ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<kramer3>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rdx,%r12
  mov    $0x3,%esi
  mov    $0x0,%eax
  call   <determinant@plt>
  movsd  D11(%rip),%xmm1
  divsd  %xmm0,%xmm1
  movsd  0x20(%rbx),%xmm4
  movsd  0x40(%rbx),%xmm3
  movsd  0x28(%rbx),%xmm2
  movsd  0x38(%rbx),%xmm5
  movsd  0x8(%rbx),%xmm7
  movsd  0x10(%rbx),%xmm6
  movapd %xmm4,%xmm0
  mulsd  %xmm3,%xmm0
  movapd %xmm2,%xmm8
  mulsd  %xmm5,%xmm8
  subsd  %xmm8,%xmm0
  mulsd  0x0(%rbp),%xmm0
  mulsd  %xmm7,%xmm3
  mulsd  %xmm6,%xmm5
  subsd  %xmm5,%xmm3
  mulsd  0x8(%rbp),%xmm3
  subsd  %xmm3,%xmm0
  mulsd  %xmm7,%xmm2
  mulsd  %xmm6,%xmm4
  subsd  %xmm4,%xmm2
  mulsd  0x10(%rbp),%xmm2
  addsd  %xmm2,%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%r12)
  movsd  0x18(%rbx),%xmm4
  movsd  0x40(%rbx),%xmm3
  movsd  0x28(%rbx),%xmm2
  movsd  0x30(%rbx),%xmm5
  movsd  (%rbx),%xmm7
  movsd  0x10(%rbx),%xmm6
  movapd %xmm4,%xmm0
  mulsd  %xmm3,%xmm0
  movapd %xmm2,%xmm8
  mulsd  %xmm5,%xmm8
  subsd  %xmm8,%xmm0
  mulsd  0x0(%rbp),%xmm0
  mulsd  %xmm7,%xmm3
  mulsd  %xmm6,%xmm5
  subsd  %xmm5,%xmm3
  mulsd  0x8(%rbp),%xmm3
  subsd  %xmm3,%xmm0
  mulsd  %xmm7,%xmm2
  mulsd  %xmm6,%xmm4
  subsd  %xmm4,%xmm2
  mulsd  0x10(%rbp),%xmm2
  addsd  %xmm2,%xmm0
  xorpd  D13(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,0x8(%r12)
  movsd  0x18(%rbx),%xmm4
  movsd  0x38(%rbx),%xmm3
  movsd  0x20(%rbx),%xmm2
  movsd  0x30(%rbx),%xmm5
  movsd  (%rbx),%xmm7
  movsd  0x8(%rbx),%xmm6
  movapd %xmm4,%xmm0
  mulsd  %xmm3,%xmm0
  movapd %xmm2,%xmm8
  mulsd  %xmm5,%xmm8
  subsd  %xmm8,%xmm0
  mulsd  0x0(%rbp),%xmm0
  mulsd  %xmm7,%xmm3
  mulsd  %xmm6,%xmm5
  subsd  %xmm5,%xmm3
  mulsd  0x8(%rbp),%xmm3
  subsd  %xmm3,%xmm0
  mulsd  %xmm7,%xmm2
  mulsd  %xmm6,%xmm4
  subsd  %xmm4,%xmm2
  mulsd  0x10(%rbp),%xmm2
  addsd  %xmm2,%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,0x10(%r12)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: label: D11
type_guessed: f64
name: parse_data
arguments: label: D13
type_guessed: byte[16]

tool: D11: 1

tool: {"D13": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
void kramer3(double A[3][3], double b[3], double x[3]) {
  double det;
  double determinant();
  det = 1.0 / determinant(A, 3);
  x[0] = (b[0] * (A[1][1] * A[2][2] - A[1][2] * A[2][1]) - b[1] * (A[0][1] * A[2][2] - A[0][2] * A[2][1]) + b[2] * (A[0][1] * A[1][2] - A[0][2] * A[1][1])) * det;
  x[1] = -(b[0] * (A[1][0] * A[2][2] - A[1][2] * A[2][0]) - b[1] * (A[0][0] * A[2][2] - A[0][2] * A[2][0]) + b[2] * (A[0][0] * A[1][2] - A[0][2] * A[1][0])) * det;
  x[2] = (b[0] * (A[1][0] * A[2][1] - A[1][1] * A[2][0]) - b[1] * (A[0][0] * A[2][1] - A[0][1] * A[2][0]) + b[2] * (A[0][0] * A[1][1] - A[0][1] * A[1][0])) * det;
}
```
user: Translate the below assembly code into C programming language:

```asm
<kramer3>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x3,%esi
  push   %rbx
  mov    %rdi,%rbx
  call   <determinant@plt>
  movsd  0x20(%rbx),%xmm4
  movsd  0x28(%rbx),%xmm2
  movsd  D11(%rip),%xmm1
  movsd  0x8(%rbx),%xmm7
  movsd  0x10(%rbx),%xmm6
  movsd  0x40(%rbx),%xmm3
  movapd %xmm2,%xmm8
  divsd  %xmm0,%xmm1
  movsd  0x38(%rbx),%xmm5
  movapd %xmm4,%xmm0
