ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<frand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movq   $0x41a7,-0x30(%rbp)
  movq   $0x7fffffff,-0x28(%rbp)
  movq   $0x1f31d,-0x20(%rbp)
  movq   $0xb14,-0x18(%rbp)
  mov    D22(%rip),%rax
  cqto
  idivq  -0x20(%rbp)
  mov    %rax,-0x10(%rbp)
  mov    D22(%rip),%rdx
  mov    -0x20(%rbp),%rax
  imul   -0x10(%rbp),%rax
  mov    %rax,%rcx
  mov    %rdx,%rax
  sub    %rcx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x30(%rbp),%rax
  imul   -0x8(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  imul   -0x10(%rbp),%rax
  mov    %rax,%rcx
  mov    %rdx,%rax
  sub    %rcx,%rax
  mov    %rax,D38(%rip)
  mov    D38(%rip),%rax
  test   %rax,%rax
  jg     L8
  mov    D38(%rip),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,D38(%rip)
L8:
  mov    D38(%rip),%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movsd  D38(%rip),%xmm1
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0x34(%rbp)
  movss  -0x34(%rbp),%xmm0
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D38
data_type: f64

tool: D38: 2147483646

assistant: ```c
float frand() {
  long a = 16807, m = 2147483647, q = 127773, r = 2836;
  long hi, lo;
  static long seed = 31207321;
  float fnumb;
  hi = seed / q;
  lo = seed - q * hi;
  seed = a * lo - r * hi;
  if (seed <= 0)
    seed = seed + m;
  fnumb = seed / 2147483646.0;
  return (fnumb);
}
```
user: What is the c source code of the assembly code below:

```asm
<frand>:
  endbr64
  mov    D0(%rip),%rcx
  movabs $0x41a705af1fe3fb79,%rdx
  mov    %rcx,%rax
  imul   %rdx
  sar    $0xf,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  imul   $0x1f31d,%rdx,%rax
  sub    %rax,%rcx
  imul   $0x41a7,%rcx,%rcx
  imul   $0xb14,%rdx,%rdx
  sub    %rdx,%rcx
  lea    0x7fffffff(%rcx),%rax
  test   %rcx,%rcx
  cmovle %rax,%rcx
  mov    %rcx,D18(%rip)
  pxor   %xmm0,%xmm0
  cvtsi2sd %rcx,%xmm0
  divsd  D18(%rip),%xmm0
  cvtsd2ss %xmm0,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: tag: D18
datatype: f64

tool: D18: 2147483646

assistant: ```c
float frand() {
  long a = 16807, m = 2147483647, q = 127773, r = 2836;
  long hi, lo;
  static long seed = 31207321;
  float fnumb;
  hi = seed / q;
  lo = seed - q * hi;
  seed = a * lo - r * hi;
  if (seed <= 0)
    seed = seed + m;
  fnumb = seed / 2147483646.0;
  return (fnumb);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<frand>:
  endbr64
  mov    D9(%rip),%rcx
  movabs $0x41a705af1fe3fb79,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0xf,%rdx
  sub    %rax,%rdx
  imul   $0x1f31d,%rdx,%rax
  imul   $0xb14,%rdx,%rdx
  sub    %rax,%rcx
  imul   $0x41a7,%rcx,%rax
  sub    %rdx,%rax
  test   %rax,%rax
  jg     L3
  add    $0x7fffffff,%rax
L3:
  pxor   %xmm0,%xmm0
  mov    %rax,D1(%rip)
  cvtsi2sd %rax,%xmm0
  divsd  D1(%rip),%xmm0
  cvtsd2ss %xmm0,%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
type_guessed: f64

tool: D1: 2147483646

assistant: ```c
float frand() {
  long a = 16807, m = 2147483647, q = 127773, r = 2836;
  long hi, lo;
  static long seed = 31207321;
  float fnumb;
  hi = seed / q;
  lo = seed - q * hi;
  seed = a * lo - r * hi;
  if (seed <= 0)
    seed = seed + m;
  fnumb = seed / 2147483646.0;
  return (fnumb);
}
```
user: What is the c source code of the assembly code below?

```asm
<frand>:
  endbr64
  mov    D26(%rip),%rcx
  movabs $0x41a705af1fe3fb79,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0xf,%rdx
  sub    %rax,%rdx
  imul   $0x1f31d,%rdx,%rax
  imul   $0xb14,%rdx,%rdx
  sub    %rax,%rcx
  imul   $0x41a7,%rcx,%rax
  sub    %rdx,%rax
  test   %rax,%rax
  jg     L9
  add    $0x7fffffff,%rax
L9:
  pxor   %xmm0,%xmm0
  mov    %rax,D19(%rip)
  cvtsi2sd %rax,%xmm0
  divsd  D19(%rip),%xmm0
  cvtsd2ss %xmm0,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D19
type_guessed: f64

