ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x3d0900,%edi
  call   <fibonacci@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D5
data_type: string

tool: D5: Answer: %d

assistant: ```c
int main(void) {
  long total = fibonacci(4000000);
  printf("Answer: %d", total);
  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    $0x3d0900,%edi
  call   <fibonacci@plt>
  mov    %rax,%rsi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D0
data_type: string

tool: D0: Answer: %d

assistant: ```c
int main(void) {
  long total = fibonacci(4000000);
  printf("Answer: %d", total);
  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    $0x3d0900,%edi
  call   <fibonacci@plt>
  lea    D3(%rip),%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D3
type: string

tool: D3: Answer: %d

assistant: ```c
int main(void) {
  long total = fibonacci(4000000);
  printf("Answer: %d", total);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3d0900,%edi
  call   <fibonacci@plt>
  lea    D5(%rip),%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D5
datatype: string

tool: D5: Answer: %d

assistant: ```c
int main(void) {
  long total = fibonacci(4000000);
  printf("Answer: %d", total);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<unterminated_text>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rsi
  mov    -0x4(%rbp),%eax
  lea    D4(%rip),%rcx
  mov    %eax,%edi
  call   <error@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D4
datatype: string

tool: D4: unmatched %%{

assistant: ```c
void unterminated_text(int lineno, char *line, char *cptr) {
  error(lineno, line, cptr, "unmatched %%{");
  exit(1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<unterminated_text>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rcx
  call   <error@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D9
type_guessed: string

tool: D9: unmatched %%{

assistant: ```c
void unterminated_text(int lineno, char *line, char *cptr) {
  error(lineno, line, cptr, "unmatched %%{");
  exit(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<unterminated_text>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rcx
  call   <error@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
datatype: string

tool: D6: unmatched %%{

assistant: ```c
void unterminated_text(int lineno, char *line, char *cptr) {
  error(lineno, line, cptr, "unmatched %%{");
  exit(1);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<unterminated_text>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rcx
  call   <error@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
type: string

tool: D8: unmatched %%{

assistant: ```c
void unterminated_text(int lineno, char *line, char *cptr) {
  error(lineno, line, cptr, "unmatched %%{");
  exit(1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<softmaxPass>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L34
L28:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L34:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L28
  movl   $0x0,-0xc(%rbp)
  jmp    L13
L39:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  divsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L13:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L39
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void softmaxPass(double *hiddenLayer, double *outputLayer, int outputSize) {
  int counter;
  double sum = 0;
  for (counter = 0; counter < outputSize; counter++) {
    sum += hiddenLayer[counter];
  }
  for (counter = 0; counter < outputSize; counter++) {
    outputLayer[counter] = hiddenLayer[counter] / sum;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<softmaxPass>:
  endbr64
  test   %edx,%edx
  jle    L23
  mov    %rdi,%rax
  lea    -0x1(%rdx),%ecx
  lea    0x8(%rdi,%rcx,8),%rcx
  pxor   %xmm1,%xmm1
L24:
  addsd  (%rax),%xmm1
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L24
  mov    %edx,%edx
  mov    $0x0,%eax
L16:
  movsd  (%rdi,%rax,8),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rsi,%rax,8)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L16
L23:
  ret
```

assistant: ```c
void softmaxPass(double *hiddenLayer, double *outputLayer, int outputSize) {
  int counter;
  double sum = 0;
  for (counter = 0; counter < outputSize; counter++) {
    sum += hiddenLayer[counter];
  }
  for (counter = 0; counter < outputSize; counter++) {
    outputLayer[counter] = hiddenLayer[counter] / sum;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<softmaxPass>:
  endbr64
  test   %edx,%edx
  jle    L24
  lea    -0x1(%rdx),%ecx
  mov    %rdi,%rax
  pxor   %xmm1,%xmm1
  lea    0x8(%rdi,%rcx,8),%rcx
  nopw   0x0(%rax,%rax,1)
L9:
  addsd  (%rax),%xmm1
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L9
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L2:
  movsd  (%rdi,%rax,8),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rsi,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L2
L24:
  ret
```

