ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<operator_priority>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  movsbl -0x14(%rbp),%eax
  cmp    $0x2f,%eax
  jg     L29
  test   %eax,%eax
  js     L21
  cmp    $0x2f,%eax
  ja     L21
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D6(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D6(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
L29:
  cmp    $0x7c,%eax
  je     L45
  jmp    L21
  addl   $0x1,-0x4(%rbp)
  addl   $0x1,-0x4(%rbp)
  addl   $0x1,-0x4(%rbp)
  addl   $0x1,-0x4(%rbp)
  addl   $0x1,-0x4(%rbp)
  addl   $0x1,-0x4(%rbp)
L45:
  addl   $0x1,-0x4(%rbp)
  addl   $0x1,-0x4(%rbp)
L21:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int operator_priority(char op) {
  int priority = 0;
  switch (op) {
  case '/':
    priority++;
  case '*':
    priority++;
  case '%':
    priority++;
  case '+':
    priority++;
  case '-':
    priority++;
  case '&':
    priority++;
  case '|':
    priority++;
  case '\0':
    priority++;
  }
  return priority;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<operator_priority>:
  endbr64
  cmp    $0x2f,%dil
  jg     L77
  test   %dil,%dil
  js     L24
  cmp    $0x2f,%dil
  ja     L79
  movzbl %dil,%edi
  lea    D6(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x1,%eax
  jmp    L78
L77:
  mov    $0x0,%eax
  cmp    $0x7c,%dil
  je     L70
  ret
  mov    $0x0,%eax
L78:
  add    $0x1,%eax
L43:
  add    $0x1,%eax
L11:
  add    $0x1,%eax
L48:
  add    $0x1,%eax
L5:
  add    $0x1,%eax
L70:
  add    $0x1,%eax
L59:
  add    $0x1,%eax
  ret
  mov    $0x0,%eax
  jmp    L43
  mov    $0x0,%eax
  jmp    L11
  mov    $0x0,%eax
  jmp    L48
  mov    $0x0,%eax
  jmp    L5
  mov    $0x0,%eax
  jmp    L59
L24:
  mov    $0x0,%eax
  ret
L79:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int operator_priority(char op) {
  int priority = 0;
  switch (op) {
  case '/':
    priority++;
  case '*':
    priority++;
  case '%':
    priority++;
  case '+':
    priority++;
  case '-':
    priority++;
  case '&':
    priority++;
  case '|':
    priority++;
  case '\0':
    priority++;
  }
  return priority;
}
```
user: What is the c source code of the assembly code below:

```asm
<operator_priority>:
  endbr64
  cmp    $0x2f,%dil
  jg     L9
  test   %dil,%dil
  js     L17
  cmp    $0x2f,%dil
  ja     L17
  lea    D6(%rip),%rdx
  movzbl %dil,%edi
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nopw   0x0(%rax,%rax,1)
L17:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  xor    %eax,%eax
  cmp    $0x7c,%dil
  sete   %al
  add    %eax,%eax
  ret
  nopl   0x0(%rax)
  mov    $0x4,%eax
  ret
  xchg   %ax,%ax
  mov    $0x5,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0x7,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0x3,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0x6,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0x1,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0x8,%eax
  ret
```

assistant: ```c
int operator_priority(char op) {
  int priority = 0;
  switch (op) {
  case '/':
    priority++;
  case '*':
    priority++;
  case '%':
    priority++;
  case '+':
    priority++;
  case '-':
    priority++;
  case '&':
    priority++;
  case '|':
    priority++;
  case '\0':
    priority++;
  }
  return priority;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<operator_priority>:
  endbr64
  cmp    $0x2f,%dil
  jg     L3
  test   %dil,%dil
  js     L24
  cmp    $0x2f,%dil
  ja     L24
  lea    D4(%rip),%rdx
  movzbl %dil,%edi
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nopw   0x0(%rax,%rax,1)
L24:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  cmp    $0x7c,%dil
  sete   %al
  add    %eax,%eax
  ret
  nopl   0x0(%rax)
  mov    $0x4,%eax
  ret
  xchg   %ax,%ax
  mov    $0x5,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0x7,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0x3,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0x6,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0x1,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0x8,%eax
  ret
```

assistant: ```c
int operator_priority(char op) {
  int priority = 0;
  switch (op) {
  case '/':
    priority++;
  case '*':
    priority++;
  case '%':
    priority++;
  case '+':
    priority++;
  case '-':
    priority++;
  case '&':
    priority++;
  case '|':
    priority++;
  case '\0':
