ylfeng/ReF-Decompile-dataset huggingface dataset:
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    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L15:
  lea    D93(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D88(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D61(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D95(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D73(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D103(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  cmp    $0x5,%eax
  ja     L15
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D65(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D65(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x0,%eax
  call   <myStrLenTD@plt>
  jmp    L26
  mov    $0x0,%eax
  call   <myStrCpyTD@plt>
  jmp    L26
  mov    $0x0,%eax
  call   <myStrCatTD@plt>
  jmp    L26
  mov    $0x0,%eax
  call   <countVowelsInStrTD@plt>
  jmp    L26
  mov    $0x0,%eax
  call   <countWordsInStrTD@plt>
  jmp    L26
  mov    $0x1,%edi
  call   <exit@plt>
  nop
L26:
  jmp    L15
```

tool_calls: name: read_data
arguments: tag: D93
type_guessed: string
name: read_data
arguments: tag: D106
type_guessed: string
name: read_data
arguments: tag: D9
type_guessed: string
name: read_data
arguments: tag: D88
type_guessed: string
name: read_data
arguments: tag: D61
type_guessed: string
name: read_data
arguments: tag: D45
type_guessed: string
name: read_data
arguments: tag: D95
type_guessed: string
name: read_data
arguments: tag: D73
type_guessed: string
name: read_data
arguments: tag: D103
type_guessed: string

tool: D93: 
******* Operations on Strings ********


tool: D106: 0.Compute the length of a string


tool: D9: 1.Copy a string from Source to Destination


tool: D88: 2.Concatinate a String with the Other


tool: D61: 3.Compute no of Vowels in a String


tool: D45: 4.Compute no of words in String


tool: D95: 5.Exit


tool: D73: Enter your choice


tool: D103: %d

assistant: ```c
main(void) {
  int choice;
  while (1) {
    printf("\n******* Operations on Strings ********\n");
    printf("0.Compute the length of a string\n");
    printf("1.Copy a string from Source to Destination\n");
    printf("2.Concatinate a String with the Other\n");
    printf("3.Compute no of Vowels in a String\n");
    printf("4.Compute no of words in String\n");
    printf("5.Exit\n");
    printf("Enter your choice\n");
    scanf("%d", &choice);
    switch (choice) {
    case 0:
      myStrLenTD();
      break;
    case 1:
      myStrCpyTD();
      break;
    case 2:
      myStrCatTD();
      break;
    case 3:
      countVowelsInStrTD();
      break;
    case 4:
      countWordsInStrTD();
      break;
    case 5:
      exit(1);
      break;
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D76(%rip),%r13
  lea    D33(%rip),%r12
  lea    D19(%rip),%rbp
  lea    D64(%rip),%rbx
  jmp    L44
  mov    $0x0,%eax
  call   <myStrLenTD@plt>
L44:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    D93(%rip),%rdi
  call   <printf@plt>
  lea    D71(%rip),%rdi
  call   <printf@plt>
  lea    D39(%rip),%rdi
  call   <printf@plt>
  lea    D38(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D5(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x5,0x4(%rsp)
  ja     L44
  mov    0x4(%rsp),%eax
  lea    D72(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x0,%eax
  call   <myStrCpyTD@plt>
  jmp    L44
  mov    $0x0,%eax
  call   <myStrCatTD@plt>
  jmp    L44
  mov    $0x0,%eax
  call   <countVowelsInStrTD@plt>
  jmp    L44
  mov    $0x0,%eax
  call   <countWordsInStrTD@plt>
  jmp    L44
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L44
```

tool_calls: name: read_data
arguments: data_label: D76
type_guessed: string
name: read_data
arguments: data_label: D33
type_guessed: string
name: read_data
arguments: data_label: D19
type_guessed: string
name: read_data
arguments: data_label: D64
type_guessed: string
name: read_data
arguments: data_label: D93
type_guessed: string
name: read_data
arguments: data_label: D71
type_guessed: string
name: read_data
arguments: data_label: D39
type_guessed: string
name: read_data
arguments: data_label: D38
type_guessed: string
name: read_data
arguments: data_label: D5
type_guessed: string

