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

```asm
<multiply243>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  test   %ebp,%ebp
  jle    L16
  mov    %rbp,%r11
  mov    %esi,%ebx
  lea    0x4(%rdx),%r10
  mov    %r8,%rax
  mov    %r11d,%edx
  mov    %r11d,%esi
  lea    0x4(%rcx),%r9
  shl    $0x2,%rbp
  and    $0xfffffffc,%esi
  shr    $0x2,%edx
  movd   %ebx,%xmm4
  xor    %r12d,%r12d
  lea    0x1(%rsi),%ecx
  lea    0x2(%rsi),%edi
  mov    %esi,-0x24(%rsp)
  movslq %esi,%r13
  lea    -0x1(%rdx),%r8d
  mov    %r11d,%esi
  lea    -0x1(%r11),%edx
  mov    %ecx,-0x20(%rsp)
  pshufd $0x0,%xmm4,%xmm2
  mov    %edi,-0x1c(%rsp)
  add    $0x1,%r8
  movslq %ecx,%r14
  mov    %edx,-0x28(%rsp)
  movdqa %xmm2,%xmm3
  shl    $0x4,%r8
  movslq %edi,%r15
  mov    %rsi,-0x10(%rsp)
  psrlq  $0x20,%xmm3
  nopl   (%rax)
L31:
  mov    %rax,%rdx
  lea    -0x4(%r10),%rcx
  lea    -0x4(%r9),%rsi
  sub    %r10,%rdx
  cmp    $0x8,%rdx
  mov    %rax,%rdx
  seta   %dil
  sub    %r9,%rdx
  cmp    $0x8,%rdx
  seta   %dl
  test   %dl,%dil
  je     L10
  xor    %edx,%edx
  cmpl   $0x2,-0x28(%rsp)
  jbe    L10
  nopl   0x0(%rax)
L38:
  movdqu (%rcx,%rdx,1),%xmm0
  movdqu (%rcx,%rdx,1),%xmm1
  psrlq  $0x20,%xmm0
  pmuludq %xmm2,%xmm1
  pmuludq %xmm3,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  movdqu (%rsi,%rdx,1),%xmm0
  paddd  %xmm1,%xmm0
  movups %xmm0,(%rax,%rdx,1)
  add    $0x10,%rdx
  cmp    %r8,%rdx
  jne    L38
  cmp    -0x24(%rsp),%r11d
  je     L56
  mov    (%rcx,%r13,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r13,4),%edx
  mov    %edx,(%rax,%r13,4)
  cmp    -0x20(%rsp),%r11d
  jle    L56
  mov    (%rcx,%r14,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r14,4),%edx
  mov    %edx,(%rax,%r14,4)
  cmp    -0x1c(%rsp),%r11d
  jle    L56
  mov    (%rcx,%r15,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r15,4),%edx
  mov    %edx,(%rax,%r15,4)
L56:
  add    $0x1,%r12d
  add    %rbp,%rax
  add    %rbp,%r10
  add    %rbp,%r9
  cmp    %r12d,%r11d
  jne    L31
L16:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L10:
  mov    %r8,-0x18(%rsp)
  mov    -0x10(%rsp),%r8
  xor    %edx,%edx
  nopl   0x0(%rax)
L63:
  mov    (%rcx,%rdx,4),%edi
  imul   %ebx,%edi
  add    (%rsi,%rdx,4),%edi
  mov    %edi,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%r8
  jne    L63
  mov    -0x18(%rsp),%r8
  jmp    L56
```

assistant: ```c
void multiply243(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<reset_control_assert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0x1,%edi
  call   <WARN_ON@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int reset_control_assert(struct reset_control *rstc) {
  WARN_ON(1);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reset_control_assert>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <WARN_ON@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int reset_control_assert(struct reset_control *rstc) {
  WARN_ON(1);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reset_control_assert>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <WARN_ON@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int reset_control_assert(struct reset_control *rstc) {
  WARN_ON(1);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<reset_control_assert>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <WARN_ON@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int reset_control_assert(struct reset_control *rstc) {
  WARN_ON(1);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<QueryRootPosition>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %rdi,-0x58(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %rdx,-0x68(%rbp)
