ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<displaySetChPitch>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %ecx,%eax
  mov    %r8d,%r10d
  mov    %r9d,%r8d
  mov    0x10(%rbp),%ecx
  mov    %dil,-0x4(%rbp)
  mov    %sil,-0x8(%rbp)
  mov    %dl,-0xc(%rbp)
  mov    %al,-0x10(%rbp)
  mov    %r10d,%eax
  mov    %al,-0x14(%rbp)
  mov    %r8d,%eax
  mov    %al,-0x18(%rbp)
  mov    %ecx,%eax
  mov    %al,-0x1c(%rbp)
  push   $0x1
  push   $0x0
  push   $0x0
  push   $0x0
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <displaySetPins@plt>
  add    $0x20,%rsp
  movsbl -0x1c(%rbp),%edi
  movsbl -0x18(%rbp),%esi
  movsbl -0x14(%rbp),%ecx
  movsbl -0x10(%rbp),%r9d
  movsbl -0xc(%rbp),%r8d
  movsbl -0x8(%rbp),%edx
  movsbl -0x4(%rbp),%eax
  push   %rdi
  push   %rsi
  push   %rcx
  push   $0x0
  mov    %edx,%ecx
  mov    %eax,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <displaySetPins@plt>
  add    $0x20,%rsp
  mov    $0x0,%eax
  call   <displayWriteData@plt>
  nop
  leave
  ret
```

assistant: ```c
void displaySetChPitch(char DB7, char DB6, char DB5, char DB4, char DB2, char DB1, char DB0) {
  displaySetPins(0, 1, 0, 0, 0, 0, 0, 0, 0, 1);
  displaySetPins(0, 0, DB7, DB6, DB5, DB4, 0, DB2, DB1, DB0);
  displayWriteData();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<displaySetChPitch>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %edx,%r13d
  mov    %ecx,%r14d
  mov    %r8d,0xc(%rsp)
  mov    %r9d,%r12d
  mov    0x50(%rsp),%r15d
  push   $0x1
  push   $0x0
  push   $0x0
  push   $0x0
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <displaySetPins@plt>
  add    $0x20,%rsp
  movsbl %bpl,%ecx
  movsbl %bl,%edx
  movsbl %r15b,%r15d
  push   %r15
  movsbl %r12b,%r12d
  push   %r12
  movsbl 0x1c(%rsp),%eax
  push   %rax
  push   $0x0
  movsbl %r14b,%r9d
  movsbl %r13b,%r8d
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <displaySetPins@plt>
  add    $0x20,%rsp
  mov    $0x0,%eax
  call   <displayWriteData@plt>
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void displaySetChPitch(char DB7, char DB6, char DB5, char DB4, char DB2, char DB1, char DB0) {
  displaySetPins(0, 1, 0, 0, 0, 0, 0, 0, 0, 1);
  displaySetPins(0, 0, DB7, DB6, DB5, DB4, 0, DB2, DB1, DB0);
  displayWriteData();
}
```
user: What is the c source code of the assembly code below:

```asm
<displaySetChPitch>:
  endbr64
  push   %r15
  push   %r14
  mov    %ecx,%r14d
  xor    %ecx,%ecx
  push   %r13
  mov    %edx,%r13d
  xor    %edx,%edx
  push   %r12
  mov    %r9d,%r12d
  xor    %r9d,%r9d
  push   %rbp
  mov    %esi,%ebp
  mov    $0x1,%esi
  movsbl %r12b,%r12d
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  sub    $0x18,%rsp
  mov    %r8d,0xc(%rsp)
  mov    0x50(%rsp),%r15d
  xor    %r8d,%r8d
  push   $0x1
  push   $0x0
  movsbl %r15b,%r15d
  push   $0x0
  push   $0x0
  call   <displaySetPins@plt>
  add    $0x20,%rsp
  movsbl %bpl,%ecx
  movsbl %bl,%edx
  push   %r15
  movsbl %r14b,%r9d
  movsbl %r13b,%r8d
  xor    %esi,%esi
  push   %r12
  movsbl 0x1c(%rsp),%eax
  xor    %edi,%edi
  push   %rax
  push   $0x0
  call   <displaySetPins@plt>
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <displayWriteData@plt>
```