tool: D76: 
******* Operations on Strings ********


tool: D33: 0.Compute the length of a string


tool: D19: 1.Copy a string from Source to Destination


tool: D64: 2.Concatinate a String with the Other


tool: D93: 3.Compute no of Vowels in a String


tool: D71: 4.Compute no of words in String


tool: D39: 5.Exit


tool: D38: Enter your choice


tool: D5: %d

assistant: ```c
main(void) {
  int choice;
  while (1) {
    printf("\n******* Operations on Strings ********\n");
    printf("0.Compute the length of a string\n");
    printf("1.Copy a string from Source to Destination\n");
    printf("2.Concatinate a String with the Other\n");
    printf("3.Compute no of Vowels in a String\n");
    printf("4.Compute no of words in String\n");
    printf("5.Exit\n");
    printf("Enter your choice\n");
    scanf("%d", &choice);
    switch (choice) {
    case 0:
      myStrLenTD();
      break;
    case 1:
      myStrCpyTD();
      break;
    case 2:
      myStrCatTD();
      break;
    case 3:
      countVowelsInStrTD();
      break;
    case 4:
      countWordsInStrTD();
      break;
    case 5:
      exit(1);
      break;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<minMax>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  movsd  %xmm0,-0x18(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  movsd  -0x20(%rbp),%xmm0
  comisd -0x18(%rbp),%xmm0
  jbe    L0
  mov    -0x8(%rbp),%rax
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x10(%rbp),%rax
  movsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  jmp    L15
L0:
  mov    -0x8(%rbp),%rax
  movsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x10(%rbp),%rax
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,(%rax)
L15:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void minMax(double *min, double *max, double a, double b) {
  if (a < b) {
    *min = a;
    *max = b;
  } else {
    *min = b;
    *max = a;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<minMax>:
  endbr64
  comisd %xmm0,%xmm1
  jbe    L1
  movq   %xmm0,%rax
  movapd %xmm1,%xmm0
  movq   %rax,%xmm1
L1:
  movsd  %xmm1,(%rdi)
  movsd  %xmm0,(%rsi)
  ret
```

assistant: ```c
void minMax(double *min, double *max, double a, double b) {
  if (a < b) {
    *min = a;
    *max = b;
  } else {
    *min = b;
    *max = a;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<minMax>:
  endbr64
  comisd %xmm0,%xmm1
  jbe    L0
  movq   %xmm0,%rax
  movapd %xmm1,%xmm0
  movq   %rax,%xmm1
L0:
  movsd  %xmm1,(%rdi)
  movsd  %xmm0,(%rsi)
  ret
```

assistant: ```c
void minMax(double *min, double *max, double a, double b) {
  if (a < b) {
    *min = a;
    *max = b;
  } else {
    *min = b;
    *max = a;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<minMax>:
  endbr64
  comisd %xmm0,%xmm1
  jbe    L3
  movq   %xmm0,%rax
  movapd %xmm1,%xmm0
  movq   %rax,%xmm1
L3:
  movsd  %xmm1,(%rdi)
  movsd  %xmm0,(%rsi)
  ret
```

assistant: ```c
void minMax(double *min, double *max, double a, double b) {
  if (a < b) {
    *min = a;
    *max = b;
  } else {
    *min = b;
    *max = a;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchrnul@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L11
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x1c(%rbp)
  jne    L13
  mov    -0x8(%rbp),%rax
  jmp    L37
L13:
  mov    $0x0,%eax
  jmp    L37
L11:
  mov    -0x8(%rbp),%rax
L37:
  leave
  ret
```

assistant: ```c
char *strchr(const char *str, int ch) {
  char *chp = strchrnul(str, ch);
  if (*chp == '\0') {
    return ch == *chp ? chp : NULL;
  }
  return chp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strchr>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  call   <strchrnul@plt>
  cmpb   $0x0,(%rax)
  je     L16
L9:
  pop    %rbx
  ret
L16:
  test   %ebx,%ebx
  mov    $0x0,%edx
  cmovne %rdx,%rax
  jmp    L9
```

assistant: ```c
char *strchr(const char *str, int ch) {
  char *chp = strchrnul(str, ch);
  if (*chp == '\0') {
    return ch == *chp ? chp : NULL;
  }
  return chp;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strchr>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  call   <strchrnul@plt>
  cmpb   $0x0,(%rax)
  jne    L0
  xor    %edx,%edx
  test   %ebx,%ebx
  cmovne %rdx,%rax
L0:
  pop    %rbx
  ret
```

