ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<all_java_tests>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_java_comments@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <test_java_comment_entities@plt>
```

assistant: ```c
void all_java_tests() {
  test_java_comments();
  test_java_comment_entities();
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<all_java_tests>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_java_comments@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <test_java_comment_entities@plt>
```

assistant: ```c
void all_java_tests() {
  test_java_comments();
  test_java_comment_entities();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <fromFile@plt>
  mov    $0x0,%eax
  call   <findBlocks@plt>
  mov    $0x0,%eax
  call   <makeCfg@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  fromFile();
  findBlocks();
  makeCfg();
  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    $0x0,%eax
  call   <fromFile@plt>
  mov    $0x0,%eax
  call   <findBlocks@plt>
  mov    $0x0,%eax
  call   <makeCfg@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  fromFile();
  findBlocks();
  makeCfg();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

assistant: ```c
int main() {
  fromFile();
  findBlocks();
  makeCfg();
  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
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <fromFile@plt>
  xor    %eax,%eax
  call   <findBlocks@plt>
  xor    %eax,%eax
  call   <makeCfg@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  fromFile();
  findBlocks();
  makeCfg();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isPrime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x9,-0x10(%rbp)
  jmp    L44
L19:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    -0x18(%rbp),%rdx
  lea    -0x4(%rdx),%rcx
  mov    $0x0,%edx
  div    %rcx
  mov    %rdx,%rax
  add    $0x2,%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x1,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L59
L64:
  mov    -0x8(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L2
  mov    -0xc(%rbp),%eax
  imul   -0x4(%rbp),%eax
  cltq
  mov    $0x0,%edx
  divq   -0x18(%rbp)
  mov    %rdx,%rax
  mov    %eax,-0xc(%rbp)
L2:
  mov    -0x4(%rbp),%eax
  imul   %eax,%eax
  cltq
  mov    $0x0,%edx
  divq   -0x18(%rbp)
  mov    %rdx,%rax
  mov    %eax,-0x4(%rbp)
  sarl   -0x8(%rbp)
L59:
  cmpl   $0x0,-0x8(%rbp)
  jne    L64
  cmpl   $0x1,-0xc(%rbp)
  je     L44
  mov    $0x0,%eax
  jmp    L32
L44:
  mov    -0x10(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  test   %eax,%eax
  jne    L19
  mov    $0x1,%eax
L32:
  leave
  ret
```

assistant: ```c
unsigned long isPrime(unsigned long val) {
  int i, p, exp, a;
  for (i = 9; i--;) {
    a = (rand() % (val - 4)) + 2;
    p = 1;
    exp = val - 1;
    while (exp) {
      if (exp & 1)
        p = (p * a) % val;
      a = (a * a) % val;
      exp >>= 1;
    }
    if (p != 1) {
      return 0;
    }
  }
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<isPrime>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x9,%ebp
  lea    -0x4(%rdi),%r12
  jmp    L16
L7:
  imul   %ecx,%ecx
  movslq %ecx,%rax
  mov    $0x0,%edx
  div    %rbx
  mov    %edx,%ecx
  sar    %esi
  je     L19
L33:
  test   $0x1,%sil
  je     L7
  imul   %ecx,%edi
  movslq %edi,%rax
  mov    $0x0,%edx
  div    %rbx
  mov    %edx,%edi
  jmp    L7
L19:
  cmp    $0x1,%edi
  jne    L61
