ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<draw_404_page>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1e0,%r8d
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0x2d0,%ecx
  mov    $0x2,%edi
  call   <bmp_fill@plt>
  mov    $0x14,%edx
  mov    $0xa,%esi
  lea    D63(%rip),%rcx
  mov    $0x70000,%edi
  call   <bmp_printf@plt>
  mov    $0x64,%edx
  mov    $0xa,%esi
  lea    D47(%rip),%rcx
  mov    $0x20000,%edi
  call   <bmp_printf@plt>
  mov    $0x78,%edx
  mov    $0xa,%esi
  lea    D19(%rip),%rcx
  mov    $0x20000,%edi
  call   <bmp_printf@plt>
  mov    $0xb4,%edx
  mov    $0xa,%esi
  lea    D9(%rip),%rcx
  mov    $0x20000,%edi
  call   <bmp_printf@plt>
  mov    $0xf0,%edx
  mov    $0xa,%esi
  lea    D3(%rip),%rcx
  mov    $0x20000,%edi
  call   <bmp_printf@plt>
  mov    $0x104,%edx
  mov    $0xa,%esi
  lea    D10(%rip),%rcx
  mov    $0x20000,%edi
  call   <bmp_printf@plt>
  lea    D14(%rip),%rcx
  mov    $0x140,%edx
  mov    $0xa,%esi
  mov    $0x20000,%edi
  add    $0x8,%rsp
  jmp    1060 <bmp_printf@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D63
type_infered: string
name: retrieve_data
arguments: data_label: D47
type_infered: string
name: retrieve_data
arguments: data_label: D19
type_infered: string
name: retrieve_data
arguments: data_label: D9
type_infered: string
name: retrieve_data
arguments: data_label: D3
type_infered: string
name: retrieve_data
arguments: data_label: D10
type_infered: string
name: retrieve_data
arguments: data_label: D14
type_infered: string

tool: D63: 404 Undocumented Feature

tool: D47: This feature is probably not yet documented.

tool: D19: After all, we are programmers, not tech writers.

tool: D9: But... you can simply try it and see what it does.

tool: D3: Then, write a short paragraph to describe it,

tool: D10: and we will include it in the user guide.

tool: D14: Thanks!

assistant: ```c
void draw_404_page() {
  bmp_fill(0x02, 0, 0, 720, 480);
  bmp_printf((0 | (((7) << 16) & (0xFFFF0000)) | ((0) & 0xFF) << 8 | ((0) & 0xFF) << 0), 10, 20, "404 Undocumented Feature");
  bmp_printf((0 | (((2) << 16) & (0xFFFF0000)) | ((0) & 0xFF) << 8 | ((0) & 0xFF) << 0), 10, 100, "This feature is probably not yet documented.");
  bmp_printf((0 | (((2) << 16) & (0xFFFF0000)) | ((0) & 0xFF) << 8 | ((0) & 0xFF) << 0), 10, 120, "After all, we are programmers, not tech writers.");
  bmp_printf((0 | (((2) << 16) & (0xFFFF0000)) | ((0) & 0xFF) << 8 | ((0) & 0xFF) << 0), 10, 180, "But... you can simply try it and see what it does.");
  bmp_printf((0 | (((2) << 16) & (0xFFFF0000)) | ((0) & 0xFF) << 8 | ((0) & 0xFF) << 0), 10, 240, "Then, write a short paragraph to describe it,");
  bmp_printf((0 | (((2) << 16) & (0xFFFF0000)) | ((0) & 0xFF) << 8 | ((0) & 0xFF) << 0), 10, 260, "and we will include it in the user guide.");
  bmp_printf((0 | (((2) << 16) & (0xFFFF0000)) | ((0) & 0xFF) << 8 | ((0) & 0xFF) << 0), 10, 320, "Thanks!");
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x2e,-0x8(%rbp)
  movl   $0x14,-0x4(%rbp)
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x1,-0x10(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L14
L18:
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0xc(%rbp),%xmm0
  movsd  D15(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rdx
  mov    -0x10(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0xc(%rbp)
L14:
  cmpl   $0x10,-0x10(%rbp)
  jle    L18
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D6
type: string
name: disassemble_data
arguments: data_tag: D15
type: f64
name: disassemble_data
arguments: data_tag: D20
type: string

