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

```asm
<main9>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0xd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
L43:
  movb   $0x0,0x10(%rsp,%rax,1)
  movb   $0x0,0x50(%rsp,%rax,1)
  movb   $0x0,0x90(%rsp,%rax,1)
  add    $0x1,%rax
  cmp    $0x32,%rax
  jne    L43
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movb   $0x0,0xf(%rsp)
  lea    0x10(%rsp),%rbx
  lea    D24(%rip),%rbp
L31:
  lea    0xf(%rsp),%rsi
  mov    $0x1,%edx
  mov    %rbp,%rdi
  call   <scanf_s@plt>
  movzbl 0xf(%rsp),%eax
  mov    %al,(%rbx)
  add    $0x1,%rbx
  cmp    $0xa,%al
  jne    L31
  mov    $0x0,%ebx
  lea    0x10(%rsp),%rbp
L5:
  movsbl (%rbx,%rbp,1),%edi
  call   <CaesarEn1@plt>
  mov    %al,0x50(%rsp,%rbx,1)
  add    $0x1,%rbx
  cmp    $0x32,%rbx
  jne    L5
  lea    0x50(%rsp),%rsi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  lea    0x50(%rsp),%rbp
L45:
  movsbl (%rbx,%rbp,1),%edi
  call   <CaesarDe1@plt>
  mov    %al,0x90(%rsp,%rbx,1)
  add    $0x1,%rbx
  cmp    $0x32,%rbx
  jne    L45
  lea    0x90(%rsp),%rsi
  lea    D30(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <fileEncrypt@plt>
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  mov    $0x0,%eax
  add    $0xd8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D17
type: string
name: unpack_data
arguments: data_tag: D24
type: string
name: unpack_data
arguments: data_tag: D6
type: string
name: unpack_data
arguments: data_tag: D30
type: string

tool: D17: Enter a Plain Text Message: To End message, Press  

tool: D24: %c

tool: D6: 
Cipher code is %s 


tool: D30: 
Plain Text is %s 




assistant: ```c
int main9() {
  char ch[50];
  char cipher[50];
  char plain[50];
  for (int i = 0; i < 50; i++) {
    ch[i] = '\0';
    cipher[i] = '\0';
    plain[i] = '\0';
  }
  printf("Enter a Plain Text Message: To End message, Press  ");
  char c = '\0';
  int j = 0;
  while (c != '\n') {
    scanf_s("%c", &c, 1);
    ch[j] = c;
    j++;
  }
  for (int i = 0; i < 50; i++) {
    cipher[i] = CaesarEn1(ch[i]);
  }
  printf("\nCipher code is %s \n", cipher);
  for (int i = 0; i < 50; i++) {
    plain[i] = CaesarDe1(cipher[i]);
  }
  printf("\nPlain Text is %s \n\n\n", plain);
  fileEncrypt();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main9>:
  endbr64
  push   %r13
  pxor   %xmm0,%xmm0
  xor    %edx,%edx
  xor    %ecx,%ecx
  push   %r12
  xor    %esi,%esi
  lea    D22(%rip),%rdi
  push   %rbp
  lea    D26(%rip),%rbp
  push   %rbx
  sub    $0xd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  mov    %dx,0x40(%rsp)
  lea    0x10(%rsp),%r12
  lea    0xf(%rsp),%r13
  mov    %cx,0x80(%rsp)
  mov    %r12,%rbx
  mov    %si,0xc0(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,0x60(%rsp)
  movaps %xmm0,0x70(%rsp)
  movaps %xmm0,0x90(%rsp)
  movaps %xmm0,0xa0(%rsp)
  movaps %xmm0,0xb0(%rsp)
  call   <printf@plt>
  movb   $0x0,0xf(%rsp)
  xchg   %ax,%ax
L4:
  mov    $0x1,%edx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <scanf_s@plt>
  movzbl 0xf(%rsp),%eax
  mov    %al,-0x1(%rbx)
  cmp    $0xa,%al
  jne    L4
  xor    %ebx,%ebx
  lea    0x50(%rsp),%rbp
  nopw   0x0(%rax,%rax,1)
L15:
  movsbl (%r12,%rbx,1),%edi
  call   <CaesarEn1@plt>
  mov    %al,0x0(%rbp,%rbx,1)
  add    $0x1,%rbx
  cmp    $0x32,%rbx
  jne    L15
  mov    %rbp,%rsi
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <printf@plt>
  lea    0x90(%rsp),%r12
  nopl   0x0(%rax,%rax,1)
L7:
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <CaesarDe1@plt>
  mov    %al,(%r12,%rbx,1)
  add    $0x1,%rbx
  cmp    $0x32,%rbx
  jne    L7
  xor    %eax,%eax
  mov    %r12,%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <fileEncrypt@plt>
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0xd8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D22
type: string
name: unpack_data
arguments: tag: D26
type: string
name: unpack_data
arguments: tag: D13
type: string
name: unpack_data
arguments: tag: D6
type: string

tool: D22: Enter a Plain Text Message: To End message, Press  

