ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<print_status_HSV>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    $0x7,%ebx
  mov    $0x80,%ebp
  lea    D16(%rip),%r13
  jmp    L7
L1:
  mov    $0x20,%edi
  call   <putchar@plt>
L6:
  sub    $0x1,%ebx
L7:
  test   %rbp,%r12
  setne  %sil
  movzbl %sil,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  shr    %rbp
  cmp    $0x2,%ebx
  je     L1
  test   %ebx,%ebx
  jne    L6
  mov    %r12,%rsi
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: label: D16
type_guessed: string
name: retrieve_data
arguments: label: D14
type_guessed: string

tool: D16: %d

tool: D14:  (%lu)


assistant: ```c
void print_status_HSV(unsigned long status) {
  unsigned int t = 8;
  unsigned int i;
  unsigned long int j = 1 << (t - 1);
  for (i = t; i > 0; --i) {
    printf("%d", (status & j) != 0);
    j >>= 1;
    if (i == 3)
      printf(" ");
  }
  printf(" (%lu)\n", status);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<print_status_HSV>:
  endbr64
  push   %r14
  lea    D10(%rip),%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    $0x8,%ebp
  push   %rbx
  mov    $0x80,%ebx
  jmp    L38
  nopw   0x0(%rax,%rax,1)
L39:
  mov    %r12d,%ebp
L38:
  xor    %esi,%esi
  test   %rbx,%r13
  lea    -0x1(%rbp),%r12d
  mov    %r14,%rdi
  setne  %sil
  xor    %eax,%eax
  shr    %rbx
  call   <printf@plt>
  cmp    $0x3,%ebp
  je     L4
  test   %r12d,%r12d
  jne    L39
  pop    %rbx
  mov    %r13,%rsi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  lea    D4(%rip),%rdi
  pop    %r13
  pop    %r14
  jmp    1070 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L4:
  mov    $0x20,%edi
  call   <putchar@plt>
  jmp    L39
```

tool_calls: name: retrieve_data
arguments: data_label: D10
type_guessed: string
name: retrieve_data
arguments: data_label: D4
type_guessed: string

tool: D10: %d

tool: D4:  (%lu)


assistant: ```c
void print_status_HSV(unsigned long status) {
  unsigned int t = 8;
  unsigned int i;
  unsigned long int j = 1 << (t - 1);
  for (i = t; i > 0; --i) {
    printf("%d", (status & j) != 0);
    j >>= 1;
    if (i == 3)
      printf(" ");
  }
  printf(" (%lu)\n", status);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_status_HSV>:
  endbr64
  push   %r14
  lea    D5(%rip),%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    $0x8,%ebp
  push   %rbx
  mov    $0x80,%ebx
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L37:
  mov    %r12d,%ebp
L15:
  xor    %esi,%esi
  test   %rbx,%r13
  lea    -0x1(%rbp),%r12d
  mov    %r14,%rdi
  setne  %sil
  xor    %eax,%eax
  shr    %rbx
  call   <printf@plt>
  cmp    $0x3,%ebp
  je     L2
  test   %r12d,%r12d
  jne    L37
  pop    %rbx
  mov    %r13,%rsi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  lea    D19(%rip),%rdi
  pop    %r13
  pop    %r14
  jmp    1070 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L2:
  mov    $0x20,%edi
  call   <putchar@plt>
  jmp    L37
```

tool_calls: name: read_data
arguments: data_label: D5
type_guessed: string
name: read_data
arguments: data_label: D19
type_guessed: string

tool: D5: %d

tool: D19:  (%lu)


assistant: ```c
void print_status_HSV(unsigned long status) {
  unsigned int t = 8;
  unsigned int i;
  unsigned long int j = 1 << (t - 1);
  for (i = t; i > 0; --i) {
    printf("%d", (status & j) != 0);
    j >>= 1;
    if (i == 3)
      printf(" ");
  }
  printf(" (%lu)\n", status);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  call   <calcularcomprimento@plt>
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  call   <area@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D12
type_infered: string
name: retrieve_data
arguments: tag: D16
type_infered: string

tool: D12: Digite o raio: 

tool: D16: %lf

assistant: ```c
main() {
  double raio;
  printf("Digite o raio: ");
  scanf("%lf", &raio);
  calcularcomprimento(raio);
  area(raio);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D6(%rip),%rdi
  call   <scanf@plt>
  movsd  (%rsp),%xmm0
  call   <calcularcomprimento@plt>
  movsd  (%rsp),%xmm0
  call   <area@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D18
type_infered: string
name: read_data
arguments: tag: D6
type_infered: string