L34:
  sub    $0x1,%ebp
  je     L18
L16:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    $0x0,%edx
  div    %r12
  lea    0x2(%rdx),%ecx
  mov    %ebx,%esi
  sub    $0x1,%esi
  je     L34
  mov    $0x1,%edi
  jmp    L33
L18:
  mov    $0x1,%eax
L12:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L61:
  mov    $0x0,%eax
  jmp    L12
```

assistant: ```c
unsigned long isPrime(unsigned long val) {
  int i, p, exp, a;
  for (i = 9; i--;) {
    a = (rand() % (val - 4)) + 2;
    p = 1;
    exp = val - 1;
    while (exp) {
      if (exp & 1)
        p = (p * a) % val;
      a = (a * a) % val;
      exp >>= 1;
    }
    if (p != 1) {
      return 0;
    }
  }
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<isPrime>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  lea    -0x1(%rdi),%r12d
  push   %rbp
  lea    -0x4(%rdi),%rbp
  push   %rbx
  mov    $0x9,%ebx
  sub    $0x8,%rsp
  xchg   %ax,%ax
L16:
  xor    %eax,%eax
  call   <rand@plt>
  xor    %edx,%edx
  div    %rbp
  lea    0x2(%rdx),%ecx
  test   %r12d,%r12d
  je     L40
  mov    %r12d,%esi
  mov    $0x1,%edi
  nopl   0x0(%rax)
L33:
  test   $0x1,%sil
  je     L12
  imul   %ecx,%edi
  xor    %edx,%edx
  movslq %edi,%rax
  div    %r13
  mov    %edx,%edi
L12:
  imul   %ecx,%ecx
  xor    %edx,%edx
  movslq %ecx,%rax
  div    %r13
  sar    %esi
  mov    %edx,%ecx
  jne    L33
  cmp    $0x1,%edi
  jne    L25
L40:
  sub    $0x1,%ebx
  jne    L16
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L25:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
unsigned long isPrime(unsigned long val) {
  int i, p, exp, a;
  for (i = 9; i--;) {
    a = (rand() % (val - 4)) + 2;
    p = 1;
    exp = val - 1;
    while (exp) {
      if (exp & 1)
        p = (p * a) % val;
      a = (a * a) % val;
      exp >>= 1;
    }
    if (p != 1) {
      return 0;
    }
  }
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<isPrime>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  lea    -0x1(%rdi),%r12d
  push   %rbp
  lea    -0x4(%rdi),%rbp
  push   %rbx
  mov    $0x9,%ebx
  sub    $0x8,%rsp
  xchg   %ax,%ax
L49:
  xor    %eax,%eax
  call   <rand@plt>
  xor    %edx,%edx
  div    %rbp
  lea    0x2(%rdx),%ecx
  test   %r12d,%r12d
  je     L32
  mov    %r12d,%esi
  mov    $0x1,%edi
  nopl   0x0(%rax)
L25:
  test   $0x1,%sil
  je     L37
  imul   %ecx,%edi
  xor    %edx,%edx
  movslq %edi,%rax
  div    %r13
  mov    %edx,%edi
L37:
  imul   %ecx,%ecx
  xor    %edx,%edx
  movslq %ecx,%rax
  div    %r13
  sar    %esi
  mov    %edx,%ecx
  jne    L25
  cmp    $0x1,%edi
  jne    L11
L32:
  sub    $0x1,%ebx
  jne    L49
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L11:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
unsigned long isPrime(unsigned long val) {
  int i, p, exp, a;
  for (i = 9; i--;) {
    a = (rand() % (val - 4)) + 2;
    p = 1;
    exp = val - 1;
    while (exp) {
      if (exp & 1)
        p = (p * a) % val;
      a = (a * a) % val;
      exp >>= 1;
    }
    if (p != 1) {
      return 0;
    }
  }
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x8(%rbp)
  jmp    L11
L15:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x4(%rbp),%eax
  sub    -0x8(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x4(%rbp),%eax
  sub    -0x8(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x9(%rbp),%eax
  mov    %al,(%rdx)
  subl   $0x1,-0x8(%rbp)
L11:
  cmpl   $0x0,-0x8(%rbp)
  jns    L15
  nop
