ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movss  D47(%rip),%xmm0
  movss  %xmm0,-0xc(%rbp)
  movss  D76(%rip),%xmm0
  movss  %xmm0,-0x8(%rbp)
  movss  D45(%rip),%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0xc(%rbp),%xmm0
  movss  %xmm0,-0x10(%rbp)
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    D15(%rip),%eax
  movd   %eax,%xmm0
  call   <fahrenheit@plt>
  movq   %xmm0,%rdx
  mov    D49(%rip),%rax
  movq   %rdx,%xmm1
  movq   %rax,%xmm0
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  jmp    L14
L6:
  mov    -0x10(%rbp),%eax
  movd   %eax,%xmm0
  call   <fahrenheit@plt>
  movq   %xmm0,%rdx
  pxor   %xmm2,%xmm2
  cvtss2sd -0x10(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rdx,%xmm1
  movq   %rax,%xmm0
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  movss  -0x10(%rbp),%xmm0
  addss  -0x4(%rbp),%xmm0
  movss  %xmm0,-0x10(%rbp)
L14:
  movss  -0x8(%rbp),%xmm0
  comiss -0x10(%rbp),%xmm0
  jae    L6
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D47
type_guessed: byte[4]
name: disassemble_data
arguments: data_tag: D76
type_guessed: byte[4]
name: disassemble_data
arguments: data_tag: D45
type_guessed: byte[4]
name: disassemble_data
arguments: data_tag: D22
type_guessed: string
name: disassemble_data
arguments: data_tag: D15
type_guessed: byte[4]
name: disassemble_data
arguments: data_tag: D49
type_guessed: byte[8]
name: disassemble_data
arguments: data_tag: D9
type_guessed: string
name: disassemble_data
arguments: data_tag: D58
type_guessed: string

tool: {"D47": [0x00,0x00,0x70,0xc2]}

tool: {"D76": [0x00,0x00,0x96,0x43]}

tool: {"D45": [0x00,0x00,0xa0,0x41]}

tool: D22: Celsius Fahrenheit

tool: {"D15": [0x33,0x93,0x88,0xc3]}

tool: {"D49": [0x66,0x66,0x66,0x66,0x66,0x12,0x71,0xc0]}

tool: D9: %7.2f %10.1f


tool: D58: %7.0f %10.2f


assistant: ```c
int main(int argc, char *argv[]) {
  float celsius;
  float lower, upper, step;
  lower = -60;
  upper = 300;
  step = 20;
  celsius = lower;
  printf("Celsius Fahrenheit\n");
  printf("%7.2f %10.1f\n", -273.15, fahrenheit(-273.15));
  while (celsius <= upper) {
    printf("%7.0f %10.2f\n", celsius, fahrenheit(celsius));
    celsius = celsius + step;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  lea    D43(%rip),%rdi
  call   <puts@plt>
  movss  D14(%rip),%xmm0
  call   <fahrenheit@plt>
  movapd %xmm0,%xmm1
  movsd  D9(%rip),%xmm0
  lea    D12(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  mov    $0x13,%ebx
  movss  D46(%rip),%xmm4
  movss  %xmm4,0xc(%rsp)
  lea    D18(%rip),%rbp
L0:
  movss  0xc(%rsp),%xmm0
  call   <fahrenheit@plt>
  movapd %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtss2sd 0xc(%rsp),%xmm0
  mov    %rbp,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  movss  D40(%rip),%xmm3
  addss  0xc(%rsp),%xmm3
  movss  %xmm3,0xc(%rsp)
  sub    $0x1,%ebx
  jne    L0
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D43
data_type: string
name: unpack_data
arguments: tag: D14
data_type: byte[4]
name: unpack_data
arguments: tag: D9
data_type: byte[8]
name: unpack_data
arguments: tag: D12
data_type: string
name: unpack_data
arguments: tag: D46
data_type: byte[4]
name: unpack_data
arguments: tag: D18
data_type: string
name: unpack_data
arguments: tag: D40
data_type: byte[4]