assistant: ```c
void softmaxPass(double *hiddenLayer, double *outputLayer, int outputSize) {
  int counter;
  double sum = 0;
  for (counter = 0; counter < outputSize; counter++) {
    sum += hiddenLayer[counter];
  }
  for (counter = 0; counter < outputSize; counter++) {
    outputLayer[counter] = hiddenLayer[counter] / sum;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<softmaxPass>:
  endbr64
  mov    %rdi,%rcx
  test   %edx,%edx
  jle    L33
  cmp    $0x1,%edx
  je     L21
  mov    %rdi,%rax
  mov    %edx,%edi
  pxor   %xmm0,%xmm0
  shr    %edi
  sub    $0x1,%edi
  shl    $0x4,%rdi
  lea    0x10(%rcx,%rdi,1),%rdi
  nop
L90:
  movsd  (%rax),%xmm1
  add    $0x10,%rax
  addsd  %xmm1,%xmm0
  movsd  -0x8(%rax),%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rdi,%rax
  jne    L90
  mov    %edx,%eax
  and    $0xfffffffe,%eax
  test   $0x1,%dl
  je     L48
L28:
  cltq
  addsd  (%rcx,%rax,8),%xmm0
L48:
  lea    0x8(%rcx),%rax
  cmp    %rax,%rsi
  je     L46
  cmp    $0x1,%edx
  je     L46
  mov    %edx,%edi
  movapd %xmm0,%xmm2
  xor    %eax,%eax
  shr    %edi
  unpcklpd %xmm2,%xmm2
  shl    $0x4,%rdi
  nopl   0x0(%rax,%rax,1)
L39:
  movupd (%rcx,%rax,1),%xmm1
  divpd  %xmm2,%xmm1
  movups %xmm1,(%rsi,%rax,1)
  add    $0x10,%rax
  cmp    %rdi,%rax
  jne    L39
  mov    %edx,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%edx
  je     L33
  movsd  (%rcx,%rax,8),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%rsi,%rax,8)
  ret
  nop
L46:
  movslq %edx,%rdx
  xor    %eax,%eax
  nopl   (%rax)
L89:
  movsd  (%rcx,%rax,8),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%rsi,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L89
L33:
  ret
L21:
  pxor   %xmm0,%xmm0
  xor    %eax,%eax
  jmp    L28
```

assistant: ```c
void softmaxPass(double *hiddenLayer, double *outputLayer, int outputSize) {
  int counter;
  double sum = 0;
  for (counter = 0; counter < outputSize; counter++) {
    sum += hiddenLayer[counter];
  }
  for (counter = 0; counter < outputSize; counter++) {
    outputLayer[counter] = hiddenLayer[counter] / sum;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L28:
  addl   $0x2,-0x4(%rbp)
  cmpl   $0x4,-0x4(%rbp)
  jle    L14
  subl   $0x1,-0x4(%rbp)
L14:
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x8(%rbp)
L11:
  cmpl   $0x4,-0x8(%rbp)
  jle    L28
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
type_infered: string

tool: D1: %i%i 

assistant: ```c
int main() {
  int x = 0;
  int y = 0;
  while (x < 5) {
    y = y + 2;
    if (y > 4)
      y = y - 1;
    printf("%i%i ", x, y);
    x = x + 1;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%ebp
  mov    $0x2,%ebx
  lea    D8(%rip),%r12
L18:
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebp
  cmp    $0x5,%ebp
  je     L1
  lea    0x2(%rbx),%eax
  add    $0x1,%ebx
  cmp    $0x4,%eax
  cmovle %eax,%ebx
  jmp    L18
L1:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
type_infered: string

tool: D8: %i%i 

assistant: ```c
int main() {
  int x = 0;
  int y = 0;
  while (x < 5) {
    y = y + 2;
    if (y > 4)
      y = y - 1;
    printf("%i%i ", x, y);
    x = x + 1;
  }
  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   %r13
  mov    $0x4,%r13d
  push   %r12
  lea    D6(%rip),%r12
  push   %rbp
  mov    $0x2,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
L10:
  mov    %ebx,%esi
  mov    %ebp,%edx
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x5,%ebx
  je     L19
  lea    0x1(%rbp),%eax
  cmp    $0x2,%ebp
  cmove  %r13d,%eax
  mov    %eax,%ebp
  jmp    L10
  nopl   0x0(%rax)
L19:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D6
type: string

tool: D6: %i%i 

assistant: ```c
int main() {
  int x = 0;
  int y = 0;
  while (x < 5) {
    y = y + 2;
    if (y > 4)
      y = y - 1;
    printf("%i%i ", x, y);
    x = x + 1;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x4,%r13d
  push   %r12
  lea    D9(%rip),%r12
  push   %rbp
  mov    $0x2,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
L2:
  mov    %ebx,%esi
  mov    %ebp,%edx
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x5,%ebx