  nop
  leave
  ret
```

assistant: ```c
void reverse(char str[]) {
  int len = strlen(str);
  char tmp;
  for (int i = len / 2; i >= 0; i--) {
    tmp = str[i];
    str[i] = str[len - i - 1];
    str[len - i - 1] = tmp;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<reverse>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  sar    %edx
  cmp    $0xffffffff,%eax
  jl     L8
  movslq %edx,%rdx
  cltq
  sub    $0x1,%rax
  sub    %rdx,%rax
  add    %rbx,%rax
L2:
  movzbl (%rbx,%rdx,1),%ecx
  movzbl (%rax),%esi
  mov    %sil,(%rbx,%rdx,1)
  mov    %cl,(%rax)
  sub    $0x1,%rdx
  add    $0x1,%rax
  test   %edx,%edx
  jns    L2
L8:
  pop    %rbx
  ret
```

assistant: ```c
void reverse(char str[]) {
  int len = strlen(str);
  char tmp;
  for (int i = len / 2; i >= 0; i--) {
    tmp = str[i];
    str[i] = str[len - i - 1];
    str[len - i - 1] = tmp;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<reverse>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  sar    %edx
  cmp    $0xffffffff,%eax
  jl     L5
  cltq
  movslq %edx,%rdx
  sub    $0x1,%rax
  sub    %rdx,%rax
  add    %rbx,%rax
  nopw   0x0(%rax,%rax,1)
L6:
  movzbl (%rax),%esi
  movzbl (%rbx,%rdx,1),%ecx
  add    $0x1,%rax
  mov    %sil,(%rbx,%rdx,1)
  sub    $0x1,%rdx
  mov    %cl,-0x1(%rax)
  test   %edx,%edx
  jns    L6
L5:
  pop    %rbx
  ret
```

assistant: ```c
void reverse(char str[]) {
  int len = strlen(str);
  char tmp;
  for (int i = len / 2; i >= 0; i--) {
    tmp = str[i];
    str[i] = str[len - i - 1];
    str[len - i - 1] = tmp;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<reverse>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  sar    %edx
  cmp    $0xffffffff,%eax
  jl     L0
  cltq
  movslq %edx,%rdx
  sub    $0x1,%rax
  sub    %rdx,%rax
  add    %rbx,%rax
  nopw   0x0(%rax,%rax,1)
L12:
  movzbl (%rax),%esi
  movzbl (%rbx,%rdx,1),%ecx
  add    $0x1,%rax
  mov    %sil,(%rbx,%rdx,1)
  sub    $0x1,%rdx
  mov    %cl,-0x1(%rax)
  test   %edx,%edx
  jns    L12
L0:
  pop    %rbx
  ret
```

assistant: ```c
void reverse(char str[]) {
  int len = strlen(str);
  char tmp;
  for (int i = len / 2; i >= 0; i--) {
    tmp = str[i];
    str[i] = str[len - i - 1];
    str[len - i - 1] = tmp;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<find>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L34
L37:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x20(%rbp)
  jne    L27
  mov    $0x1,%eax
  jmp    L58
L27:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x20(%rbp)
  jle    L57
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L34
L57:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
L34:
  mov    -0x8(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jge    L37
  mov    $0x0,%eax
L58:
  pop    %rbp
  ret
```

assistant: ```c
int find(int *l, int a_size, int num) {
  int mid, low, high;
  low = 0;
  high = a_size - 1;
  while (high >= low) {
    mid = (low + high) / 2;
    if (l[mid] == num)
      return 1;
    else if (l[mid] < num) {
      low = mid + 1;
    } else {
      high = mid - 1;
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<find>:
  endbr64
  sub    $0x1,%esi
  js     L50
  mov    $0x0,%r8d
  jmp    L44
L51:
  lea    -0x1(%rax),%esi
L34:
  cmp    %esi,%r8d
  jg     L36
L44:
  lea    (%r8,%rsi,1),%ecx
  mov    %ecx,%eax
  shr    $0x1f,%eax
  add    %ecx,%eax
  sar    %eax
  movslq %eax,%rcx
  mov    (%rdi,%rcx,4),%ecx
  cmp    %edx,%ecx
