ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Auth>:
  endbr64
  push   %r13
  lea    D25(%rip),%rdi
  lea    D24(%rip),%r13
  push   %r12
  push   %rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  lea    0x30(%rsp),%r12
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  lea    0x7(%rsp),%rdx
  mov    %r12,%rcx
  mov    %rbp,%rsi
  movabs $0x64726f7773736150,%rax
  lea    0x2(%rsp),%rdi
  movb   $0x0,0x6(%rsp)
  movl   $0x48545541,0x2(%rsp)
  mov    %rax,0x7(%rsp)
  movb   $0x0,0xf(%rsp)
  call   <SendMessages@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x50,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D25
datatype: string
name: disassemble_data
arguments: data_tag: D24
datatype: string
name: disassemble_data
arguments: data_tag: D3
datatype: string

tool: D25: Username: 

tool: D24: %s

tool: D3: Password: 

assistant: ```c
void Auth() {
  char username[20];
  char password[20];
  printf("Username: ");
  scanf("%s", username);
  printf("Password: ");
  scanf("%s", password);
  char operation[] = "AUTH";
  char key[] = "Password";
  SendMessages(operation, username, key, password);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Auth>:
  endbr64
  push   %r13
  lea    D7(%rip),%rdi
  lea    D5(%rip),%r13
  push   %r12
  push   %rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  lea    0x30(%rsp),%r12
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  lea    D21(%rip),%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  lea    0x7(%rsp),%rdx
  mov    %r12,%rcx
  mov    %rbp,%rsi
  movabs $0x64726f7773736150,%rax
  lea    0x2(%rsp),%rdi
  movb   $0x0,0x6(%rsp)
  movl   $0x48545541,0x2(%rsp)
  mov    %rax,0x7(%rsp)
  movb   $0x0,0xf(%rsp)
  call   <SendMessages@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x50,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D7
type: string
name: disassemble_data
arguments: label: D5
type: string
name: disassemble_data
arguments: label: D21
type: string

tool: D7: Username: 

tool: D5: %s

tool: D21: Password: 

assistant: ```c
void Auth() {
  char username[20];
  char password[20];
  printf("Username: ");
  scanf("%s", username);
  printf("Password: ");
  scanf("%s", password);
  char operation[] = "AUTH";
  char key[] = "Password";
  SendMessages(operation, username, key, password);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void swap(int permutation[], int idx1, int idx2) {
  int tmp;
  tmp = permutation[idx1];
  permutation[idx1] = permutation[idx2];
  permutation[idx2] = tmp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<swap>:
  endbr64
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rcx
  mov    (%rcx),%esi
  movslq %edx,%rdx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
  ret
```

assistant: ```c
void swap(int permutation[], int idx1, int idx2) {
  int tmp;
  tmp = permutation[idx1];
  permutation[idx1] = permutation[idx2];
  permutation[idx2] = tmp;
}
```
user: Translate the below assembly code into C programming language:

```asm
<swap>:
  endbr64
  movslq %esi,%rsi
  movslq %edx,%rdx
  lea    (%rdi,%rsi,4),%rcx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rcx),%esi
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
  ret
```

assistant: ```c
void swap(int permutation[], int idx1, int idx2) {
  int tmp;
  tmp = permutation[idx1];
  permutation[idx1] = permutation[idx2];
  permutation[idx2] = tmp;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<swap>:
  endbr64
  movslq %esi,%rsi
  movslq %edx,%rdx
  lea    (%rdi,%rsi,4),%rcx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rcx),%esi
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
  ret
```

assistant: ```c
void swap(int permutation[], int idx1, int idx2) {
  int tmp;
  tmp = permutation[idx1];
  permutation[idx1] = permutation[idx2];
  permutation[idx2] = tmp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<choose>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  js     L41
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L8
L41:
  mov    $0x0,%eax
  jmp    L22
L8:
  cmpl   $0x0,-0x18(%rbp)
  je     L55
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jne    L29
L55:
  mov    $0x1,%eax
  jmp    L22
L29:
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jle    L83
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,-0x18(%rbp)
L83:
  movl   $0x1,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L79
L34:
  mov    -0x14(%rbp),%eax
  sub    -0x4(%rbp),%eax
  imul   -0x8(%rbp),%eax
  mov    -0x4(%rbp),%edx
  lea    0x1(%rdx),%ecx
  cltd
  idiv   %ecx
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L79:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L34
  mov    -0x8(%rbp),%eax
L22:
  pop    %rbp
  ret
```