tool: D43: Celsius Fahrenheit

tool: {"D14": [0x33,0x93,0x88,0xc3]}

tool: {"D9": [0x66,0x66,0x66,0x66,0x66,0x12,0x71,0xc0]}

tool: D12: %7.2f %10.1f


tool: {"D46": [0x00,0x00,0x70,0xc2]}

tool: D18: %7.0f %10.2f


tool: {"D40": [0x00,0x00,0xa0,0x41]}

assistant: ```c
int main(int argc, char *argv[]) {
  float celsius;
  float lower, upper, step;
  lower = -60;
  upper = 300;
  step = 20;
  celsius = lower;
  printf("Celsius Fahrenheit\n");
  printf("%7.2f %10.1f\n", -273.15, fahrenheit(-273.15));
  while (celsius <= upper) {
    printf("%7.0f %10.2f\n", celsius, fahrenheit(celsius));
    celsius = celsius + step;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D16(%rip),%rdi
  lea    D49(%rip),%rbp
  push   %rbx
  mov    $0x13,%ebx
  sub    $0x18,%rsp
  call   <puts@plt>
  movss  D12(%rip),%xmm0
  call   <fahrenheit@plt>
  lea    D65(%rip),%rdi
  mov    $0x2,%eax
  movapd %xmm0,%xmm1
  movsd  D13(%rip),%xmm0
  call   <printf@plt>
  movss  D9(%rip),%xmm2
  nopl   0x0(%rax)
L2:
  movaps %xmm2,%xmm0
  movss  %xmm2,0xc(%rsp)
  call   <fahrenheit@plt>
  movss  0xc(%rsp),%xmm2
  mov    %rbp,%rdi
  mov    $0x2,%eax
  movapd %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtss2sd %xmm2,%xmm0
  call   <printf@plt>
  movss  0xc(%rsp),%xmm2
  addss  D8(%rip),%xmm2
  sub    $0x1,%ebx
  jne    L2
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D16
data_type: string
name: read_data
arguments: data_tag: D49
data_type: string
name: read_data
arguments: data_tag: D12
data_type: byte[4]
name: read_data
arguments: data_tag: D65
data_type: string
name: read_data
arguments: data_tag: D13
data_type: byte[8]
name: read_data
arguments: data_tag: D9
data_type: byte[4]
name: read_data
arguments: data_tag: D8
data_type: byte[4]

tool: D16: Celsius Fahrenheit

tool: D49: %7.0f %10.2f


tool: {"D12": [0x33,0x93,0x88,0xc3]}

tool: D65: %7.2f %10.1f


tool: {"D13": [0x66,0x66,0x66,0x66,0x66,0x12,0x71,0xc0]}

tool: {"D9": [0x00,0x00,0x70,0xc2]}

tool: {"D8": [0x00,0x00,0xa0,0x41]}

assistant: ```c
int main(int argc, char *argv[]) {
  float celsius;
  float lower, upper, step;
  lower = -60;
  upper = 300;
  step = 20;
  celsius = lower;
  printf("Celsius Fahrenheit\n");
  printf("%7.2f %10.1f\n", -273.15, fahrenheit(-273.15));
  while (celsius <= upper) {
    printf("%7.0f %10.2f\n", celsius, fahrenheit(celsius));
    celsius = celsius + step;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D10(%rip),%rdi
  lea    D23(%rip),%rbp
  push   %rbx
  mov    $0x13,%ebx
  sub    $0x18,%rsp
  call   <puts@plt>
  movss  D8(%rip),%xmm0
  call   <fahrenheit@plt>
  lea    D22(%rip),%rdi
  mov    $0x2,%eax
  movapd %xmm0,%xmm1
  movsd  D17(%rip),%xmm0
  call   <printf@plt>
  movss  D65(%rip),%xmm2
  nopl   0x0(%rax)
L8:
  movaps %xmm2,%xmm0
  movss  %xmm2,0xc(%rsp)
  call   <fahrenheit@plt>
  movss  0xc(%rsp),%xmm2
  mov    %rbp,%rdi
  mov    $0x2,%eax
  movapd %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtss2sd %xmm2,%xmm0
  call   <printf@plt>
  movss  0xc(%rsp),%xmm2
  addss  D25(%rip),%xmm2
  sub    $0x1,%ebx
  jne    L8
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D10
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D23
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D8
type_guessed: byte[4]
name: extract_symbolic_data
arguments: data_label: D22
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D17
type_guessed: byte[8]
name: extract_symbolic_data
arguments: data_label: D65
type_guessed: byte[4]
name: extract_symbolic_data
arguments: data_label: D25
type_guessed: byte[4]