  mov    %rcx,-0x70(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rdx
  mov    -0x60(%rbp),%rcx
  mov    -0x58(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <XGetWindowAttributes@plt>
  test   %rax,%rax
  jne    L36
  mov    $0x1,%eax
  jmp    L12
L36:
  mov    -0x20(%rbp),%rax
  cmp    %rax,-0x60(%rbp)
  jne    L14
  mov    -0x68(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x70(%rbp),%rax
  movl   $0x0,(%rax)
  mov    $0x0,%eax
  jmp    L12
L14:
  mov    -0x18(%rbp),%edx
  mov    -0x68(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x14(%rbp),%edx
  mov    -0x70(%rbp),%rax
  mov    %edx,(%rax)
  lea    -0x3c(%rbp),%r8
  lea    -0x28(%rbp),%rdi
  lea    -0x30(%rbp),%rcx
  lea    -0x38(%rbp),%rdx
  mov    -0x60(%rbp),%rsi
  mov    -0x58(%rbp),%rax
  mov    %r8,%r9
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <XQueryTree@plt>
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <XFree@plt>
  mov    -0x30(%rbp),%rsi
  lea    -0x40(%rbp),%rcx
  lea    -0x44(%rbp),%rdx
  mov    -0x58(%rbp),%rax
  mov    %rax,%rdi
  call   <QueryRootPosition>
  mov    -0x68(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x44(%rbp),%eax
  add    %eax,%edx
  mov    -0x68(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x70(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x40(%rbp),%eax
  add    %eax,%edx
  mov    -0x70(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
L12:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L39
  call   <__stack_chk_fail@plt>
L39:
  leave
  ret
```

assistant: ```c
int QueryRootPosition(Display *dpy, Window win, int *x, int *y) {
  int X, Y;
  unsigned int nchildren;
  XWindowAttributes win_att;
  Window root, parent, *children;
  if (XGetWindowAttributes(dpy, win, &win_att) == 0)
    return (1);
  if (win == win_att.root) {
    *x = 0;
    *y = 0;
    return (0);
  }
  *x = win_att.x;
  *y = win_att.y;
  XQueryTree(dpy, win, &root, &parent, &children, &nchildren);
  XFree(children);
  QueryRootPosition(dpy, parent, &X, &Y);
  *x += X;
  *y += Y;
  return (0);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<QueryRootPosition>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r13
  mov    %rcx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rdx
  call   <XGetWindowAttributes@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  je     L12
  cmp    %rbx,0x30(%rsp)
  je     L14
  mov    0x38(%rsp),%eax
  mov    %eax,0x0(%r13)
  mov    0x3c(%rsp),%eax
  mov    %eax,(%r12)
  lea    0x20(%rsp),%rcx
  lea    0x18(%rsp),%rdx
  lea    0x14(%rsp),%r9
  lea    0x28(%rsp),%r8
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <XQueryTree@plt>
  mov    0x28(%rsp),%rdi
  call   <XFree@plt>
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  mov    0x20(%rsp),%rsi
  mov    %rbp,%rdi
  call   <QueryRootPosition>
  mov    0xc(%rsp),%eax
  add    %eax,0x0(%r13)
  mov    0x10(%rsp),%eax
  add    %eax,(%r12)
  mov    $0x0,%eax
L12:
  mov    0x48(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L35
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  movl   $0x0,0x0(%r13)
  movl   $0x0,(%r12)
  mov    $0x0,%eax
  jmp    L12
L35:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int QueryRootPosition(Display *dpy, Window win, int *x, int *y) {
  int X, Y;
  unsigned int nchildren;
  XWindowAttributes win_att;
  Window root, parent, *children;
  if (XGetWindowAttributes(dpy, win, &win_att) == 0)
    return (1);
  if (win == win_att.root) {
    *x = 0;
    *y = 0;
    return (0);
  }
  *x = win_att.x;
  *y = win_att.y;
  XQueryTree(dpy, win, &root, &parent, &children, &nchildren);
  XFree(children);
