ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ae_wait>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    %rdx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdi
  mov    $0xc,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  mov    %ebp,0xc(%rsp)
  test   $0x1,%bl
  je     L22
  mov    D38(%rip),%rax
  mov    (%rax),%eax
  or     %eax,0x10(%rsp)
L22:
  test   $0x2,%bl
  je     L10
  mov    D39(%rip),%rax
  mov    (%rax),%eax
  or     %eax,0x10(%rsp)
L10:
  lea    0xc(%rsp),%rdi
  mov    %r12,%rdx
  mov    $0x1,%esi
  call   <poll@plt>
  cmp    $0x1,%eax
  je     L34
L40:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L34:
  mov    0x14(%rsp),%edx
  mov    D38(%rip),%rax
  mov    %edx,%ebx
  and    (%rax),%ebx
  setne  %al
  movzbl %al,%eax
  mov    D39(%rip),%rcx
  mov    %edx,%esi
  and    (%rcx),%esi
  mov    %eax,%ecx
  or     $0x2,%ecx
  test   %esi,%esi
  cmovne %ecx,%eax
  mov    D44(%rip),%rcx
  mov    %edx,%esi
  and    (%rcx),%esi
  mov    %eax,%ecx
  or     $0x2,%ecx
  test   %esi,%esi
  cmovne %ecx,%eax
  mov    D49(%rip),%rcx
  and    (%rcx),%edx
  mov    %eax,%ecx
  or     $0x2,%ecx
  test   %edx,%edx
  cmovne %ecx,%eax
  jmp    L40
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ae_wait(int fd, int mask, long long milliseconds) {
  struct pollfd pfd;
  int retmask = 0, retval;
  memset(&pfd, 0, sizeof(pfd));
  pfd.fd = fd;
  if (mask & 1)
    pfd.events |= POLLIN;
  if (mask & 2)
    pfd.events |= POLLOUT;
  if ((retval = poll(&pfd, 1, milliseconds)) == 1) {
    if (pfd.revents & POLLIN)
      retmask |= 1;
    if (pfd.revents & POLLOUT)
      retmask |= 2;
    if (pfd.revents & POLLERR)
      retmask |= 2;
    if (pfd.revents & POLLHUP)
      retmask |= 2;
    return retmask;
  } else {
    return retval;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ae_wait>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  mov    $0xc,%edx
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  xor    %esi,%esi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r13
  mov    %r13,%rdi
  call   <memset@plt>
  mov    %ebp,0xc(%rsp)
  test   $0x1,%bl
  je     L13
  mov    D19(%rip),%rax
  mov    (%rax),%eax
  or     %eax,0x10(%rsp)
L13:
  and    $0x2,%ebx
  je     L38
  mov    D22(%rip),%rax
  mov    (%rax),%eax
  or     %eax,0x10(%rsp)
L38:
  mov    %r12,%rdx
  mov    $0x1,%esi
  mov    %r13,%rdi
  call   <poll@plt>
  cmp    $0x1,%eax
  jne    L7
  mov    0x14(%rsp),%edx
  mov    D19(%rip),%rax
  mov    D22(%rip),%rcx
  mov    %edx,%ebx
  and    (%rax),%ebx
  setne  %al
  mov    (%rcx),%esi
  movzbl %al,%eax
  and    %edx,%esi
  mov    %eax,%ecx
  or     $0x2,%ecx
  test   %esi,%esi
  cmovne %ecx,%eax
  mov    D10(%rip),%rcx
  mov    (%rcx),%esi
  mov    %eax,%ecx
  or     $0x2,%ecx
  and    %edx,%esi
  test   %esi,%esi
  cmovne %ecx,%eax
  mov    D56(%rip),%rcx
  and    (%rcx),%edx
  mov    %eax,%ecx
  or     $0x2,%ecx
  test   %edx,%edx
  cmovne %ecx,%eax
L7:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ae_wait(int fd, int mask, long long milliseconds) {
  struct pollfd pfd;
  int retmask = 0, retval;
  memset(&pfd, 0, sizeof(pfd));
  pfd.fd = fd;
  if (mask & 1)
    pfd.events |= POLLIN;
  if (mask & 2)
    pfd.events |= POLLOUT;
  if ((retval = poll(&pfd, 1, milliseconds)) == 1) {
    if (pfd.revents & POLLIN)
      retmask |= 1;
    if (pfd.revents & POLLOUT)
      retmask |= 2;
    if (pfd.revents & POLLERR)
      retmask |= 2;
    if (pfd.revents & POLLHUP)
      retmask |= 2;
    return retmask;
  } else {
    return retval;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ae_wait>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  mov    $0xc,%edx
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  xor    %esi,%esi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r13
  mov    %r13,%rdi
  call   <memset@plt>
  mov    %ebp,0xc(%rsp)
  test   $0x1,%bl
  je     L34
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  or     %eax,0x10(%rsp)
L34:
  and    $0x2,%ebx
  je     L31
  mov    D16(%rip),%rax
  mov    (%rax),%eax
  or     %eax,0x10(%rsp)
L31:
  mov    %r12,%rdx
  mov    $0x1,%esi
  mov    %r13,%rdi
  call   <poll@plt>
  cmp    $0x1,%eax
  jne    L16
  mov    0x14(%rsp),%edx
  mov    D2(%rip),%rax
  mov    D16(%rip),%rcx
  mov    %edx,%ebx
  and    (%rax),%ebx
  setne  %al
  mov    (%rcx),%esi
  movzbl %al,%eax
  and    %edx,%esi
  mov    %eax,%ecx
  or     $0x2,%ecx
  test   %esi,%esi
  cmovne %ecx,%eax
  mov    D55(%rip),%rcx
  mov    (%rcx),%esi
  mov    %eax,%ecx
  or     $0x2,%ecx
  and    %edx,%esi
  test   %esi,%esi
  cmovne %ecx,%eax
  mov    D39(%rip),%rcx
  and    (%rcx),%edx
  mov    %eax,%ecx
  or     $0x2,%ecx
  test   %edx,%edx
  cmovne %ecx,%eax
L16:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ae_wait(int fd, int mask, long long milliseconds) {
  struct pollfd pfd;
  int retmask = 0, retval;
  memset(&pfd, 0, sizeof(pfd));
  pfd.fd = fd;
  if (mask & 1)
    pfd.events |= POLLIN;
  if (mask & 2)
    pfd.events |= POLLOUT;
  if ((retval = poll(&pfd, 1, milliseconds)) == 1) {
    if (pfd.revents & POLLIN)
      retmask |= 1;
    if (pfd.revents & POLLOUT)
      retmask |= 2;
    if (pfd.revents & POLLERR)
      retmask |= 2;
    if (pfd.revents & POLLHUP)
      retmask |= 2;
