ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r14
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D55(%rip),%rdi
  call   <puts@plt>
  lea    0x20(%rsp),%rsi
  lea    D30(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D44(%rip),%rdi
  call   <puts@plt>
  lea    0x18(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movsd  0x20(%rsp),%xmm2
  movapd %xmm2,%xmm6
  mulsd  D1(%rip),%xmm6
  movq   %xmm6,%r14
  movsd  0x18(%rsp),%xmm1
  movapd %xmm6,%xmm0
  call   <fun@plt>
  movq   %xmm0,%rbx
  movsd  0x18(%rsp),%xmm1
  movq   %r14,%xmm0
  call   <deriv@plt>
  movq   %rbx,%xmm1
  divsd  %xmm0,%xmm1
  movq   %r14,%xmm0
  subsd  %xmm1,%xmm0
  pxor   %xmm7,%xmm7
  movsd  %xmm7,0x8(%rsp)
L11:
  movsd  0x20(%rsp),%xmm2
  movsd  0x18(%rsp),%xmm1
  call   <next@plt>
  movsd  D32(%rip),%xmm3
  addsd  0x8(%rsp),%xmm3
  movsd  %xmm3,0x8(%rsp)
  movsd  D49(%rip),%xmm5
  comisd %xmm3,%xmm5
  ja     L11
  lea    D15(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %r14
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D55
data_type: string
name: extract_symbolic_data
arguments: data_tag: D30
data_type: string
name: extract_symbolic_data
arguments: data_tag: D44
data_type: string
name: extract_symbolic_data
arguments: data_tag: D1
data_type: byte[8]
name: extract_symbolic_data
arguments: data_tag: D32
data_type: byte[8]
name: extract_symbolic_data
arguments: data_tag: D49
data_type: byte[8]
name: extract_symbolic_data
arguments: data_tag: D15
data_type: string

tool: D55: Please enter the value of c

tool: D30: %lf

tool: D44: Please enter the value of n

tool: {"D1": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: {"D32": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: {"D49": [0x00,0x00,0x00,0x00,0x00,0x6a,0xf8,0x40]}

tool: D15: The root of the funcation is %f


assistant: ```c
main() {
  double x_in, n, c, x_nxt, i, tm;
  printf("Please enter the value of c\n");
  scanf("%lf", &c);
  printf("Please enter the value of n\n");
  scanf("%lf", &n);
  x_in = c / 2;
  x_nxt = x_in - fun(x_in, n, c) / deriv(x_in, n);
  for (i = 0; i < 100000; i++) {
    x_nxt = next(x_nxt, n, c);
  }
  printf("The root of the funcation is %f\n", x_nxt);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D32(%rip),%rdi
  lea    D44(%rip),%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x20(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D21(%rip),%rdi
  call   <puts@plt>
  lea    0x18(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movsd  0x20(%rsp),%xmm2
  movsd  D69(%rip),%xmm3
  movsd  0x18(%rsp),%xmm1
  mulsd  %xmm2,%xmm3
  movapd %xmm3,%xmm0
  movsd  %xmm3,(%rsp)
  call   <fun@plt>
  movsd  (%rsp),%xmm3
  movsd  0x18(%rsp),%xmm1
  movsd  %xmm0,0x8(%rsp)
  movapd %xmm3,%xmm0
  call   <deriv@plt>
  movsd  0x8(%rsp),%xmm2
  movsd  (%rsp),%xmm3
  movapd %xmm2,%xmm1
  divsd  %xmm0,%xmm1
  subsd  %xmm1,%xmm3
  movapd %xmm3,%xmm0
  pxor   %xmm3,%xmm3
  nopl   0x0(%rax)
L14:
  movsd  0x20(%rsp),%xmm2
  movsd  0x18(%rsp),%xmm1
  movsd  %xmm3,(%rsp)
  call   <next@plt>
  movsd  (%rsp),%xmm3
  addsd  D27(%rip),%xmm3
  movsd  D64(%rip),%xmm4
  comisd %xmm3,%xmm4
  ja     L14
  lea    D57(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D32
type: string
name: disassemble_data
arguments: data_tag: D44
type: string
name: disassemble_data
arguments: data_tag: D21
type: string
name: disassemble_data
arguments: data_tag: D69
type: byte[8]
name: disassemble_data
arguments: data_tag: D27
type: byte[8]
name: disassemble_data
arguments: data_tag: D64
type: byte[8]
name: disassemble_data
arguments: data_tag: D57
type: string