tool: D26: %c

tool: D13: 
Cipher code is %s 


tool: D6: 
Plain Text is %s 




assistant: ```c
int main9() {
  char ch[50];
  char cipher[50];
  char plain[50];
  for (int i = 0; i < 50; i++) {
    ch[i] = '\0';
    cipher[i] = '\0';
    plain[i] = '\0';
  }
  printf("Enter a Plain Text Message: To End message, Press  ");
  char c = '\0';
  int j = 0;
  while (c != '\n') {
    scanf_s("%c", &c, 1);
    ch[j] = c;
    j++;
  }
  for (int i = 0; i < 50; i++) {
    cipher[i] = CaesarEn1(ch[i]);
  }
  printf("\nCipher code is %s \n", cipher);
  for (int i = 0; i < 50; i++) {
    plain[i] = CaesarDe1(cipher[i]);
  }
  printf("\nPlain Text is %s \n\n\n", plain);
  fileEncrypt();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main9>:
  endbr64
  push   %r13
  pxor   %xmm0,%xmm0
  xor    %edx,%edx
  xor    %ecx,%ecx
  push   %r12
  xor    %esi,%esi
  lea    D7(%rip),%rdi
  push   %rbp
  lea    D27(%rip),%rbp
  push   %rbx
  sub    $0xd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  mov    %dx,0x40(%rsp)
  lea    0x10(%rsp),%r12
  lea    0xf(%rsp),%r13
  mov    %cx,0x80(%rsp)
  mov    %r12,%rbx
  mov    %si,0xc0(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,0x60(%rsp)
  movaps %xmm0,0x70(%rsp)
  movaps %xmm0,0x90(%rsp)
  movaps %xmm0,0xa0(%rsp)
  movaps %xmm0,0xb0(%rsp)
  call   <printf@plt>
  movb   $0x0,0xf(%rsp)
  xchg   %ax,%ax
L27:
  mov    $0x1,%edx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <scanf_s@plt>
  movzbl 0xf(%rsp),%eax
  mov    %al,-0x1(%rbx)
  cmp    $0xa,%al
  jne    L27
  xor    %ebx,%ebx
  lea    0x50(%rsp),%rbp
  nopw   0x0(%rax,%rax,1)
L39:
  movsbl (%r12,%rbx,1),%edi
  call   <CaesarEn1@plt>
  mov    %al,0x0(%rbp,%rbx,1)
  add    $0x1,%rbx
  cmp    $0x32,%rbx
  jne    L39
  mov    %rbp,%rsi
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <printf@plt>
  lea    0x90(%rsp),%r12
  nopl   0x0(%rax,%rax,1)
L7:
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <CaesarDe1@plt>
  mov    %al,(%r12,%rbx,1)
  add    $0x1,%rbx
  cmp    $0x32,%rbx
  jne    L7
  xor    %eax,%eax
  mov    %r12,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <fileEncrypt@plt>
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0xd8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
type: string
name: extract_symbolic_data
arguments: data_tag: D27
type: string
name: extract_symbolic_data
arguments: data_tag: D14
type: string
name: extract_symbolic_data
arguments: data_tag: D2
type: string

tool: D7: Enter a Plain Text Message: To End message, Press  

tool: D27: %c

tool: D14: 
Cipher code is %s 


tool: D2: 
Plain Text is %s 




assistant: ```c
int main9() {
  char ch[50];
  char cipher[50];
  char plain[50];
  for (int i = 0; i < 50; i++) {
    ch[i] = '\0';
    cipher[i] = '\0';
    plain[i] = '\0';
  }
  printf("Enter a Plain Text Message: To End message, Press  ");
  char c = '\0';
  int j = 0;
  while (c != '\n') {
    scanf_s("%c", &c, 1);
    ch[j] = c;
    j++;
  }
  for (int i = 0; i < 50; i++) {
    cipher[i] = CaesarEn1(ch[i]);
  }
  printf("\nCipher code is %s \n", cipher);
  for (int i = 0; i < 50; i++) {
    plain[i] = CaesarDe1(cipher[i]);
  }
  printf("\nPlain Text is %s \n\n\n", plain);
  fileEncrypt();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<seven>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  jmp    L2
L11:
  addl   $0x1,-0x4(%rbp)
L2:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jne    L11
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

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

tool: D2: %d is now equal to %d


assistant: ```c
void seven(int a, const int b) {
  while (a != b) {
    a++;
  }
  printf("%d is now equal to %d\n", a, b);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<seven>:
  endbr64
  sub    $0x8,%rsp
  mov    %esi,%edx
  cmp    %esi,%edi
  je     L5
L23:
  add    $0x1,%edi
  cmp    %edi,%edx
  jne    L23
L10:
  mov    %edi,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
L5:
  mov    %esi,%edi
  jmp    L10
```