  QueryRootPosition(dpy, parent, &X, &Y);
  *x += X;
  *y += Y;
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<QueryRootPosition>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rdx
  call   <XGetWindowAttributes@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  je     L26
  cmp    %rbp,0x30(%rsp)
  je     L3
  mov    0x38(%rsp),%eax
  lea    0x20(%rsp),%rcx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  lea    0x18(%rsp),%rdx
  lea    0x14(%rsp),%r9
  mov    %eax,0x0(%r13)
  mov    0x3c(%rsp),%eax
  lea    0x28(%rsp),%r8
  mov    %eax,(%rbx)
  call   <XQueryTree@plt>
  mov    0x28(%rsp),%rdi
  call   <XFree@plt>
  mov    0x20(%rsp),%rsi
  lea    0x10(%rsp),%rcx
  mov    %r12,%rdi
  lea    0xc(%rsp),%rdx
  call   <QueryRootPosition>
  mov    0xc(%rsp),%eax
  add    %eax,0x0(%r13)
  mov    0x10(%rsp),%eax
  add    %eax,(%rbx)
  xor    %eax,%eax
L26:
  mov    0x48(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L37
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L3:
  movl   $0x0,0x0(%r13)
  xor    %eax,%eax
  movl   $0x0,(%rbx)
  jmp    L26
L37:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int QueryRootPosition(Display *dpy, Window win, int *x, int *y) {
  int X, Y;
  unsigned int nchildren;
  XWindowAttributes win_att;
  Window root, parent, *children;
  if (XGetWindowAttributes(dpy, win, &win_att) == 0)
    return (1);
  if (win == win_att.root) {
    *x = 0;
    *y = 0;
    return (0);
  }
  *x = win_att.x;
  *y = win_att.y;
  XQueryTree(dpy, win, &root, &parent, &children, &nchildren);
  XFree(children);
  QueryRootPosition(dpy, parent, &X, &Y);
  *x += X;
  *y += Y;
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<QueryRootPosition>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rdx
  call   <XGetWindowAttributes@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  je     L16
  cmp    %rbp,0x30(%rsp)
  je     L14
  mov    0x38(%rsp),%eax
  lea    0x20(%rsp),%rcx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  lea    0x18(%rsp),%rdx
  lea    0x14(%rsp),%r9
  mov    %eax,0x0(%r13)
  mov    0x3c(%rsp),%eax
  lea    0x28(%rsp),%r8
  mov    %eax,(%rbx)
  call   <XQueryTree@plt>
  mov    0x28(%rsp),%rdi
  call   <XFree@plt>
  mov    0x20(%rsp),%rsi
  lea    0x10(%rsp),%rcx
  mov    %r12,%rdi
  lea    0xc(%rsp),%rdx
  call   <QueryRootPosition>
  mov    0xc(%rsp),%eax
  add    %eax,0x0(%r13)
  mov    0x10(%rsp),%eax
  add    %eax,(%rbx)
  xor    %eax,%eax
L16:
  mov    0x48(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L35
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L14:
  movl   $0x0,0x0(%r13)
  xor    %eax,%eax
  movl   $0x0,(%rbx)
  jmp    L16
L35:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int QueryRootPosition(Display *dpy, Window win, int *x, int *y) {
  int X, Y;
  unsigned int nchildren;
  XWindowAttributes win_att;
  Window root, parent, *children;
  if (XGetWindowAttributes(dpy, win, &win_att) == 0)
    return (1);
  if (win == win_att.root) {
    *x = 0;
    *y = 0;
    return (0);
  }
  *x = win_att.x;
  *y = win_att.y;
  XQueryTree(dpy, win, &root, &parent, &children, &nchildren);
  XFree(children);
  QueryRootPosition(dpy, parent, &X, &Y);
  *x += X;
  *y += Y;
  return (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sax_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L3:
  mov    -0x4(%rbp),%eax
  shl    $0x5,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  shr    $0x2,%eax
  add    %eax,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %edx,%eax
  xor    %eax,-0x4(%rbp)
  addq   $0x1,-0x18(%rbp)
L17:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L3
