ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x14(%rbp),%eax
  imul   %eax,%edx
  mov    -0x14(%rbp),%eax
  imul   %edx,%eax
  add    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x14(%rbp),%eax
  imul   %edx,%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  mov    -0x10(%rbp),%edx
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    -0xc(%rbp),%edx
  mov    %eax,%esi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D28
data_type: string
name: read_data
arguments: data_tag: D33
data_type: string
name: read_data
arguments: data_tag: D12
data_type: string
name: read_data
arguments: data_tag: D35
data_type: string
name: read_data
arguments: data_tag: D7
data_type: string

tool: D28: Numero inteiro: 

tool: D33: %d

tool: D12: Sucessor do triplo de %d: %d


tool: D35: Antecessor do dobro de %d: %d


tool: D7: Soma: %d


assistant: ```c
int main() {
  int num, suc_triplo, ant_dobro;
  printf("Numero inteiro: ");
  scanf("%d", &num);
  suc_triplo = (num * num * num) + 1;
  ant_dobro = (num * num) - 1;
  printf("Sucessor do triplo de %d: %d\n", num, suc_triplo);
  printf("Antecessor do dobro de %d: %d\n", num, ant_dobro);
  printf("Soma: %d\n", suc_triplo + ant_dobro);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D2(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D39(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  mov    %esi,%ebp
  imul   %esi,%ebp
  mov    %esi,%ebx
  imul   %ebp,%ebx
  add    $0x1,%ebx
  sub    $0x1,%ebp
  mov    %ebx,%edx
  lea    D34(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%edx
  mov    0x4(%rsp),%esi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    (%rbx,%rbp,1),%esi
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D2
type_infered: string
name: parse_data
arguments: tag: D39
type_infered: string
name: parse_data
arguments: tag: D34
type_infered: string
name: parse_data
arguments: tag: D4
type_infered: string
name: parse_data
arguments: tag: D14
type_infered: string

tool: D2: Numero inteiro: 

tool: D39: %d

tool: D34: Sucessor do triplo de %d: %d


tool: D4: Antecessor do dobro de %d: %d


tool: D14: Soma: %d


assistant: ```c
int main() {
  int num, suc_triplo, ant_dobro;
  printf("Numero inteiro: ");
  scanf("%d", &num);
  suc_triplo = (num * num * num) + 1;
  ant_dobro = (num * num) - 1;
  printf("Sucessor do triplo de %d: %d\n", num, suc_triplo);
  printf("Antecessor do dobro de %d: %d\n", num, ant_dobro);
  printf("Soma: %d\n", suc_triplo + ant_dobro);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D17(%rip),%rdi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D10(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  lea    D42(%rip),%rdi
  xor    %eax,%eax
  mov    %esi,%ebp
  mov    %esi,%ebx
  imul   %esi,%ebp
  imul   %ebp,%ebx
  sub    $0x1,%ebp
  add    $0x1,%ebx
  mov    %ebx,%edx
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  xor    %eax,%eax
  mov    %ebp,%edx
  lea    D37(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  lea    (%rbx,%rbp,1),%esi
  lea    D26(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D17
type_guessed: string
name: disassemble_data
arguments: label: D10
type_guessed: string
name: disassemble_data
arguments: label: D42
type_guessed: string
name: disassemble_data
arguments: label: D37
type_guessed: string
name: disassemble_data
arguments: label: D26
type_guessed: string

tool: D17: Numero inteiro: 

tool: D10: %d

tool: D42: Sucessor do triplo de %d: %d


tool: D37: Antecessor do dobro de %d: %d


tool: D26: Soma: %d


assistant: ```c
int main() {
  int num, suc_triplo, ant_dobro;
  printf("Numero inteiro: ");
  scanf("%d", &num);
  suc_triplo = (num * num * num) + 1;
  ant_dobro = (num * num) - 1;
  printf("Sucessor do triplo de %d: %d\n", num, suc_triplo);
  printf("Antecessor do dobro de %d: %d\n", num, ant_dobro);
  printf("Soma: %d\n", suc_triplo + ant_dobro);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D15(%rip),%rdi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D39(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  lea    D32(%rip),%rdi
  xor    %eax,%eax
  mov    %esi,%ebp
  mov    %esi,%ebx
  imul   %esi,%ebp
  imul   %ebp,%ebx
  sub    $0x1,%ebp
  add    $0x1,%ebx
  mov    %ebx,%edx
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  xor    %eax,%eax
  mov    %ebp,%edx
  lea    D23(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  lea    (%rbx,%rbp,1),%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D15
type_infered: string
name: unpack_data
arguments: label: D39
type_infered: string
name: unpack_data
arguments: label: D32
type_infered: string
name: unpack_data
arguments: label: D23
type_infered: string
name: unpack_data
arguments: label: D7
type_infered: string