    return retmask;
  } else {
    return retval;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  add    $0x136ff696,%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L25
L46:
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  mov    %ax,-0x1a(%rbp)
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  add    $0x6,%rax
  lea    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  shl    $0x3,%rdx
  add    %rcx,%rdx
  movzwl (%rax),%eax
  mov    %ax,(%rdx)
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  lea    0x6(%rax),%rdx
  movzwl -0x1a(%rbp),%eax
  mov    %ax,(%rdx)
  addq   $0x2,-0x18(%rbp)
L25:
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  cmp    %rax,-0x18(%rbp)
  jb     L46
  movq   $0x0,-0x18(%rbp)
  jmp    L28
L33:
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  addq   $0x1,-0x18(%rbp)
L28:
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  cmp    %rax,-0x18(%rbp)
  jb     L33
  mov    -0x10(%rbp),%rax
  lea    -0x2b2feb28(%rax),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy11;
  {
    state[0UL] = (input[0UL] + 51238316UL) + 274866410UL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      copy11 = *((unsigned short *)(&state[local1]) + 0);
      *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 3);
      *((unsigned short *)(&state[local1]) + 3) = copy11;
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[0UL] = state[local1] + state[0UL];
      local1++;
    }
    output[0UL] = state[0UL] - 724560680UL;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    (%rdi),%rax
  add    $0x136ff696,%rax
  mov    %rax,(%rsp)
  mov    0x8(%rdi),%rdx
  test   %rdx,%rdx
  je     L4
  mov    %rsp,%rax
  lea    -0x8(,%rdx,8),%rdi
  and    $0xfffffffffffffff0,%rdi
  lea    0x10(%rsp),%rcx
  add    %rcx,%rdi
L25:
  movzwl (%rax),%edx
  movzwl 0x6(%rax),%ecx
  mov    %cx,(%rax)
  mov    %dx,0x6(%rax)
  add    $0x10,%rax
  cmp    %rdi,%rax
  jne    L25
  shlq   (%rsp)
L4:
  mov    (%rsp),%rax
  sub    $0x2b2feb28,%rax
  mov    %rax,(%rsi)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x18,%rsp
  ret
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy11;
  {
    state[0UL] = (input[0UL] + 51238316UL) + 274866410UL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      copy11 = *((unsigned short *)(&state[local1]) + 0);
      *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 3);
      *((unsigned short *)(&state[local1]) + 3) = copy11;
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[0UL] = state[local1] + state[0UL];
      local1++;
    }
    output[0UL] = state[0UL] - 724560680UL;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  mov    0x8(%rdi),%rdx
  lea    0x136ff696(%rax),%rcx
  test   %rdx,%rdx
  je     L18
  xor    %eax,%eax
  nopl   0x0(%rax)
L12:
  add    $0x2,%rax
  cmp    %rax,%rdx
  ja     L12
  add    %rcx,%rcx
L18:
  sub    $0x2b2feb28,%rcx
  mov    %rcx,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy11;
  {
    state[0UL] = (input[0UL] + 51238316UL) + 274866410UL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      copy11 = *((unsigned short *)(&state[local1]) + 0);
      *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 3);
      *((unsigned short *)(&state[local1]) + 3) = copy11;
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[0UL] = state[local1] + state[0UL];
      local1++;
    }
    output[0UL] = state[0UL] - 724560680UL;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  mov    0x8(%rdi),%rdx
  lea    0x136ff696(%rax),%rcx
  test   %rdx,%rdx
  je     L7
  xor    %eax,%eax
  nopl   0x0(%rax)
L12:
  add    $0x2,%rax
  cmp    %rax,%rdx
  ja     L12
  add    %rcx,%rcx
L7:
  sub    $0x2b2feb28,%rcx
  mov    %rcx,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy11;
  {
    state[0UL] = (input[0UL] + 51238316UL) + 274866410UL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      copy11 = *((unsigned short *)(&state[local1]) + 0);
      *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 3);
      *((unsigned short *)(&state[local1]) + 3) = copy11;
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[0UL] = state[local1] + state[0UL];
      local1++;
    }
    output[0UL] = state[0UL] - 724560680UL;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<isprime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x1,-0x8(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L34
  cmpq   $0x1,-0x18(%rbp)
  jne    L35
L34:
  mov    $0x0,%eax
  jmp    L55
L53:
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x8(%rbp)
  mov    %rdx,%rax
  test   %rax,%rax
  jne    L35
  mov    $0x0,%eax
  jmp    L55
L35:
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jb     L53
  mov    $0x1,%eax
L55:
  pop    %rbp
  ret
```

assistant: ```c
bool isprime(long long unsigned int num) {
  long long unsigned int i = 1;
  if (num <= 0 || num == 1)
    return 0;
  while (++i < num)
    if (num % i == 0)
      return 0;
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<isprime>:
  endbr64
  mov    $0x0,%eax
  mov    $0x1,%ecx
  cmp    $0x1,%rdi
  jbe    L26
L12:
  add    $0x1,%rcx
  cmp    %rcx,%rdi
  je     L13
  mov    %rdi,%rax
  mov    $0x0,%edx
  div    %rcx
  test   %rdx,%rdx
  jne    L12
  mov    $0x0,%eax
L26:
  ret
L13:
  mov    $0x1,%eax
  ret
```