    priority++;
  }
  return priority;
}
```
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    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x18(%rbp)
  jmp    L8
L60:
  mov    -0x1c(%rbp),%eax
  mov    -0x18(%rbp),%edx
  add    $0x1,%edx
  sub    %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L64
L57:
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L64:
  mov    -0x14(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jl     L57
  movl   $0x0,-0x14(%rbp)
  jmp    L39
L18:
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L39:
  mov    -0x14(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L18
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L8:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jl     L60
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L53
  call   <__stack_chk_fail@plt>
L53:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D1
data_type: string
name: read_data
arguments: label: D39
data_type: string
name: read_data
arguments: label: D22
data_type: string
name: read_data
arguments: label: D6
data_type: string
name: read_data
arguments: label: D4
data_type: string

tool: D1: Cuantos niveles?


tool: D39: %d

tool: D22:  

tool: D6: *

tool: D4: 


assistant: ```c
int main() {
  int iter, i, j;
  printf("Cuantos niveles?\n");
  scanf("%d", &iter);
  for (i = 0; i < iter; i++) {
    int esp = iter - (i + 1);
    int ast = iter - esp;
    for (j = 0; j < esp; j++) {
      printf(" ");
    }
    for (j = 0; j < ast; j++) {
      printf("*");
    }
    printf("\n");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D41(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D35(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ebp
  test   %ebp,%ebp
  jle    L51
  mov    $0x0,%r15d
  lea    D16(%rip),%r14
  lea    D4(%rip),%r13
L21:
  mov    %r15d,%r12d
  add    $0x1,%r15d
  sub    %r15d,%ebp
  test   %ebp,%ebp
  jle    L65
  mov    $0x0,%ebx
L5:
  mov    %r14,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L5
L65:
  test   %r15d,%r15d
  jle    L25
  mov    $0x0,%ebx
L34:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,%r12d
  jne    L34
L25:
  lea    D26(%rip),%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%ebp
  cmp    %ebp,%r15d
  jl     L21
L51:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L62
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L62:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D41
type_guessed: string
name: disassemble_data
arguments: data_tag: D35
type_guessed: string
name: disassemble_data
arguments: data_tag: D16
type_guessed: string
name: disassemble_data
arguments: data_tag: D4
type_guessed: string
name: disassemble_data
arguments: data_tag: D26
type_guessed: string

tool: D41: Cuantos niveles?


tool: D35: %d

tool: D16:  

tool: D4: *

tool: D26: 


assistant: ```c
int main() {
  int iter, i, j;
  printf("Cuantos niveles?\n");
  scanf("%d", &iter);
  for (i = 0; i < iter; i++) {
    int esp = iter - (i + 1);
    int ast = iter - esp;
    for (j = 0; j < esp; j++) {
      printf(" ");
    }
    for (j = 0; j < ast; j++) {
      printf("*");
    }
    printf("\n");
  }
  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
  push   %r15
  lea    D43(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D22(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ebx
  test   %ebx,%ebx
  jle    L3
  xor    %r14d,%r14d
  lea    D7(%rip),%r13
  lea    D28(%rip),%r12
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %r14d,%ebp
  add    $0x1,%r14d
  sub    %r14d,%ebx
  test   %ebx,%ebx
  jle    L39
  xor    %r15d,%r15d
  nopl   0x0(%rax)
L55:
  mov    %r12,%rdi
  add    $0x1,%r15d
  call   <printf@plt>
  cmp    %r15d,%ebx
  jne    L55
L39:
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L20:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,%ebp
  jne    L20
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%ebx
  cmp    %ebx,%r14d
  jl     L5
L3:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L57
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L57:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D43
type_guessed: string
name: parse_data
arguments: label: D22
type_guessed: string
name: parse_data
arguments: label: D7
type_guessed: string
name: parse_data
arguments: label: D28
type_guessed: string
name: parse_data
arguments: label: D11
type_guessed: string

tool: D43: Cuantos niveles?