assistant: ```c
char *strchr(const char *str, int ch) {
  char *chp = strchrnul(str, ch);
  if (*chp == '\0') {
    return ch == *chp ? chp : NULL;
  }
  return chp;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strchr>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  call   <strchrnul@plt>
  cmpb   $0x0,(%rax)
  jne    L3
  xor    %edx,%edx
  test   %ebx,%ebx
  cmovne %rdx,%rax
L3:
  pop    %rbx
  ret
```

assistant: ```c
char *strchr(const char *str, int ch) {
  char *chp = strchrnul(str, ch);
  if (*chp == '\0') {
    return ch == *chp ? chp : NULL;
  }
  return chp;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<kfifo_reset>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <__kfifo_reset@plt>
  nop
  leave
  ret
```

assistant: ```c
void kfifo_reset(struct kfifo *fifo) {
  unsigned long flags;
  __kfifo_reset(fifo);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<kfifo_reset>:
  endbr64
  sub    $0x8,%rsp
  call   <__kfifo_reset@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void kfifo_reset(struct kfifo *fifo) {
  unsigned long flags;
  __kfifo_reset(fifo);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<kfifo_reset>:
  endbr64
  jmp    1050 <__kfifo_reset@plt>
```

assistant: ```c
void kfifo_reset(struct kfifo *fifo) {
  unsigned long flags;
  __kfifo_reset(fifo);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<kfifo_reset>:
  endbr64
  jmp    1050 <__kfifo_reset@plt>
```

assistant: ```c
void kfifo_reset(struct kfifo *fifo) {
  unsigned long flags;
  __kfifo_reset(fifo);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<array_ref_len>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L7
L66:
  cmpl   $0x5b,-0xc(%rbp)
  jne    L7
  addl   $0x1,-0x10(%rbp)
L7:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  je     L37
  cmpl   $0x5d,-0xc(%rbp)
  jne    L66
  subl   $0x1,-0x10(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  jne    L66
L37:
  cmpl   $0x0,-0xc(%rbp)
  jne    L49
  mov    $0x0,%eax
  jmp    L53
L49:
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
L53:
  pop    %rbp
  ret
```

assistant: ```c
int array_ref_len(cp) const char *cp;
{
  const char *s = cp;
  int c;
  int depth = 0;
  while ((c = *s++) && (c != ']' || --depth))
    if (c == '[')
      depth++;
  if (!c)
    return 0;
  return s - cp;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<array_ref_len>:
  endbr64
  lea    0x1(%rdi),%rdx
  movsbl (%rdi),%eax
  mov    $0x0,%ecx
  test   %eax,%eax
  jne    L38
  ret
L4:
  cmp    $0x5b,%eax
  sete   %al
  movzbl %al,%eax
  add    %eax,%ecx
L31:
  add    $0x1,%rdx
  movsbl -0x1(%rdx),%eax
  test   %eax,%eax
  je     L36
L38:
  cmp    $0x5d,%eax
  jne    L4
  sub    $0x1,%ecx
  jne    L31
  mov    %edx,%eax
  sub    %edi,%eax
L36:
  ret
```

assistant: ```c
int array_ref_len(cp) const char *cp;
{
  const char *s = cp;
  int c;
  int depth = 0;
  while ((c = *s++) && (c != ']' || --depth))
    if (c == '[')
      depth++;
  if (!c)
    return 0;
  return s - cp;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<array_ref_len>:
  endbr64
  movsbl (%rdi),%eax
  lea    0x1(%rdi),%rdx
  xor    %ecx,%ecx
  test   %eax,%eax
  jne    L35
  jmp    L9
  nopl   0x0(%rax,%rax,1)
L10:
  cmp    $0x5b,%eax
  sete   %al
  movzbl %al,%eax
  add    %eax,%ecx
L47:
  movsbl (%rdx),%eax
  add    $0x1,%rdx
  test   %eax,%eax
  je     L18
L35:
  cmp    $0x5d,%eax
  jne    L10
  sub    $0x1,%ecx
  jne    L47
  mov    %edx,%eax
  sub    %edi,%eax
L18:
  ret
L9:
  ret
```

assistant: ```c
int array_ref_len(cp) const char *cp;
{
  const char *s = cp;
  int c;
  int depth = 0;
  while ((c = *s++) && (c != ']' || --depth))
    if (c == '[')
      depth++;
  if (!c)
    return 0;
  return s - cp;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<array_ref_len>:
  endbr64
  movsbl (%rdi),%eax
  lea    0x1(%rdi),%rdx
  xor    %ecx,%ecx
  test   %eax,%eax
  jne    L35
  jmp    L8
  nopl   0x0(%rax,%rax,1)
L23:
  cmp    $0x5b,%eax
  sete   %al
  movzbl %al,%eax
  add    %eax,%ecx
L5:
  movsbl (%rdx),%eax
  add    $0x1,%rdx
  test   %eax,%eax
  je     L14
L35:
  cmp    $0x5d,%eax
  jne    L23
  sub    $0x1,%ecx
  jne    L5
  mov    %edx,%eax
  sub    %edi,%eax
L14:
  ret
L8:
  ret
```

