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

```asm
<maxHeapSort>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%ebx
  call   <buildMaxHeap@plt>
  lea    D26(%rip),%rdi
  call   <puts@plt>
  test   %ebx,%ebx
  jle    L7
  mov    %rbp,%r12
  lea    -0x1(%rbx),%eax
  lea    0x4(%rbp,%rax,4),%r14
  lea    D5(%rip),%r13
L30:
  mov    (%r12),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%r12
  cmp    %r14,%r12
  jne    L30
L7:
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    $0x1,%ebx
  jle    L36
  movslq %ebx,%rbx
  lea    D27(%rip),%r12
L24:
  mov    0x0(%rbp),%esi
  mov    -0x4(%rbp,%rbx,4),%eax
  mov    %eax,0x0(%rbp)
  mov    %esi,-0x4(%rbp,%rbx,4)
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1(%rbx),%esi
  mov    %rbp,%rdi
  call   <buildMaxHeap@plt>
  sub    $0x1,%rbx
  cmp    $0x1,%ebx
  jg     L24
L36:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: label: D26
type: string
name: parse_data
arguments: label: D5
type: string
name: parse_data
arguments: label: D27
type: string

tool: D26: build max root heap

tool: D5: %d 

tool: D27: %d  

assistant: ```c
void maxHeapSort(int A[], int len) {
  int i, temp;
  buildMaxHeap(A, len);
  printf("build max root heap\n");
  for (i = 0; i < len; i++)
    printf("%d ", A[i]);
  printf("\n");
  for (i = len; i > 1; i--) {
    temp = A[0];
    A[0] = A[i - 1];
    A[i - 1] = temp;
    printf("%d  ", A[i - 1]);
    buildMaxHeap(A, i - 1);
  }
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<maxHeapSort>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %esi,%rbx
  mov    %ebx,%esi
  call   <buildMaxHeap@plt>
  lea    D20(%rip),%rdi
  call   <puts@plt>
  test   %ebx,%ebx
  jle    L17
  lea    -0x1(%rbx),%eax
  mov    %rbp,%r12
  lea    D26(%rip),%r13
  lea    0x4(%rbp,%rax,4),%r14
  nop
L8:
  mov    (%r12),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%r12
  call   <printf@plt>
  cmp    %r14,%r12
  jne    L8
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    $0x1,%ebx
  jle    L9
  lea    D3(%rip),%r12
  nopl   (%rax)
L49:
  mov    -0x4(%rbp,%rbx,4),%eax
  mov    0x0(%rbp),%esi
  mov    %r12,%rdi
  mov    %eax,0x0(%rbp)
  xor    %eax,%eax
  mov    %esi,-0x4(%rbp,%rbx,4)
  call   <printf@plt>
  lea    -0x1(%rbx),%esi
  mov    %rbp,%rdi
  sub    $0x1,%rbx
  call   <buildMaxHeap@plt>
  cmp    $0x1,%ebx
  jg     L49
L9:
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1080 <putchar@plt>
  xchg   %ax,%ax
L17:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L9
```

tool_calls: name: retrieve_data
arguments: data_label: D20
type_infered: string
name: retrieve_data
arguments: data_label: D26
type_infered: string
name: retrieve_data
arguments: data_label: D3
type_infered: string