tool: D22: %d

tool: D7: *

tool: D28:  

tool: D11: 


assistant: ```c
int main() {
  int iter, i, j;
  printf("Cuantos niveles?\n");
  scanf("%d", &iter);
  for (i = 0; i < iter; i++) {
    int esp = iter - (i + 1);
    int ast = iter - esp;
    for (j = 0; j < esp; j++) {
      printf(" ");
    }
    for (j = 0; j < ast; j++) {
      printf("*");
    }
    printf("\n");
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  lea    D44(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D17(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ebx
  test   %ebx,%ebx
  jle    L23
  xor    %r14d,%r14d
  lea    D23(%rip),%r13
  lea    D13(%rip),%r12
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %r14d,%ebp
  add    $0x1,%r14d
  sub    %r14d,%ebx
  test   %ebx,%ebx
  jle    L58
  xor    %r15d,%r15d
  nopl   0x0(%rax)
L32:
  mov    %r12,%rdi
  add    $0x1,%r15d
  call   <printf@plt>
  cmp    %r15d,%ebx
  jne    L32
L58:
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L33:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,%ebp
  jne    L33
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%ebx
  cmp    %ebx,%r14d
  jl     L8
L23:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D44
type: string
name: disassemble_data
arguments: data_tag: D17
type: string
name: disassemble_data
arguments: data_tag: D23
type: string
name: disassemble_data
arguments: data_tag: D13
type: string
name: disassemble_data
arguments: data_tag: D5
type: string

tool: D44: Cuantos niveles?


tool: D17: %d

tool: D23: *

tool: D13:  

tool: D5: 


assistant: ```c
int main() {
  int iter, i, j;
  printf("Cuantos niveles?\n");
  scanf("%d", &iter);
  for (i = 0; i < iter; i++) {
    int esp = iter - (i + 1);
    int ast = iter - esp;
    for (j = 0; j < esp; j++) {
      printf(" ");
    }
    for (j = 0; j < ast; j++) {
      printf("*");
    }
    printf("\n");
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<flvlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L64
L39:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdelimiter@plt>
  test   %rax,%rax
  jne    L45
  addl   $0x1,-0x8(%rbp)
L43:
  addl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L64
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdelimiter@plt>
  test   %rax,%rax
  je     L43
  jmp    L64
L45:
  addl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L64
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdelimiter@plt>
  test   %rax,%rax
  jne    L45
L64:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L39
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int flvlen(const char *FORMULAV) {
  int r = 0;
  int i = 0;
  while (FORMULAV[i]) {
    if (!isdelimiter(FORMULAV[i])) {
      ++r;
      do {
        ++i;
      } while (FORMULAV[i] && !isdelimiter(FORMULAV[i]));
    } else {
      do {
        ++i;
      } while (FORMULAV[i] && isdelimiter(FORMULAV[i]));
    }
  }
  return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<flvlen>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L56
  mov    $0x0,%ebx
  mov    $0x0,%r12d
  jmp    L17
L45:
  lea    0x1(%rbx),%r13d
  movslq %r13d,%r13
L13:
  mov    %r13d,%ebx
  movzbl 0x0(%rbp,%r13,1),%edi
  test   %dil,%dil
  je     L37
  movsbl %dil,%edi
  call   <isdelimiter@plt>
  add    $0x1,%r13
  test   %rax,%rax
  jne    L13
L66:
  movslq %ebx,%rax
  movzbl 0x0(%rbp,%rax,1),%edi
  test   %dil,%dil
  je     L37
L17:
  movsbl %dil,%edi
  call   <isdelimiter@plt>
  test   %rax,%rax
  jne    L45
  add    $0x1,%r12d