  je     L49
  jge    L51
  lea    0x1(%rax),%r8d
  jmp    L34
L36:
  mov    $0x0,%eax
  ret
L50:
  mov    $0x0,%eax
  ret
L49:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int find(int *l, int a_size, int num) {
  int mid, low, high;
  low = 0;
  high = a_size - 1;
  while (high >= low) {
    mid = (low + high) / 2;
    if (l[mid] == num)
      return 1;
    else if (l[mid] < num) {
      low = mid + 1;
    } else {
      high = mid - 1;
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<find>:
  endbr64
  sub    $0x1,%esi
  js     L39
  xor    %ecx,%ecx
  jmp    L55
  nopl   (%rax)
L27:
  lea    0x1(%rax),%ecx
  cmp    %ecx,%esi
  jl     L39
L55:
  lea    (%rsi,%rcx,1),%eax
  sar    %eax
  movslq %eax,%r8
  cmp    %edx,(%rdi,%r8,4)
  je     L51
  jl     L27
  lea    -0x1(%rax),%esi
  cmp    %ecx,%esi
  jge    L55
L39:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L51:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int find(int *l, int a_size, int num) {
  int mid, low, high;
  low = 0;
  high = a_size - 1;
  while (high >= low) {
    mid = (low + high) / 2;
    if (l[mid] == num)
      return 1;
    else if (l[mid] < num) {
      low = mid + 1;
    } else {
      high = mid - 1;
    }
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<find>:
  endbr64
  sub    $0x1,%esi
  js     L2
  xor    %ecx,%ecx
  nopl   0x0(%rax,%rax,1)
L30:
  lea    (%rsi,%rcx,1),%eax
  sar    %eax
  movslq %eax,%r8
  cmp    %edx,(%rdi,%r8,4)
  je     L49
L56:
  jge    L4
  lea    0x1(%rax),%ecx
  cmp    %esi,%ecx
  jle    L30
L2:
  xor    %eax,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  lea    -0x1(%rax),%esi
  cmp    %ecx,%esi
  jl     L2
  lea    (%rsi,%rcx,1),%eax
  sar    %eax
  movslq %eax,%r8
  cmp    %edx,(%rdi,%r8,4)
  jne    L56
L49:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int find(int *l, int a_size, int num) {
  int mid, low, high;
  low = 0;
  high = a_size - 1;
  while (high >= low) {
    mid = (low + high) / 2;
    if (l[mid] == num)
      return 1;
    else if (l[mid] < num) {
      low = mid + 1;
    } else {
      high = mid - 1;
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<GPIOIntUnregister>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <GPIOGetIntNumber@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <IntDisable@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <IntUnregister@plt>
  nop
  leave
  ret
```

assistant: ```c
void GPIOIntUnregister(unsigned long ulPort) {
  ;
  ulPort = GPIOGetIntNumber(ulPort);
  IntDisable(ulPort);
  IntUnregister(ulPort);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<GPIOIntUnregister>:
  endbr64
  push   %rbx
  call   <GPIOGetIntNumber@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <IntDisable@plt>
  mov    %rbx,%rdi
  call   <IntUnregister@plt>
  pop    %rbx
  ret
```

assistant: ```c
void GPIOIntUnregister(unsigned long ulPort) {
  ;
  ulPort = GPIOGetIntNumber(ulPort);
  IntDisable(ulPort);
  IntUnregister(ulPort);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<GPIOIntUnregister>:
  endbr64
  push   %rbp
  call   <GPIOGetIntNumber@plt>
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <IntDisable@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1080 <IntUnregister@plt>
```

assistant: ```c
void GPIOIntUnregister(unsigned long ulPort) {
  ;
  ulPort = GPIOGetIntNumber(ulPort);
  IntDisable(ulPort);
