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

```asm
<print_graph>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  movslq %esi,%r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  push   %rbx
  mov    %r15,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    0xf(,%r15,4),%rax
  mov    %rsp,%rsi
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L7
L37:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L37
L7:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L9
L50:
  mov    %rsp,%r13
  mov    %edx,%ecx
  mov    %ebx,%esi
  mov    %r14,%rdi
  mov    %r13,%rdx
  call   <calc_bars@plt>
  test   %ebx,%ebx
  jle    L26
  xor    %ebx,%ebx
  lea    D5(%rip),%r12
  nop
L39:
  mov    (%r14,%rbx,4),%ecx
  mov    %ebx,%esi
  test   %ecx,%ecx
  jle    L55
  mov    %ebx,%edx
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x0(%r13,%rbx,4),%edi
  call   <print_bar@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
L55:
  add    $0x1,%rbx
  cmp    %r15,%rbx
  jne    L39
L26:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nop
L9:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L50
L53:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
type_infered: string

tool: D5: '%c' (%3d) %4d 

assistant: ```c
void print_graph(int values[], int size, int max_bar_size) {
  int bars[size];
  calc_bars(values, size, bars, max_bar_size);
  int i;
  for (i = 0; i < size; i++) {
    if (values[i] > 0) {
      printf("'%c' (%3d) %4d ", i, i, values[i]);
      print_bar(bars[i]);
      putchar('\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
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L28:
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D82(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D86(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D83(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  cmp    $0x6,%eax
  ja     L61
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D37(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D37(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x0,%eax
  call   <push1@plt>
  jmp    L61
  mov    $0x0,%eax
  call   <push2@plt>
  jmp    L61
  mov    $0x0,%eax
  call   <pop1@plt>
  jmp    L61
  mov    $0x0,%eax
  call   <pop2@plt>
  jmp    L61
  mov    $0x0,%eax
  call   <display@plt>
  nop
L61:
  mov    -0xc(%rbp),%eax
  cmp    $0x6,%eax
  jne    L28
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L57
  call   <__stack_chk_fail@plt>
L57:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D54
type: string
name: retrieve_data
arguments: data_tag: D35
type: string
name: retrieve_data
arguments: data_tag: D82
type: string
name: retrieve_data
arguments: data_tag: D16
type: string
name: retrieve_data
arguments: data_tag: D52
type: string
name: retrieve_data
arguments: data_tag: D86
type: string
name: retrieve_data
arguments: data_tag: D83
type: string

tool: D54: 
Push in Stack 1


tool: D35: Push in stack 2


tool: D82: Pop from stack 1


tool: D16: Pop from stack 2


tool: D52: Display both Stack


tool: D86: Exit


tool: D83: %d

assistant: ```c
void main() {
  int n;
  do {
    printf("\nPush in Stack 1\n");
    printf("Push in stack 2\n");
    printf("Pop from stack 1\n");
    printf("Pop from stack 2\n");
    printf("Display both Stack\n");
    printf("Exit\n");
    scanf("%d", &n);
    switch (n) {
    case 1: {
      push1();
      break;
    }
    case 2: {
      push2();
      break;
    }
    case 3: {
      pop1();
      break;
    }
    case 4: {
      pop2();
      break;
    }
    case 5: {
      display();
      break;
    }
    case 6: {
      break;
    }
    }
  } while (n != 6);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D9(%rip),%r13
  lea    D0(%rip),%r12
  lea    D12(%rip),%rbp
  lea    D24(%rip),%rbx
  jmp    L7
  mov    $0x0,%eax
  call   <push1@plt>
L39:
  cmpl   $0x6,0x4(%rsp)
  je     L71
L7:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    D14(%rip),%rdi
  call   <printf@plt>
  lea    D60(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D41(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x5,0x4(%rsp)
  ja     L39
  mov    0x4(%rsp),%eax
  lea    D61(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x0,%eax
  call   <push2@plt>
  jmp    L39
  mov    $0x0,%eax
  call   <pop1@plt>
  jmp    L39
  mov    $0x0,%eax
  call   <pop2@plt>
  jmp    L39
  mov    $0x0,%eax
  call   <display@plt>
  jmp    L39
L71:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D9
type_infered: string
name: parse_data
arguments: data_label: D0
type_infered: string
name: parse_data
arguments: data_label: D12
type_infered: string
name: parse_data
arguments: data_label: D24
type_infered: string
name: parse_data
arguments: data_label: D14
type_infered: string
name: parse_data
arguments: data_label: D60
type_infered: string
name: parse_data
arguments: data_label: D41
type_infered: string

tool: D9: 
Push in Stack 1


tool: D0: Push in stack 2