assistant: ```c
void displaySetChPitch(char DB7, char DB6, char DB5, char DB4, char DB2, char DB1, char DB0) {
  displaySetPins(0, 1, 0, 0, 0, 0, 0, 0, 0, 1);
  displaySetPins(0, 0, DB7, DB6, DB5, DB4, 0, DB2, DB1, DB0);
  displayWriteData();
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<displaySetChPitch>:
  endbr64
  push   %r15
  push   %r14
  mov    %ecx,%r14d
  xor    %ecx,%ecx
  push   %r13
  mov    %edx,%r13d
  xor    %edx,%edx
  push   %r12
  mov    %r9d,%r12d
  xor    %r9d,%r9d
  push   %rbp
  mov    %esi,%ebp
  mov    $0x1,%esi
  movsbl %r12b,%r12d
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  sub    $0x18,%rsp
  mov    %r8d,0xc(%rsp)
  mov    0x50(%rsp),%r15d
  xor    %r8d,%r8d
  push   $0x1
  push   $0x0
  movsbl %r15b,%r15d
  push   $0x0
  push   $0x0
  call   <displaySetPins@plt>
  add    $0x20,%rsp
  movsbl %bpl,%ecx
  movsbl %bl,%edx
  push   %r15
  movsbl %r14b,%r9d
  movsbl %r13b,%r8d
  xor    %esi,%esi
  push   %r12
  movsbl 0x1c(%rsp),%eax
  xor    %edi,%edi
  push   %rax
  push   $0x0
  call   <displaySetPins@plt>
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <displayWriteData@plt>
```

assistant: ```c
void displaySetChPitch(char DB7, char DB6, char DB5, char DB4, char DB2, char DB1, char DB0) {
  displaySetPins(0, 1, 0, 0, 0, 0, 0, 0, 0, 1);
  displaySetPins(0, 0, DB7, DB6, DB5, DB4, 0, DB2, DB1, DB0);
  displayWriteData();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ft_toupper>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <ft_islower@plt>
  test   %rax,%rax
  je     L19
  mov    -0x4(%rbp),%eax
  sub    $0x20,%eax
  jmp    L16
L19:
  mov    -0x4(%rbp),%eax
L16:
  leave
  ret
```

assistant: ```c
int ft_toupper(int c) {
  if (ft_islower(c))
    return (c - 32);
  else
    return (c);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ft_toupper>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <ft_islower@plt>
  mov    %rax,%rdx
  lea    -0x20(%rbx),%eax
  test   %rdx,%rdx
  cmovne %eax,%ebx
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int ft_toupper(int c) {
  if (ft_islower(c))
    return (c - 32);
  else
    return (c);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ft_toupper>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  call   <ft_islower@plt>
  mov    %rax,%r8
  lea    -0x20(%r12),%eax
  test   %r8,%r8
  cmovne %eax,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int ft_toupper(int c) {
  if (ft_islower(c))
    return (c - 32);
  else
    return (c);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_toupper>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  call   <ft_islower@plt>
  mov    %rax,%r8
  lea    -0x20(%r12),%eax
  test   %r8,%r8
  cmovne %eax,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int ft_toupper(int c) {
  if (ft_islower(c))
    return (c - 32);
  else
    return (c);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L14
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L44
L14:
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L34
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L44
L34:
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <is_prime@plt>
  mov    $0x0,%eax
  jmp    L44
L44:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L48
  call   <__stack_chk_fail@plt>
L48:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D30
type_infered: string
name: retrieve_data
arguments: data_tag: D42
type_infered: string
name: retrieve_data
arguments: data_tag: D32
type_infered: string
name: retrieve_data
arguments: data_tag: D23
type_infered: string

tool: D30: Is your number less than 0? If yes - input 1; If no - input any other number: 