  IntUnregister(ulPort);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<GPIOIntUnregister>:
  endbr64
  push   %rbp
  call   <GPIOGetIntNumber@plt>
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <IntDisable@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1080 <IntUnregister@plt>
```

assistant: ```c
void GPIOIntUnregister(unsigned long ulPort) {
  ;
  ulPort = GPIOGetIntNumber(ulPort);
  IntDisable(ulPort);
  IntUnregister(ulPort);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<init_power_hashmap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x1,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  movl   $0x0,-0xc(%rbp)
  jmp    L19
L8:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdx
  shl    $0x8,%rdx
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rcx
  movabs $0x89705f3112a28fe5,%rdx
  mov    %rcx,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x1d,%rax
  imul   $0x3b9aca07,%rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rdx
  mov    %edx,(%rax)
  addl   $0x1,-0xc(%rbp)
L19:
  mov    -0x14(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jg     L8
  mov    -0x20(%rbp),%rax
  leave
  ret
```

assistant: ```c
unsigned *init_power_hashmap(int N, unsigned *power_hashmap) {
  int i;
  unsigned long long power = 1;
  power_hashmap = (unsigned *)malloc(sizeof(unsigned) * N);
  power_hashmap[0] = power;
  for (i = 0; i <= N - 1; i++) {
    power *= 257;
    power %= 1000000007;
    power_hashmap[i] = power;
  }
  return power_hashmap;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<init_power_hashmap>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  lea    0x0(,%rdi,4),%edi
  call   <malloc@plt>
  mov    %rax,%r9
  movl   $0x1,(%rax)
  test   %ebx,%ebx
  jle    L19
  mov    %rax,%rsi
  mov    %ebx,%ebx
  lea    (%rax,%rbx,4),%r8
  mov    $0x1,%edx
  movabs $0x89705f3112a28fe5,%rdi
L11:
  mov    %rdx,%rcx
  shl    $0x8,%rcx
  add    %rdx,%rcx
  mov    %rcx,%rax
  mul    %rdi
  shr    $0x1d,%rdx
  imul   $0x3b9aca07,%rdx,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    %ecx,(%rsi)
  add    $0x4,%rsi
  cmp    %r8,%rsi
  jne    L11
L19:
  mov    %r9,%rax
  pop    %rbx
  ret
```

assistant: ```c
unsigned *init_power_hashmap(int N, unsigned *power_hashmap) {
  int i;
  unsigned long long power = 1;
  power_hashmap = (unsigned *)malloc(sizeof(unsigned) * N);
  power_hashmap[0] = power;
  for (i = 0; i <= N - 1; i++) {
    power *= 257;
    power %= 1000000007;
    power_hashmap[i] = power;
  }
  return power_hashmap;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<init_power_hashmap>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  lea    0x0(,%rbx,4),%edi
  call   <malloc@plt>
  movl   $0x1,(%rax)
  mov    %rax,%r9
  test   %ebx,%ebx
  jle    L9
  mov    %rax,%rsi
  lea    (%rax,%rbx,4),%r8
  mov    $0x1,%edx
  movabs $0x89705f3112a28fe5,%rdi
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    %rdx,%rcx
  add    $0x4,%rsi
  shl    $0x8,%rcx
  add    %rdx,%rcx
  mov    %rcx,%rax
  mul    %rdi
  shr    $0x1d,%rdx
  imul   $0x3b9aca07,%rdx,%rax
  sub    %rax,%rcx
  mov    %ecx,-0x4(%rsi)
  mov    %rcx,%rdx
  cmp    %r8,%rsi
  jne    L14
L9:
  mov    %r9,%rax
  pop    %rbx
  ret
```

assistant: ```c
unsigned *init_power_hashmap(int N, unsigned *power_hashmap) {
  int i;
  unsigned long long power = 1;
  power_hashmap = (unsigned *)malloc(sizeof(unsigned) * N);
  power_hashmap[0] = power;
  for (i = 0; i <= N - 1; i++) {
    power *= 257;
