((*array_depth) == 0) {
        pop_node_stack(stack, stack_size);
      }
    }
  } else if ((parent == 0x00000004 || parent == 0x00000005) && stack_top == 0x00000008) {
    (*array_depth)--;
    pop_node_stack(stack, stack_size);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<fileOpener>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  lea    D15(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    -0x8(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L11
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L19
L11:
  mov    $0x1,%eax
L19:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D15
datatype: string
name: disassemble_data
arguments: tag: D12
datatype: string

tool: D15: r

tool: D12: Error: cannot open file.


assistant: ```c
int fileOpener(FILE **file, char *fileName) {
  *file = fopen(fileName, "r");
  if (*file == NULL) {
    printf("Error: cannot open file.\n");
    return -1;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fileOpener>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  lea    D8(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,(%rbx)
  mov    $0x1,%edx
  test   %rax,%rax
  je     L3
L16:
  mov    %edx,%eax
  pop    %rbx
  ret
L3:
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%edx
  jmp    L16
```

tool_calls: name: disassemble_data
arguments: tag: D8
type_guessed: string
name: disassemble_data
arguments: tag: D11
type_guessed: string

tool: D8: r

tool: D11: Error: cannot open file.


assistant: ```c
int fileOpener(FILE **file, char *fileName) {
  *file = fopen(fileName, "r");
  if (*file == NULL) {
    printf("Error: cannot open file.\n");
    return -1;
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fileOpener>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  lea    D15(%rip),%rsi
  call   <fopen@plt>
  mov    $0x1,%r8d
  mov    %rax,(%rbx)
  test   %rax,%rax
  je     L16
L18:
  mov    %r8d,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%r8d
  jmp    L18
```

tool_calls: name: retrieve_data
arguments: data_tag: D15
datatype: string
name: retrieve_data
arguments: data_tag: D4
datatype: string

tool: D15: r

tool: D4: Error: cannot open file.


assistant: ```c
int fileOpener(FILE **file, char *fileName) {
  *file = fopen(fileName, "r");
  if (*file == NULL) {
    printf("Error: cannot open file.\n");
    return -1;
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fileOpener>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  lea    D16(%rip),%rsi
  call   <fopen@plt>
  mov    $0x1,%r8d
  mov    %rax,(%rbx)
  test   %rax,%rax
  je     L17
L4:
  mov    %r8d,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%r8d
  jmp    L4
```

tool_calls: name: extract_symbolic_data
arguments: label: D16
datatype: string
name: extract_symbolic_data
arguments: label: D8
datatype: string

tool: D16: r

tool: D8: Error: cannot open file.


assistant: ```c
int fileOpener(FILE **file, char *fileName) {
  *file = fopen(fileName, "r");
  if (*file == NULL) {
    printf("Error: cannot open file.\n");
    return -1;
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getfn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rcx
  mov    %eax,%edx
  lea    D2(%rip),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    D0(%rip),%rax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D2
datatype: string

tool: D2: run%03d.%s

assistant: ```c
char *getfn(int id, char *suffix) {
  static char fn[1000];
  sprintf(fn, "run%03d.%s", id, suffix);
  return (fn);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<getfn>:
  endbr64
  push   %rbx
  mov    %edi,%edx
  mov    %rsi,%rcx
  lea    D2(%rip),%rsi
  lea    D19(%rip),%rbx
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D2
type_guessed: string

tool: D2: run%03d.%s

assistant: ```c
char *getfn(int id, char *suffix) {
  static char fn[1000];
  sprintf(fn, "run%03d.%s", id, suffix);
  return (fn);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getfn>:
  endbr64
  push   %r12
  lea    D17(%rip),%r12
  mov    %edi,%edx
  mov    %rsi,%rcx
  mov    %r12,%rdi
  lea    D7(%rip),%rsi
  call   <sprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
type_guessed: string

tool: D7: run%03d.%s

assistant: ```c
char *getfn(int id, char *suffix) {
  static char fn[1000];
  sprintf(fn, "run%03d.%s", id, suffix);
  return (fn);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getfn>:
  endbr64
  push   %r12
  lea    D16(%rip),%r12
  mov    %edi,%edx
  mov    %rsi,%rcx
  mov    %r12,%rdi
  lea    D0(%rip),%rsi
  call   <sprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D0
data_type: string

tool: D0: run%03d.%s

assistant: ```c
char *getfn(int id, char *suffix) {
  static char fn[1000];
  sprintf(fn, "run%03d.%s", id, suffix);
  return (fn);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<scan_known>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L52
L43:
  addq   $0x1,-0x8(%rbp)
L52:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L61
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jle    L43
  jmp    L61
L24:
  addq   $0x1,-0x8(%rbp)
L61:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L22
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L24
L22:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L32
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <syntax@plt>
L32:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *scan_known(char *from, char *kn) {
  while (*from && *from <= ' ')
    from++;
  while (*kn && *kn++ == *from)
    from++;
  if (*kn)
    syntax(from);
  return from;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<scan_known>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%eax
  test   %al,%al
  je     L68
  cmp    $0x20,%al
  jg     L68
L49:
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  test   %al,%al
  je     L68
  cmp    $0x20,%al
  jle    L49
L68:
  movzbl (%rsi),%eax
  test   %al,%al
  je     L40
L30:
  add    $0x1,%rsi
  cmp    %al,(%rbx)
  jne    L89
  add    $0x1,%rbx
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L30
L40:
  mov    %rbx,%rax
L80:
  pop    %rbx
  ret
L89:
  mov    %rbx,%rax
  cmpb   $0x0,(%rsi)
  je     L80
  mov    %rbx,%rdi
  call   <syntax@plt>
  mov    %rbx,%rax
  jmp    L80
```