tool: D18: Digite o raio: 

tool: D6: %lf

assistant: ```c
main() {
  double raio;
  printf("Digite o raio: ");
  scanf("%lf", &raio);
  calcularcomprimento(raio);
  area(raio);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D17(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D8(%rip),%rdi
  call   <scanf@plt>
  movsd  (%rsp),%xmm0
  call   <calcularcomprimento@plt>
  movsd  (%rsp),%xmm0
  call   <area@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D17
type: string
name: extract_symbolic_data
arguments: data_label: D8
type: string

tool: D17: Digite o raio: 

tool: D8: %lf

assistant: ```c
main() {
  double raio;
  printf("Digite o raio: ");
  scanf("%lf", &raio);
  calcularcomprimento(raio);
  area(raio);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D3(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D10(%rip),%rdi
  call   <scanf@plt>
  movsd  (%rsp),%xmm0
  call   <calcularcomprimento@plt>
  movsd  (%rsp),%xmm0
  call   <area@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D3
type_guessed: string
name: read_data
arguments: label: D10
type_guessed: string

tool: D3: Digite o raio: 

tool: D10: %lf

assistant: ```c
main() {
  double raio;
  printf("Digite o raio: ");
  scanf("%lf", &raio);
  calcularcomprimento(raio);
  area(raio);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D67(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  lea    D81(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%ecx
  mov    $0x1,%edx
  mov    $0x1,%esi
  lea    D81(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x8,%ecx
  mov    $0x4,%edx
  mov    $0x2,%esi
  lea    D81(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1b,%ecx
  mov    $0x9,%edx
  mov    $0x3,%esi
  lea    D81(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x40,%ecx
  mov    $0x10,%edx
  mov    $0x4,%esi
  lea    D99(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x7d,%ecx
  mov    $0x19,%edx
  mov    $0x5,%esi
  lea    D99(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xd8,%ecx
  mov    $0x24,%edx
  mov    $0x6,%esi
  lea    D99(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x157,%ecx
  mov    $0x31,%edx
  mov    $0x7,%esi
  lea    D99(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x200,%ecx
  mov    $0x40,%edx
  mov    $0x8,%esi
  lea    D99(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x2d9,%ecx
  mov    $0x51,%edx
  mov    $0x9,%esi
  lea    D99(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x3e8,%ecx
  mov    $0x64,%edx
  mov    $0xa,%esi
  lea    D84(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D67
data_type: string
name: disassemble_data
arguments: label: D81
data_type: string
name: disassemble_data
arguments: label: D99
data_type: string
name: disassemble_data
arguments: label: D84
data_type: string

tool: D67: numero  Quadrado  Cubo

tool: D81: %d       %d         %d   


tool: D99: %d       %d        %d   


tool: D84: %d      %d       %d   


assistant: ```c
int main(void) {
  printf("numero  Quadrado  Cubo\n");
  printf("%d       %d         %d   \n", 0, 0, 0);
  printf("%d       %d         %d   \n", 1, 1 * 1, 1 * 1 * 1);
  printf("%d       %d         %d   \n", 2, 2 * 2, 2 * 2 * 2);
  printf("%d       %d         %d   \n", 3, 3 * 3, 3 * 3 * 3);
  printf("%d       %d        %d   \n", 4, 4 * 4, 4 * 4 * 4);
  printf("%d       %d        %d   \n", 5, 5 * 5, 5 * 5 * 5);
  printf("%d       %d        %d   \n", 6, 6 * 6, 6 * 6 * 6);
  printf("%d       %d        %d   \n", 7, 7 * 7, 7 * 7 * 7);
  printf("%d       %d        %d   \n", 8, 8 * 8, 8 * 8 * 8);
  printf("%d       %d        %d   \n", 9, 9 * 9, 9 * 9 * 9);
  printf("%d      %d       %d   \n", 10, 10 * 10, 10 * 10 * 10);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbx
  lea    D2(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  lea    D27(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%ecx
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x8,%ecx
  mov    $0x4,%edx
  mov    $0x2,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1b,%ecx
  mov    $0x9,%edx
  mov    $0x3,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x40,%ecx
  mov    $0x10,%edx
  mov    $0x4,%esi
  lea    D18(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x7d,%ecx
  mov    $0x19,%edx
  mov    $0x5,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xd8,%ecx
  mov    $0x24,%edx
  mov    $0x6,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x157,%ecx
  mov    $0x31,%edx
  mov    $0x7,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x200,%ecx
  mov    $0x40,%edx
  mov    $0x8,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x2d9,%ecx
  mov    $0x51,%edx
  mov    $0x9,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x3e8,%ecx
  mov    $0x64,%edx
  mov    $0xa,%esi
  lea    D32(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: tag: D2
data_type: string
name: parse_data
arguments: tag: D27
data_type: string
name: parse_data
arguments: tag: D18
data_type: string
name: parse_data
arguments: tag: D32
data_type: string