assistant: ```c
int choose(int n, int k) {
  if (k < 0 || k > n)
    return 0;
  else if (k == 0 || k == n)
    return 1;
  else {
    if (k > n - k)
      k = n - k;
    int c = 1;
    int i;
    for (i = 0; i < k; i++)
      c = c * (n - i) / (i + 1);
    return c;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<choose>:
  endbr64
  test   %esi,%esi
  js     L20
  cmp    %edi,%esi
  jg     L20
  test   %esi,%esi
  je     L45
  cmp    %edi,%esi
  je     L45
  mov    %edi,%eax
  sub    %esi,%eax
  cmp    %esi,%eax
  cmovle %eax,%esi
  test   %esi,%esi
  jle    L3
  mov    $0x0,%ecx
  mov    $0x1,%eax
L37:
  mov    %edi,%edx
  sub    %ecx,%edx
  imul   %edx,%eax
  add    $0x1,%ecx
  cltd
  idiv   %ecx
  cmp    %esi,%ecx
  jne    L37
  ret
L20:
  mov    $0x0,%eax
  ret
L45:
  mov    $0x1,%eax
  ret
L3:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int choose(int n, int k) {
  if (k < 0 || k > n)
    return 0;
  else if (k == 0 || k == n)
    return 1;
  else {
    if (k > n - k)
      k = n - k;
    int c = 1;
    int i;
    for (i = 0; i < k; i++)
      c = c * (n - i) / (i + 1);
    return c;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<choose>:
  endbr64
  test   %esi,%esi
  js     L28
  cmp    %edi,%esi
  jg     L28
  test   %esi,%esi
  je     L42
  cmp    %edi,%esi
  je     L42
  mov    %edi,%eax
  sub    %esi,%eax
  cmp    %esi,%eax
  cmovle %eax,%esi
  xor    %ecx,%ecx
  mov    $0x1,%eax
  nopl   0x0(%rax)
L27:
  mov    %edi,%edx
  sub    %ecx,%edx
  add    $0x1,%ecx
  imul   %edx,%eax
  cltd
  idiv   %ecx
  cmp    %esi,%ecx
  jne    L27
  ret
  nopw   0x0(%rax,%rax,1)
L42:
  mov    $0x1,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L28:
  xor    %eax,%eax
  ret
```

assistant: ```c
int choose(int n, int k) {
  if (k < 0 || k > n)
    return 0;
  else if (k == 0 || k == n)
    return 1;
  else {
    if (k > n - k)
      k = n - k;
    int c = 1;
    int i;
    for (i = 0; i < k; i++)
      c = c * (n - i) / (i + 1);
    return c;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<choose>:
  endbr64
  test   %esi,%esi
  js     L38
  cmp    %edi,%esi
  jg     L38
  test   %esi,%esi
  je     L41
  cmp    %edi,%esi
  je     L41
  mov    %edi,%eax
  sub    %esi,%eax
  cmp    %esi,%eax
  cmovle %eax,%esi
  xor    %ecx,%ecx
  mov    $0x1,%eax
  nopl   0x0(%rax)
L17:
  mov    %edi,%edx
  sub    %ecx,%edx
  add    $0x1,%ecx
  imul   %edx,%eax
  cltd
  idiv   %ecx
  cmp    %esi,%ecx
  jne    L17
  ret
  nopw   0x0(%rax,%rax,1)
L41:
  mov    $0x1,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L38:
  xor    %eax,%eax
  ret
```

assistant: ```c
int choose(int n, int k) {
  if (k < 0 || k > n)
    return 0;
  else if (k == 0 || k == n)
    return 1;
  else {
    if (k > n - k)
      k = n - k;
    int c = 1;
    int i;
    for (i = 0; i < k; i++)
      c = c * (n - i) / (i + 1);
    return c;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```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)
  jmp    L16
L10:
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf_s@plt>
  addl   $0x1,-0xc(%rbp)
L16:
  cmpl   $0x9,-0xc(%rbp)
  jle    L10
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D20
datatype: string
name: parse_data
arguments: label: D12
datatype: string
name: parse_data
arguments: label: D10
datatype: string

tool: D20: Enter the grade: 

tool: D12: %d

tool: D10: pause

assistant: ```c
main() {
  int i = 0, student;
  while (i < 10) {
    printf("Enter the grade: ");
    scanf_s("%d", &student);
    i++;
  }
  system("pause");
}
```
user: 请将以下的汇编代码转换成C语言代码：