tool: D19: 2147483646

assistant: ```c
float frand() {
  long a = 16807, m = 2147483647, q = 127773, r = 2836;
  long hi, lo;
  static long seed = 31207321;
  float fnumb;
  hi = seed / q;
  lo = seed - q * hi;
  seed = a * lo - r * hi;
  if (seed <= 0)
    seed = seed + m;
  fnumb = seed / 2147483646.0;
  return (fnumb);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<testBorder>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x5,-0x4(%rbp)
  je     L33
  cmpl   $0x5,-0x4(%rbp)
  jg     L58
  cmpl   $0x4,-0x4(%rbp)
  jg     L58
  cmpl   $0x3,-0x4(%rbp)
  jge    L99
  cmpl   $0x1,-0x4(%rbp)
  jg     L29
  cmpl   $0x0,-0x4(%rbp)
  jns    L131
  jmp    L58
L29:
  cmpl   $0x2,-0x4(%rbp)
  je     L96
  jmp    L58
L131:
  cmpl   $0x0,-0x8(%rbp)
  je     L52
  mov    $0x1,%eax
  jmp    L72
L52:
  mov    $0x2,%eax
  jmp    L72
L96:
  mov    $0x4,%eax
  jmp    L72
L99:
  cmpl   $0x0,-0x8(%rbp)
  je     L36
  mov    $0x2,%eax
  jmp    L72
L36:
  mov    $0x1,%eax
  jmp    L72
L33:
  mov    $0x4,%eax
  jmp    L72
L58:
  mov    $0x0,%eax
L72:
  pop    %rbp
  ret
```

assistant: ```c
int testBorder(int pozice, int leftright) {
  switch (pozice) {
  case 0:
  case 1:
    return (leftright ? 1 : 2);
    break;
  case 2:
    return 4;
    break;
  case 3:
  case 4:
    return (leftright ? 2 : 1);
    break;
  case 5:
    return 4;
    break;
  default:
    return 0;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<testBorder>:
  endbr64
  cmp    $0x4,%edi
  jg     L14
  cmp    $0x2,%edi
  jg     L11
  mov    $0x4,%eax
  cmp    $0x1,%edi
  jg     L15
  cmp    $0x1,%esi
  mov    $0x1,%eax
  adc    $0x0,%eax
  test   %edi,%edi
  mov    $0x0,%edx
  cmovs  %edx,%eax
  ret
L14:
  cmp    $0x5,%edi
  sete   %al
  movzbl %al,%eax
  shl    $0x2,%eax
  ret
L11:
  cmp    $0x1,%esi
  mov    $0x1,%eax
  sbb    $0xffffffff,%eax
L15:
  ret
```

assistant: ```c
int testBorder(int pozice, int leftright) {
  switch (pozice) {
  case 0:
  case 1:
    return (leftright ? 1 : 2);
    break;
  case 2:
    return 4;
    break;
  case 3:
  case 4:
    return (leftright ? 2 : 1);
    break;
  case 5:
    return 4;
    break;
  default:
    return 0;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<testBorder>:
  endbr64
  cmp    $0x4,%edi
  jg     L17
  cmp    $0x2,%edi
  jg     L0
  mov    $0x4,%eax
  je     L29
  cmp    $0x1,%esi
  mov    $0x1,%eax
  adc    $0x0,%eax
  xor    %edx,%edx
  test   %edi,%edi
  cmovs  %edx,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L17:
  xor    %eax,%eax
  cmp    $0x5,%edi
  sete   %al
  shl    $0x2,%eax
  ret
  nopl   0x0(%rax)
L0:
  cmp    $0x1,%esi
  mov    $0x1,%eax
  sbb    $0xffffffff,%eax
L29:
  ret
```

assistant: ```c
int testBorder(int pozice, int leftright) {
  switch (pozice) {
  case 0:
  case 1:
    return (leftright ? 1 : 2);
    break;
  case 2:
    return 4;
    break;
  case 3:
  case 4:
    return (leftright ? 2 : 1);
    break;
  case 5:
    return 4;
    break;
  default:
    return 0;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<testBorder>:
  endbr64
  cmp    $0x4,%edi
  jg     L21
  cmp    $0x2,%edi
  jg     L12
  mov    $0x4,%eax
  je     L23
  cmp    $0x1,%esi
  mov    $0x1,%eax
  adc    $0x0,%eax
  xor    %edx,%edx
  test   %edi,%edi
  cmovs  %edx,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L21:
  xor    %eax,%eax
  cmp    $0x5,%edi
  sete   %al
  shl    $0x2,%eax
  ret
  nopl   0x0(%rax)
L12:
  cmp    $0x1,%esi
  mov    $0x1,%eax
  sbb    $0xffffffff,%eax
L23:
  ret
```