  mulsd  %xmm3,%xmm0
  mulsd  %xmm5,%xmm8
  mulsd  %xmm7,%xmm3
  mulsd  %xmm6,%xmm5
  mulsd  %xmm6,%xmm4
  mulsd  %xmm7,%xmm2
  subsd  %xmm8,%xmm0
  mulsd  0x0(%rbp),%xmm0
  subsd  %xmm5,%xmm3
  mulsd  0x8(%rbp),%xmm3
  subsd  %xmm4,%xmm2
  mulsd  0x10(%rbp),%xmm2
  subsd  %xmm3,%xmm0
  addsd  %xmm2,%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%r12)
  movsd  0x18(%rbx),%xmm4
  movsd  0x28(%rbx),%xmm2
  movsd  (%rbx),%xmm7
  movsd  0x10(%rbx),%xmm6
  movsd  0x40(%rbx),%xmm3
  movsd  0x30(%rbx),%xmm5
  movapd %xmm4,%xmm0
  movapd %xmm2,%xmm8
  mulsd  %xmm6,%xmm4
  mulsd  %xmm5,%xmm8
  mulsd  %xmm3,%xmm0
  mulsd  %xmm6,%xmm5
  mulsd  %xmm7,%xmm3
  mulsd  %xmm7,%xmm2
  subsd  %xmm8,%xmm0
  mulsd  0x0(%rbp),%xmm0
  subsd  %xmm5,%xmm3
  mulsd  0x8(%rbp),%xmm3
  subsd  %xmm4,%xmm2
  mulsd  0x10(%rbp),%xmm2
  subsd  %xmm3,%xmm0
  addsd  %xmm2,%xmm0
  xorpd  D0(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,0x8(%r12)
  movsd  0x18(%rbx),%xmm4
  movsd  (%rbx),%xmm7
  movsd  0x20(%rbx),%xmm2
  movsd  0x38(%rbx),%xmm3
  movsd  0x30(%rbx),%xmm5
  movsd  0x8(%rbx),%xmm6
  movapd %xmm4,%xmm0
  mulsd  %xmm3,%xmm0
  movapd %xmm2,%xmm8
  mulsd  %xmm5,%xmm8
  mulsd  %xmm7,%xmm3
  mulsd  %xmm6,%xmm5
  mulsd  %xmm7,%xmm2
  mulsd  %xmm6,%xmm4
  subsd  %xmm8,%xmm0
  mulsd  0x0(%rbp),%xmm0
  subsd  %xmm5,%xmm3
  mulsd  0x8(%rbp),%xmm3
  subsd  %xmm4,%xmm2
  mulsd  0x10(%rbp),%xmm2
  subsd  %xmm3,%xmm0
  addsd  %xmm2,%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,0x10(%r12)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: label: D11
type_guessed: f64
name: unpack_data
arguments: label: D0
type_guessed: byte[16]

tool: D11: 1

tool: {"D0": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
void kramer3(double A[3][3], double b[3], double x[3]) {
  double det;
  double determinant();
  det = 1.0 / determinant(A, 3);
  x[0] = (b[0] * (A[1][1] * A[2][2] - A[1][2] * A[2][1]) - b[1] * (A[0][1] * A[2][2] - A[0][2] * A[2][1]) + b[2] * (A[0][1] * A[1][2] - A[0][2] * A[1][1])) * det;
  x[1] = -(b[0] * (A[1][0] * A[2][2] - A[1][2] * A[2][0]) - b[1] * (A[0][0] * A[2][2] - A[0][2] * A[2][0]) + b[2] * (A[0][0] * A[1][2] - A[0][2] * A[1][0])) * det;
  x[2] = (b[0] * (A[1][0] * A[2][1] - A[1][1] * A[2][0]) - b[1] * (A[0][0] * A[2][1] - A[0][1] * A[2][0]) + b[2] * (A[0][0] * A[1][1] - A[0][1] * A[1][0])) * det;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<kramer3>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x3,%esi
  push   %rbx
  mov    %rdi,%rbx
  call   <determinant@plt>
  movsd  0x20(%rbx),%xmm4
  movsd  0x28(%rbx),%xmm2
  movsd  D16(%rip),%xmm1
  movsd  0x8(%rbx),%xmm7
  movsd  0x10(%rbx),%xmm6
  movsd  0x40(%rbx),%xmm3
  movapd %xmm2,%xmm8
  divsd  %xmm0,%xmm1
  movsd  0x38(%rbx),%xmm5
  movapd %xmm4,%xmm0
  mulsd  %xmm3,%xmm0
  mulsd  %xmm5,%xmm8
  mulsd  %xmm7,%xmm3
  mulsd  %xmm6,%xmm5
  mulsd  %xmm6,%xmm4
  mulsd  %xmm7,%xmm2
  subsd  %xmm8,%xmm0
  mulsd  0x0(%rbp),%xmm0
  subsd  %xmm5,%xmm3
  mulsd  0x8(%rbp),%xmm3
  subsd  %xmm4,%xmm2
  mulsd  0x10(%rbp),%xmm2
  subsd  %xmm3,%xmm0
  addsd  %xmm2,%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%r12)
  movsd  0x18(%rbx),%xmm4
  movsd  0x28(%rbx),%xmm2
  movsd  (%rbx),%xmm7
  movsd  0x10(%rbx),%xmm6