```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
  mov    $0xa,%ebx
  lea    D21(%rip),%r13
  lea    0x4(%rsp),%r12
  lea    D23(%rip),%rbp
L7:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf_s@plt>
  sub    $0x1,%ebx
  jne    L7
  lea    D9(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D21
data_type: string
name: extract_symbolic_data
arguments: data_label: D23
data_type: string
name: extract_symbolic_data
arguments: data_label: D9
data_type: string

tool: D21: Enter the grade: 

tool: D23: %d

tool: D9: pause

assistant: ```c
main() {
  int i = 0, student;
  while (i < 10) {
    printf("Enter the grade: ");
    scanf_s("%d", &student);
    i++;
  }
  system("pause");
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r13
  lea    D29(%rip),%r13
  push   %r12
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  cs nopw 0x0(%rax,%rax,1)
L16:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf_s@plt>
  sub    $0x1,%ebx
  jne    L16
  lea    D7(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D29
data_type: string
name: disassemble_data
arguments: tag: D6
data_type: string
name: disassemble_data
arguments: tag: D7
data_type: string

tool: D29: Enter the grade: 

tool: D6: %d

tool: D7: pause

assistant: ```c
main() {
  int i = 0, student;
  while (i < 10) {
    printf("Enter the grade: ");
    scanf_s("%d", &student);
    i++;
  }
  system("pause");
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  lea    D29(%rip),%r13
  push   %r12
  push   %rbp
  lea    D10(%rip),%rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  cs nopw 0x0(%rax,%rax,1)
L11:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf_s@plt>
  sub    $0x1,%ebx
  jne    L11
  lea    D14(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D29
type: string
name: parse_data
arguments: data_label: D10
type: string
name: parse_data
arguments: data_label: D14
type: string

tool: D29: Enter the grade: 

tool: D10: %d

tool: D14: pause

assistant: ```c
main() {
  int i = 0, student;
  while (i < 10) {
    printf("Enter the grade: ");
    scanf_s("%d", &student);
    i++;
  }
  system("pause");
}
```
user: Translate the below assembly code into C programming language:

```asm
<wep_seedkeygen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L14
L19:
  mov    -0x14(%rbp),%eax
  imul   $0x343fd,%eax,%eax
  mov    %eax,-0x14(%rbp)
  addl   $0x269ec3,-0x14(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x14(%rbp),%edx
  sar    $0x10,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x4(%rbp)
L14:
  cmpl   $0x13,-0x4(%rbp)
  jle    L19
  nop
  pop    %rbp
  ret
```

assistant: ```c
void wep_seedkeygen(int val, u_char *keys) {
  int i;
  for (i = 0; i < (5 * 4); i++) {
    val *= 0x343fd;
    val += 0x269ec3;
    keys[i] = val >> 16;
  }
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<wep_seedkeygen>:
  endbr64
  lea    0x50(%rsi),%rdx
L0:
  imul   $0x343fd,%edi,%edi
  add    $0x269ec3,%edi
  mov    %edi,%eax
  sar    $0x10,%eax
  mov    %eax,(%rsi)
  add    $0x4,%rsi
  cmp    %rdx,%rsi
  jne    L0
  ret
```

assistant: ```c
void wep_seedkeygen(int val, u_char *keys) {
  int i;
  for (i = 0; i < (5 * 4); i++) {
    val *= 0x343fd;
    val += 0x269ec3;
    keys[i] = val >> 16;
  }
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<wep_seedkeygen>:
  endbr64
  lea    0x50(%rsi),%rdx
  nopl   0x0(%rax,%rax,1)
L7:
  imul   $0x343fd,%edi,%edi
  add    $0x4,%rsi
  add    $0x269ec3,%edi
  mov    %edi,%eax
  sar    $0x10,%eax
  mov    %eax,-0x4(%rsi)
  cmp    %rdx,%rsi
  jne    L7
  ret
```

assistant: ```c
void wep_seedkeygen(int val, u_char *keys) {
  int i;
  for (i = 0; i < (5 * 4); i++) {
    val *= 0x343fd;
    val += 0x269ec3;
    keys[i] = val >> 16;
  }
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<wep_seedkeygen>:
  endbr64
  lea    0x50(%rsi),%rdx
  nopl   0x0(%rax,%rax,1)
L6:
  imul   $0x343fd,%edi,%edi
  add    $0x4,%rsi
  add    $0x269ec3,%edi
  mov    %edi,%eax
  sar    $0x10,%eax
  mov    %eax,-0x4(%rsi)
  cmp    %rsi,%rdx
  jne    L6
  ret
```