tool: D42: %d

tool: D32: -1

tool: D23: 
Insert your number - 

assistant: ```c
int main(void) {
  int x, flag;
  flag = 0;
  printf("Is your number less than 0? If yes - input 1; If no - input any other number: ");
  if (scanf("%d", &flag) != 1) {
    printf("-1");
    return;
  }
  printf("\nInsert your number - ");
  if (scanf("%d", &x) != 1) {
    printf("-1");
    return;
  }
  is_prime(x, flag);
}
```
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
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    D49(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D57(%rip),%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L49
  lea    D33(%rip),%rdi
  call   <printf@plt>
L32:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L40
  add    $0x18,%rsp
  ret
L49:
  lea    D54(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D57(%rip),%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L7
  lea    D33(%rip),%rdi
  call   <printf@plt>
  jmp    L32
L7:
  mov    0x4(%rsp),%esi
  mov    (%rsp),%edi
  call   <is_prime@plt>
  mov    $0x0,%eax
  jmp    L32
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D49
type_infered: string
name: retrieve_data
arguments: data_tag: D57
type_infered: string
name: retrieve_data
arguments: data_tag: D33
type_infered: string
name: retrieve_data
arguments: data_tag: D54
type_infered: string

tool: D49: Is your number less than 0? If yes - input 1; If no - input any other number: 

tool: D57: %d

tool: D33: -1

tool: D54: 
Insert your number - 

assistant: ```c
int main(void) {
  int x, flag;
  flag = 0;
  printf("Is your number less than 0? If yes - input 1; If no - input any other number: ");
  if (scanf("%d", &flag) != 1) {
    printf("-1");
    return;
  }
  printf("\nInsert your number - ");
  if (scanf("%d", &x) != 1) {
    printf("-1");
    return;
  }
  is_prime(x, flag);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D18(%rip),%rdi
  lea    D6(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L5
  lea    D21(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  sub    $0x1,%eax
  je     L32
L5:
  lea    D35(%rip),%rdi
  call   <printf@plt>
L39:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x10,%rsp
  pop    %rbp
  ret
L32:
  mov    0x4(%rsp),%esi
  mov    (%rsp),%edi
  call   <is_prime@plt>
  xor    %eax,%eax
  jmp    L39
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D18
type: string
name: retrieve_data
arguments: data_tag: D6
type: string
name: retrieve_data
arguments: data_tag: D21
type: string
name: retrieve_data
arguments: data_tag: D35
type: string

tool: D18: Is your number less than 0? If yes - input 1; If no - input any other number: 

tool: D6: %d

tool: D21: 
Insert your number - 

tool: D35: -1

assistant: ```c
int main(void) {
  int x, flag;
  flag = 0;
  printf("Is your number less than 0? If yes - input 1; If no - input any other number: ");
  if (scanf("%d", &flag) != 1) {
    printf("-1");
    return;
  }
  printf("\nInsert your number - ");
  if (scanf("%d", &x) != 1) {
    printf("-1");
    return;
  }
  is_prime(x, flag);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D19(%rip),%rdi
  lea    D18(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L1
  lea    D34(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  sub    $0x1,%eax
  je     L4
L1:
  lea    D17(%rip),%rdi
  call   <printf@plt>
L22:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L25
  add    $0x10,%rsp
  pop    %rbp
  ret
L4:
  mov    0x4(%rsp),%esi
  mov    (%rsp),%edi
  call   <is_prime@plt>
  xor    %eax,%eax
  jmp    L22
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D19
type_infered: string
name: read_data
arguments: data_label: D18
type_infered: string
name: read_data
arguments: data_label: D34
type_infered: string
name: read_data
arguments: data_label: D17
type_infered: string

tool: D19: Is your number less than 0? If yes - input 1; If no - input any other number: 

tool: D18: %d

tool: D34: 
Insert your number - 

tool: D17: -1

assistant: ```c
int main(void) {
  int x, flag;
  flag = 0;
  printf("Is your number less than 0? If yes - input 1; If no - input any other number: ");
  if (scanf("%d", &flag) != 1) {
    printf("-1");
    return;
  }
  printf("\nInsert your number - ");
  if (scanf("%d", &x) != 1) {
    printf("-1");
    return;
  }
  is_prime(x, flag);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<tests>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <cmps@plt>
  mov    $0x0,%eax
  call   <nanz@plt>
  mov    $0x0,%eax
  call   <adds@plt>
  mov    $0x0,%eax
  call   <muls@plt>
  mov    $0x0,%eax
  call   <divs@plt>
  mov    $0x0,%eax
  call   <cvts@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void tests(void) {
  cmps();
  nanz();
  adds();
  muls();
  divs();
  cvts();
}
```
user: Translate the below assembly code into C programming language:

```asm
<tests>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <cmps@plt>
  mov    $0x0,%eax
  call   <nanz@plt>
  mov    $0x0,%eax
  call   <adds@plt>
  mov    $0x0,%eax
  call   <muls@plt>
  mov    $0x0,%eax
  call   <divs@plt>
  mov    $0x0,%eax
  call   <cvts@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void tests(void) {
  cmps();
  nanz();
  adds();
  muls();
  divs();
  cvts();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<tests>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <cmps@plt>
  xor    %eax,%eax
  call   <nanz@plt>
  xor    %eax,%eax
  call   <adds@plt>
  xor    %eax,%eax
  call   <muls@plt>
  xor    %eax,%eax
  call   <divs@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10c0 <cvts@plt>
```