tool: D6:  ounces  cost

tool: D15: 100

tool: D20: %5d   $%4.2f


assistant: ```c
int main() {
  const int FIRST_OZ = 46;
  const int NEXT_OZ = 20;
  int ounces, cost;
  printf(" ounces  cost\n");
  for (ounces = 1, cost = FIRST_OZ; ounces <= 16; ounces++, cost += NEXT_OZ)
    printf("%5d   $%4.2f\n", ounces, cost / 100.0);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    D4(%rip),%rdi
  call   <puts@plt>
  mov    $0x2e,%ebp
  mov    $0x1,%ebx
  lea    D19(%rip),%r12
L3:
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebp,%xmm0
  divsd  D16(%rip),%xmm0
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  add    $0x14,%ebp
  cmp    $0x11,%ebx
  jne    L3
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D4
type_guessed: string
name: read_data
arguments: label: D19
type_guessed: string
name: read_data
arguments: label: D16
type_guessed: f64

tool: D4:  ounces  cost

tool: D19: %5d   $%4.2f


tool: D16: 100

assistant: ```c
int main() {
  const int FIRST_OZ = 46;
  const int NEXT_OZ = 20;
  int ounces, cost;
  printf(" ounces  cost\n");
  for (ounces = 1, cost = FIRST_OZ; ounces <= 16; ounces++, cost += NEXT_OZ)
    printf("%5d   $%4.2f\n", ounces, cost / 100.0);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  lea    D15(%rip),%rdi
  lea    D29(%rip),%r12
  push   %rbp
  mov    $0x2e,%ebp
  push   %rbx
  mov    $0x1,%ebx
  call   <puts@plt>
  nopl   (%rax)
L9:
  mov    %ebx,%esi
  pxor   %xmm0,%xmm0
  mov    %r12,%rdi
  mov    $0x1,%eax
  add    $0x1,%ebx
  cvtsi2sd %ebp,%xmm0
  add    $0x14,%ebp
  divsd  D20(%rip),%xmm0
  call   <printf@plt>
  cmp    $0x11,%ebx
  jne    L9
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D15
data_type: string
name: extract_symbolic_data
arguments: data_tag: D29
data_type: string
name: extract_symbolic_data
arguments: data_tag: D20
data_type: f64

tool: D15:  ounces  cost

tool: D29: %5d   $%4.2f


tool: D20: 100

assistant: ```c
int main() {
  const int FIRST_OZ = 46;
  const int NEXT_OZ = 20;
  int ounces, cost;
  printf(" ounces  cost\n");
  for (ounces = 1, cost = FIRST_OZ; ounces <= 16; ounces++, cost += NEXT_OZ)
    printf("%5d   $%4.2f\n", ounces, cost / 100.0);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  lea    D24(%rip),%rdi
  lea    D12(%rip),%r12
  push   %rbp
  mov    $0x2e,%ebp
  push   %rbx
  mov    $0x1,%ebx
  call   <puts@plt>
  nopl   (%rax)
L6:
  mov    %ebx,%esi
  pxor   %xmm0,%xmm0
  mov    %r12,%rdi
  mov    $0x1,%eax
  add    $0x1,%ebx
  cvtsi2sd %ebp,%xmm0
  add    $0x14,%ebp
  divsd  D18(%rip),%xmm0
  call   <printf@plt>
  cmp    $0x11,%ebx
  jne    L6
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: tag: D24
type: string
name: unpack_data
arguments: tag: D12
type: string
name: unpack_data
arguments: tag: D18
type: f64