tool: D2: numero  Quadrado  Cubo

tool: D27: %d       %d         %d   


tool: D18: %d       %d        %d   


tool: D32: %d      %d       %d   


assistant: ```c
int main(void) {
  printf("numero  Quadrado  Cubo\n");
  printf("%d       %d         %d   \n", 0, 0, 0);
  printf("%d       %d         %d   \n", 1, 1 * 1, 1 * 1 * 1);
  printf("%d       %d         %d   \n", 2, 2 * 2, 2 * 2 * 2);
  printf("%d       %d         %d   \n", 3, 3 * 3, 3 * 3 * 3);
  printf("%d       %d        %d   \n", 4, 4 * 4, 4 * 4 * 4);
  printf("%d       %d        %d   \n", 5, 5 * 5, 5 * 5 * 5);
  printf("%d       %d        %d   \n", 6, 6 * 6, 6 * 6 * 6);
  printf("%d       %d        %d   \n", 7, 7 * 7, 7 * 7 * 7);
  printf("%d       %d        %d   \n", 8, 8 * 8, 8 * 8 * 8);
  printf("%d       %d        %d   \n", 9, 9 * 9, 9 * 9 * 9);
  printf("%d      %d       %d   \n", 10, 10 * 10, 10 * 10 * 10);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D26(%rip),%rdi
  lea    D7(%rip),%rbp
  call   <puts@plt>
  mov    %rbp,%rdi
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%ecx
  xor    %eax,%eax
  mov    $0x1,%edx
  mov    $0x1,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x8,%ecx
  xor    %eax,%eax
  mov    $0x4,%edx
  mov    $0x2,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1b,%ecx
  xor    %eax,%eax
  lea    D23(%rip),%rbp
  mov    $0x9,%edx
  mov    $0x3,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x40,%ecx
  xor    %eax,%eax
  mov    $0x10,%edx
  mov    $0x4,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x7d,%ecx
  xor    %eax,%eax
  mov    $0x19,%edx
  mov    $0x5,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0xd8,%ecx
  xor    %eax,%eax
  mov    $0x24,%edx
  mov    $0x6,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x157,%ecx
  xor    %eax,%eax
  mov    $0x31,%edx
  mov    $0x7,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x200,%ecx
  xor    %eax,%eax
  mov    $0x40,%edx
  mov    $0x8,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x2d9,%ecx
  xor    %eax,%eax
  mov    $0x51,%edx
  mov    $0x9,%esi
  call   <printf@plt>
  mov    $0x3e8,%ecx
  xor    %eax,%eax
  mov    $0x64,%edx
  mov    $0xa,%esi
  lea    D32(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D26
type_infered: string
name: parse_data
arguments: tag: D7
type_infered: string
name: parse_data
arguments: tag: D23
type_infered: string
name: parse_data
arguments: tag: D32
type_infered: string