assistant: ```c
bool isprime(long long unsigned int num) {
  long long unsigned int i = 1;
  if (num <= 0 || num == 1)
    return 0;
  while (++i < num)
    if (num % i == 0)
      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
  xor    %eax,%eax
  mov    $0x1,%ecx
  cmp    $0x1,%rdi
  ja     L5
  jmp    L31
  nopl   0x0(%rax,%rax,1)
L37:
  mov    %rdi,%rax
  xor    %edx,%edx
  div    %rcx
  test   %rdx,%rdx
  je     L38
L5:
  add    $0x1,%rcx
  cmp    %rcx,%rdi
  jne    L37
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax)
L38:
  xor    %eax,%eax
L31:
  ret
```

assistant: ```c
bool isprime(long long unsigned int num) {
  long long unsigned int i = 1;
  if (num <= 0 || num == 1)
    return 0;
  while (++i < num)
    if (num % i == 0)
      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
  xor    %eax,%eax
  mov    $0x1,%ecx
  cmp    $0x1,%rdi
  ja     L34
  jmp    L38
  nopl   0x0(%rax,%rax,1)
L20:
  mov    %rdi,%rax
  xor    %edx,%edx
  div    %rcx
  test   %rdx,%rdx
  je     L28
L34:
  add    $0x1,%rcx
  cmp    %rcx,%rdi
  jne    L20
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax)
L28:
  xor    %eax,%eax
L38:
  ret
```