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int sax_hash(const char *key) {
  unsigned int h = 0;
  while (*key) {
    h ^= (h << 5) + (h >> 2) + (unsigned char)*key;
    ++key;
  }
  return h;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sax_hash>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L17
  mov    $0x0,%ecx
L2:
  mov    %ecx,%eax
  shl    $0x5,%eax
  mov    %ecx,%esi
  shr    $0x2,%esi
  add    %esi,%eax
  movzbl %dl,%edx
  add    %edx,%eax
  xor    %eax,%ecx
  add    $0x1,%rdi
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L2
L26:
  mov    %ecx,%eax
  ret
L17:
  mov    $0x0,%ecx
  jmp    L26
```

assistant: ```c
unsigned int sax_hash(const char *key) {
  unsigned int h = 0;
  while (*key) {
    h ^= (h << 5) + (h >> 2) + (unsigned char)*key;
    ++key;
  }
  return h;
}
```
user: Translate the below assembly code into C programming language:

```asm
<sax_hash>:
  endbr64
  movzbl (%rdi),%edx
  xor    %r8d,%r8d
  test   %dl,%dl
  je     L3
  xchg   %ax,%ax
L11:
  mov    %r8d,%eax
  mov    %r8d,%ecx
  add    $0x1,%rdi
  shl    $0x5,%eax
  shr    $0x2,%ecx
  add    %ecx,%eax
  add    %edx,%eax
  movzbl (%rdi),%edx
  xor    %eax,%r8d
  test   %dl,%dl
  jne    L11
L3:
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int sax_hash(const char *key) {
  unsigned int h = 0;
  while (*key) {
    h ^= (h << 5) + (h >> 2) + (unsigned char)*key;
    ++key;
  }
  return h;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<sax_hash>:
  endbr64
  movzbl (%rdi),%edx
  xor    %r8d,%r8d
  test   %dl,%dl
  je     L19
  xchg   %ax,%ax
L8:
  mov    %r8d,%eax
  mov    %r8d,%ecx
  add    $0x1,%rdi
  shl    $0x5,%eax
  shr    $0x2,%ecx
  add    %ecx,%eax
  add    %edx,%eax
  movzbl (%rdi),%edx
  xor    %eax,%r8d
  test   %dl,%dl
  jne    L8
L19:
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int sax_hash(const char *key) {
  unsigned int h = 0;
  while (*key) {
    h ^= (h << 5) + (h >> 2) + (unsigned char)*key;
    ++key;
  }
  return h;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<byteSwapDouble>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movsd  %xmm0,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movzbl -0x11(%rbp),%eax
  mov    %al,-0x10(%rbp)
  movzbl -0x12(%rbp),%eax
  mov    %al,-0xf(%rbp)
  movzbl -0x13(%rbp),%eax
  mov    %al,-0xe(%rbp)
  movzbl -0x14(%rbp),%eax
  mov    %al,-0xd(%rbp)
  movzbl -0x15(%rbp),%eax
  mov    %al,-0xc(%rbp)
  movzbl -0x16(%rbp),%eax
  mov    %al,-0xb(%rbp)
  movzbl -0x17(%rbp),%eax
  mov    %al,-0xa(%rbp)
  movzbl -0x18(%rbp),%eax
  mov    %al,-0x9(%rbp)
  movsd  -0x10(%rbp),%xmm0
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double byteSwapDouble(double a) {
  union {
    double whole;
    unsigned char bytes[4];
  } u, v;
  u.whole = a;
  v.bytes[0] = u.bytes[7];
  v.bytes[1] = u.bytes[6];
  v.bytes[2] = u.bytes[5];
  v.bytes[3] = u.bytes[4];
  v.bytes[4] = u.bytes[3];
  v.bytes[5] = u.bytes[2];
  v.bytes[6] = u.bytes[1];
  v.bytes[7] = u.bytes[0];
  return v.whole;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<byteSwapDouble>:
  endbr64
  movq   %xmm0,%rdx
  movq   %xmm0,%rcx
  shr    $0x38,%rcx
  mov    $0x0,%eax
  mov    %cl,%al
  movq   %xmm0,%rcx
  shr    $0x30,%rcx
  mov    %cl,%ah
  movq   %xmm0,%rcx
  shr    $0x18,%rcx
  and    $0xff0000,%ecx
  and    $0xffffffffff00ffff,%rax
  or     %rcx,%rax
  movq   %xmm0,%rcx
  shr    $0x8,%rcx
  and    $0xff000000,%ecx
  movabs $0xffffffff00ffffff,%rsi
  and    %rsi,%rax
  or     %rcx,%rax
  movq   %xmm0,%rcx
  shr    $0x18,%rcx
  movzbl %cl,%ecx
  shl    $0x20,%rcx
  movabs $0xffffff00ffffffff,%rsi
  and    %rsi,%rax
  or     %rcx,%rax
  movq   %xmm0,%rcx