tool_calls: name: read_data
arguments: data_label: D0
type_infered: string

tool: D0: %d is now equal to %d


assistant: ```c
void seven(int a, const int b) {
  while (a != b) {
    a++;
  }
  printf("%d is now equal to %d\n", a, b);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<seven>:
  endbr64
  mov    %esi,%edx
  lea    D3(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: label: D3
type: string

tool: D3: %d is now equal to %d


assistant: ```c
void seven(int a, const int b) {
  while (a != b) {
    a++;
  }
  printf("%d is now equal to %d\n", a, b);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<seven>:
  endbr64
  mov    %esi,%edx
  lea    D0(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
type: string

tool: D0: %d is now equal to %d


assistant: ```c
void seven(int a, const int b) {
  while (a != b) {
    a++;
  }
  printf("%d is now equal to %d\n", a, b);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bzero1_null>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movq   $0x0,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <bzero@plt>
  nop
  leave
  ret
```

assistant: ```c
void bzero1_null() {
  char *a = 0;
  bzero(a, 10);
}
```
user: What is the c source code of the assembly code below:

```asm
<bzero1_null>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%esi
  mov    $0x0,%edi
  call   <bzero@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void bzero1_null() {
  char *a = 0;
  bzero(a, 10);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bzero1_null>:
  endbr64
  mov    $0xa,%esi
  xor    %edi,%edi
  jmp    1050 <bzero@plt>
```