tool: D24:  ounces  cost

tool: D12: %5d   $%4.2f


tool: D18: 100

assistant: ```c
int main() {
  const int FIRST_OZ = 46;
  const int NEXT_OZ = 20;
  int ounces, cost;
  printf(" ounces  cost\n");
  for (ounces = 1, cost = FIRST_OZ; ounces <= 16; ounces++, cost += NEXT_OZ)
    printf("%5d   $%4.2f\n", ounces, cost / 100.0);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bubble_decrease>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jae    L10
  mov    -0x20(%rbp),%rax
  sub    -0x18(%rbp),%rax
  sar    $0x2,%rax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L40
L11:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jge    L42
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
L42:
  addl   $0x1,-0x8(%rbp)
L40:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L11
  mov    -0x20(%rbp),%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <bubble_decrease>
  jmp    L8
L10:
  nop
L8:
  leave
  ret
```

assistant: ```c
void bubble_decrease(int *front, int *end) {
  if (front >= end)
    return;
  int size = end - front;
  int i;
  for (i = 0; i < size; ++i)
    if (*(front + i) < *(front + i + 1))
      swap(front + i, front + i + 1);
  bubble_decrease(front, end - 1);
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<bubble_decrease>:
  endbr64
  cmp    %rsi,%rdi
  jae    L22
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    %rsi,%rax
  sub    %rdi,%rax
  sar    $0x2,%rax
  test   %eax,%eax
  jle    L48
  mov    %rdi,%rbx
  lea    -0x1(%rax),%eax
  lea    0x4(%rdi,%rax,4),%r12
  jmp    L2
L28:
  cmp    %rbx,%r12
  je     L48
L2:
  mov    %rbx,%rdi
  add    $0x4,%rbx
  mov    0x4(%rdi),%eax
  cmp    %eax,(%rdi)
  jge    L28
  mov    %rbx,%rsi
  call   <swap@plt>
  jmp    L28
L48:
  lea    -0x4(%rbp),%rsi
  mov    %r13,%rdi
  call   <bubble_decrease>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L22:
  ret
```

assistant: ```c
void bubble_decrease(int *front, int *end) {
  if (front >= end)
    return;
  int size = end - front;
  int i;
  for (i = 0; i < size; ++i)
    if (*(front + i) < *(front + i + 1))
      swap(front + i, front + i + 1);
  bubble_decrease(front, end - 1);
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<bubble_decrease>:
  endbr64
  cmp    %rdi,%rsi
  jbe    L7
  push   %r14
  lea    0x4(%rdi),%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  nopl   0x0(%rax,%rax,1)
L32:
  mov    %r12,%rax
  sub    %r13,%rax
  sar    $0x2,%rax
  test   %eax,%eax
  jle    L58
  sub    $0x1,%eax
  mov    %r13,%rbx
  lea    (%r14,%rax,4),%rbp
  jmp    L43
  nopw   0x0(%rax,%rax,1)
L12:
  cmp    %rbp,%rbx
  je     L58
L43:
  mov    %rbx,%rdi
  add    $0x4,%rbx
  mov    0x4(%rdi),%eax
  cmp    %eax,(%rdi)
  jge    L12
  mov    %rbx,%rsi
  call   <swap@plt>
  cmp    %rbp,%rbx
  jne    L43
L58:
  sub    $0x4,%r12
  cmp    %r13,%r12
  ja     L32
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L7:
  ret
```

assistant: ```c
void bubble_decrease(int *front, int *end) {
  if (front >= end)
    return;
  int size = end - front;
  int i;
  for (i = 0; i < size; ++i)
    if (*(front + i) < *(front + i + 1))
      swap(front + i, front + i + 1);
  bubble_decrease(front, end - 1);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bubble_decrease>:
  endbr64
  cmp    %rdi,%rsi
  jbe    L43
  push   %r14
  lea    0x4(%rdi),%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %r12,%rax
  sub    %r13,%rax
  sar    $0x2,%rax
  test   %eax,%eax
  jle    L51
  sub    $0x1,%eax
  mov    %r13,%rbx
  lea    (%r14,%rax,4),%rbp
  jmp    L16
  nopw   0x0(%rax,%rax,1)
L21:
  cmp    %rbp,%rbx
  je     L51
L16:
  mov    %rbx,%rdi
  add    $0x4,%rbx
  mov    0x4(%rdi),%eax
  cmp    %eax,(%rdi)
  jge    L21
  mov    %rbx,%rsi
  call   <swap@plt>
  cmp    %rbp,%rbx
  jne    L16
L51:
  sub    $0x4,%r12
  cmp    %r13,%r12
  ja     L6
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L43:
  ret
```