tool: D32: Please enter the value of c

tool: D44: %lf

tool: D21: Please enter the value of n

tool: {"D69": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: {"D27": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: {"D64": [0x00,0x00,0x00,0x00,0x00,0x6a,0xf8,0x40]}

tool: D57: The root of the funcation is %f


assistant: ```c
main() {
  double x_in, n, c, x_nxt, i, tm;
  printf("Please enter the value of c\n");
  scanf("%lf", &c);
  printf("Please enter the value of n\n");
  scanf("%lf", &n);
  x_in = c / 2;
  x_nxt = x_in - fun(x_in, n, c) / deriv(x_in, n);
  for (i = 0; i < 100000; i++) {
    x_nxt = next(x_nxt, n, c);
  }
  printf("The root of the funcation is %f\n", x_nxt);
}
```
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
  lea    D58(%rip),%rdi
  lea    D37(%rip),%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x20(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D13(%rip),%rdi
  call   <puts@plt>
  lea    0x18(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movsd  0x20(%rsp),%xmm2
  movsd  D46(%rip),%xmm3
  movsd  0x18(%rsp),%xmm1
  mulsd  %xmm2,%xmm3
  movapd %xmm3,%xmm0
  movsd  %xmm3,(%rsp)
  call   <fun@plt>
  movsd  (%rsp),%xmm3
  movsd  0x18(%rsp),%xmm1
  movsd  %xmm0,0x8(%rsp)
  movapd %xmm3,%xmm0
  call   <deriv@plt>
  movsd  0x8(%rsp),%xmm2
  movsd  (%rsp),%xmm3
  movapd %xmm2,%xmm1
  divsd  %xmm0,%xmm1
  subsd  %xmm1,%xmm3
  movapd %xmm3,%xmm0
  pxor   %xmm3,%xmm3
  nopl   0x0(%rax)
L6:
  movsd  0x20(%rsp),%xmm2
  movsd  0x18(%rsp),%xmm1
  movsd  %xmm3,(%rsp)
  call   <next@plt>
  movsd  (%rsp),%xmm3
  addsd  D68(%rip),%xmm3
  movsd  D29(%rip),%xmm4
  comisd %xmm3,%xmm4
  ja     L6
  lea    D22(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D58
datatype: string
name: extract_symbolic_data
arguments: data_label: D37
datatype: string
name: extract_symbolic_data
arguments: data_label: D13
datatype: string
name: extract_symbolic_data
arguments: data_label: D46
datatype: byte[8]
name: extract_symbolic_data
arguments: data_label: D68
datatype: byte[8]
name: extract_symbolic_data
arguments: data_label: D29
datatype: byte[8]
name: extract_symbolic_data
arguments: data_label: D22
datatype: string

tool: D58: Please enter the value of c

tool: D37: %lf

tool: D13: Please enter the value of n

tool: {"D46": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: {"D68": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: {"D29": [0x00,0x00,0x00,0x00,0x00,0x6a,0xf8,0x40]}

tool: D22: The root of the funcation is %f


assistant: ```c
main() {
  double x_in, n, c, x_nxt, i, tm;
  printf("Please enter the value of c\n");
  scanf("%lf", &c);
  printf("Please enter the value of n\n");
  scanf("%lf", &n);
  x_in = c / 2;
  x_nxt = x_in - fun(x_in, n, c) / deriv(x_in, n);
  for (i = 0; i < 100000; i++) {
    x_nxt = next(x_nxt, n, c);
  }
  printf("The root of the funcation is %f\n", x_nxt);
}
```
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    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <c_escape_tests@plt>
  mov    $0x0,%eax
  call   <case_convert_tests@plt>
  mov    $0x0,%eax
  call   <char_class_tests@plt>