tool: D10: Celsius Fahrenheit

tool: D23: %7.0f %10.2f


tool: {"D8": [0x33,0x93,0x88,0xc3]}

tool: D22: %7.2f %10.1f


tool: {"D17": [0x66,0x66,0x66,0x66,0x66,0x12,0x71,0xc0]}

tool: {"D65": [0x00,0x00,0x70,0xc2]}

tool: {"D25": [0x00,0x00,0xa0,0x41]}

assistant: ```c
int main(int argc, char *argv[]) {
  float celsius;
  float lower, upper, step;
  lower = -60;
  upper = 300;
  step = 20;
  celsius = lower;
  printf("Celsius Fahrenheit\n");
  printf("%7.2f %10.1f\n", -273.15, fahrenheit(-273.15));
  while (celsius <= upper) {
    printf("%7.0f %10.2f\n", celsius, fahrenheit(celsius));
    celsius = celsius + step;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<next_permutation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  sub    $0x1,%edx
  mov    %edx,(%rax)
  mov    -0x24(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  imul   %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  movl   $0x1,-0x24(%rbp)
  jmp    L15
L12:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x24(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x24(%rbp)
L15:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L12
  mov    -0x24(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int next_permutation(int *o, int *p, int k) {
  p[k]--;
  int j = k % 2 * p[k];
  int tmp = o[j];
  o[j] = o[k];
  o[k] = tmp;
  for (k = 1; p[k] == 0; k++)
    p[k] = k;
  return k;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<next_permutation>:
  endbr64
  mov    %rdi,%rcx
  movslq %edx,%rdi
  shl    $0x2,%rdi
  lea    (%rsi,%rdi,1),%rax
  mov    (%rax),%r10d
  lea    -0x1(%r10),%r8d
  mov    %r8d,(%rax)
  mov    %edx,%r9d
  shr    $0x1f,%r9d
  lea    (%rdx,%r9,1),%eax
  and    $0x1,%eax
  sub    %r9d,%eax
  imul   %r8d,%eax
  cltq
  lea    (%rcx,%rax,4),%rax
  mov    (%rax),%edx
  add    %rdi,%rcx
  mov    (%rcx),%edi
  mov    %edi,(%rax)
  mov    %edx,(%rcx)
  lea    0x4(%rsi),%rax
  cmpl   $0x0,0x4(%rsi)
  jne    L22
  mov    $0x1,%edx
L17:
  mov    %edx,(%rax)
  add    $0x1,%edx
  add    $0x4,%rax
  cmpl   $0x0,(%rax)
  je     L17
L27:
  mov    %edx,%eax
  ret
L22:
  mov    $0x1,%edx
  jmp    L27
```

assistant: ```c
int next_permutation(int *o, int *p, int k) {
  p[k]--;
  int j = k % 2 * p[k];
  int tmp = o[j];
  o[j] = o[k];
  o[k] = tmp;
  for (k = 1; p[k] == 0; k++)
    p[k] = k;
  return k;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<next_permutation>:
  endbr64
  mov    %rdi,%rcx
  movslq %edx,%rdi
  mov    %edx,%r9d
  shl    $0x2,%rdi
  shr    $0x1f,%r9d
  lea    (%rsi,%rdi,1),%rax
  mov    (%rax),%r10d
  lea    -0x1(%r10),%r8d
  mov    %r8d,(%rax)
  lea    (%rdx,%r9,1),%eax
  and    $0x1,%eax
  sub    %r9d,%eax
  imul   %r8d,%eax
  mov    $0x1,%r8d
  cltq
  lea    (%rcx,%rax,4),%rax
  add    %rdi,%rcx
  mov    (%rax),%edx
  mov    (%rcx),%edi
  mov    %edi,(%rax)
  lea    0x4(%rsi),%rax
  mov    %edx,(%rcx)
  mov    0x4(%rsi),%ecx
  test   %ecx,%ecx
  jne    L13
  nopl   0x0(%rax,%rax,1)
L17:
  mov    %r8d,(%rax)
  mov    0x4(%rax),%edx
  add    $0x4,%rax
  add    $0x1,%r8d
  test   %edx,%edx
  je     L17
L13:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int next_permutation(int *o, int *p, int k) {
  p[k]--;
  int j = k % 2 * p[k];
  int tmp = o[j];
  o[j] = o[k];
  o[k] = tmp;
  for (k = 1; p[k] == 0; k++)
    p[k] = k;
  return k;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<next_permutation>:
  endbr64
  mov    %rdi,%rcx
  movslq %edx,%rdi
  mov    %edx,%r9d
  shl    $0x2,%rdi
  shr    $0x1f,%r9d
  lea    (%rsi,%rdi,1),%rax
  mov    (%rax),%r10d
  lea    -0x1(%r10),%r8d
  mov    %r8d,(%rax)
  lea    (%rdx,%r9,1),%eax
  and    $0x1,%eax
  sub    %r9d,%eax
  imul   %r8d,%eax
  mov    $0x1,%r8d
  cltq
  lea    (%rcx,%rax,4),%rax
  add    %rdi,%rcx
  mov    (%rax),%edx
  mov    (%rcx),%edi
  mov    %edi,(%rax)
  lea    0x4(%rsi),%rax
  mov    %edx,(%rcx)
  mov    0x4(%rsi),%ecx
  test   %ecx,%ecx
  jne    L11
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %r8d,(%rax)
  mov    0x4(%rax),%edx
  add    $0x4,%rax
  add    $0x1,%r8d
  test   %edx,%edx
  je     L1
L11:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int next_permutation(int *o, int *p, int k) {
  p[k]--;
  int j = k % 2 * p[k];
  int tmp = o[j];
  o[j] = o[k];
  o[k] = tmp;
  for (k = 1; p[k] == 0; k++)
    p[k] = k;
  return k;
}
```
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   $0x1,-0xc(%rbp)
  movl   $0x2,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <add@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
type_infered: string