tool: D15: Numero inteiro: 

tool: D39: %d

tool: D32: Sucessor do triplo de %d: %d


tool: D23: Antecessor do dobro de %d: %d


tool: D7: Soma: %d


assistant: ```c
int main() {
  int num, suc_triplo, ant_dobro;
  printf("Numero inteiro: ");
  scanf("%d", &num);
  suc_triplo = (num * num * num) + 1;
  ant_dobro = (num * num) - 1;
  printf("Sucessor do triplo de %d: %d\n", num, suc_triplo);
  printf("Antecessor do dobro de %d: %d\n", num, ant_dobro);
  printf("Soma: %d\n", suc_triplo + ant_dobro);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L10:
  mov    $0x0,%eax
  call   <N@plt>
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L12:
  cmpl   $0x9,-0x4(%rbp)
  jle    L10
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D0
type: string

tool: D0: 


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 0; i < 10; i++) {
    N();
    printf("\n");
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0xa,%ebx
  lea    D5(%rip),%rbp
L1:
  mov    $0x0,%eax
  call   <N@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L1
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D5
data_type: string

tool: D5: 


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 0; i < 10; i++) {
    N();
    printf("\n");
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L1:
  xor    %eax,%eax
  call   <N@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L1
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D6
data_type: string

tool: D6: 


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 0; i < 10; i++) {
    N();
    printf("\n");
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D4(%rip),%rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L7:
  xor    %eax,%eax
  call   <N@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L7
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D4
type_infered: string

tool: D4: 


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 0; i < 10; i++) {
    N();
    printf("\n");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<uc_punctuation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <uc_program_current_character@plt>
  sub    $0x41,%eax
  cmp    $0x10,%eax
  ja     L3
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D13(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D13(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x2e,%eax
  jmp    L58
  mov    $0x21,%eax
  jmp    L58
  mov    $0x3f,%eax
  jmp    L58
  mov    $0x2c,%eax
  jmp    L58
  mov    $0x27,%eax
  jmp    L58
L3:
  mov    $0x0,%eax
L58:
  leave
  ret
```

assistant: ```c
char uc_punctuation(struct uc_program *program) {
  switch (uc_program_current_character(program)) {
  case 'P':
    return '.';
  case 'E':
    return '!';
  case 'Q':
    return '?';
  case 'C':
    return ',';
  case 'A':
    return '\'';
  default:
    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
<uc_punctuation>:
  endbr64
  sub    $0x8,%rsp
  call   <uc_program_current_character@plt>
  sub    $0x41,%eax
  cmp    $0x10,%eax
  ja     L24
  mov    %eax,%eax
  lea    D7(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x21,%eax
  jmp    L10
  mov    $0x3f,%eax
L10:
  add    $0x8,%rsp
  ret
  mov    $0x2c,%eax
  jmp    L10
  mov    $0x27,%eax
  jmp    L10
L24:
  mov    $0x0,%eax
  jmp    L10
  mov    $0x2e,%eax
  jmp    L10
```

assistant: ```c
char uc_punctuation(struct uc_program *program) {
  switch (uc_program_current_character(program)) {
  case 'P':
    return '.';
  case 'E':
    return '!';
  case 'Q':
    return '?';
  case 'C':
    return ',';
  case 'A':
    return '\'';
  default:
    return 0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<uc_punctuation>:
  endbr64
  sub    $0x8,%rsp
  call   <uc_program_current_character@plt>
  xor    %r8d,%r8d
  sub    $0x41,%eax
  cmp    $0x10,%eax
  ja     L7
  lea    D4(%rip),%rdx
  movzbl (%rdx,%rax,1),%r8d
L7:
  mov    %r8d,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D4: '

assistant: ```c
char uc_punctuation(struct uc_program *program) {
  switch (uc_program_current_character(program)) {
  case 'P':
    return '.';
  case 'E':
    return '!';
  case 'Q':
    return '?';
  case 'C':
    return ',';
  case 'A':
    return '\'';
  default:
    return 0;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<uc_punctuation>:
  endbr64
  sub    $0x8,%rsp
  call   <uc_program_current_character@plt>
  xor    %r8d,%r8d
  sub    $0x41,%eax
  cmp    $0x10,%eax
  ja     L2
  lea    D3(%rip),%rdx
  movzbl (%rdx,%rax,1),%r8d
L2:
  mov    %r8d,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D3: '