assistant: ```c
void bubble_decrease(int *front, int *end) {
  if (front >= end)
    return;
  int size = end - front;
  int i;
  for (i = 0; i < size; ++i)
    if (*(front + i) < *(front + i + 1))
      swap(front + i, front + i + 1);
  bubble_decrease(front, end - 1);
  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
<read_long_number>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    $0x6,%esi
  mov    %rax,%rdi
  call   <read_n10@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x6,-0x14(%rbp)
  jne    L75
  mov    $0x0,%eax
  jmp    L11
L75:
  cmpl   $0x0,-0x14(%rbp)
  jne    L26
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <read_bit@plt>
  test   %rax,%rax
  jne    L1
  mov    $0x1,%eax
  jmp    L11
L1:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <read_bit@plt>
  test   %rax,%rax
  je     L47
  mov    $0x2,%eax
  jmp    L11
L47:
  mov    $0x3,%eax
  jmp    L11
L26:
  mov    -0x14(%rbp),%eax
  add    $0x2,%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  lea    -0x1(%rax),%ebx
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <read_number@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
L11:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
unsigned read_long_number(struct state *state) {
  unsigned order = read_n10(state, 6);
  if (order == 6)
    return 0;
  if (order == 0) {
    if (!read_bit(state))
      return 1;
    return read_bit(state) ? 2 : 3;
  }
  return (1 << (order + 2)) - 1 - read_number(state, order + 1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<read_long_number>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x6,%esi
  call   <read_n10@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  cmp    $0x6,%ebx
  je     L38
  test   %ebx,%ebx
  jne    L15
  mov    %rbp,%rdi
  call   <read_bit@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  je     L38
  mov    %rbp,%rdi
  call   <read_bit@plt>
  cmp    $0x1,%rax
  mov    $0x2,%eax
  adc    $0x0,%eax
  jmp    L38
L15:
  lea    0x1(%rbx),%esi
  mov    %rbp,%rdi
  call   <read_number@plt>
  mov    %eax,%edx
  lea    0x2(%rbx),%ecx
  mov    $0x1,%eax
  shl    %cl,%eax
  sub    $0x1,%eax
  sub    %edx,%eax
L38:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned read_long_number(struct state *state) {
  unsigned order = read_n10(state, 6);
  if (order == 6)
    return 0;
  if (order == 0) {
    if (!read_bit(state))
      return 1;
    return read_bit(state) ? 2 : 3;
  }
  return (1 << (order + 2)) - 1 - read_number(state, order + 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_long_number>:
  endbr64
  push   %rbp
  mov    $0x6,%esi
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <read_n10@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  cmp    $0x6,%ebx
  je     L9
  test   %ebx,%ebx
  jne    L12
  mov    %rbp,%rdi
  call   <read_bit@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  jne    L13
L9:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L12:
  lea    0x1(%rbx),%esi
  mov    %rbp,%rdi
  call   <read_number@plt>
  lea    0x2(%rbx),%ecx
  add    $0x8,%rsp
  mov    %eax,%r8d
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  shl    %cl,%eax
  sub    $0x1,%eax
  sub    %r8d,%eax
  ret
  nopl   (%rax)
L13:
  mov    %rbp,%rdi
  call   <read_bit@plt>
  cmp    $0x1,%rax
  mov    $0x2,%eax
  adc    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned read_long_number(struct state *state) {
  unsigned order = read_n10(state, 6);
  if (order == 6)
    return 0;
  if (order == 0) {
    if (!read_bit(state))
      return 1;
    return read_bit(state) ? 2 : 3;
  }
  return (1 << (order + 2)) - 1 - read_number(state, order + 1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<read_long_number>:
  endbr64
  push   %rbp
  mov    $0x6,%esi
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <read_n10@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  cmp    $0x6,%ebx
  je     L13
  test   %ebx,%ebx
  jne    L14
  mov    %rbp,%rdi
  call   <read_bit@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  jne    L11
L13:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L14:
  lea    0x1(%rbx),%esi
  mov    %rbp,%rdi
  call   <read_number@plt>
  lea    0x2(%rbx),%ecx
  add    $0x8,%rsp
  mov    %eax,%r8d
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  shl    %cl,%eax
  sub    $0x1,%eax
  sub    %r8d,%eax
  ret
  nopl   (%rax)
L11:
  mov    %rbp,%rdi
  call   <read_bit@plt>
  cmp    $0x1,%rax
  mov    $0x2,%eax
  adc    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned read_long_number(struct state *state) {
  unsigned order = read_n10(state, 6);
  if (order == 6)
    return 0;
  if (order == 0) {
    if (!read_bit(state))
      return 1;
    return read_bit(state) ? 2 : 3;
  }
  return (1 << (order + 2)) - 1 - read_number(state, order + 1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<randomPtr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <random@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *randomPtr(void) {
  long int x = random();
  return (void *)x;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<randomPtr>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <random@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *randomPtr(void) {
  long int x = random();
  return (void *)x;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<randomPtr>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <random@plt>
```

assistant: ```c
void *randomPtr(void) {
  long int x = random();
  return (void *)x;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<randomPtr>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <random@plt>
```