  movsd  0x40(%rbx),%xmm3
  movsd  0x30(%rbx),%xmm5
  movapd %xmm4,%xmm0
  movapd %xmm2,%xmm8
  mulsd  %xmm6,%xmm4
  mulsd  %xmm5,%xmm8
  mulsd  %xmm3,%xmm0
  mulsd  %xmm6,%xmm5
  mulsd  %xmm7,%xmm3
  mulsd  %xmm7,%xmm2
  subsd  %xmm8,%xmm0
  mulsd  0x0(%rbp),%xmm0
  subsd  %xmm5,%xmm3
  mulsd  0x8(%rbp),%xmm3
  subsd  %xmm4,%xmm2
  mulsd  0x10(%rbp),%xmm2
  subsd  %xmm3,%xmm0
  addsd  %xmm2,%xmm0
  xorpd  D11(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,0x8(%r12)
  movsd  0x18(%rbx),%xmm4
  movsd  (%rbx),%xmm7
  movsd  0x20(%rbx),%xmm2
  movsd  0x38(%rbx),%xmm3
  movsd  0x30(%rbx),%xmm5
  movsd  0x8(%rbx),%xmm6
  movapd %xmm4,%xmm0
  mulsd  %xmm3,%xmm0
  movapd %xmm2,%xmm8
  mulsd  %xmm5,%xmm8
  mulsd  %xmm7,%xmm3
  mulsd  %xmm6,%xmm5
  mulsd  %xmm7,%xmm2
  mulsd  %xmm6,%xmm4
  subsd  %xmm8,%xmm0
  mulsd  0x0(%rbp),%xmm0
  subsd  %xmm5,%xmm3
  mulsd  0x8(%rbp),%xmm3
  subsd  %xmm4,%xmm2
  mulsd  0x10(%rbp),%xmm2
  subsd  %xmm3,%xmm0
  addsd  %xmm2,%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,0x10(%r12)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D16
type_infered: f64
name: parse_data
arguments: tag: D11
type_infered: byte[16]

tool: D16: 1

tool: {"D11": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
void kramer3(double A[3][3], double b[3], double x[3]) {
  double det;
  double determinant();
  det = 1.0 / determinant(A, 3);
  x[0] = (b[0] * (A[1][1] * A[2][2] - A[1][2] * A[2][1]) - b[1] * (A[0][1] * A[2][2] - A[0][2] * A[2][1]) + b[2] * (A[0][1] * A[1][2] - A[0][2] * A[1][1])) * det;
  x[1] = -(b[0] * (A[1][0] * A[2][2] - A[1][2] * A[2][0]) - b[1] * (A[0][0] * A[2][2] - A[0][2] * A[2][0]) + b[2] * (A[0][0] * A[1][2] - A[0][2] * A[1][0])) * det;
  x[2] = (b[0] * (A[1][0] * A[2][1] - A[1][1] * A[2][0]) - b[1] * (A[0][0] * A[2][1] - A[0][1] * A[2][0]) + b[2] * (A[0][0] * A[1][1] - A[0][1] * A[1][0])) * det;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xrop3_214_1_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x24(%rbp)
  mov    %al,-0x28(%rbp)
  movl   $0xff,-0x1c(%rbp)
  movzbl -0x24(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x24(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x18(%rbp)
  movzbl -0x28(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x28(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  and    -0x18(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  or     -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  xor    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  xor    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%eax
  not    %eax
  and    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  not    %eax
  and    %eax,%edx
  mov    -0x1c(%rbp),%eax
  and    -0x18(%rbp),%eax
  or     %eax,%edx
  mov    -0x1c(%rbp),%eax
  and    -0x14(%rbp),%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xrop3_214_1_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  unsigned stk4;
  stk4 = T & S;
  stk3 = D | stk4;
  stk2 = S ^ stk3;
  stk1 = T ^ stk2;
  return (stk1 & (~S) & (~T)) | (D & S) | (D & T);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<xrop3_214_1_1>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%eax
  mov    %eax,%esi
  shl    $0x8,%esi
  or     %eax,%esi