  mov    $0x0,%eax
  call   <elide_tests@plt>
  mov    $0x0,%eax
  call   <quote_tests@plt>
  mov    $0x0,%eax
  call   <transform_tests@plt>
  mov    $0x0,%eax
  call   <uint8_convert_tests@plt>
  mov    $0x0,%eax
  call   <url_encode_tests@plt>
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  c_escape_tests();
  case_convert_tests();
  char_class_tests();
  elide_tests();
  quote_tests();
  transform_tests();
  uint8_convert_tests();
  url_encode_tests();
  return EXIT_SUCCESS;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <c_escape_tests@plt>
  mov    $0x0,%eax
  call   <case_convert_tests@plt>
  mov    $0x0,%eax
  call   <char_class_tests@plt>
  mov    $0x0,%eax
  call   <elide_tests@plt>
  mov    $0x0,%eax
  call   <quote_tests@plt>
  mov    $0x0,%eax
  call   <transform_tests@plt>
  mov    $0x0,%eax
  call   <uint8_convert_tests@plt>
  mov    $0x0,%eax
  call   <url_encode_tests@plt>
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  c_escape_tests();
  case_convert_tests();
  char_class_tests();
  elide_tests();
  quote_tests();
  transform_tests();
  uint8_convert_tests();
  url_encode_tests();
  return EXIT_SUCCESS;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <c_escape_tests@plt>
  xor    %eax,%eax
  call   <case_convert_tests@plt>
  xor    %eax,%eax
  call   <char_class_tests@plt>
  xor    %eax,%eax
  call   <elide_tests@plt>
  xor    %eax,%eax
  call   <quote_tests@plt>
  xor    %eax,%eax
  call   <transform_tests@plt>
  xor    %eax,%eax
  call   <uint8_convert_tests@plt>
  xor    %eax,%eax
  call   <url_encode_tests@plt>
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  c_escape_tests();
  case_convert_tests();
  char_class_tests();
  elide_tests();
  quote_tests();
  transform_tests();
  uint8_convert_tests();
  url_encode_tests();
  return EXIT_SUCCESS;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <c_escape_tests@plt>
  xor    %eax,%eax
  call   <case_convert_tests@plt>
  xor    %eax,%eax
  call   <char_class_tests@plt>
  xor    %eax,%eax
  call   <elide_tests@plt>
  xor    %eax,%eax
  call   <quote_tests@plt>
  xor    %eax,%eax
  call   <transform_tests@plt>
  xor    %eax,%eax
  call   <uint8_convert_tests@plt>
  xor    %eax,%eax
  call   <url_encode_tests@plt>
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  c_escape_tests();
  case_convert_tests();
  char_class_tests();
  elide_tests();
  quote_tests();
  transform_tests();
  uint8_convert_tests();
  url_encode_tests();
  return EXIT_SUCCESS;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rsi
  movq   $0x1,-0x18(%rbp)
  mov    $0x2,%edx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%edi
  mov    $0x0,%edx
  div    %rdi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L0:
  cmp    %rdx,%rsp
  je     L13
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L0
L13:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L4
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L4:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x10(%rbp)
  mov    $0x2,%eax
  mov    %rsi,%rsp
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