assistant: ```c
void tests(void) {
  cmps();
  nanz();
  adds();
  muls();
  divs();
  cvts();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<tests>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <cmps@plt>
  xor    %eax,%eax
  call   <nanz@plt>
  xor    %eax,%eax
  call   <adds@plt>
  xor    %eax,%eax
  call   <muls@plt>
  xor    %eax,%eax
  call   <divs@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10c0 <cvts@plt>
```

assistant: ```c
void tests(void) {
  cmps();
  nanz();
  adds();
  muls();
  divs();
  cvts();
}
```
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    $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
L1:
  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    L16
L3:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  lea    D18(%rip),%rax
  mov    %rax,%rsi
  lea    D12(%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    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L16:
  cmpl   $0x2,-0x14(%rbp)
  jle    L3
  addl   $0x1,-0x18(%rbp)
L28:
  cmpl   $0x6,-0x18(%rbp)
  jle    L1
  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: extract_symbolic_data
arguments: data_tag: D18
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D12
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D28
type_guessed: string

tool: D18: sminabs

tool: D12: # c::%s:len=%d


tool: D28: 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", "sminabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```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    D28(%rip),%r15
  jmp    L12
L11:
  add    $0x1,%r14d
  add    $0x3,%r12d
  cmp    $0x15,%r12d
  je     L16
L12:
  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
L35:
  add    $0x1,%ebx
  mov    %ebp,%edx
  mov    %r15,%rsi
  lea    D16(%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    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L35
  jmp    L11
L16:
  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: parse_data
arguments: data_tag: D28
type_infered: string
name: parse_data
arguments: data_tag: D16
type_infered: string
name: parse_data
arguments: data_tag: D26
type_infered: string

tool: D28: sminabs

tool: D16: # c::%s:len=%d


tool: D26: 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", "sminabs", 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
  xor    %edi,%edi
  lea    D9(%rip),%r15
  push   %r14
  lea    D12(%rip),%r14
  push   %r13
  lea    D14(%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)
L8:
  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
L11:
  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    L11
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L8
  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: retrieve_data
arguments: label: D9
datatype: string
name: retrieve_data
arguments: label: D12
datatype: string
name: retrieve_data
arguments: label: D14
datatype: string

tool: D9: sminabs

tool: D12: # c::%s:len=%d


tool: D14: 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", "sminabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D17(%rip),%r15
  push   %r14
  lea    D9(%rip),%r14
  push   %r13
  lea    D7(%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)
L9:
  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
L5:
  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    L5
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L9
  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: disassemble_data
arguments: label: D17
type_guessed: string
name: disassemble_data
arguments: label: D9
type_guessed: string
name: disassemble_data
arguments: label: D7
type_guessed: string

tool: D17: sminabs

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


tool: D7: 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", "sminabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```