tool: D12: Pop from stack 1


tool: D24: Pop from stack 2


tool: D14: Display both Stack


tool: D60: Exit


tool: D41: %d

assistant: ```c
void main() {
  int n;
  do {
    printf("\nPush in Stack 1\n");
    printf("Push in stack 2\n");
    printf("Pop from stack 1\n");
    printf("Pop from stack 2\n");
    printf("Display both Stack\n");
    printf("Exit\n");
    scanf("%d", &n);
    switch (n) {
    case 1: {
      push1();
      break;
    }
    case 2: {
      push2();
      break;
    }
    case 3: {
      pop1();
      break;
    }
    case 4: {
      pop2();
      break;
    }
    case 5: {
      display();
      break;
    }
    case 6: {
      break;
    }
    }
  } while (n != 6);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  lea    D32(%rip),%r15
  push   %r14
  lea    D57(%rip),%r14
  push   %r13
  lea    D7(%rip),%r13
  push   %r12
  lea    D78(%rip),%r12
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  lea    D13(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rax
  mov    %rax,0x8(%rsp)
  cs nopw 0x0(%rax,%rax,1)
L59:
  mov    %r15,%rdi
  call   <printf@plt>
  mov    %r14,%rdi
  call   <printf@plt>
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    D79(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rsi
  lea    D46(%rip),%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  cmp    $0x5,%eax
  ja     L46
  mov    %eax,%edx
  movslq (%rbx,%rdx,4),%rdx
  add    %rbx,%rdx
  notrack jmp *%rdx
  nopw   0x0(%rax,%rax,1)
  xor    %eax,%eax
  call   <display@plt>
  mov    0x14(%rsp),%eax
L46:
  cmp    $0x6,%eax
  jne    L59
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
  xor    %eax,%eax
  call   <pop2@plt>
  mov    0x14(%rsp),%eax
  jmp    L46
  nopl   (%rax)
  xor    %eax,%eax
  call   <pop1@plt>
  mov    0x14(%rsp),%eax
  jmp    L46
  nopl   (%rax)
  xor    %eax,%eax
  call   <push1@plt>
  mov    0x14(%rsp),%eax
  jmp    L46
  nopl   (%rax)
  xor    %eax,%eax
  call   <push2@plt>
  mov    0x14(%rsp),%eax
  jmp    L46
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D32
type_infered: string
name: retrieve_data
arguments: tag: D57
type_infered: string
name: retrieve_data
arguments: tag: D7
type_infered: string
name: retrieve_data
arguments: tag: D78
type_infered: string
name: retrieve_data
arguments: tag: D8
type_infered: string
name: retrieve_data
arguments: tag: D79
type_infered: string
name: retrieve_data
arguments: tag: D46
type_infered: string

tool: D32: 
Push in Stack 1


tool: D57: Push in stack 2


tool: D7: Pop from stack 1


tool: D78: Pop from stack 2


tool: D8: Display both Stack


tool: D79: Exit


tool: D46: %d

assistant: ```c
void main() {
  int n;
  do {
    printf("\nPush in Stack 1\n");
    printf("Push in stack 2\n");
    printf("Pop from stack 1\n");
    printf("Pop from stack 2\n");
    printf("Display both Stack\n");
    printf("Exit\n");
    scanf("%d", &n);
    switch (n) {
    case 1: {
      push1();
      break;
    }
    case 2: {
      push2();
      break;
    }
    case 3: {
      pop1();
      break;
    }
    case 4: {
      pop2();
      break;
    }
    case 5: {
      display();
      break;
    }
    case 6: {
      break;
    }
    }
  } while (n != 6);
}
```
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   %r15
  lea    D33(%rip),%r15
  push   %r14
  lea    D37(%rip),%r14
  push   %r13
  lea    D50(%rip),%r13
  push   %r12
  lea    D56(%rip),%r12
  push   %rbp
  lea    D74(%rip),%rbp
  push   %rbx
  lea    D22(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rax
  mov    %rax,0x8(%rsp)
  cs nopw 0x0(%rax,%rax,1)
L10:
  mov    %r15,%rdi
  call   <printf@plt>
  mov    %r14,%rdi
  call   <printf@plt>
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    D60(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rsi
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  cmp    $0x5,%eax
  ja     L5
  mov    %eax,%edx
  movslq (%rbx,%rdx,4),%rdx
  add    %rbx,%rdx
  notrack jmp *%rdx
  nopw   0x0(%rax,%rax,1)
  xor    %eax,%eax
  call   <display@plt>
  mov    0x14(%rsp),%eax
L5:
  cmp    $0x6,%eax
  jne    L10
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L49
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
  xor    %eax,%eax
  call   <pop2@plt>
  mov    0x14(%rsp),%eax
  jmp    L5
  nopl   (%rax)
  xor    %eax,%eax
  call   <pop1@plt>
  mov    0x14(%rsp),%eax
  jmp    L5
  nopl   (%rax)
  xor    %eax,%eax
  call   <push1@plt>
  mov    0x14(%rsp),%eax
  jmp    L5
  nopl   (%rax)
  xor    %eax,%eax
  call   <push2@plt>
  mov    0x14(%rsp),%eax
  jmp    L5
L49:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D33
datatype: string
name: unpack_data
arguments: label: D37
datatype: string
name: unpack_data
arguments: label: D50
datatype: string
name: unpack_data
arguments: label: D56
datatype: string
name: unpack_data
arguments: label: D74
datatype: string
name: unpack_data
arguments: label: D60
datatype: string
name: unpack_data
arguments: label: D16
datatype: string