assistant: ```c
char uc_punctuation(struct uc_program *program) {
  switch (uc_program_current_character(program)) {
  case 'P':
    return '.';
  case 'E':
    return '!';
  case 'Q':
    return '?';
  case 'C':
    return ',';
  case 'A':
    return '\'';
  default:
    return 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<lockon>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  nop
L3:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L3
  mov    -0x8(%rbp),%rax
  movl   $0x1,(%rax)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int lockon(int *block) {
  while (*block)
    ;
  *block = 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<lockon>:
  endbr64
  mov    (%rdi),%eax
L7:
  test   %eax,%eax
  jne    L7
  movl   $0x1,(%rdi)
  ret
```

assistant: ```c
int lockon(int *block) {
  while (*block)
    ;
  *block = 1;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<lockon>:
  endbr64
  mov    (%rdi),%eax
  cs nopw 0x0(%rax,%rax,1)
L4:
  test   %eax,%eax
  jne    L4
  movl   $0x1,(%rdi)
  ret
```

assistant: ```c
int lockon(int *block) {
  while (*block)
    ;
  *block = 1;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<lockon>:
  endbr64
  mov    (%rdi),%eax
  test   %eax,%eax
  je     L14
L11:
  jmp    L11
  nopl   0x0(%rax)
L14:
  movl   $0x1,(%rdi)
  xor    %eax,%eax
  ret
```

assistant: ```c
int lockon(int *block) {
  while (*block)
    ;
  *block = 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<insort_index>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  movl   $0x1,-0x14(%rbp)
  jmp    L13
L44:
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L22
L25:
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x30(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  subl   $0x1,-0x10(%rbp)
L22:
  cmpl   $0x0,-0x10(%rbp)
  jle    L7
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  comisd -0x8(%rbp),%xmm0
  ja     L25
L7:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x14(%rbp)
L13:
  mov    -0x14(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L44
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void insort_index(double *a, int *index, int len) {
  int i, j;
  double tempd;
  int tempi;
  for (i = 1; i < len; i++) {
    j = i;
    tempd = a[j];
    tempi = index[j];
    while (j > 0 && ((a[j - 1]) > (tempd))) {
      a[j] = a[j - 1];
      index[j] = index[j - 1];
      j--;
    }
    a[j] = tempd;
    index[j] = tempi;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<insort_index>:
  endbr64
  cmp    $0x1,%edx
  jle    L53
  mov    %edx,%r9d
  mov    $0x1,%ecx
  jmp    L30
L68:
  movslq %edx,%rdx
  movsd  %xmm1,(%rdi,%rdx,8)
  mov    %r8d,(%rsi,%rdx,4)
  add    $0x1,%rcx
  cmp    %r9,%rcx
  je     L53
L30:
  mov    %ecx,%edx
  movsd  (%rdi,%rcx,8),%xmm1
  mov    (%rsi,%rcx,4),%r8d
  mov    %rcx,%rax
  test   %ecx,%ecx
  jle    L68
L10:
  mov    %eax,%edx
  movsd  -0x8(%rdi,%rax,8),%xmm0
  comisd %xmm1,%xmm0
  jbe    L68
  movsd  %xmm0,(%rdi,%rax,8)
  mov    -0x4(%rsi,%rax,4),%edx
  mov    %edx,(%rsi,%rax,4)
  lea    -0x1(%rax),%edx
  sub    $0x1,%rax
  test   %eax,%eax
  jg     L10
  jmp    L68
L53:
  ret
```