  je     L9
  lea    0x1(%rbp),%eax
  cmp    $0x2,%ebp
  cmove  %r13d,%eax
  mov    %eax,%ebp
  jmp    L2
  nopl   0x0(%rax)
L9:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D9
type_guessed: string

tool: D9: %i%i 

assistant: ```c
int main() {
  int x = 0;
  int y = 0;
  while (x < 5) {
    y = y + 2;
    if (y > 4)
      y = y - 1;
    printf("%i%i ", x, y);
    x = x + 1;
  }
  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    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L30
L21:
  mov    -0x10(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  imul   %eax,%eax
  imul   -0xc(%rbp),%eax
  add    %eax,-0x14(%rbp)
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x10(%rbp)
L30:
  cmpl   $0x0,-0x10(%rbp)
  jne    L21
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jne    L13
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L24
L13:
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L24:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D15
type: string
name: extract_symbolic_data
arguments: data_label: D35
type: string
name: extract_symbolic_data
arguments: data_label: D28
type: string
name: extract_symbolic_data
arguments: data_label: D18
type: string

tool: D15: Enter an integer


tool: D35: %d

tool: D28: Entered number is an armstrong number.


tool: D18: Entered number is not an armstrong number.


assistant: ```c
int main() {
  int number, sum = 0, temp, remainder;
  printf("Enter an integer\n");
  scanf("%d", &number);
  temp = number;
  while (temp != 0) {
    remainder = temp % 10;
    sum = sum + remainder * remainder * remainder;
    temp = temp / 10;
  }
  if (number == sum)
    printf("Entered number is an armstrong number.\n");
  else
    printf("Entered number is not an armstrong number.\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D36(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D28(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r8d
  test   %r8d,%r8d
  je     L34
  mov    %r8d,%edx
  mov    $0x0,%edi
L18:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%esi
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,%edx
  imul   %eax,%edx
  imul   %edx,%eax
  add    %eax,%edi
  mov    %esi,%edx
  test   %esi,%esi
  jne    L18
  cmp    %edi,%r8d
  je     L34
  lea    D39(%rip),%rdi
  call   <printf@plt>
L16:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L34:
  lea    D13(%rip),%rdi
  call   <printf@plt>
  jmp    L16
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D36
datatype: string
name: disassemble_data
arguments: label: D28
datatype: string
name: disassemble_data
arguments: label: D39
datatype: string
name: disassemble_data
arguments: label: D13
datatype: string

tool: D36: Enter an integer


tool: D28: %d

tool: D39: Entered number is not an armstrong number.


tool: D13: Entered number is an armstrong number.


assistant: ```c
int main() {
  int number, sum = 0, temp, remainder;
  printf("Enter an integer\n");
  scanf("%d", &number);
  temp = number;
  while (temp != 0) {
    remainder = temp % 10;
    sum = sum + remainder * remainder * remainder;
    temp = temp / 10;
  }
  if (number == sum)
    printf("Entered number is an armstrong number.\n");
  else
    printf("Entered number is not an armstrong number.\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
  sub    $0x18,%rsp
  lea    D28(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r8d
  test   %r8d,%r8d
  je     L23
  mov    %r8d,%edx
  xor    %edi,%edi
  nopl   0x0(%rax)
L35:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%esi
  mov    %eax,%ecx
  mov    %edx,%eax
  add    %esi,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  imul   %eax,%edx
  imul   %edx,%eax
  mov    %ecx,%edx
  add    %eax,%edi
  test   %ecx,%ecx
  jne    L35
  cmp    %edi,%r8d
  je     L23
  lea    D33(%rip),%rdi
  call   <printf@plt>
L41:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L23:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L41
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D28
type: string
name: extract_symbolic_data
arguments: label: D1