tool: D20: build max root heap

tool: D26: %d 

tool: D3: %d  

assistant: ```c
void maxHeapSort(int A[], int len) {
  int i, temp;
  buildMaxHeap(A, len);
  printf("build max root heap\n");
  for (i = 0; i < len; i++)
    printf("%d ", A[i]);
  printf("\n");
  for (i = len; i > 1; i--) {
    temp = A[0];
    A[0] = A[i - 1];
    A[i - 1] = temp;
    printf("%d  ", A[i - 1]);
    buildMaxHeap(A, i - 1);
  }
  printf("\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<maxHeapSort>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %esi,%rbx
  mov    %ebx,%esi
  call   <buildMaxHeap@plt>
  lea    D4(%rip),%rdi
  call   <puts@plt>
  test   %ebx,%ebx
  jle    L41
  lea    -0x1(%rbx),%eax
  mov    %rbp,%r12
  lea    D12(%rip),%r13
  lea    0x4(%rbp,%rax,4),%r14
  nop
L24:
  mov    (%r12),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%r12
  call   <printf@plt>
  cmp    %r12,%r14
  jne    L24
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    $0x1,%ebx
  jle    L5
  lea    D18(%rip),%r12
  nopl   (%rax)
L2:
  mov    -0x4(%rbp,%rbx,4),%eax
  mov    0x0(%rbp),%esi
  mov    %r12,%rdi
  mov    %eax,0x0(%rbp)
  xor    %eax,%eax
  mov    %esi,-0x4(%rbp,%rbx,4)
  call   <printf@plt>
  lea    -0x1(%rbx),%esi
  mov    %rbp,%rdi
  sub    $0x1,%rbx
  call   <buildMaxHeap@plt>
  cmp    $0x1,%ebx
  jg     L2
L5:
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1080 <putchar@plt>
  xchg   %ax,%ax
L41:
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L5
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
datatype: string
name: disassemble_data
arguments: data_tag: D12
datatype: string
name: disassemble_data
arguments: data_tag: D18
datatype: string