  lea    0x1(%rbx),%r13d
  movslq %r13d,%r13
L82:
  mov    %r13d,%ebx
  movzbl 0x0(%rbp,%r13,1),%edi
  test   %dil,%dil
  je     L37
  movsbl %dil,%edi
  call   <isdelimiter@plt>
  add    $0x1,%r13
  test   %rax,%rax
  je     L82
  jmp    L66
L56:
  mov    $0x0,%r12d
L37:
  mov    %r12d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int flvlen(const char *FORMULAV) {
  int r = 0;
  int i = 0;
  while (FORMULAV[i]) {
    if (!isdelimiter(FORMULAV[i])) {
      ++r;
      do {
        ++i;
      } while (FORMULAV[i] && !isdelimiter(FORMULAV[i]));
    } else {
      do {
        ++i;
      } while (FORMULAV[i] && isdelimiter(FORMULAV[i]));
    }
  }
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<flvlen>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L95
  xor    %ebp,%ebp
  xor    %r14d,%r14d
  call   <isdelimiter@plt>
  test   %rax,%rax
  jne    L13
  nopw   0x0(%rax,%rax,1)
L20:
  movslq %ebp,%rax
  lea    0x1(%rbp),%r12d
  add    $0x1,%r14d
  lea    0x1(%r13,%rax,1),%rax
  sub    %eax,%r12d
  jmp    L71
  nopl   (%rax)
L31:
  call   <isdelimiter@plt>
  mov    %rax,%r8
  lea    0x1(%rbx),%rax
  test   %r8,%r8
  jne    L8
L71:
  movsbl (%rax),%edi
  lea    (%r12,%rax,1),%ebp
  mov    %rax,%rbx
  test   %dil,%dil
  jne    L31
L80:
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L8:
  movsbl (%rbx),%edi
  test   %dil,%dil
  je     L80
  call   <isdelimiter@plt>
  test   %rax,%rax
  je     L20
L13:
  movslq %ebp,%rax
  lea    0x1(%rbp),%r12d
  lea    0x1(%r13,%rax,1),%rax
  sub    %eax,%r12d
  jmp    L98
  nopl   0x0(%rax,%rax,1)
L103:
  call   <isdelimiter@plt>
  mov    %rax,%r8
  lea    0x1(%rbx),%rax
  test   %r8,%r8
  je     L8
L98:
  movsbl (%rax),%edi
  lea    (%r12,%rax,1),%ebp
  mov    %rax,%rbx
  test   %dil,%dil
  jne    L103
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L95:
  xor    %r14d,%r14d
  jmp    L80
```

assistant: ```c
int flvlen(const char *FORMULAV) {
  int r = 0;
  int i = 0;
  while (FORMULAV[i]) {
    if (!isdelimiter(FORMULAV[i])) {
      ++r;
      do {
        ++i;
      } while (FORMULAV[i] && !isdelimiter(FORMULAV[i]));
    } else {
      do {
        ++i;
      } while (FORMULAV[i] && isdelimiter(FORMULAV[i]));
    }
  }
  return r;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<flvlen>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L52
  xor    %ebp,%ebp
  xor    %r14d,%r14d
  call   <isdelimiter@plt>
  test   %rax,%rax
  jne    L100
  nopw   0x0(%rax,%rax,1)
L40:
  movslq %ebp,%rax
  lea    0x1(%rbp),%r12d
  add    $0x1,%r14d
  lea    0x1(%r13,%rax,1),%rax
  sub    %eax,%r12d
  jmp    L4
  nopl   (%rax)
L53:
  call   <isdelimiter@plt>
  mov    %rax,%r8
  lea    0x1(%rbx),%rax
  test   %r8,%r8
  jne    L78
L4:
  movsbl (%rax),%edi
  lea    (%r12,%rax,1),%ebp
  mov    %rax,%rbx
  test   %dil,%dil
  jne    L53
L75:
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L78:
  movsbl (%rbx),%edi
  test   %dil,%dil
  je     L75
  call   <isdelimiter@plt>
  test   %rax,%rax
  je     L40
L100:
  movslq %ebp,%rax
  lea    0x1(%rbp),%r12d
  lea    0x1(%r13,%rax,1),%rax
  sub    %eax,%r12d
  jmp    L10
  nopl   0x0(%rax,%rax,1)
L2:
  call   <isdelimiter@plt>
  mov    %rax,%r8
  lea    0x1(%rbx),%rax
  test   %r8,%r8
  je     L78
L10:
  movsbl (%rax),%edi
  lea    (%r12,%rax,1),%ebp
  mov    %rax,%rbx
  test   %dil,%dil
  jne    L2
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L52:
  xor    %r14d,%r14d
  jmp    L75
```

assistant: ```c
int flvlen(const char *FORMULAV) {
  int r = 0;
  int i = 0;