assistant: ```c
void *randomPtr(void) {
  long int x = random();
  return (void *)x;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void swap(int *a, int *b) {
  int w = *a;
  *a = *b;
  *b = w;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<swap>:
  endbr64
  mov    (%rdi),%eax
  mov    (%rsi),%edx
  mov    %edx,(%rdi)
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void swap(int *a, int *b) {
  int w = *a;
  *a = *b;
  *b = w;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<swap>:
  endbr64
  mov    (%rdi),%eax
  mov    (%rsi),%edx
  mov    %edx,(%rdi)
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void swap(int *a, int *b) {
  int w = *a;
  *a = *b;
  *b = w;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<swap>:
  endbr64
  mov    (%rdi),%eax
  mov    (%rsi),%edx
  mov    %edx,(%rdi)
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void swap(int *a, int *b) {
  int w = *a;
  *a = *b;
  *b = w;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<get_nibble_value>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x2f,-0x4(%rbp)
  jle    L4
  cmpb   $0x39,-0x4(%rbp)
  jg     L4
  movzbl -0x4(%rbp),%eax
  sub    $0x30,%eax
  mov    %al,-0x4(%rbp)
L4:
  cmpb   $0x60,-0x4(%rbp)
  jle    L54
  cmpb   $0x66,-0x4(%rbp)
  jg     L54
  movzbl -0x4(%rbp),%eax
  sub    $0x6b,%eax
  mov    %al,-0x4(%rbp)
L54:
  cmpb   $0x40,-0x4(%rbp)
  jle    L12
  cmpb   $0x46,-0x4(%rbp)
  jg     L12
  movzbl -0x4(%rbp),%eax
  sub    $0x4b,%eax
  mov    %al,-0x4(%rbp)
L12:
  movsbl -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int get_nibble_value(char c) {
  if ((c >= '0') && (c <= '9')) {
    c = c - '0';
  }
  if ((c >= 'a') && (c <= 'f')) {
    c = c - 'a' - (char)10;
  }
  if ((c >= 'A') && (c <= 'F')) {
    c = c - 'A' - (char)10;
  }
  return (int)c;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<get_nibble_value>:
  endbr64
  mov    %edi,%eax
  lea    -0x30(%rdi),%edx
  cmp    $0x9,%dl
  ja     L47
  mov    %edx,%eax
L48:
  movsbl %al,%eax
  ret
L47:
  lea    -0x61(%rdi),%edx
  cmp    $0x5,%dl
  ja     L14
  lea    -0x6b(%rdi),%eax
  jmp    L48
L14:
  lea    -0x41(%rdi),%edx
  cmp    $0x5,%dl
  ja     L48
  lea    -0x4b(%rdi),%eax
  jmp    L48
```

assistant: ```c
int get_nibble_value(char c) {
  if ((c >= '0') && (c <= '9')) {
    c = c - '0';
  }
  if ((c >= 'a') && (c <= 'f')) {
    c = c - 'a' - (char)10;
  }
  if ((c >= 'A') && (c <= 'F')) {
    c = c - 'A' - (char)10;
  }
  return (int)c;
}
```
user: What is the c source code of the assembly code below?

```asm
<get_nibble_value>:
  endbr64
  lea    -0x30(%rdi),%edx
  mov    %edi,%eax
  cmp    $0x9,%dl
  ja     L0
  mov    %edx,%eax
L26:
  movsbl %al,%eax
  ret
  nopl   0x0(%rax)
L0:
  lea    -0x61(%rdi),%edx
  cmp    $0x5,%dl
  ja     L6
  lea    -0x6b(%rdi),%eax
  movsbl %al,%eax
  ret
L6:
  lea    -0x41(%rdi),%edx
  cmp    $0x5,%dl
  ja     L26
  lea    -0x4b(%rdi),%eax
  movsbl %al,%eax
  ret
```