tool: D33: 
Push in Stack 1


tool: D37: Push in stack 2


tool: D50: Pop from stack 1


tool: D56: Pop from stack 2


tool: D74: Display both Stack


tool: D60: Exit


tool: D16: %d

assistant: ```c
void main() {
  int n;
  do {
    printf("\nPush in Stack 1\n");
    printf("Push in stack 2\n");
    printf("Pop from stack 1\n");
    printf("Pop from stack 2\n");
    printf("Display both Stack\n");
    printf("Exit\n");
    scanf("%d", &n);
    switch (n) {
    case 1: {
      push1();
      break;
    }
    case 2: {
      push2();
      break;
    }
    case 3: {
      pop1();
      break;
    }
    case 4: {
      pop2();
      break;
    }
    case 5: {
      display();
      break;
    }
    case 6: {
      break;
    }
    }
  } while (n != 6);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<CRCResultRead>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x44030000,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <ASSERT@plt>
  cmpl   $0x0,-0x8(%rbp)
  je     L16
  mov    -0x4(%rbp),%eax
  add    $0x418,%eax
  cltq
  mov    (%rax),%eax
  jmp    L0
L16:
  mov    -0x4(%rbp),%eax
  add    $0x410,%eax
  cltq
  mov    (%rax),%eax
L0:
  leave
  ret
```

assistant: ```c
uint32_t CRCResultRead(uint32_t ui32Base, bool bPPResult) {
  ASSERT(ui32Base == 0x44030000);
  if (bPPResult) {
    return ((*((volatile uint32_t *)(ui32Base + 0x00000418))));
  } else {
    return ((*((volatile uint32_t *)(ui32Base + 0x00000410))));
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<CRCResultRead>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  cmp    $0x44030000,%edi
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  test   %ebp,%ebp
  je     L9
  add    $0x418,%ebx
  movslq %ebx,%rbx
  mov    (%rbx),%eax
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  add    $0x410,%ebx
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  jmp    L17
```