type: string
name: extract_symbolic_data
arguments: label: D33
type: string
name: extract_symbolic_data
arguments: label: D6
type: string

tool: D28: Enter an integer


tool: D1: %d

tool: D33: Entered number is not an armstrong number.


tool: D6: Entered number is an armstrong number.


assistant: ```c
int main() {
  int number, sum = 0, temp, remainder;
  printf("Enter an integer\n");
  scanf("%d", &number);
  temp = number;
  while (temp != 0) {
    remainder = temp % 10;
    sum = sum + remainder * remainder * remainder;
    temp = temp / 10;
  }
  if (number == sum)
    printf("Entered number is an armstrong number.\n");
  else
    printf("Entered number is not an armstrong number.\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D26(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r8d
  test   %r8d,%r8d
  je     L13
  mov    %r8d,%edx
  xor    %edi,%edi
  nopl   0x0(%rax)
L20:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%esi
  mov    %eax,%ecx
  mov    %edx,%eax
  add    %esi,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  imul   %eax,%edx
  imul   %edx,%eax
  mov    %ecx,%edx
  add    %eax,%edi
  test   %ecx,%ecx
  jne    L20
  cmp    %edi,%r8d
  je     L13
  lea    D9(%rip),%rdi
  call   <printf@plt>
L33:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L13:
  lea    D24(%rip),%rdi
  call   <printf@plt>
  jmp    L33
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D26
data_type: string
name: parse_data
arguments: label: D13
data_type: string
name: parse_data
arguments: label: D9
data_type: string
name: parse_data
arguments: label: D24
data_type: string

tool: D26: Enter an integer


tool: D13: %d

tool: D9: Entered number is not an armstrong number.


tool: D24: Entered number is an armstrong number.


assistant: ```c
int main() {
  int number, sum = 0, temp, remainder;
  printf("Enter an integer\n");
  scanf("%d", &number);
  temp = number;
  while (temp != 0) {
    remainder = temp % 10;
    sum = sum + remainder * remainder * remainder;
    temp = temp / 10;
  }
  if (number == sum)
    printf("Entered number is an armstrong number.\n");
  else
    printf("Entered number is not an armstrong number.\n");
  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    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x2e2e2e2e2e2e2e2e,%rax
  mov    %rax,-0x70(%rbp)
  movw   $0x2e2e,-0x68(%rbp)
  mov    %rax,-0x66(%rbp)
  movw   $0x2e2e,-0x5e(%rbp)
  mov    %rax,-0x5c(%rbp)
  movw   $0x2e2e,-0x54(%rbp)
  mov    %rax,-0x52(%rbp)
  movw   $0x2e2e,-0x4a(%rbp)
  mov    %rax,-0x48(%rbp)
  movw   $0x2e2e,-0x40(%rbp)
  mov    %rax,-0x3e(%rbp)
  movw   $0x2e2e,-0x36(%rbp)
  mov    %rax,-0x34(%rbp)
  movw   $0x2e2e,-0x2c(%rbp)
  mov    %rax,-0x2a(%rbp)
  movw   $0x2e2e,-0x22(%rbp)
  mov    %rax,-0x20(%rbp)
  movw   $0x2e2e,-0x18(%rbp)
  mov    %rax,-0x16(%rbp)
  movw   $0x2e2e,-0xe(%rbp)
  mov    $0x0,%eax
  call   <walk_path@plt>
  mov    $0x0,%eax
  call   <print_path@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int main(void) {
  char path[10][10] = {{'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}};
  walk_path();
  print_path();
  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    $0x0,%eax
  call   <walk_path@plt>
  mov    $0x0,%eax
  call   <print_path@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  char path[10][10] = {{'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}};
  walk_path();
  print_path();
  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
  xor    %eax,%eax
  call   <walk_path@plt>
  xor    %eax,%eax
  call   <print_path@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  char path[10][10] = {{'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}};
  walk_path();
  print_path();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <walk_path@plt>
  xor    %eax,%eax
  call   <print_path@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  char path[10][10] = {{'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}, {'.', '.', '.', '.', '.', '.', '.', '.', '.', '.'}};
  walk_path();
  print_path();
  return 0;
}
```