tool: D4: build max root heap

tool: D12: %d 

tool: D18: %d  

assistant: ```c
void maxHeapSort(int A[], int len) {
  int i, temp;
  buildMaxHeap(A, len);
  printf("build max root heap\n");
  for (i = 0; i < len; i++)
    printf("%d ", A[i]);
  printf("\n");
  for (i = len; i > 1; i--) {
    temp = A[0];
    A[0] = A[i - 1];
    A[i - 1] = temp;
    printf("%d  ", A[i - 1]);
    buildMaxHeap(A, i - 1);
  }
  printf("\n");
}
```
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
  lea    -0x3d0000(%rsp),%r11
L35:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L35
  sub    $0x930,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x3d0910(%rbp),%rax
  mov    %rax,%rdi
  call   <init_arr@plt>
  lea    -0x3d0924(%rbp),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x3d091c(%rbp)
  jmp    L23
L13:
  lea    -0x3d0920(%rbp),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x3d0920(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x3d0910(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x3d0910(%rbp,%rax,4)
  addl   $0x1,-0x3d091c(%rbp)
L23:
  mov    -0x3d0924(%rbp),%eax
  cmp    %eax,-0x3d091c(%rbp)
  jl     L13
  movl   $0x0,-0x3d091c(%rbp)
  jmp    L1
L14:
  mov    -0x3d091c(%rbp),%eax
  cltq
  mov    -0x3d0910(%rbp,%rax,4),%eax
  mov    %eax,-0x3d0914(%rbp)
  movl   $0x0,-0x3d0918(%rbp)
  jmp    L70
L16:
  mov    -0x3d091c(%rbp),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x3d0918(%rbp)
L70:
  mov    -0x3d0918(%rbp),%eax
  cmp    -0x3d0914(%rbp),%eax
  jl     L16
  addl   $0x1,-0x3d091c(%rbp)
L1:
  cmpl   $0xf4240,-0x3d091c(%rbp)
  jle    L14
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L58
  call   <__stack_chk_fail@plt>
L58:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D25
data_type: string
name: retrieve_data
arguments: data_tag: D23
data_type: string

tool: D25: %d

tool: D23: %d


assistant: ```c
int main() {
  int arr[1000000 + 1];
  init_arr(arr);
  int cases;
  scanf("%d", &cases);
  int i, n;
  for (i = 0; i < cases; i++) {
    scanf("%d", &n);
    arr[n]++;
  }
  int j, cur_val;
  for (i = 0; i <= 1000000; i++) {
    cur_val = arr[i];
    for (j = 0; j < cur_val; j++) {
      printf("%d\n", i);
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x3d0000(%rsp),%r11
L58:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L58
  sub    $0x920,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3d0918(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  call   <init_arr@plt>
  lea    0x8(%rsp),%rsi
  lea    D21(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x8(%rsp)
  jle    L4
  mov    $0x0,%ebx
  lea    D21(%rip),%rbp
L30:
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movslq 0xc(%rsp),%rax
  addl   $0x1,0x10(%rsp,%rax,4)
  add    $0x1,%ebx
  cmp    %ebx,0x8(%rsp)
  jg     L30
L4:
  mov    $0x0,%r14d
  lea    D28(%rip),%r13
  jmp    L80
L46:
  add    $0x1,%r14
  cmp    $0xf4241,%r14
  je     L39
L80:
  mov    %r14d,%r12d
  mov    0x10(%rsp,%r14,4),%ebp
  test   %ebp,%ebp
  jle    L46
  mov    $0x0,%ebx
L87:
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L87
  jmp    L46
L39:
  mov    0x3d0918(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L82
  mov    $0x0,%eax
  add    $0x3d0920,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L82:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D21
type_infered: string
name: parse_data
arguments: tag: D28
type_infered: string

tool: D21: %d

tool: D28: %d


assistant: ```c
int main() {
  int arr[1000000 + 1];
  init_arr(arr);
  int cases;
  scanf("%d", &cases);
  int i, n;
  for (i = 0; i < cases; i++) {
    scanf("%d", &n);
    arr[n]++;
  }
  int j, cur_val;
  for (i = 0; i <= 1000000; i++) {
    cur_val = arr[i];
    for (j = 0; j < cur_val; j++) {
      printf("%d\n", i);
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x3d0000(%rsp),%r11
L61:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L61
  sub    $0x928,%rsp
  lea    D1(%rip),%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x3d0918(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r15
  mov    %r15,%rdi
  call   <init_arr@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  test   %eax,%eax
  jle    L23
  xor    %ebx,%ebx
  lea    0xc(%rsp),%r12
L9:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <scanf@plt>
  movslq 0xc(%rsp),%rax
  addl   $0x1,0x10(%rsp,%rax,4)
  cmp    %ebx,0x8(%rsp)
  jg     L9
L23:
  xor    %r14d,%r14d
  lea    D12(%rip),%r13
  nopl   0x0(%rax)
L17:
  mov    (%r15,%r14,4),%ebp
  mov    %r14d,%r12d
  test   %ebp,%ebp
  jle    L31