assistant: ```c
main() {
  char x[(1, 2)];
  return sizeof(x);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

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

assistant: ```c
main() {
  char x[(1, 2)];
  return sizeof(x);
}
```
user: Translate the below assembly code into C programming language:

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

assistant: ```c
main() {
  char x[(1, 2)];
  return sizeof(x);
}
```
user: 下面的汇编指令对应的C源代码是什么:

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

assistant: ```c
main() {
  char x[(1, 2)];
  return sizeof(x);
}
```
user: What is the c source code of the assembly code below?

```asm
<sum2d>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L33
L37:
  movl   $0x0,-0x8(%rbp)
  jmp    L30
L8:
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  add    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L30:
  cmpl   $0x3,-0x8(%rbp)
  jle    L8
  addl   $0x1,-0xc(%rbp)
L33:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L37
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int sum2d(int ar[][4], int rows) {
  int r;
  int c;
  int tot = 0;
  for (r = 0; r < rows; r++) {
    for (c = 0; c < 4; c++) {
      tot += ar[r][c];
    }
  }
  return tot;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sum2d>:
  endbr64
  test   %esi,%esi
  jle    L10
  mov    %rdi,%rdx
  lea    -0x1(%rsi),%eax
  shl    $0x4,%rax
  lea    0x10(%rdi,%rax,1),%rcx
  mov    $0x0,%eax
L13:
  add    (%rdx),%eax
  add    0x4(%rdx),%eax
  add    0x8(%rdx),%eax
  add    0xc(%rdx),%eax
  add    $0x10,%rdx
  cmp    %rcx,%rdx
  jne    L13
  ret
L10:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int sum2d(int ar[][4], int rows) {
  int r;
  int c;
  int tot = 0;
  for (r = 0; r < rows; r++) {
    for (c = 0; c < 4; c++) {
      tot += ar[r][c];
    }
  }
  return tot;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sum2d>:
  endbr64
  test   %esi,%esi
  jle    L13
  lea    -0x1(%rsi),%eax
  shl    $0x4,%rax
  lea    0x10(%rdi,%rax,1),%rdx
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L4:
  add    (%rdi),%eax
  add    $0x10,%rdi
  add    -0xc(%rdi),%eax
  add    -0x8(%rdi),%eax
  add    -0x4(%rdi),%eax
  cmp    %rdx,%rdi
  jne    L4
  ret
L13:
  xor    %eax,%eax
  ret
```

assistant: ```c
int sum2d(int ar[][4], int rows) {
  int r;
  int c;
  int tot = 0;
  for (r = 0; r < rows; r++) {
    for (c = 0; c < 4; c++) {
      tot += ar[r][c];
    }
  }
  return tot;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<sum2d>:
  endbr64
  test   %esi,%esi
  jle    L5
  lea    -0x1(%rsi),%eax
  pxor   %xmm0,%xmm0
  shl    $0x4,%rax
  lea    0x10(%rdi,%rax,1),%rax
  nopl   0x0(%rax,%rax,1)
L4:
  movdqu (%rdi),%xmm2
  add    $0x10,%rdi
  paddd  %xmm2,%xmm0
  cmp    %rax,%rdi
  jne    L4
  movdqa %xmm0,%xmm1
  psrldq $0x8,%xmm1
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  ret
L5:
  xor    %eax,%eax
  ret
```

assistant: ```c
int sum2d(int ar[][4], int rows) {
  int r;
  int c;
  int tot = 0;
  for (r = 0; r < rows; r++) {
    for (c = 0; c < 4; c++) {
      tot += ar[r][c];
    }
  }
  return tot;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x5,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D8
datatype: string

tool: D8: %d


assistant: ```c
int main() {
  1 + 2;
  printf("%d\n", 1 < 2 ? 5 : 8);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D0: %d


assistant: ```c
int main() {
  1 + 2;
  printf("%d\n", 1 < 2 ? 5 : 8);
  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
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D2
type: string

tool: D2: %d


assistant: ```c
int main() {
  1 + 2;
  printf("%d\n", 1 < 2 ? 5 : 8);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D4
type_guessed: string