    power %= 1000000007;
    power_hashmap[i] = power;
  }
  return power_hashmap;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<init_power_hashmap>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  lea    0x0(,%rbx,4),%edi
  call   <malloc@plt>
  movl   $0x1,(%rax)
  mov    %rax,%r9
  test   %ebx,%ebx
  jle    L15
  mov    %rax,%rsi
  lea    (%rax,%rbx,4),%r8
  mov    $0x1,%edx
  movabs $0x89705f3112a28fe5,%rdi
  cs nopw 0x0(%rax,%rax,1)
L13:
  mov    %rdx,%rcx
  add    $0x4,%rsi
  shl    $0x8,%rcx
  add    %rdx,%rcx
  mov    %rcx,%rax
  mul    %rdi
  shr    $0x1d,%rdx
  imul   $0x3b9aca07,%rdx,%rax
  sub    %rax,%rcx
  mov    %ecx,-0x4(%rsi)
  mov    %rcx,%rdx
  cmp    %rsi,%r8
  jne    L13
L15:
  mov    %r9,%rax
  pop    %rbx
  ret
```

assistant: ```c
unsigned *init_power_hashmap(int N, unsigned *power_hashmap) {
  int i;
  unsigned long long power = 1;
  power_hashmap = (unsigned *)malloc(sizeof(unsigned) * N);
  power_hashmap[0] = power;
  for (i = 0; i <= N - 1; i++) {
    power *= 257;
    power %= 1000000007;
    power_hashmap[i] = power;
  }
  return power_hashmap;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x7f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x7e4(%rbp)
  lea    -0x7f0(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x7e0(%rbp),%eax
  mov    %eax,-0x7ec(%rbp)
  mov    -0x7e0(%rbp),%eax
  mov    %eax,-0x7e8(%rbp)
  jmp    L27
L10:
  lea    -0x7e0(%rbp),%rdx
  mov    -0x7e4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x7e4(%rbp)
L27:
  mov    -0x7f0(%rbp),%eax
  cmp    %eax,-0x7e4(%rbp)
  jl     L10
  mov    -0x7f0(%rbp),%esi
  lea    -0x7e8(%rbp),%rcx
  lea    -0x7ec(%rbp),%rdx
  lea    -0x7e0(%rbp),%rax
  mov    %rax,%rdi
  call   <minmax@plt>
  mov    -0x7ec(%rbp),%edx
  mov    -0x7e8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
data_type: string
name: retrieve_data
arguments: tag: D12
data_type: string

tool: D2: %d

tool: D12: %d


assistant: ```c
int main() {
  int arr[500];
  int size, counter = 0, min, max;
  scanf("%d", &size);
  min = arr[0];
  max = arr[0];
  while (counter < size) {
    scanf("%d", &arr[counter]);
    counter++;
  }
  minmax(arr, size, &min, &max);
  printf("%d\n", min + max);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x7f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x7e8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%eax
  mov    %eax,0x8(%rsp)
  mov    %eax,0xc(%rsp)
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L0
  lea    0x10(%rsp),%rbp
  mov    $0x0,%ebx
  lea    D14(%rip),%r12
L19:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    0x4(%rsp),%esi
  add    $0x4,%rbp
  cmp    %ebx,%esi
  jg     L19
L0:
  lea    0xc(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  lea    0x10(%rsp),%rdi
  call   <minmax@plt>
  mov    0xc(%rsp),%esi
  add    0x8(%rsp),%esi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    0x7e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x7f0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D14
type_guessed: string
name: retrieve_data
arguments: data_label: D10
type_guessed: string

tool: D14: %d

tool: D10: %d


assistant: ```c
int main() {
  int arr[500];
  int size, counter = 0, min, max;
  scanf("%d", &size);
  min = arr[0];
  max = arr[0];
  while (counter < size) {
    scanf("%d", &arr[counter]);
    counter++;
  }
  minmax(arr, size, &min, &max);
  printf("%d\n", min + max);
  return 0;
}
```