  xor    %ebx,%ebx
  nopl   (%rax)
L51:
  mov    %r12d,%esi
  mov    %r13,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L51
L31:
  add    $0x1,%r14
  cmp    $0xf4241,%r14
  jne    L17
  mov    0x3d0918(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L57
  add    $0x3d0928,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L57:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D1
datatype: string
name: extract_symbolic_data
arguments: data_label: D12
datatype: string

tool: D1: %d

tool: D12: %d


assistant: ```c
int main() {
  int arr[1000000 + 1];
  init_arr(arr);
  int cases;
  scanf("%d", &cases);
  int i, n;
  for (i = 0; i < cases; i++) {
    scanf("%d", &n);
    arr[n]++;
  }
  int j, cur_val;
  for (i = 0; i <= 1000000; i++) {
    cur_val = arr[i];
    for (j = 0; j < cur_val; j++) {
      printf("%d\n", i);
    }
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x3d0000(%rsp),%r11
L27:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L27
  sub    $0x928,%rsp
  lea    D15(%rip),%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x3d0918(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r15
  mov    %r15,%rdi
  call   <init_arr@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  test   %eax,%eax
  jle    L44
  xor    %ebx,%ebx
  lea    0xc(%rsp),%r12
L47:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <scanf@plt>
  movslq 0xc(%rsp),%rax
  addl   $0x1,0x10(%rsp,%rax,4)
  cmp    %ebx,0x8(%rsp)
  jg     L47
L44:
  xor    %r14d,%r14d
  lea    D4(%rip),%r13
  nopl   0x0(%rax)
L1:
  mov    (%r15,%r14,4),%ebp
  mov    %r14d,%r12d
  test   %ebp,%ebp
  jle    L37
  xor    %ebx,%ebx
  nopl   (%rax)
L52:
  mov    %r12d,%esi
  mov    %r13,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L52
L37:
  add    $0x1,%r14
  cmp    $0xf4241,%r14
  jne    L1
  mov    0x3d0918(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L62
  add    $0x3d0928,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L62:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D15
data_type: string
name: retrieve_data
arguments: label: D4
data_type: string

tool: D15: %d

tool: D4: %d


assistant: ```c
int main() {
  int arr[1000000 + 1];
  init_arr(arr);
  int cases;
  scanf("%d", &cases);
  int i, n;
  for (i = 0; i < cases; i++) {
    scanf("%d", &n);
    arr[n]++;
  }
  int j, cur_val;
  for (i = 0; i <= 1000000; i++) {
    cur_val = arr[i];
    for (j = 0; j < cur_val; j++) {
      printf("%d\n", i);
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<get_key>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xa(%rbp),%r8
  lea    -0xc(%rbp),%rdi
  lea    -0xe(%rbp),%rcx
  lea    -0x10(%rbp),%rdx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %r8,%r9
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <find_key@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
void *get_key(struct cube *cube, int key) {
  short w, x, y, z;
  return find_key(cube, key, &w, &x, &y, &z);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_key>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x2(%rsp),%rcx
  mov    %rsp,%rdx
  lea    0x6(%rsp),%r9
  lea    0x4(%rsp),%r8
  call   <find_key@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *get_key(struct cube *cube, int key) {
  short w, x, y, z;
  return find_key(cube, key, &w, &x, &y, &z);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_key>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsp,%rdx
  lea    0x2(%rsp),%rcx
  lea    0x6(%rsp),%r9
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r8
  call   <find_key@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *get_key(struct cube *cube, int key) {
  short w, x, y, z;
  return find_key(cube, key, &w, &x, &y, &z);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<get_key>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsp,%rdx
  lea    0x2(%rsp),%rcx
  lea    0x6(%rsp),%r9
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r8
  call   <find_key@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *get_key(struct cube *cube, int key) {
  short w, x, y, z;
  return find_key(cube, key, &w, &x, &y, &z);
}
```
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)
  movq   $0x0,-0x10(%rbp)
  movq   $0x2,-0x8(%rbp)
  jmp    L23
L9:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <is_equal@plt>
  test   %rax,%rax
  je     L4
  mov    -0x8(%rbp),%rax
  add    %rax,-0x10(%rbp)
L4:
  addq   $0x1,-0x8(%rbp)
L23:
  cmpq   $0xf4240,-0x8(%rbp)
  jbe    L9
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
type_guessed: string