tool: D1: %d



assistant: ```c
int main(void) {
  int a = 1, b = 2;
  int c = add(a, b);
  printf("%d\n\n", c);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <add@plt>
  mov    %eax,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D5
type_infered: string

tool: D5: %d



assistant: ```c
int main(void) {
  int a = 1, b = 2;
  int c = add(a, b);
  printf("%d\n\n", c);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <add@plt>
  lea    D3(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D3
datatype: string

tool: D3: %d



assistant: ```c
int main(void) {
  int a = 1, b = 2;
  int c = add(a, b);
  printf("%d\n\n", c);
  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
  sub    $0x8,%rsp
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <add@plt>
  lea    D5(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D5
type_infered: string

tool: D5: %d



assistant: ```c
int main(void) {
  int a = 1, b = 2;
  int c = add(a, b);
  printf("%d\n\n", c);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Josephusproblem4>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D30(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D2(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    $0x10,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  movl   $0x1,(%rax)
  cmpl   $0x1,(%rsp)
  jle    L62
  mov    %rax,%rbx
  mov    $0x2,%ebp
L4:
  mov    $0x10,%edi
  call   <malloc@plt>
  mov    %rbx,%rdx
  mov    %rax,%rbx
  mov    %rax,0x8(%rdx)
  mov    %ebp,(%rax)
  mov    %r12,0x8(%rax)
  add    $0x1,%ebp
  mov    (%rsp),%ecx
  cmp    %ebp,%ecx
  jge    L4
  cmp    $0x1,%ecx
  jle    L42
  mov    0x4(%rsp),%esi
  lea    -0x1(%rsi),%edx
  jmp    L5
L65:
  mov    0x8(%rbx),%rbx
  add    $0x1,%eax
  cmp    %eax,%edx
  jne    L65
L40:
  mov    0x8(%rbx),%rax
  mov    0x8(%rax),%rax
  mov    %rax,0x8(%rbx)
  sub    $0x1,%ecx
  cmp    $0x1,%ecx
  je     L42
L5:
  mov    $0x0,%eax
  cmp    $0x1,%esi
  jg     L65
  jmp    L40
L62:
  mov    %rax,%rbx
L42:
  mov    (%rbx),%esi
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D30
type_guessed: string
name: extract_symbolic_data
arguments: label: D2
type_guessed: string
name: extract_symbolic_data
arguments: label: D26
type_guessed: string
name: extract_symbolic_data
arguments: label: D10
type_guessed: string