assistant: ```c
int array_ref_len(cp) const char *cp;
{
  const char *s = cp;
  int c;
  int depth = 0;
  while ((c = *s++) && (c != ']' || --depth))
    if (c == '[')
      depth++;
  if (!c)
    return 0;
  return s - cp;
}
```
user: What is the c source code of the assembly code below?

```asm
<removeArquivo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x24(%rbp),%edx
  lea    -0x20(%rbp),%rax
  lea    D0(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <remove@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
data_type: string

tool: D0: rodada-%d.txt

assistant: ```c
void removeArquivo(int numeroArq) {
  char arqRodada[20];
  sprintf(arqRodada, "rodada-%d.txt", numeroArq);
  remove(arqRodada);
}
```
user: What is the c source code of the assembly code below:

```asm
<removeArquivo>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %edi,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    D0(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rdi
  call   <remove@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x20,%rsp
  pop    %rbx
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
type_infered: string

tool: D0: rodada-%d.txt

assistant: ```c
void removeArquivo(int numeroArq) {
  char arqRodada[20];
  sprintf(arqRodada, "rodada-%d.txt", numeroArq);
  remove(arqRodada);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<removeArquivo>:
  endbr64
  push   %rbp
  mov    %edi,%edx
  lea    D7(%rip),%rsi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <remove@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x20,%rsp
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D7
type: string

tool: D7: rodada-%d.txt

assistant: ```c
void removeArquivo(int numeroArq) {
  char arqRodada[20];
  sprintf(arqRodada, "rodada-%d.txt", numeroArq);
  remove(arqRodada);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<removeArquivo>:
  endbr64
  push   %rbp
  mov    %edi,%edx
  lea    D4(%rip),%rsi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <remove@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x20,%rsp
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D4
type: string

tool: D4: rodada-%d.txt

assistant: ```c
void removeArquivo(int numeroArq) {
  char arqRodada[20];
  sprintf(arqRodada, "rodada-%d.txt", numeroArq);
  remove(arqRodada);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<btlza_ctxindex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  sar    $0x10,%eax
  imul   $0xfb,%eax,%eax
  sar    $0x8,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shl    $0x10,%eax
  xor    -0x14(%rbp),%eax
  and    -0x18(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int btlza_ctxindex(int ctx, int mask) {
  int i;
  i = ((ctx >> 16) * 251) >> 8;
  return ((ctx ^ (i << 16)) & mask);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<btlza_ctxindex>:
  endbr64
  mov    %edi,%eax
  sar    $0x10,%eax
  imul   $0xfb,%eax,%eax
  sar    $0x8,%eax
  shl    $0x10,%eax
  xor    %edi,%eax
  and    %esi,%eax
  ret
```

assistant: ```c
int btlza_ctxindex(int ctx, int mask) {
  int i;
  i = ((ctx >> 16) * 251) >> 8;
  return ((ctx ^ (i << 16)) & mask);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<btlza_ctxindex>:
  endbr64
  mov    %edi,%eax
  sar    $0x10,%eax
  imul   $0xfb,%eax,%eax
  sar    $0x8,%eax
  shl    $0x10,%eax
  xor    %edi,%eax
  and    %esi,%eax
  ret
```