assistant: ```c
char *scan_known(char *from, char *kn) {
  while (*from && *from <= ' ')
    from++;
  while (*kn && *kn++ == *from)
    from++;
  if (*kn)
    syntax(from);
  return from;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<scan_known>:
  endbr64
  push   %r12
  movzbl (%rdi),%eax
  mov    %rdi,%r12
  test   %al,%al
  jne    L16
  jmp    L11
  nopw   0x0(%rax,%rax,1)
L59:
  movzbl 0x1(%r12),%eax
  add    $0x1,%r12
  test   %al,%al
  je     L11
L16:
  cmp    $0x20,%al
  jle    L59
L11:
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L36
  jmp    L74
  nopl   0x0(%rax,%rax,1)
L69:
  add    $0x1,%r12
  test   %al,%al
  je     L74
L36:
  add    $0x1,%rsi
  mov    %eax,%edx
  movzbl (%rsi),%eax
  cmp    %dl,(%r12)
  je     L69
  test   %al,%al
  je     L74
  mov    %r12,%rdi
  call   <syntax@plt>
L74:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *scan_known(char *from, char *kn) {
  while (*from && *from <= ' ')
    from++;
  while (*kn && *kn++ == *from)
    from++;
  if (*kn)
    syntax(from);
  return from;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<scan_known>:
  endbr64
  push   %r12
  movzbl (%rdi),%eax
  mov    %rdi,%r12
  test   %al,%al
  jne    L33
  jmp    L57
  nopw   0x0(%rax,%rax,1)
L85:
  movzbl 0x1(%r12),%eax
  add    $0x1,%r12
  test   %al,%al
  je     L57
L33:
  cmp    $0x20,%al
  jle    L85
L57:
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L52
  jmp    L62
  nopl   0x0(%rax,%rax,1)
L24:
  add    $0x1,%r12
  test   %al,%al
  je     L62
L52:
  add    $0x1,%rsi
  mov    %eax,%edx
  movzbl (%rsi),%eax
  cmp    %dl,(%r12)
  je     L24
  test   %al,%al
  je     L62
  mov    %r12,%rdi
  call   <syntax@plt>
L62:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *scan_known(char *from, char *kn) {
  while (*from && *from <= ' ')
    from++;
  while (*kn && *kn++ == *from)
    from++;
  if (*kn)
    syntax(from);
  return from;
}
```