assistant: ```c
void bzero1_null() {
  char *a = 0;
  bzero(a, 10);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bzero1_null>:
  endbr64
  mov    $0xa,%esi
  xor    %edi,%edi
  jmp    1050 <bzero@plt>
```

assistant: ```c
void bzero1_null() {
  char *a = 0;
  bzero(a, 10);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <load_menu@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  load_menu();
  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    $0x0,%eax
  call   <load_menu@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  load_menu();
  return 0;
}
```
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
  xor    %eax,%eax
  call   <load_menu@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  load_menu();
  return 0;
}
```
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   <load_menu@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  load_menu();
  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
<InfoMaster>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    %ecx,-0x14(%rbp)
  mov    %r8d,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D18
datatype: string
name: read_data
arguments: tag: D17
datatype: string
name: read_data
arguments: tag: D8
datatype: string
name: read_data
arguments: tag: D31
datatype: string

tool: D18: >> Image output		 :	[%s]


tool: D17: >> Viewport dimension	:	[%dx%d]


tool: D8: >> Parallel computing MPI :	[OK]

tool: D31: >> Processus/lines :		[%d/%d]


assistant: ```c
void InfoMaster(char *img_name, int w, int h, int n_proc, int n_lines) {
  printf(">> Image output		 :	[%s]\n", img_name);
  printf(">> Viewport dimension	:	[%dx%d]\n", w, h);
  printf(">> Parallel computing MPI :	[OK]\n");
  printf(">> Processus/lines :		[%d/%d]\n", n_proc, n_lines);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<InfoMaster>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%r12d
  mov    %edx,%r13d
  mov    %ecx,%ebx
  mov    %r8d,%ebp
  mov    %rdi,%rsi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r13d,%edx
  mov    %r12d,%esi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D34(%rip),%rdi
  call   <puts@plt>
  mov    %ebp,%edx
  mov    %ebx,%esi
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D4
datatype: string
name: extract_symbolic_data
arguments: label: D16
datatype: string
name: extract_symbolic_data
arguments: label: D34
datatype: string
name: extract_symbolic_data
arguments: label: D10
datatype: string

tool: D4: >> Image output		 :	[%s]


tool: D16: >> Viewport dimension	:	[%dx%d]


tool: D34: >> Parallel computing MPI :	[OK]

tool: D10: >> Processus/lines :		[%d/%d]


assistant: ```c
void InfoMaster(char *img_name, int w, int h, int n_proc, int n_lines) {
  printf(">> Image output		 :	[%s]\n", img_name);
  printf(">> Viewport dimension	:	[%dx%d]\n", w, h);
  printf(">> Parallel computing MPI :	[OK]\n");
  printf(">> Processus/lines :		[%d/%d]\n", n_proc, n_lines);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<InfoMaster>:
  endbr64
  push   %r14
  xor    %eax,%eax
  mov    %edx,%r14d
  push   %r13
  mov    %esi,%r13d
  mov    %rdi,%rsi
  lea    D19(%rip),%rdi
  push   %r12
  mov    %r8d,%r12d
  push   %rbp
  mov    %ecx,%ebp
  sub    $0x8,%rsp
  call   <printf@plt>
  mov    %r14d,%edx
  mov    %r13d,%esi
  xor    %eax,%eax
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    D31(%rip),%rdi
  call   <puts@plt>
  add    $0x8,%rsp
  mov    %r12d,%edx
  mov    %ebp,%esi
  lea    D34(%rip),%rdi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1070 <printf@plt>
```

tool_calls: name: read_data
arguments: data_tag: D19
datatype: string
name: read_data
arguments: data_tag: D8
datatype: string
name: read_data
arguments: data_tag: D31
datatype: string
name: read_data
arguments: data_tag: D34
datatype: string

tool: D19: >> Image output		 :	[%s]


tool: D8: >> Viewport dimension	:	[%dx%d]


tool: D31: >> Parallel computing MPI :	[OK]

tool: D34: >> Processus/lines :		[%d/%d]


assistant: ```c
void InfoMaster(char *img_name, int w, int h, int n_proc, int n_lines) {
  printf(">> Image output		 :	[%s]\n", img_name);
  printf(">> Viewport dimension	:	[%dx%d]\n", w, h);
  printf(">> Parallel computing MPI :	[OK]\n");
  printf(">> Processus/lines :		[%d/%d]\n", n_proc, n_lines);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<InfoMaster>:
  endbr64
  push   %r14
  xor    %eax,%eax
  mov    %edx,%r14d
  push   %r13
  mov    %esi,%r13d
  mov    %rdi,%rsi
  lea    D1(%rip),%rdi
  push   %r12
  mov    %r8d,%r12d
  push   %rbp
  mov    %ecx,%ebp
  sub    $0x8,%rsp
  call   <printf@plt>
  mov    %r14d,%edx
  mov    %r13d,%esi
  xor    %eax,%eax
  lea    D10(%rip),%rdi
  call   <printf@plt>
  lea    D29(%rip),%rdi
  call   <puts@plt>
  add    $0x8,%rsp
  mov    %r12d,%edx
  mov    %ebp,%esi
  lea    D15(%rip),%rdi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1070 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
data_type: string
name: extract_symbolic_data
arguments: data_tag: D10
data_type: string
name: extract_symbolic_data
arguments: data_tag: D29
data_type: string
name: extract_symbolic_data
arguments: data_tag: D15
data_type: string

tool: D1: >> Image output		 :	[%s]


tool: D10: >> Viewport dimension	:	[%dx%d]


tool: D29: >> Parallel computing MPI :	[OK]

tool: D15: >> Processus/lines :		[%d/%d]


assistant: ```c
void InfoMaster(char *img_name, int w, int h, int n_proc, int n_lines) {
  printf(">> Image output		 :	[%s]\n", img_name);
  printf(">> Viewport dimension	:	[%dx%d]\n", w, h);
  printf(">> Parallel computing MPI :	[OK]\n");
  printf(">> Processus/lines :		[%d/%d]\n", n_proc, n_lines);
}
```
user: What is the c source code of the assembly code below?