tool: D30: 
									Enter the number of players: 

tool: D2: %d

tool: D26: 
							Enter the value of k (every kth player gets eliminated): 

tool: D10: 
									     The winner is player %d

assistant: ```c
void Josephusproblem4() {
  struct josephus {
    int playerid;
    struct josephus *next;
  };
  struct josephus *start4, *ptr, *newnode;
  int n, k, i, count;
  printf("\n\t\t\t\t\t\t\t\t\tEnter the number of players: ");
  scanf("%d", &n);
  printf("\n\t\t\t\t\t\t\tEnter the value of k (every kth player gets eliminated): ");
  scanf("%d", &k);
  start4 = (struct josephus *)malloc(sizeof(struct josephus));
  start4->playerid = 1;
  ptr = start4;
  for (i = 2; i <= n; i++) {
    newnode = (struct josephus *)malloc(sizeof(struct josephus));
    ptr->next = newnode;
    newnode->playerid = i;
    newnode->next = start4;
    ptr = newnode;
  }
  for (count = n; count > 1; count--) {
    for (i = 0; i < k - 1; ++i)
      ptr = ptr->next;
    ptr->next = ptr->next->next;
  }
  printf("\n\t\t\t\t\t\t\t\t\t     The winner is player %d", ptr->playerid);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<Josephusproblem4>:
  endbr64
  push   %r12
  lea    D26(%rip),%rdi
  push   %rbp
  lea    D24(%rip),%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D8(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    $0x10,%edi
  call   <malloc@plt>
  cmpl   $0x1,(%rsp)
  movl   $0x1,(%rax)
  jle    L60
  mov    %rax,%rbx
  mov    %rax,%rbp
  mov    $0x2,%r12d
  nopl   0x0(%rax,%rax,1)
L52:
  mov    $0x10,%edi
  call   <malloc@plt>
  mov    %rbp,%rdx
  mov    (%rsp),%ecx
  mov    %r12d,%esi
  mov    %rax,0x8(%rdx)
  mov    %rax,%rbp
  mov    %r12d,(%rax)
  add    $0x1,%r12d
  mov    %rbx,0x8(%rax)
  cmp    %r12d,%ecx
  jge    L52
  cmp    $0x1,%ecx
  jle    L38
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  cs nopw 0x0(%rax,%rax,1)
L41:
  xor    %eax,%eax
  test   %edx,%edx
  jle    L32
  cs nopw 0x0(%rax,%rax,1)
L57:
  add    $0x1,%eax
  mov    %rbx,%rbp
  mov    0x8(%rbx),%rbx
  cmp    %edx,%eax
  jne    L57
L32:
  mov    0x8(%rbx),%rbx
  sub    $0x1,%ecx
  mov    %rbx,0x8(%rbp)
  cmp    $0x1,%ecx
  jne    L41
  mov    0x0(%rbp),%esi
L38:
  xor    %eax,%eax
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L60:
  mov    $0x1,%esi
  jmp    L38
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D26
data_type: string
name: extract_symbolic_data
arguments: label: D24
data_type: string
name: extract_symbolic_data
arguments: label: D8
data_type: string
name: extract_symbolic_data
arguments: label: D15
data_type: string

tool: D26: 
									Enter the number of players: 

tool: D24: %d

tool: D8: 
							Enter the value of k (every kth player gets eliminated): 