assistant: ```c
uint32_t CRCResultRead(uint32_t ui32Base, bool bPPResult) {
  ASSERT(ui32Base == 0x44030000);
  if (bPPResult) {
    return ((*((volatile uint32_t *)(ui32Base + 0x00000418))));
  } else {
    return ((*((volatile uint32_t *)(ui32Base + 0x00000410))));
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<CRCResultRead>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  cmp    $0x44030000,%edi
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  test   %ebp,%ebp
  je     L6
  add    $0x418,%ebx
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  add    $0x410,%ebx
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
uint32_t CRCResultRead(uint32_t ui32Base, bool bPPResult) {
  ASSERT(ui32Base == 0x44030000);
  if (bPPResult) {
    return ((*((volatile uint32_t *)(ui32Base + 0x00000418))));
  } else {
    return ((*((volatile uint32_t *)(ui32Base + 0x00000410))));
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<CRCResultRead>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  cmp    $0x44030000,%edi
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  test   %ebp,%ebp
  je     L2
  add    $0x418,%ebx
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  add    $0x410,%ebx
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
uint32_t CRCResultRead(uint32_t ui32Base, bool bPPResult) {
  ASSERT(ui32Base == 0x44030000);
  if (bPPResult) {
    return ((*((volatile uint32_t *)(ui32Base + 0x00000418))));
  } else {
    return ((*((volatile uint32_t *)(ui32Base + 0x00000410))));
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Whole>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  sub    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int Whole(int N) {
  int num;
  num = ((N / 10) * 10);
  return (num);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Whole>:
  endbr64
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edi
  sub    %edi,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  ret
```

assistant: ```c
int Whole(int N) {
  int num;
  num = ((N / 10) * 10);
  return (num);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<Whole>:
  endbr64
  movslq %edi,%rax
  sar    $0x1f,%edi
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sub    %edi,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  ret
```

assistant: ```c
int Whole(int N) {
  int num;
  num = ((N / 10) * 10);
  return (num);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<Whole>:
  endbr64
  movslq %edi,%rax
  sar    $0x1f,%edi
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sub    %edi,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  ret
```

assistant: ```c
int Whole(int N) {
  int num;
  num = ((N / 10) * 10);
  return (num);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L28
L10:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0xf4240,%eax
  cltd
  idiv   %ecx
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L4
L24:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  lea    D29(%rip),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%edi
  call   <print_results@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L4:
  cmpl   $0x2,-0x14(%rbp)
  jle    L24
  addl   $0x1,-0x18(%rbp)
L28:
  cmpl   $0x6,-0x18(%rbp)
  jle    L10
  mov    -0x1c(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D29
type_infered: string
name: parse_data
arguments: data_label: D10
type_infered: string
name: parse_data
arguments: data_label: D17
type_infered: string

tool: D29: sstdevpn

tool: D10: # c::%s:len=%d


tool: D17: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "sstdevpn", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  mov    $0x3,%r12d
  mov    $0x1,%r14d
  lea    D3(%rip),%r15
  jmp    L13
L21:
  add    $0x1,%r14d
  add    $0x3,%r12d
  cmp    $0x15,%r12d
  je     L11
L13:
  lea    -0x3(%r12),%ebx
  mov    %r14d,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ebp
  lea    -0x1(%r14),%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0xf4240,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r13d
L23:
  add    $0x1,%ebx
  mov    %ebp,%edx
  mov    %r15,%rsi
  lea    D21(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %r13d,%edi
  call   <benchmark@plt>
  mov    %r13d,%edi
  call   <print_results@plt>
  mov    %ebx,%esi
  lea    D24(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L23
  jmp    L21
L11:
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
data_type: string
name: disassemble_data
arguments: data_tag: D21
data_type: string
name: disassemble_data
arguments: data_tag: D24
data_type: string

tool: D3: sstdevpn

tool: D21: # c::%s:len=%d


tool: D24: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "sstdevpn", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D10(%rip),%r15
  push   %r14
  lea    D9(%rip),%r14
  push   %r13
  lea    D12(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L13:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0xf4240,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L2:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L2
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L13
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: tag: D10
type: string
name: read_data
arguments: tag: D9
type: string
name: read_data
arguments: tag: D12
type: string

tool: D10: sstdevpn

tool: D9: # c::%s:len=%d


tool: D12: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "sstdevpn", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```