```asm
<ifwin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  jmp    L35
L80:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0x800,%eax
  jne    L31
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L2:
  lea    -0xd(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L36
L43:
  nop
L36:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L43
  movzbl -0xd(%rbp),%eax
  cmp    $0x79,%al
  jne    L52
  movzbl -0xd(%rbp),%eax
  cmp    $0x6e,%al
  je     L15
L52:
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L2
L15:
  movb   $0x79,-0xd(%rbp)
  mov    $0x0,%eax
  jmp    L74
L31:
  addl   $0x1,-0xc(%rbp)
L35:
  cmpl   $0xf,-0xc(%rbp)
  jle    L80
  mov    $0xffffffff,%eax
L74:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L47
  call   <__stack_chk_fail@plt>
L47:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D13
type_guessed: string
name: read_data
arguments: data_label: D9
type_guessed: string
name: read_data
arguments: data_label: D24
type_guessed: string

tool: D13: You Got 2048,Do you want to continue?

tool: D9: %c

tool: D24: Please input y or n.

assistant: ```c
int ifwin(int *f) {
  char a;
  for (int i = 0; i < 16; i++) {
    if (*(f + i) == 2048) {
      puts("You Got 2048,Do you want to continue?");
      while (1) {
        scanf("%c", &a);
        while (getchar() != '\n')
          continue;
        if (a != 'y' || a != 'n')
          puts("Please input y or n.");
        else
          break;
      }
      return (a = 'y') ? 0 : 1;
    }
  }
  return -1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ifwin>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rdi,%rax
  lea    0x40(%rdi),%rdx
L3:
  cmpl   $0x800,(%rax)
  je     L38
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L3
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  mov    $0xffffffff,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L38:
  lea    D7(%rip),%rdi
  call   <puts@plt>
  lea    0x7(%rsp),%r12
  lea    D8(%rip),%rbp
  lea    D3(%rip),%rbx
L6:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
L7:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L7
  mov    %rbx,%rdi
  call   <puts@plt>
  jmp    L6
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D7
type_infered: string
name: parse_data
arguments: data_tag: D8
type_infered: string
name: parse_data
arguments: data_tag: D3
type_infered: string

tool: D7: You Got 2048,Do you want to continue?

tool: D8: %c

tool: D3: Please input y or n.

assistant: ```c
int ifwin(int *f) {
  char a;
  for (int i = 0; i < 16; i++) {
    if (*(f + i) == 2048) {
      puts("You Got 2048,Do you want to continue?");
      while (1) {
        scanf("%c", &a);
        while (getchar() != '\n')
          continue;
        if (a != 'y' || a != 'n')
          puts("Please input y or n.");
        else
          break;
      }
      return (a = 'y') ? 0 : 1;
    }
  }
  return -1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ifwin>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x40(%rdi),%rax
L41:
  cmpl   $0x800,(%rdi)
  je     L3
  add    $0x4,%rdi
  cmp    %rax,%rdi
  jne    L41
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  add    $0x10,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  lea    D22(%rip),%rdi
  lea    0x7(%rsp),%r12
  call   <puts@plt>
  lea    D11(%rip),%rbp
  lea    D16(%rip),%rbx
  xchg   %ax,%ax
L8:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
L33:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L33
  mov    %rbx,%rdi
  call   <puts@plt>
  jmp    L8
L44:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D22
data_type: string
name: extract_symbolic_data
arguments: data_tag: D11
data_type: string
name: extract_symbolic_data
arguments: data_tag: D16
data_type: string

tool: D22: You Got 2048,Do you want to continue?

tool: D11: %c

tool: D16: Please input y or n.

assistant: ```c
int ifwin(int *f) {
  char a;
  for (int i = 0; i < 16; i++) {
    if (*(f + i) == 2048) {
      puts("You Got 2048,Do you want to continue?");
      while (1) {
        scanf("%c", &a);
        while (getchar() != '\n')
          continue;
        if (a != 'y' || a != 'n')
          puts("Please input y or n.");
        else
          break;
      }
      return (a = 'y') ? 0 : 1;
    }
  }
  return -1;
}
```