tool: D0: Sum: %llu


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long long sum = 0;
  for (unsigned long long i = 2; i <= 1000000; i++)
    if (is_equal(i))
      sum += i;
  printf("Sum: %llu\n", sum);
  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    $0x8,%rsp
  mov    $0x2,%ebx
  mov    $0x0,%ebp
L5:
  mov    %rbx,%rdi
  call   <is_equal@plt>
  mov    %rax,%rdx
  lea    0x0(%rbp,%rbx,1),%rax
  test   %rdx,%rdx
  cmovne %rax,%rbp
  add    $0x1,%rbx
  cmp    $0xf4241,%rbx
  jne    L5
  mov    %rbp,%rsi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

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

tool: D8: Sum: %llu


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long long sum = 0;
  for (unsigned long long i = 2; i <= 1000000; i++)
    if (is_equal(i))
      sum += i;
  printf("Sum: %llu\n", sum);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x2,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L9:
  mov    %rbx,%rdi
  call   <is_equal@plt>
  mov    %rax,%r8
  lea    0x0(%rbp,%rbx,1),%rax
  test   %r8,%r8
  cmovne %rax,%rbp
  add    $0x1,%rbx
  cmp    $0xf4241,%rbx
  jne    L9
  mov    %rbp,%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D3
type_guessed: string

tool: D3: Sum: %llu


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long long sum = 0;
  for (unsigned long long i = 2; i <= 1000000; i++)
    if (is_equal(i))
      sum += i;
  printf("Sum: %llu\n", sum);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x2,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L2:
  mov    %rbx,%rdi
  call   <is_equal@plt>
  mov    %rax,%r8
  lea    0x0(%rbp,%rbx,1),%rax
  test   %r8,%r8
  cmovne %rax,%rbp
  add    $0x1,%rbx
  cmp    $0xf4241,%rbx
  jne    L2
  mov    %rbp,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D2
data_type: string

tool: D2: Sum: %llu


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long long sum = 0;
  for (unsigned long long i = 2; i <= 1000000; i++)
    if (is_equal(i))
      sum += i;
  printf("Sum: %llu\n", sum);
  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
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  movq   $0x4,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  mov    %rdx,%rcx
  mov    $0x4,%edx
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D9
data_type: string

tool: D9: n = %d, n as %zd bytes: all ints have %zd bytes.


assistant: ```c
int main(int argc, char const *argv[]) {
  int n = 0;
  size_t intsize;
  intsize = sizeof(int);
  printf("n = %d, n as %zd bytes: all ints have %zd bytes.\n", n, sizeof n, intsize);
  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    $0x4,%ecx
  mov    $0x4,%edx
  mov    $0x0,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D5: n = %d, n as %zd bytes: all ints have %zd bytes.


assistant: ```c
int main(int argc, char const *argv[]) {
  int n = 0;
  size_t intsize;
  intsize = sizeof(int);
  printf("n = %d, n as %zd bytes: all ints have %zd bytes.\n", n, sizeof n, intsize);
  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    $0x4,%ecx
  mov    $0x4,%edx
  xor    %esi,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type_guessed: string

tool: D1: n = %d, n as %zd bytes: all ints have %zd bytes.


assistant: ```c
int main(int argc, char const *argv[]) {
  int n = 0;
  size_t intsize;
  intsize = sizeof(int);
  printf("n = %d, n as %zd bytes: all ints have %zd bytes.\n", n, sizeof n, intsize);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%ecx
  mov    $0x4,%edx
  xor    %esi,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D7
type_infered: string

tool: D7: n = %d, n as %zd bytes: all ints have %zd bytes.


assistant: ```c
int main(int argc, char const *argv[]) {
  int n = 0;
  size_t intsize;
  intsize = sizeof(int);
  printf("n = %d, n as %zd bytes: all ints have %zd bytes.\n", n, sizeof n, intsize);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x7,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <fibo@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xd,-0x4(%rbp)
  je     L3
  mov    $0x0,%eax
  call   <__VERIFIER_error@plt>
L3:
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int x = 7;
  int result = fibo(x);
  if (result != 13) {
    __VERIFIER_error();
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x7,%edi
  call   <fibo@plt>
  cmp    $0xd,%eax
  jne    L2
L15:
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L2:
  mov    $0x0,%eax
  call   <__VERIFIER_error@plt>
  jmp    L15
```

assistant: ```c
int main() {
  int x = 7;
  int result = fibo(x);
  if (result != 13) {
    __VERIFIER_error();
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x7,%edi
  call   <fibo@plt>
  cmp    $0xd,%eax
  je     L3
  xor    %eax,%eax
  call   <__VERIFIER_error@plt>
L3:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x = 7;
  int result = fibo(x);
  if (result != 13) {
    __VERIFIER_error();
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x7,%edi
  call   <fibo@plt>
  cmp    $0xd,%eax
  je     L8
  xor    %eax,%eax
  call   <__VERIFIER_error@plt>
L8:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x = 7;
  int result = fibo(x);
  if (result != 13) {
    __VERIFIER_error();
  }
  return 0;
}
```