  mov    %edi,%edx
  and    %esi,%edx
  or     $0xff,%dl
  mov    %edi,%eax
  xor    %esi,%eax
  xor    %eax,%edx
  mov    %edx,%eax
  xor    $0xff,%al
  or     %esi,%edi
  and    %edi,%eax
  xor    %edx,%eax
  ret
```

assistant: ```c
unsigned xrop3_214_1_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  unsigned stk4;
  stk4 = T & S;
  stk3 = D | stk4;
  stk2 = S ^ stk3;
  stk1 = T ^ stk2;
  return (stk1 & (~S) & (~T)) | (D & S) | (D & T);
}
```
user: Translate the below assembly code into C programming language:

```asm
<xrop3_214_1_1>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%eax
  mov    %eax,%esi
  mov    %edi,%edx
  shl    $0x8,%esi
  or     %eax,%esi
  mov    %edi,%eax
  and    %esi,%edx
  xor    %esi,%eax
  or     %esi,%edi
  or     $0xff,%dl
  xor    %eax,%edx
  mov    %edx,%eax
  xor    $0xff,%al
  and    %edi,%eax
  xor    %edx,%eax
  ret
```

assistant: ```c
unsigned xrop3_214_1_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  unsigned stk4;
  stk4 = T & S;
  stk3 = D | stk4;
  stk2 = S ^ stk3;
  stk1 = T ^ stk2;
  return (stk1 & (~S) & (~T)) | (D & S) | (D & T);
}
```
user: What is the c source code of the assembly code below:

```asm
<xrop3_214_1_1>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%eax
  mov    %eax,%esi
  mov    %edi,%edx
  shl    $0x8,%esi
  or     %eax,%esi
  mov    %edi,%eax
  and    %esi,%edx
  xor    %esi,%eax
  or     %esi,%edi
  or     $0xff,%dl
  xor    %eax,%edx
  mov    %edx,%eax
  xor    $0xff,%al
  and    %edi,%eax
  xor    %edx,%eax
  ret
```

assistant: ```c
unsigned xrop3_214_1_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  unsigned stk4;
  stk4 = T & S;
  stk3 = D | stk4;
  stk2 = S ^ stk3;
  stk1 = T ^ stk2;
  return (stk1 & (~S) & (~T)) | (D & S) | (D & T);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0xc(%rbp)
  movl   $0x3e8,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <somaMultiplos3e7@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D5
datatype: string

tool: D5: Soma dos multiplos de 3 e 7, de %d ate %d: %d 


assistant: ```c
int main() {
  int inicio = 0, fim = 1000, soma = 0;
  soma = somaMultiplos3e7(inicio, fim);
  printf("Soma dos multiplos de 3 e 7, de %d ate %d: %d \n", inicio, fim, soma);
  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
  sub    $0x8,%rsp
  mov    $0x3e8,%esi
  mov    $0x0,%edi
  call   <somaMultiplos3e7@plt>
  mov    %eax,%ecx
  mov    $0x3e8,%edx
  mov    $0x0,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
type_guessed: string

tool: D9: Soma dos multiplos de 3 e 7, de %d ate %d: %d 


assistant: ```c
int main() {
  int inicio = 0, fim = 1000, soma = 0;
  soma = somaMultiplos3e7(inicio, fim);
  printf("Soma dos multiplos de 3 e 7, de %d ate %d: %d \n", inicio, fim, soma);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3e8,%esi
  xor    %edi,%edi
  call   <somaMultiplos3e7@plt>
  mov    $0x3e8,%edx
  xor    %esi,%esi
  lea    D1(%rip),%rdi
  mov    %eax,%ecx
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D1
type_infered: string

tool: D1: Soma dos multiplos de 3 e 7, de %d ate %d: %d 


assistant: ```c
int main() {
  int inicio = 0, fim = 1000, soma = 0;
  soma = somaMultiplos3e7(inicio, fim);