tool: D26: numero  Quadrado  Cubo

tool: D7: %d       %d         %d   


tool: D23: %d       %d        %d   


tool: D32: %d      %d       %d   


assistant: ```c
int main(void) {
  printf("numero  Quadrado  Cubo\n");
  printf("%d       %d         %d   \n", 0, 0, 0);
  printf("%d       %d         %d   \n", 1, 1 * 1, 1 * 1 * 1);
  printf("%d       %d         %d   \n", 2, 2 * 2, 2 * 2 * 2);
  printf("%d       %d         %d   \n", 3, 3 * 3, 3 * 3 * 3);
  printf("%d       %d        %d   \n", 4, 4 * 4, 4 * 4 * 4);
  printf("%d       %d        %d   \n", 5, 5 * 5, 5 * 5 * 5);
  printf("%d       %d        %d   \n", 6, 6 * 6, 6 * 6 * 6);
  printf("%d       %d        %d   \n", 7, 7 * 7, 7 * 7 * 7);
  printf("%d       %d        %d   \n", 8, 8 * 8, 8 * 8 * 8);
  printf("%d       %d        %d   \n", 9, 9 * 9, 9 * 9 * 9);
  printf("%d      %d       %d   \n", 10, 10 * 10, 10 * 10 * 10);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  lea    D0(%rip),%rdi
  lea    D3(%rip),%rbp
  call   <puts@plt>
  mov    %rbp,%rdi
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%ecx
  xor    %eax,%eax
  mov    $0x1,%edx
  mov    $0x1,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x8,%ecx
  xor    %eax,%eax
  mov    $0x4,%edx
  mov    $0x2,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1b,%ecx
  xor    %eax,%eax
  lea    D23(%rip),%rbp
  mov    $0x9,%edx
  mov    $0x3,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x40,%ecx
  xor    %eax,%eax
  mov    $0x10,%edx
  mov    $0x4,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x7d,%ecx
  xor    %eax,%eax
  mov    $0x19,%edx
  mov    $0x5,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0xd8,%ecx
  xor    %eax,%eax
  mov    $0x24,%edx
  mov    $0x6,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x157,%ecx
  xor    %eax,%eax
  mov    $0x31,%edx
  mov    $0x7,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x200,%ecx
  xor    %eax,%eax
  mov    $0x40,%edx
  mov    $0x8,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x2d9,%ecx
  xor    %eax,%eax
  mov    $0x51,%edx
  mov    $0x9,%esi
  call   <printf@plt>
  mov    $0x3e8,%ecx
  xor    %eax,%eax
  mov    $0x64,%edx
  mov    $0xa,%esi
  lea    D31(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D0
type: string
name: retrieve_data
arguments: data_tag: D3
type: string
name: retrieve_data
arguments: data_tag: D23
type: string
name: retrieve_data
arguments: data_tag: D31
type: string

tool: D0: numero  Quadrado  Cubo

tool: D3: %d       %d         %d   


tool: D23: %d       %d        %d   


tool: D31: %d      %d       %d   


assistant: ```c
int main(void) {
  printf("numero  Quadrado  Cubo\n");
  printf("%d       %d         %d   \n", 0, 0, 0);
  printf("%d       %d         %d   \n", 1, 1 * 1, 1 * 1 * 1);
  printf("%d       %d         %d   \n", 2, 2 * 2, 2 * 2 * 2);
  printf("%d       %d         %d   \n", 3, 3 * 3, 3 * 3 * 3);
  printf("%d       %d        %d   \n", 4, 4 * 4, 4 * 4 * 4);
  printf("%d       %d        %d   \n", 5, 5 * 5, 5 * 5 * 5);
  printf("%d       %d        %d   \n", 6, 6 * 6, 6 * 6 * 6);
  printf("%d       %d        %d   \n", 7, 7 * 7, 7 * 7 * 7);
  printf("%d       %d        %d   \n", 8, 8 * 8, 8 * 8 * 8);
  printf("%d       %d        %d   \n", 9, 9 * 9, 9 * 9 * 9);
  printf("%d      %d       %d   \n", 10, 10 * 10, 10 * 10 * 10);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<truncQuad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x8(%rbp)
  movsd  %xmm1,-0x10(%rbp)
  pxor   %xmm0,%xmm0
  ucomisd -0x10(%rbp),%xmm0
  jp     L38
  pxor   %xmm0,%xmm0
  ucomisd -0x10(%rbp),%xmm0
  jne    L38
  pxor   %xmm0,%xmm0
  jmp    L16
L38:
  movsd  -0x10(%rbp),%xmm1
  movsd  D45(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  D26(%rip),%xmm2
  movapd %xmm0,%xmm1
  divsd  %xmm2,%xmm1
  movsd  -0x8(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jb     L32
  pxor   %xmm0,%xmm0
  jmp    L16
L32:
  movsd  -0x10(%rbp),%xmm1
  movsd  D45(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  D26(%rip),%xmm1
  movapd %xmm0,%xmm2
  divsd  %xmm1,%xmm2
  movsd  -0x8(%rbp),%xmm0
  movapd %xmm0,%xmm1
  divsd  %xmm2,%xmm1
  movsd  D19(%rip),%xmm0
  subsd  %xmm1,%xmm0
L16:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D45
data_type: byte[8]
name: disassemble_data
arguments: data_label: D26
data_type: byte[8]
name: disassemble_data
arguments: data_label: D19
data_type: byte[8]