assistant: ```c
void wep_seedkeygen(int val, u_char *keys) {
  int i;
  for (i = 0; i < (5 * 4); i++) {
    val *= 0x343fd;
    val += 0x269ec3;
    keys[i] = val >> 16;
  }
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ADCSequenceOverflow>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  cltq
  and    %rdx,%rax
  pop    %rbp
  ret
```

assistant: ```c
long ADCSequenceOverflow(unsigned long ulBase, unsigned long ulSequenceNum) {
  ;
  ;
  return ((*((volatile unsigned long *)(ulBase + 0x00000010))) & (1 << ulSequenceNum));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ADCSequenceOverflow>:
  endbr64
  mov    0x10(%rdi),%rdx
  mov    $0x1,%eax
  mov    %esi,%ecx
  shl    %cl,%eax
  cltq
  and    %rdx,%rax
  ret
```

assistant: ```c
long ADCSequenceOverflow(unsigned long ulBase, unsigned long ulSequenceNum) {
  ;
  ;
  return ((*((volatile unsigned long *)(ulBase + 0x00000010))) & (1 << ulSequenceNum));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ADCSequenceOverflow>:
  endbr64
  mov    %esi,%ecx
  mov    $0x1,%eax
  mov    0x10(%rdi),%rdx
  shl    %cl,%eax
  cltq
  and    %rdx,%rax
  ret
```

assistant: ```c
long ADCSequenceOverflow(unsigned long ulBase, unsigned long ulSequenceNum) {
  ;
  ;
  return ((*((volatile unsigned long *)(ulBase + 0x00000010))) & (1 << ulSequenceNum));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ADCSequenceOverflow>:
  endbr64
  mov    %esi,%ecx
  mov    $0x1,%eax
  mov    0x10(%rdi),%rdx
  shl    %cl,%eax
  cltq
  and    %rdx,%rax
  ret
```

assistant: ```c
long ADCSequenceOverflow(unsigned long ulBase, unsigned long ulSequenceNum) {
  ;
  ;
  return ((*((volatile unsigned long *)(ulBase + 0x00000010))) & (1 << ulSequenceNum));
}
```
user: Translate the below assembly code into C programming language:

```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    L19
L14:
  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    L13
L11:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  lea    D27(%rip),%rax
  mov    %rax,%rsi
  lea    D7(%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    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L13:
  cmpl   $0x2,-0x14(%rbp)
  jle    L11
  addl   $0x1,-0x18(%rbp)
L19:
  cmpl   $0x6,-0x18(%rbp)
  jle    L14
  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: unpack_data
arguments: tag: D27
type: string
name: unpack_data
arguments: tag: D7
type: string
name: unpack_data
arguments: tag: D12
type: string

tool: D27: dsumkbn

tool: D7: # 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", "dsumkbn", 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    D0(%rip),%r15
  jmp    L22
L19:
  add    $0x1,%r14d
  add    $0x3,%r12d
  cmp    $0x15,%r12d
  je     L12
L22:
  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
L13:
  add    $0x1,%ebx
  mov    %ebp,%edx
  mov    %r15,%rsi
  lea    D24(%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    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L13
  jmp    L19
L12:
  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: label: D0
type_guessed: string
name: parse_data
arguments: label: D24
type_guessed: string
name: parse_data
arguments: label: D23
type_guessed: string

tool: D0: dsumkbn

tool: D24: # c::%s:len=%d


tool: D23: 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", "dsumkbn", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
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
  xor    %edi,%edi
  lea    D29(%rip),%r15
  push   %r14
  lea    D23(%rip),%r14
  push   %r13
  lea    D6(%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)
L10:
  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    L10
  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: tag: D29
type_infered: string
name: disassemble_data
arguments: tag: D23
type_infered: string
name: disassemble_data
arguments: tag: D6
type_infered: string

tool: D29: dsumkbn

tool: D23: # c::%s:len=%d


tool: D6: 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", "dsumkbn", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```