  printf("Soma dos multiplos de 3 e 7, de %d ate %d: %d \n", inicio, fim, soma);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3e8,%esi
  xor    %edi,%edi
  call   <somaMultiplos3e7@plt>
  mov    $0x3e8,%edx
  xor    %esi,%esi
  lea    D1(%rip),%rdi
  mov    %eax,%ecx
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D1
datatype: string

tool: D1: Soma dos multiplos de 3 e 7, de %d ate %d: %d 


assistant: ```c
int main() {
  int inicio = 0, fim = 1000, soma = 0;
  soma = somaMultiplos3e7(inicio, fim);
  printf("Soma dos multiplos de 3 e 7, de %d ate %d: %d \n", inicio, fim, soma);
  return 0;
}
```
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    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x9,-0x4(%rbp)
  jg     L0
  mov    $0x0,%eax
  call   <rand@plt>
  jmp    L13
L0:
  mov    $0x0,%eax
L13:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int x = rand();
  if (x < 10) {
    return rand();
  } else {
    return 0;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  mov    $0x0,%eax
  cmp    $0x9,%edx
  jle    L14
L18:
  add    $0x8,%rsp
  ret
L14:
  call   <rand@plt>
  jmp    L18
```

assistant: ```c
int main(int argc, char **argv) {
  int x = rand();
  if (x < 10) {
    return rand();
  } else {
    return 0;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <rand@plt>
  cmp    $0x9,%eax
  jle    L7
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L7:
  xor    %eax,%eax
  pop    %rdx
  jmp    1050 <rand@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  int x = rand();
  if (x < 10) {
    return rand();
  } else {
    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
  xor    %eax,%eax
  call   <rand@plt>
  cmp    $0x9,%eax
  jle    L1
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L1:
  xor    %eax,%eax
  pop    %rdx
  jmp    1050 <rand@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  int x = rand();
  if (x < 10) {
    return rand();
  } else {
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<page45>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D88(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D71(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D80(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D63(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xa,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D47
type_guessed: string
name: unpack_data
arguments: data_label: D88
type_guessed: string
name: unpack_data
arguments: data_label: D22
type_guessed: string
name: unpack_data
arguments: data_label: D71
type_guessed: string
name: unpack_data
arguments: data_label: D24
type_guessed: string
name: unpack_data
arguments: data_label: D80
type_guessed: string
name: unpack_data
arguments: data_label: D63
type_guessed: string
name: unpack_data
arguments: data_label: D30
type_guessed: string
name: unpack_data
arguments: data_label: D11
type_guessed: string

tool: D47: "Hrank Hrank," snorts the alarm in your living quarters.  Something is up.


tool: D88: You look at the monitor above the bathroom mirror and see the message you have


tool: D22: been waiting for all these years.  "ATTENTION TROUBLESHOOTER, YOU ARE BEING


tool: D71: ACTIVATED. PLEASE REPORT IMMEDIATELY TO MISSION ASSIGNMENT ROOM A17/GAMMA/LB22.