assistant: ```c
void insort_index(double *a, int *index, int len) {
  int i, j;
  double tempd;
  int tempi;
  for (i = 1; i < len; i++) {
    j = i;
    tempd = a[j];
    tempi = index[j];
    while (j > 0 && ((a[j - 1]) > (tempd))) {
      a[j] = a[j - 1];
      index[j] = index[j - 1];
      j--;
    }
    a[j] = tempd;
    index[j] = tempi;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<insort_index>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  cmp    $0x1,%edx
  jle    L36
  mov    %rdi,%r9
  mov    %rsi,%rbp
  lea    0x8(%rdi),%r11
  mov    $0x1,%edx
  lea    0x4(%rsi),%r10
L6:
  movsd  (%r11),%xmm1
  mov    (%r10),%esi
  mov    %r10,%rcx
  mov    %r11,%rax
  xchg   %ax,%ax
L16:
  movsd  -0x8(%rax),%xmm0
  mov    %rax,%r8
  mov    %rcx,%rdi
  comisd %xmm1,%xmm0
  jbe    L29
  movsd  %xmm0,(%rax)
  mov    -0x4(%rcx),%eax
  mov    %eax,(%rcx)
  lea    -0x8(%r8),%rax
  lea    -0x4(%rdi),%rcx
  cmp    %rax,%r9
  jne    L16
  mov    %rbp,%rdi
  mov    %r9,%r8
L29:
  add    $0x1,%edx
  movsd  %xmm1,(%r8)
  add    $0x8,%r11
  add    $0x4,%r10
  mov    %esi,(%rdi)
  cmp    %edx,%ebx
  jne    L6
L36:
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void insort_index(double *a, int *index, int len) {
  int i, j;
  double tempd;
  int tempi;
  for (i = 1; i < len; i++) {
    j = i;
    tempd = a[j];
    tempi = index[j];
    while (j > 0 && ((a[j - 1]) > (tempd))) {
      a[j] = a[j - 1];
      index[j] = index[j - 1];
      j--;
    }
    a[j] = tempd;
    index[j] = tempi;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<insort_index>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  cmp    $0x1,%edx
  jle    L30
  mov    %rdi,%r9
  mov    %rsi,%rbp
  lea    0x8(%rdi),%r11
  mov    $0x1,%edx
  lea    0x4(%rsi),%r10
L8:
  movsd  (%r11),%xmm1
  mov    (%r10),%esi
  mov    %r10,%rcx
  mov    %r11,%rax
  xchg   %ax,%ax
L37:
  movsd  -0x8(%rax),%xmm0
  mov    %rax,%r8
  mov    %rcx,%rdi
  comisd %xmm1,%xmm0
  jbe    L7
  movsd  %xmm0,(%rax)
  mov    -0x4(%rcx),%eax
  mov    %eax,(%rcx)
  lea    -0x8(%r8),%rax
  lea    -0x4(%rdi),%rcx
  cmp    %r9,%rax
  jne    L37
  mov    %rbp,%rdi
  mov    %r9,%r8
L7:
  add    $0x1,%edx
  movsd  %xmm1,(%r8)
  add    $0x8,%r11
  add    $0x4,%r10
  mov    %esi,(%rdi)
  cmp    %edx,%ebx
  jne    L8
L30:
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void insort_index(double *a, int *index, int len) {
  int i, j;
  double tempd;
  int tempi;
  for (i = 1; i < len; i++) {
    j = i;
    tempd = a[j];
    tempi = index[j];
    while (j > 0 && ((a[j - 1]) > (tempd))) {
      a[j] = a[j - 1];
      index[j] = index[j - 1];
      j--;
    }
    a[j] = tempd;
    index[j] = tempi;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x3,%eax
  mov    $0x0,%ecx
  cltd
  idiv   %ecx
  mov    %edx,%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D0
type_infered: string

tool: D0: %d


assistant: ```c
int main() {
  printf("%d\n", 3 % 0);
  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
  mov    $0x3,%eax
  mov    $0x0,%ecx
  cltd
  idiv   %ecx
  mov    %edx,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
datatype: string

tool: D5: %d


assistant: ```c
int main() {
  printf("%d\n", 3 % 0);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  ud2
```

assistant: ```c
int main() {
  printf("%d\n", 3 % 0);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  ud2
```

assistant: ```c
int main() {
  printf("%d\n", 3 % 0);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<read_macro_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
L59:
  mov    -0x28(%rbp),%rdx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <xgetline@plt>
  test   %rax,%rax
  je     L81
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L66
  subl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L66:
  mov    -0xc(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x5c,%al
  jne    L41
  subl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x1c(%rbp),%eax
  sub    -0xc(%rbp),%eax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rdi
  mov    -0x28(%rbp),%rdx
  mov    $0x2,%ecx
  mov    %eax,%esi
  call   <read_macro_line>
  add    %eax,-0xc(%rbp)
L41:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L37
L9:
  addq   $0x1,-0x8(%rbp)
L37:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L9
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x23,%al
  je     L32
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L7
L32:
  cmpl   $0x2,-0x20(%rbp)
  je     L7
  jmp    L59
L7:
  mov    -0xc(%rbp),%eax
  jmp    L75
L81:
  mov    $0x0,%eax
L75:
  leave
  ret
```

assistant: ```c
int read_macro_line(char *buf, int maxlen, FILE *confp, int flags) {
  char *p;
  int len;
repeat:
  if (xgetline(buf, maxlen, confp)) {
    len = strlen(buf);
    if (buf[len - 1] == '\n') {
      buf[--len] = '\0';
    }
    if (buf[len - 1] == '\\') {
      buf[--len] = '\0';
      len += read_macro_line(buf + len, maxlen - len, confp, (2));
    }
    p = buf;
    while (isspace(*p)) {
      p++;
    }
    if ((*p == '#' || *p == '\0') && flags != (2))
      goto repeat;
    return len;
  }
  return 0;
}
```