tool: D15: 
									     The winner is player %d

assistant: ```c
void Josephusproblem4() {
  struct josephus {
    int playerid;
    struct josephus *next;
  };
  struct josephus *start4, *ptr, *newnode;
  int n, k, i, count;
  printf("\n\t\t\t\t\t\t\t\t\tEnter the number of players: ");
  scanf("%d", &n);
  printf("\n\t\t\t\t\t\t\tEnter the value of k (every kth player gets eliminated): ");
  scanf("%d", &k);
  start4 = (struct josephus *)malloc(sizeof(struct josephus));
  start4->playerid = 1;
  ptr = start4;
  for (i = 2; i <= n; i++) {
    newnode = (struct josephus *)malloc(sizeof(struct josephus));
    ptr->next = newnode;
    newnode->playerid = i;
    newnode->next = start4;
    ptr = newnode;
  }
  for (count = n; count > 1; count--) {
    for (i = 0; i < k - 1; ++i)
      ptr = ptr->next;
    ptr->next = ptr->next->next;
  }
  printf("\n\t\t\t\t\t\t\t\t\t     The winner is player %d", ptr->playerid);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Josephusproblem4>:
  endbr64
  push   %r12
  lea    D12(%rip),%rdi
  push   %rbp
  lea    D11(%rip),%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D32(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    $0x10,%edi
  call   <malloc@plt>
  cmpl   $0x1,(%rsp)
  movl   $0x1,(%rax)
  jle    L1
  mov    %rax,%rbx
  mov    %rax,%rbp
  mov    $0x2,%r12d
  nopl   0x0(%rax,%rax,1)
L13:
  mov    $0x10,%edi
  call   <malloc@plt>
  mov    %rbp,%rdx
  mov    (%rsp),%ecx
  mov    %r12d,%esi
  mov    %rax,0x8(%rdx)
  mov    %rax,%rbp
  mov    %r12d,(%rax)
  add    $0x1,%r12d
  mov    %rbx,0x8(%rax)
  cmp    %r12d,%ecx
  jge    L13
  cmp    $0x1,%ecx
  jle    L29
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  cs nopw 0x0(%rax,%rax,1)
L59:
  xor    %eax,%eax
  test   %edx,%edx
  jle    L2
  cs nopw 0x0(%rax,%rax,1)
L53:
  add    $0x1,%eax
  mov    %rbx,%rbp
  mov    0x8(%rbx),%rbx
  cmp    %edx,%eax
  jne    L53
L2:
  mov    0x8(%rbx),%rbx
  sub    $0x1,%ecx
  mov    %rbx,0x8(%rbp)
  cmp    $0x1,%ecx
  jne    L59
  mov    0x0(%rbp),%esi
L29:
  xor    %eax,%eax
  lea    D23(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L70
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  mov    $0x1,%esi
  jmp    L29
L70:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D12
type_infered: string
name: parse_data
arguments: data_label: D11
type_infered: string
name: parse_data
arguments: data_label: D32
type_infered: string
name: parse_data
arguments: data_label: D23
type_infered: string

tool: D12: 
									Enter the number of players: 

tool: D11: %d

tool: D32: 
							Enter the value of k (every kth player gets eliminated): 

tool: D23: 
									     The winner is player %d

assistant: ```c
void Josephusproblem4() {
  struct josephus {
    int playerid;
    struct josephus *next;
  };
  struct josephus *start4, *ptr, *newnode;
  int n, k, i, count;
  printf("\n\t\t\t\t\t\t\t\t\tEnter the number of players: ");
  scanf("%d", &n);
  printf("\n\t\t\t\t\t\t\tEnter the value of k (every kth player gets eliminated): ");
  scanf("%d", &k);
  start4 = (struct josephus *)malloc(sizeof(struct josephus));
  start4->playerid = 1;
  ptr = start4;
  for (i = 2; i <= n; i++) {
    newnode = (struct josephus *)malloc(sizeof(struct josephus));
    ptr->next = newnode;
    newnode->playerid = i;
    newnode->next = start4;
    ptr = newnode;
  }
  for (count = n; count > 1; count--) {
    for (i = 0; i < k - 1; ++i)
      ptr = ptr->next;
    ptr->next = ptr->next->next;
  }
  printf("\n\t\t\t\t\t\t\t\t\t     The winner is player %d", ptr->playerid);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<jround_up>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  sub    $0x1,%rax
  add    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cqto
  idivq  -0x10(%rbp)
  mov    -0x8(%rbp),%rax
  sub    %rdx,%rax
  pop    %rbp
  ret
```

assistant: ```c
long jround_up(a, b)
long a;
long b;
{
  a += b - 1;
  return a - (a % b);
}
```