tool: D24: THANK YOU. THE COMPUTER IS YOUR FRIEND."  When you arrive at mission


tool: D80: assignment room A17-gamma/LB22 you are given your previous clone's


tool: D63: remaining possessions and notebook.  You puzzle through your predecessor's


tool: D30: cryptic notes, managing to decipher enough to lead you to the tube station and


tool: D11: the tube car to GDH7-beta.


assistant: ```c
page45() {
  printf("\"Hrank Hrank,\" snorts the alarm in your living quarters.  Something is up.\n");
  printf("You look at the monitor above the bathroom mirror and see the message you have\n");
  printf("been waiting for all these years.  \"ATTENTION TROUBLESHOOTER, YOU ARE BEING\n");
  printf("ACTIVATED. PLEASE REPORT IMMEDIATELY TO MISSION ASSIGNMENT ROOM A17/GAMMA/LB22.\n");
  printf("THANK YOU. THE COMPUTER IS YOUR FRIEND.\"  When you arrive at mission\n");
  printf("assignment room A17-gamma/LB22 you are given your previous clone\'s\n");
  printf("remaining possessions and notebook.  You puzzle through your predecessor\'s\n");
  printf("cryptic notes, managing to decipher enough to lead you to the tube station and\n");
  printf("the tube car to GDH7-beta.\n");
  return 10;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<page45>:
  endbr64
  sub    $0x8,%rsp
  lea    D20(%rip),%rdi
  call   <printf@plt>
  lea    D32(%rip),%rdi
  call   <printf@plt>
  lea    D23(%rip),%rdi
  call   <printf@plt>
  lea    D59(%rip),%rdi
  call   <printf@plt>
  lea    D80(%rip),%rdi
  call   <printf@plt>
  lea    D13(%rip),%rdi
  call   <printf@plt>
  lea    D5(%rip),%rdi
  call   <printf@plt>
  lea    D43(%rip),%rdi
  call   <printf@plt>
  lea    D83(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D20
type: string
name: parse_data
arguments: label: D32
type: string
name: parse_data
arguments: label: D23
type: string
name: parse_data
arguments: label: D59
type: string
name: parse_data
arguments: label: D80
type: string
name: parse_data
arguments: label: D13
type: string
name: parse_data
arguments: label: D5
type: string
name: parse_data
arguments: label: D43
type: string
name: parse_data
arguments: label: D83
type: string

tool: D20: "Hrank Hrank," snorts the alarm in your living quarters.  Something is up.


tool: D32: You look at the monitor above the bathroom mirror and see the message you have


tool: D23: been waiting for all these years.  "ATTENTION TROUBLESHOOTER, YOU ARE BEING


tool: D59: ACTIVATED. PLEASE REPORT IMMEDIATELY TO MISSION ASSIGNMENT ROOM A17/GAMMA/LB22.


tool: D80: THANK YOU. THE COMPUTER IS YOUR FRIEND."  When you arrive at mission


tool: D13: assignment room A17-gamma/LB22 you are given your previous clone's


tool: D5: remaining possessions and notebook.  You puzzle through your predecessor's


tool: D43: cryptic notes, managing to decipher enough to lead you to the tube station and


tool: D83: the tube car to GDH7-beta.


assistant: ```c
page45() {
  printf("\"Hrank Hrank,\" snorts the alarm in your living quarters.  Something is up.\n");
  printf("You look at the monitor above the bathroom mirror and see the message you have\n");
  printf("been waiting for all these years.  \"ATTENTION TROUBLESHOOTER, YOU ARE BEING\n");
  printf("ACTIVATED. PLEASE REPORT IMMEDIATELY TO MISSION ASSIGNMENT ROOM A17/GAMMA/LB22.\n");
  printf("THANK YOU. THE COMPUTER IS YOUR FRIEND.\"  When you arrive at mission\n");
  printf("assignment room A17-gamma/LB22 you are given your previous clone\'s\n");
  printf("remaining possessions and notebook.  You puzzle through your predecessor\'s\n");
  printf("cryptic notes, managing to decipher enough to lead you to the tube station and\n");
  printf("the tube car to GDH7-beta.\n");
  return 10;
}
```