  while (FORMULAV[i]) {
    if (!isdelimiter(FORMULAV[i])) {
      ++r;
      do {
        ++i;
      } while (FORMULAV[i] && !isdelimiter(FORMULAV[i]));
    } else {
      do {
        ++i;
      } while (FORMULAV[i] && isdelimiter(FORMULAV[i]));
    }
  }
  return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<imprimeVector>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L3
L18:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L3:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L18
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

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

tool: D4: %i 

assistant: ```c
void imprimeVector(int *minimos, int nCol) {
  int i = 0;
  for (i = 0; i < nCol; i++) {
    printf("%i ", *(minimos + i));
  }
  printf("\n");
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<imprimeVector>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L14
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%r12
  lea    D2(%rip),%rbp
L4:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L4
L14:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D2
datatype: string

tool: D2: %i 

assistant: ```c
void imprimeVector(int *minimos, int nCol) {
  int i = 0;
  for (i = 0; i < nCol; i++) {
    printf("%i ", *(minimos + i));
  }
  printf("\n");
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<imprimeVector>:
  endbr64
  test   %esi,%esi
  jle    L2
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x4(%rdi,%rax,4),%r12
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  xchg   %ax,%ax
L9:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L9
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  nopl   0x0(%rax,%rax,1)
L2:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D8
data_type: string

tool: D8: %i 

assistant: ```c
void imprimeVector(int *minimos, int nCol) {
  int i = 0;
  for (i = 0; i < nCol; i++) {
    printf("%i ", *(minimos + i));
  }
  printf("\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<imprimeVector>:
  endbr64
  test   %esi,%esi
  jle    L8
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x4(%rdi,%rax,4),%r12
  push   %rbp
  lea    D3(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  xchg   %ax,%ax
L19:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L19
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  nopl   0x0(%rax,%rax,1)
L8:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
data_type: string

tool: D3: %i 

assistant: ```c
void imprimeVector(int *minimos, int nCol) {
  int i = 0;
  for (i = 0; i < nCol; i++) {
    printf("%i ", *(minimos + i));
  }
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<selinux_relabel_path>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void selinux_relabel_path(const char *path, const char *inside_path) {
  (void)path;
  (void)inside_path;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<selinux_relabel_path>:
  endbr64
  ret
```

assistant: ```c
void selinux_relabel_path(const char *path, const char *inside_path) {
  (void)path;
  (void)inside_path;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<selinux_relabel_path>:
  endbr64
  ret
```

assistant: ```c
void selinux_relabel_path(const char *path, const char *inside_path) {
  (void)path;
  (void)inside_path;
}
```
user: Translate the below assembly code into C programming language:

```asm
<selinux_relabel_path>:
  endbr64
  ret
```

assistant: ```c
void selinux_relabel_path(const char *path, const char *inside_path) {
  (void)path;
  (void)inside_path;
}
```