assistant: ```c
bool isprime(long long unsigned int num) {
  long long unsigned int i = 1;
  if (num <= 0 || num == 1)
    return 0;
  while (++i < num)
    if (num % i == 0)
      return 0;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<__test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x1,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  je     L27
  cmpl   $0x0,-0x4(%rbp)
  je     L27
  mov    $0x1,%eax
  jmp    L5
L27:
  mov    $0x0,%eax
L5:
  pop    %rbp
  ret
```

assistant: ```c
int __test() {
  bool x = true;
  bool y = false;
  if (x && y) {
    return 1;
  } else {
    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
<__test>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int __test() {
  bool x = true;
  bool y = false;
  if (x && y) {
    return 1;
  } else {
    return 0;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<__test>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int __test() {
  bool x = true;
  bool y = false;
  if (x && y) {
    return 1;
  } else {
    return 0;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<__test>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int __test() {
  bool x = true;
  bool y = false;
  if (x && y) {
    return 1;
  } else {
    return 0;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bpf_obj_get>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <ptr_to_u64@plt>
  mov    %eax,-0xc(%rbp)
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  lea    -0xc(%rbp),%rdx
  mov    $0x4,%ecx
  mov    %eax,%esi
  mov    $0x141,%edi
  call   <syscall@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
int bpf_obj_get(const char *pathname) {
  union bpf_attr attr;
  attr.pathname = ptr_to_u64((void *)pathname);
  return syscall(321, BPF_OBJ_GET, &attr, sizeof(attr));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<bpf_obj_get>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <ptr_to_u64@plt>
  mov    %eax,0x4(%rsp)
  lea    0x4(%rsp),%rdx
  mov    $0x4,%ecx
  mov    D4(%rip),%rax
  mov    (%rax),%esi
  mov    $0x141,%edi
  call   <syscall@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bpf_obj_get(const char *pathname) {
  union bpf_attr attr;
  attr.pathname = ptr_to_u64((void *)pathname);
  return syscall(321, BPF_OBJ_GET, &attr, sizeof(attr));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bpf_obj_get>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <ptr_to_u64@plt>
  lea    0x4(%rsp),%rdx
  mov    $0x4,%ecx
  mov    $0x141,%edi
  mov    %eax,0x4(%rsp)
  mov    D4(%rip),%rax
  mov    (%rax),%esi
  call   <syscall@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bpf_obj_get(const char *pathname) {
  union bpf_attr attr;
  attr.pathname = ptr_to_u64((void *)pathname);
  return syscall(321, BPF_OBJ_GET, &attr, sizeof(attr));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bpf_obj_get>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <ptr_to_u64@plt>
  lea    0x4(%rsp),%rdx
  mov    $0x4,%ecx
  mov    $0x141,%edi
  mov    %eax,0x4(%rsp)
  mov    D5(%rip),%rax
  mov    (%rax),%esi
  call   <syscall@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bpf_obj_get(const char *pathname) {
  union bpf_attr attr;
  attr.pathname = ptr_to_u64((void *)pathname);
  return syscall(321, BPF_OBJ_GET, &attr, sizeof(attr));
}
```
user: What is the c source code of the assembly code below?

```asm
<ck_m>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D21(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L32
  mov    $0x0,%eax
  jmp    L19
L32:
  mov    -0x8(%rbp),%rax
  lea    D13(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L3
  mov    $0x1,%eax
  jmp    L19
L3:
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x2,%eax
L19:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D21
type_guessed: string
name: extract_symbolic_data
arguments: label: D13
type_guessed: string
name: extract_symbolic_data
arguments: label: D27
type_guessed: string

tool: D21: -MIC

tool: D13: -MICFS

tool: D27: Program Mode Error !

assistant: ```c
int ck_m(char *mode) {
  if (strcmp(mode, "-MIC") == 0) {
    return 0;
  } else if (strcmp(mode, "-MICFS") == 0) {
    return 1;
  } else {
    puts("Program Mode Error !");
    return 2;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ck_m>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D29(%rip),%rsi
  call   <strcmp@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  jne    L5
L18:
  pop    %rbx
  ret
L5:
  lea    D28(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  je     L18
  lea    D19(%rip),%rdi
  call   <puts@plt>
  mov    $0x2,%eax
  jmp    L18
```