assistant: ```c
int btlza_ctxindex(int ctx, int mask) {
  int i;
  i = ((ctx >> 16) * 251) >> 8;
  return ((ctx ^ (i << 16)) & mask);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<btlza_ctxindex>:
  endbr64
  mov    %edi,%eax
  sar    $0x10,%eax
  imul   $0xfb,%eax,%eax
  sar    $0x8,%eax
  shl    $0x10,%eax
  xor    %edi,%eax
  and    %esi,%eax
  ret
```

assistant: ```c
int btlza_ctxindex(int ctx, int mask) {
  int i;
  i = ((ctx >> 16) * 251) >> 8;
  return ((ctx ^ (i << 16)) & mask);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<Act_SetOut>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <ArgText@plt>
  mov    %eax,%edi
  call   <TextOf@plt>
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <Me@plt>
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  mov    %edx,%edi
  call   <SetOutMsg@plt>
  nop
  leave
  ret
```

assistant: ```c
void Act_SetOut(void) {
  char *x = TextOf(ArgText());
  SetOutMsg(Me(), x);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Act_SetOut>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <ArgText@plt>
  mov    %eax,%edi
  call   <TextOf@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  call   <Me@plt>
  mov    %eax,%edi
  mov    %rbx,%rsi
  call   <SetOutMsg@plt>
  pop    %rbx
  ret
```

assistant: ```c
void Act_SetOut(void) {
  char *x = TextOf(ArgText());
  SetOutMsg(Me(), x);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Act_SetOut>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <ArgText@plt>
  mov    %eax,%edi
  call   <TextOf@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <Me@plt>
  mov    %rbp,%rsi
  pop    %rbp
  mov    %eax,%edi
  jmp    10a0 <SetOutMsg@plt>
```

assistant: ```c
void Act_SetOut(void) {
  char *x = TextOf(ArgText());
  SetOutMsg(Me(), x);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<Act_SetOut>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <ArgText@plt>
  mov    %eax,%edi
  call   <TextOf@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <Me@plt>
  mov    %rbp,%rsi
  pop    %rbp
  mov    %eax,%edi
  jmp    10a0 <SetOutMsg@plt>
```

assistant: ```c
void Act_SetOut(void) {
  char *x = TextOf(ArgText());
  SetOutMsg(Me(), x);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<displyMatrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  mov    -0x28(%rbp),%ebx
  movslq %ebx,%rax
  sub    $0x1,%rax
  mov    %rax,-0x18(%rbp)
  movslq %ebx,%rax
  mov    %rax,%r8
  mov    $0x0,%r9d
  movl   $0x0,-0x20(%rbp)
  jmp    L21
L20:
  movl   $0x0,-0x1c(%rbp)
  jmp    L18
L19:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  movslq %ebx,%rax
  imul   %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x1c(%rbp)
L18:
  mov    -0x1c(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L19
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x20(%rbp)
L21:
  mov    -0x20(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L20
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
type_infered: string

tool: D5: %2i

assistant: ```c
int displyMatrix(int nRows, int nCols, int matrix[nRows][nCols]) {
  int i, j;
  for (i = 0; i < nRows; ++i) {
    for (j = 0; j < nCols; ++j) {
      printf("%2i", matrix[i][j]);
    }
    printf("\n");
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<displyMatrix>:
  endbr64
  test   %edi,%edi
  jle    L39
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%r15d
  mov    %esi,%r14d
  movslq %esi,%rax
  shl    $0x2,%rax
  mov    %rax,(%rsp)
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdx,%rax,4),%rbp
  mov    $0x0,%r13d
  not    %rax
  shl    $0x2,%rax
  mov    %rax,0x8(%rsp)
  lea    D3(%rip),%r12
  jmp    L28
L24:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L24
L36:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r13d
  add    (%rsp),%rbp
  cmp    %r13d,%r15d
  je     L47
L28:
  mov    0x8(%rsp),%rax
  lea    (%rax,%rbp,1),%rbx
  test   %r14d,%r14d
  jg     L24
  jmp    L36
L47:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L39:
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
type_infered: string

tool: D3: %2i

assistant: ```c
int displyMatrix(int nRows, int nCols, int matrix[nRows][nCols]) {
  int i, j;
  for (i = 0; i < nRows; ++i) {
    for (j = 0; j < nCols; ++j) {
      printf("%2i", matrix[i][j]);
    }
    printf("\n");
  }
}
```