tool: D4: %d


assistant: ```c
int main() {
  1 + 2;
  printf("%d\n", 1 < 2 ? 5 : 8);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<isInteger>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L26
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  je     L26
  mov    $0x0,%eax
  jmp    L87
L26:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L30
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L35
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L30
L35:
  mov    $0x0,%eax
  jmp    L87
L30:
  addq   $0x1,-0x8(%rbp)
  jmp    L61
L25:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L99
  mov    $0x0,%eax
  jmp    L87
L99:
  addq   $0x1,-0x8(%rbp)
L61:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L25
  mov    $0x1,%eax
L87:
  leave
  ret
```

assistant: ```c
int isInteger(char *buf) {
  if (isdigit(*buf) == 0 && *buf != '-') {
    return 0;
  }
  if (isdigit(*buf) == 0 && (*(buf + 1) == '\0' || isdigit(*(buf + 1)) == 0)) {
    return 0;
  }
  buf++;
  while (*buf != '\0') {
    if (isdigit(*buf) == 0) {
      return 0;
    }
    buf++;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<isInteger>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  movsbl (%rdi),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L86
  cmpb   $0x2d,0x0(%rbp)
  jne    L0
L86:
  movsbl 0x0(%rbp),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L52
  movzbl 0x1(%rbp),%edi
  test   %dil,%dil
  je     L0
  movsbl %dil,%edi
  call   <isdigit@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  je     L0
L52:
  lea    0x1(%rbp),%rbx
  movzbl 0x1(%rbp),%edi
  test   %dil,%dil
  je     L31
L13:
  movsbl %dil,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L93
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L13
  mov    $0x1,%eax
  jmp    L0
L31:
  mov    $0x1,%eax
  jmp    L0
L93:
  mov    $0x0,%eax
L0:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int isInteger(char *buf) {
  if (isdigit(*buf) == 0 && *buf != '-') {
    return 0;
  }
  if (isdigit(*buf) == 0 && (*(buf + 1) == '\0' || isdigit(*(buf + 1)) == 0)) {
    return 0;
  }
  buf++;
  while (*buf != '\0') {
    if (isdigit(*buf) == 0) {
      return 0;
    }
    buf++;
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isInteger>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  call   <isdigit@plt>
  movsbl 0x0(%rbp),%edi
  test   %rax,%rax
  jne    L38
  cmp    $0x2d,%dil
  je     L38
L55:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L38:
  call   <isdigit@plt>
  test   %rax,%rax
  je     L70
L93:
  movsbl 0x1(%rbp),%edi
  lea    0x1(%rbp),%rbx
  test   %dil,%dil
  jne    L73
  jmp    L56
  nopl   0x0(%rax)
L41:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L56
L73:
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L41
  jmp    L55
  nopl   0x0(%rax)
L70:
  movsbl 0x1(%rbp),%edi
  test   %dil,%dil
  je     L55
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L93
  jmp    L55
  nopl   (%rax)
L56:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int isInteger(char *buf) {
  if (isdigit(*buf) == 0 && *buf != '-') {
    return 0;
  }
  if (isdigit(*buf) == 0 && (*(buf + 1) == '\0' || isdigit(*(buf + 1)) == 0)) {
    return 0;
  }
  buf++;
  while (*buf != '\0') {
    if (isdigit(*buf) == 0) {
      return 0;
    }
    buf++;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<isInteger>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  call   <isdigit@plt>
  movsbl 0x0(%rbp),%edi
  test   %rax,%rax
  jne    L21
  cmp    $0x2d,%dil
  je     L21
L73:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L21:
  call   <isdigit@plt>
  test   %rax,%rax
  je     L11
L0:
  movsbl 0x1(%rbp),%edi
  lea    0x1(%rbp),%rbx
  test   %dil,%dil
  jne    L19
  jmp    L91
  nopl   0x0(%rax)
L50:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L91
L19:
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L50
  jmp    L73
  nopl   0x0(%rax)
L11:
  movsbl 0x1(%rbp),%edi
  test   %dil,%dil
  je     L73
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L0
  jmp    L73
  nopl   (%rax)
L91:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int isInteger(char *buf) {
  if (isdigit(*buf) == 0 && *buf != '-') {
    return 0;
  }
  if (isdigit(*buf) == 0 && (*(buf + 1) == '\0' || isdigit(*(buf + 1)) == 0)) {
    return 0;
  }
  buf++;
  while (*buf != '\0') {
    if (isdigit(*buf) == 0) {
      return 0;
    }
    buf++;
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<tbx_chksum_type_valid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jle    L12
  cmpl   $0x4,-0x4(%rbp)
  jg     L12
  mov    $0x1,%eax
  jmp    L26
L12:
  mov    $0x0,%eax
L26:
  pop    %rbp
  ret
```

assistant: ```c
int tbx_chksum_type_valid(int type) {
  if ((type > 0) && (type < 5))
    return (1);
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<tbx_chksum_type_valid>:
  endbr64
  sub    $0x1,%edi
  cmp    $0x3,%edi
  setbe  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int tbx_chksum_type_valid(int type) {
  if ((type > 0) && (type < 5))
    return (1);
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<tbx_chksum_type_valid>:
  endbr64
  sub    $0x1,%edi
  xor    %eax,%eax
  cmp    $0x3,%edi
  setbe  %al
  ret
```

assistant: ```c
int tbx_chksum_type_valid(int type) {
  if ((type > 0) && (type < 5))
    return (1);
  return (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<tbx_chksum_type_valid>:
  endbr64
  sub    $0x1,%edi
  xor    %eax,%eax
  cmp    $0x3,%edi
  setbe  %al
  ret
```

assistant: ```c
int tbx_chksum_type_valid(int type) {
  if ((type > 0) && (type < 5))
    return (1);
  return (0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<cgiAppendSForHttps>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <cgiServerHttpsIsOn@plt>
  test   %rax,%rax
  je     L14
  lea    D7(%rip),%rax
  jmp    L1
L14:
  lea    D8(%rip),%rax
L1:
  pop    %rbp
  ret
```

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

tool: D7: s

tool: 

assistant: ```c
char *cgiAppendSForHttps() {
  if (cgiServerHttpsIsOn())
    return "s";
  return "";
}
```