tool: {"D45": [0x00,0x00,0x00,0x00,0x00,0x00,0x22,0x40]}

tool: {"D26": [0x00,0x00,0x00,0x00,0x00,0x00,0x10,0x40]}

tool: {"D19": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double truncQuad(double epsilon, double thr) {
  if (thr == 0) {
    return 0;
  }
  if (epsilon >= thr * 9 / 4) {
    return 0;
  }
  return 1 - (epsilon / (thr * 9 / 4));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<truncQuad>:
  endbr64
  ucomisd D23(%rip),%xmm1
  jp     L18
  pxor   %xmm2,%xmm2
  je     L15
L18:
  mulsd  D4(%rip),%xmm1
  mulsd  D10(%rip),%xmm1
  pxor   %xmm2,%xmm2
  comisd %xmm1,%xmm0
  jae    L15
  divsd  %xmm1,%xmm0
  movsd  D7(%rip),%xmm2
  subsd  %xmm0,%xmm2
L15:
  movapd %xmm2,%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D23
type: byte[8]
name: disassemble_data
arguments: data_tag: D4
type: byte[8]
name: disassemble_data
arguments: data_tag: D10
type: byte[8]
name: disassemble_data
arguments: data_tag: D7
type: byte[8]

tool: {"D23": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D4": [0x00,0x00,0x00,0x00,0x00,0x00,0x22,0x40]}

tool: {"D10": [0x00,0x00,0x00,0x00,0x00,0x00,0xd0,0x3f]}

tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double truncQuad(double epsilon, double thr) {
  if (thr == 0) {
    return 0;
  }
  if (epsilon >= thr * 9 / 4) {
    return 0;
  }
  return 1 - (epsilon / (thr * 9 / 4));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<truncQuad>:
  endbr64
  pxor   %xmm3,%xmm3
  ucomisd %xmm3,%xmm1
  jp     L27
  movapd %xmm3,%xmm2
  je     L16
L27:
  mulsd  D25(%rip),%xmm1
  movapd %xmm3,%xmm2
  mulsd  D23(%rip),%xmm1
  comisd %xmm1,%xmm0
  jae    L16
  divsd  %xmm1,%xmm0
  movsd  D8(%rip),%xmm2
  subsd  %xmm0,%xmm2
L16:
  movapd %xmm2,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: data_label: D25
datatype: byte[8]
name: parse_data
arguments: data_label: D23
datatype: byte[8]
name: parse_data
arguments: data_label: D8
datatype: byte[8]

tool: {"D25": [0x00,0x00,0x00,0x00,0x00,0x00,0x22,0x40]}

tool: {"D23": [0x00,0x00,0x00,0x00,0x00,0x00,0xd0,0x3f]}

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double truncQuad(double epsilon, double thr) {
  if (thr == 0) {
    return 0;
  }
  if (epsilon >= thr * 9 / 4) {
    return 0;
  }
  return 1 - (epsilon / (thr * 9 / 4));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<truncQuad>:
  endbr64
  pxor   %xmm3,%xmm3
  ucomisd %xmm3,%xmm1
  jp     L22
  movapd %xmm3,%xmm2
  je     L8
L22:
  mulsd  D15(%rip),%xmm1
  movapd %xmm3,%xmm2
  mulsd  D10(%rip),%xmm1
  comisd %xmm1,%xmm0
  jae    L8
  divsd  %xmm1,%xmm0
  movsd  D11(%rip),%xmm2
  subsd  %xmm0,%xmm2
L8:
  movapd %xmm2,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: data_label: D15
datatype: byte[8]
name: parse_data
arguments: data_label: D10
datatype: byte[8]
name: parse_data
arguments: data_label: D11
datatype: byte[8]

tool: {"D15": [0x00,0x00,0x00,0x00,0x00,0x00,0x22,0x40]}

tool: {"D10": [0x00,0x00,0x00,0x00,0x00,0x00,0xd0,0x3f]}

tool: {"D11": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double truncQuad(double epsilon, double thr) {
  if (thr == 0) {
    return 0;
  }
  if (epsilon >= thr * 9 / 4) {
    return 0;
  }
  return 1 - (epsilon / (thr * 9 / 4));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x10,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  movl   $0x1,(%rax)
  mov    -0x8(%rbp),%rax
  movl   $0x2,(%rax)
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int *a = (int *)malloc(16);
  a[0] = 0;
  a[0] = 1;
  a[0] = 2;
  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    $0x10,%edi
  call   <malloc@plt>
  movl   $0x2,(%rax)
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int *a = (int *)malloc(16);
  a[0] = 0;
  a[0] = 1;
  a[0] = 2;
  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    $0x10,%edi
  call   <malloc@plt>
  movl   $0x2,(%rax)
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int *a = (int *)malloc(16);
  a[0] = 0;
  a[0] = 1;
  a[0] = 2;
  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
  mov    $0x10,%edi
  call   <malloc@plt>
  movl   $0x2,(%rax)
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int *a = (int *)malloc(16);
  a[0] = 0;
  a[0] = 1;
  a[0] = 2;
  return 0;
}
```