  shr    $0x10,%rcx
  movzbl %cl,%ecx
  shl    $0x28,%rcx
  movabs $0xffff00ffffffffff,%rsi
  and    %rsi,%rax
  or     %rcx,%rax
  movzbl %dh,%ecx
  shl    $0x30,%rcx
  movabs $0xff00ffffffffffff,%rsi
  and    %rsi,%rax
  or     %rcx,%rax
  shl    $0x38,%rdx
  movabs $0xffffffffffffff,%rcx
  and    %rcx,%rax
  or     %rdx,%rax
  movq   %rax,%xmm0
  ret
```

assistant: ```c
double byteSwapDouble(double a) {
  union {
    double whole;
    unsigned char bytes[4];
  } u, v;
  u.whole = a;
  v.bytes[0] = u.bytes[7];
  v.bytes[1] = u.bytes[6];
  v.bytes[2] = u.bytes[5];
  v.bytes[3] = u.bytes[4];
  v.bytes[4] = u.bytes[3];
  v.bytes[5] = u.bytes[2];
  v.bytes[6] = u.bytes[1];
  v.bytes[7] = u.bytes[0];
  return v.whole;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<byteSwapDouble>:
  endbr64
  movq   %xmm0,%rcx
  xor    %eax,%eax
  movq   %xmm0,%rdx
  movabs $0xffffffff00ffffff,%rsi
  shr    $0x38,%rcx
  mov    %cl,%al
  movq   %xmm0,%rcx
  shr    $0x30,%rcx
  mov    %cl,%ah
  movq   %xmm0,%rcx
  shr    $0x18,%rcx
  and    $0xffffffffff00ffff,%rax
  and    $0xff0000,%ecx
  or     %rcx,%rax
  movq   %xmm0,%rcx
  shr    $0x8,%rcx
  and    %rsi,%rax
  movabs $0xffffff00ffffffff,%rsi
  and    $0xff000000,%ecx
  or     %rcx,%rax
  movq   %xmm0,%rcx
  shr    $0x18,%rcx
  and    %rsi,%rax
  movabs $0xffff00ffffffffff,%rsi
  movzbl %cl,%ecx
  shl    $0x20,%rcx
  or     %rcx,%rax
  movq   %xmm0,%rcx
  shr    $0x10,%rcx
  and    %rsi,%rax
  movabs $0xff00ffffffffffff,%rsi
  movzbl %cl,%ecx
  shl    $0x28,%rcx
  or     %rcx,%rax
  movzbl %dh,%ecx
  shl    $0x38,%rdx
  shl    $0x30,%rcx
  and    %rsi,%rax
  or     %rcx,%rax
  movabs $0xffffffffffffff,%rcx
  and    %rcx,%rax
  or     %rdx,%rax
  movq   %rax,%xmm0
  ret
```

assistant: ```c
double byteSwapDouble(double a) {
  union {
    double whole;
    unsigned char bytes[4];
  } u, v;
  u.whole = a;
  v.bytes[0] = u.bytes[7];
  v.bytes[1] = u.bytes[6];
  v.bytes[2] = u.bytes[5];
  v.bytes[3] = u.bytes[4];
  v.bytes[4] = u.bytes[3];
  v.bytes[5] = u.bytes[2];
  v.bytes[6] = u.bytes[1];
  v.bytes[7] = u.bytes[0];
  return v.whole;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<byteSwapDouble>:
  endbr64
  movq   %xmm0,%rdx
  push   %rbx
  movq   %xmm0,%r9
  movq   %xmm0,%r8
  movzbl %dl,%eax
  movzbl %dh,%ebx
  shr    $0x10,%r9
  movq   %xmm0,%rdi
  shl    $0x8,%rax
  movzbl %r9b,%r9d
  shr    $0x18,%r8
  movq   %xmm0,%rsi
  or     %rbx,%rax
  movzbl %r8b,%r8d
  shr    $0x20,%rdi
  movq   %xmm0,%rcx
  shl    $0x8,%rax
  movzbl %dil,%edi
  shr    $0x28,%rsi
  pop    %rbx
  shr    $0x30,%rcx
  or     %r9,%rax
  movzbl %sil,%esi
  shr    $0x38,%rdx
  shl    $0x8,%rax
  movzbl %cl,%ecx
  or     %r8,%rax
  shl    $0x8,%rax
  or     %rdi,%rax
  shl    $0x8,%rax
  or     %rsi,%rax
  shl    $0x8,%rax
  or     %rcx,%rax
  shl    $0x8,%rax
  or     %rdx,%rax
  movq   %rax,%xmm0
  ret
```

assistant: ```c
double byteSwapDouble(double a) {
  union {
    double whole;
    unsigned char bytes[4];
  } u, v;
  u.whole = a;
  v.bytes[0] = u.bytes[7];
  v.bytes[1] = u.bytes[6];
  v.bytes[2] = u.bytes[5];
  v.bytes[3] = u.bytes[4];
  v.bytes[4] = u.bytes[3];
  v.bytes[5] = u.bytes[2];
  v.bytes[6] = u.bytes[1];
  v.bytes[7] = u.bytes[0];
  return v.whole;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<reapchild>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D11(%rip),%rax
  mov    (%rax),%eax
  mov    D12(%rip),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <signal@plt>
  mov    $0x0,%edi
  call   <wait@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void reapchild() {
  signal(SIGCHLD, reapchild);