tool_calls: name: unpack_data
arguments: data_label: D29
type_infered: string
name: unpack_data
arguments: data_label: D28
type_infered: string
name: unpack_data
arguments: data_label: D19
type_infered: string

tool: D29: -MIC

tool: D28: -MICFS

tool: D19: Program Mode Error !

assistant: ```c
int ck_m(char *mode) {
  if (strcmp(mode, "-MIC") == 0) {
    return 0;
  } else if (strcmp(mode, "-MICFS") == 0) {
    return 1;
  } else {
    puts("Program Mode Error !");
    return 2;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ck_m>:
  endbr64
  push   %rbp
  lea    D0(%rip),%rsi
  mov    %rdi,%rbp
  call   <strcmp@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L5
L4:
  pop    %rbp
  ret
L5:
  lea    D26(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  je     L4
  lea    D22(%rip),%rdi
  call   <puts@plt>
  mov    $0x2,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D0
type_guessed: string
name: unpack_data
arguments: data_label: D26
type_guessed: string
name: unpack_data
arguments: data_label: D22
type_guessed: string

tool: D0: -MIC

tool: D26: -MICFS

tool: D22: Program Mode Error !

assistant: ```c
int ck_m(char *mode) {
  if (strcmp(mode, "-MIC") == 0) {
    return 0;
  } else if (strcmp(mode, "-MICFS") == 0) {
    return 1;
  } else {
    puts("Program Mode Error !");
    return 2;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ck_m>:
  endbr64
  push   %rbp
  lea    D8(%rip),%rsi
  mov    %rdi,%rbp
  call   <strcmp@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L15
L5:
  pop    %rbp
  ret
L15:
  lea    D9(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  je     L5
  lea    D17(%rip),%rdi
  call   <puts@plt>
  mov    $0x2,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D8
data_type: string
name: read_data
arguments: data_tag: D9
data_type: string
name: read_data
arguments: data_tag: D17
data_type: string

tool: D8: -MIC

tool: D9: -MICFS

tool: D17: Program Mode Error !

assistant: ```c
int ck_m(char *mode) {
  if (strcmp(mode, "-MIC") == 0) {
    return 0;
  } else if (strcmp(mode, "-MICFS") == 0) {
    return 1;
  } else {
    puts("Program Mode Error !");
    return 2;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  imul   -0x14(%rbp),%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  sub    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <bar@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int foo(int a, int b) {
  int c;
  c = bar(b - a, b * a);
  return c;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%eax
  mov    %esi,%edx
  imul   %edi,%edx
  mov    %esi,%edi
  sub    %eax,%edi
  mov    %edx,%esi
  call   <bar@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int foo(int a, int b) {
  int c;
  c = bar(b - a, b * a);
  return c;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<foo>:
  endbr64
  mov    %esi,%r8d
  mov    %edi,%eax
  imul   %edi,%r8d
  mov    %esi,%edi
  sub    %eax,%edi
  mov    %r8d,%esi
  jmp    1050 <bar@plt>
```

assistant: ```c
int foo(int a, int b) {
  int c;
  c = bar(b - a, b * a);
  return c;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<foo>:
  endbr64
  mov    %esi,%r8d
  mov    %edi,%eax
  imul   %edi,%r8d
  mov    %esi,%edi
  sub    %eax,%edi
  mov    %r8d,%esi
  jmp    1050 <bar@plt>
```

assistant: ```c
int foo(int a, int b) {
  int c;
  c = bar(b - a, b * a);
  return c;
}
```