assistant: ```c
int get_nibble_value(char c) {
  if ((c >= '0') && (c <= '9')) {
    c = c - '0';
  }
  if ((c >= 'a') && (c <= 'f')) {
    c = c - 'a' - (char)10;
  }
  if ((c >= 'A') && (c <= 'F')) {
    c = c - 'A' - (char)10;
  }
  return (int)c;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<get_nibble_value>:
  endbr64
  lea    -0x30(%rdi),%edx
  mov    %edi,%eax
  cmp    $0x9,%dl
  ja     L24
  mov    %edx,%eax
L28:
  movsbl %al,%eax
  ret
  nopl   0x0(%rax)
L24:
  lea    -0x61(%rdi),%edx
  cmp    $0x5,%dl
  ja     L19
  lea    -0x6b(%rdi),%eax
  movsbl %al,%eax
  ret
L19:
  lea    -0x41(%rdi),%edx
  cmp    $0x5,%dl
  ja     L28
  lea    -0x4b(%rdi),%eax
  movsbl %al,%eax
  ret
```

assistant: ```c
int get_nibble_value(char c) {
  if ((c >= '0') && (c <= '9')) {
    c = c - '0';
  }
  if ((c >= 'a') && (c <= 'f')) {
    c = c - 'a' - (char)10;
  }
  if ((c >= 'A') && (c <= 'F')) {
    c = c - 'A' - (char)10;
  }
  return (int)c;
}
```
user: What is the c source code of the assembly code below?

```asm
<list_splice>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <list_empty@plt>
  test   %eax,%eax
  jne    L8
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <__list_splice@plt>
L8:
  nop
  leave
  ret
```

assistant: ```c
void list_splice(struct list_head *list, struct list_head *head) {
  if (!list_empty(list))
    __list_splice(list, head);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<list_splice>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <list_empty@plt>
  test   %eax,%eax
  je     L8
L19:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <__list_splice@plt>
  jmp    L19
```

assistant: ```c
void list_splice(struct list_head *list, struct list_head *head) {
  if (!list_empty(list))
    __list_splice(list, head);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<list_splice>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <list_empty@plt>
  test   %eax,%eax
  je     L0
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <__list_splice@plt>
```

assistant: ```c
void list_splice(struct list_head *list, struct list_head *head) {
  if (!list_empty(list))
    __list_splice(list, head);
}
```