  wait(NULL);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<reapchild>:
  endbr64
  sub    $0x8,%rsp
  mov    D5(%rip),%rsi
  mov    D18(%rip),%rax
  mov    (%rax),%edi
  call   <signal@plt>
  mov    $0x0,%edi
  call   <wait@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void reapchild() {
  signal(SIGCHLD, reapchild);
  wait(NULL);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<reapchild>:
  endbr64
  sub    $0x8,%rsp
  mov    D12(%rip),%rax
  mov    D1(%rip),%rsi
  mov    (%rax),%edi
  call   <signal@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <wait@plt>
```

assistant: ```c
void reapchild() {
  signal(SIGCHLD, reapchild);
  wait(NULL);
}
```
user: Translate the below assembly code into C programming language:

```asm
<reapchild>:
  endbr64
  sub    $0x8,%rsp
  mov    D18(%rip),%rax
  mov    D6(%rip),%rsi
  mov    (%rax),%edi
  call   <signal@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <wait@plt>
```

assistant: ```c
void reapchild() {
  signal(SIGCHLD, reapchild);
  wait(NULL);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  sub    $0x6a,%eax
  mov    %al,-0x9(%rbp)
  movb   $0x0,-0xa(%rbp)
  jmp    L21
L22:
  movb   $0x0,-0xb(%rbp)
  jmp    L32
L19:
  movzbl -0x9(%rbp),%edx
  movzbl -0xb(%rbp),%eax
  add    %edx,%edx
  cltq
  mov    %dl,-0x9(%rbp,%rax,1)
  movzbl -0xb(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0xb(%rbp)
L32:
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    %al,-0xb(%rbp)
  jb     L19
  addb   $0x2,-0xa(%rbp)
L21:
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    %al,-0xa(%rbp)
  jb     L22
  movzbl -0x9(%rbp),%eax
  lea    -0x72(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L49
  call   <__stack_chk_fail@plt>
L49:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local2;
  unsigned char local1;
  {
    state[0UL] = (input[0UL] + 51238316UL) + (unsigned char)234;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      local2 = 0UL;
      while (local2 < input[1UL]) {
        state[local2] = state[0UL] + state[0UL];
        local2++;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + (unsigned char)142;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  lea    -0x6a(%rax),%edx
  movzbl 0x1(%rdi),%ecx
  mov    $0x0,%edi
  test   %cl,%cl
  je     L0
L9:
  mov    $0x0,%eax
L3:
  add    %edx,%edx
  add    $0x1,%eax
  cmp    %cl,%al
  jne    L3
  add    $0x2,%edi
  cmp    %cl,%dil
  jb     L9
L0:
  sub    $0x72,%edx
  mov    %dl,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local2;
  unsigned char local1;
  {
    state[0UL] = (input[0UL] + 51238316UL) + (unsigned char)234;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      local2 = 0UL;
      while (local2 < input[1UL]) {
        state[local2] = state[0UL] + state[0UL];
        local2++;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + (unsigned char)142;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  movzbl 0x1(%rdi),%ecx
  xor    %edi,%edi
  lea    -0x6a(%rax),%edx
  test   %cl,%cl
  je     L26
  nopl   0x0(%rax)
L1:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L4:
  add    $0x1,%eax
  add    %edx,%edx
  cmp    %cl,%al
  jne    L4
  add    $0x2,%edi
  cmp    %cl,%dil
  jb     L1
L26:
  sub    $0x72,%edx
  mov    %dl,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local2;
  unsigned char local1;
  {
    state[0UL] = (input[0UL] + 51238316UL) + (unsigned char)234;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      local2 = 0UL;
      while (local2 < input[1UL]) {
        state[local2] = state[0UL] + state[0UL];
        local2++;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + (unsigned char)142;
  }
}
```