assistant: ```c
int testBorder(int pozice, int leftright) {
  switch (pozice) {
  case 0:
  case 1:
    return (leftright ? 1 : 2);
    break;
  case 2:
    return 4;
    break;
  case 3:
  case 4:
    return (leftright ? 2 : 1);
    break;
  case 5:
    return 4;
    break;
  default:
    return 0;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parent_dir_name_is>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <parent_dir_name@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L3
  mov    $0x0,%eax
  jmp    L12
L3:
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0xc(%rbp),%eax
L12:
  leave
  ret
```

assistant: ```c
int parent_dir_name_is(const char *path, const char *str) {
  char *parent_dir;
  int result;
  parent_dir = parent_dir_name(path);
  if (!parent_dir)
    return 0;
  result = !strcmp(parent_dir, str);
  free(parent_dir);
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<parent_dir_name_is>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbp
  call   <parent_dir_name@plt>
  mov    $0x0,%r12d
  test   %rax,%rax
  je     L2
  mov    %rax,%rbx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  sete   %r12b
  movzbl %r12b,%r12d
  mov    %rbx,%rdi
  call   <free@plt>
L2:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int parent_dir_name_is(const char *path, const char *str) {
  char *parent_dir;
  int result;
  parent_dir = parent_dir_name(path);
  if (!parent_dir)
    return 0;
  result = !strcmp(parent_dir, str);
  free(parent_dir);
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<parent_dir_name_is>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  call   <parent_dir_name@plt>
  test   %rax,%rax
  je     L7
  mov    %rax,%rdi
  mov    %r12,%rsi
  mov    %rax,%rbp
  xor    %r13d,%r13d
  call   <strcmp@plt>
  mov    %rbp,%rdi
  test   %eax,%eax
  sete   %r13b
  call   <free@plt>
L7:
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int parent_dir_name_is(const char *path, const char *str) {
  char *parent_dir;
  int result;
  parent_dir = parent_dir_name(path);
  if (!parent_dir)
    return 0;
  result = !strcmp(parent_dir, str);
  free(parent_dir);
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<parent_dir_name_is>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  call   <parent_dir_name@plt>
  test   %rax,%rax
  je     L4
  mov    %rax,%rdi
  mov    %r12,%rsi
  mov    %rax,%rbp
  xor    %r13d,%r13d
  call   <strcmp@plt>
  mov    %rbp,%rdi
  test   %eax,%eax
  sete   %r13b
  call   <free@plt>
L4:
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int parent_dir_name_is(const char *path, const char *str) {
  char *parent_dir;
  int result;
  parent_dir = parent_dir_name(path);
  if (!parent_dir)
    return 0;
  result = !strcmp(parent_dir, str);
  free(parent_dir);
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<data_fragmenter>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    %r8,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jge    L0
  mov    -0x28(%rbp),%rax
  sub    -0x18(%rbp),%rax
  cmp    %rax,-0x10(%rbp)
  jle    L21
  mov    -0x28(%rbp),%rax
  sub    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
L21:
  mov    -0x18(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%rax
  jmp    L26
L0:
  mov    $0x0,%eax
L26:
  leave
  ret
```

assistant: ```c
uint8_t data_fragmenter(uint8_t fragment[], uint8_t fragSize, uint8_t offset, uint8_t data[], uint8_t dataSize) {
  if (offset < dataSize) {
    if (dataSize - offset < fragSize) {
      fragSize = dataSize - offset;
    }
    memcpy(fragment, &data[offset], fragSize);
    return fragSize;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<data_fragmenter>:
  endbr64
  push   %rbx
  mov    $0x0,%ebx
  cmp    %r8,%rdx
  jl     L15
L3:
  mov    %rbx,%rax
  pop    %rbx
  ret
L15:
  sub    %rdx,%r8
  cmp    %rsi,%r8
  mov    %rsi,%rbx
  cmovle %r8,%rbx
  lea    (%rcx,%rdx,8),%rsi
  mov    %rbx,%rdx
  call   <memcpy@plt>
  jmp    L3
```

assistant: ```c
uint8_t data_fragmenter(uint8_t fragment[], uint8_t fragSize, uint8_t offset, uint8_t data[], uint8_t dataSize) {
  if (offset < dataSize) {
    if (dataSize - offset < fragSize) {
      fragSize = dataSize - offset;
    }
    memcpy(fragment, &data[offset], fragSize);
    return fragSize;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<data_fragmenter>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  cmp    %r8,%rdx
  jl     L2
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L2:
  sub    %rdx,%r8
  cmp    %rsi,%r8
  cmovle %r8,%rsi
  mov    %rsi,%r12
  lea    (%rcx,%rdx,8),%rsi
  mov    %r12,%rdx
  call   <memcpy@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
uint8_t data_fragmenter(uint8_t fragment[], uint8_t fragSize, uint8_t offset, uint8_t data[], uint8_t dataSize) {
  if (offset < dataSize) {
    if (dataSize - offset < fragSize) {
      fragSize = dataSize - offset;
    }
    memcpy(fragment, &data[offset], fragSize);
    return fragSize;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<data_fragmenter>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  cmp    %r8,%rdx
  jl     L5
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L5:
  sub    %rdx,%r8
  cmp    %rsi,%r8
  cmovle %r8,%rsi
  mov    %rsi,%r12
  lea    (%rcx,%rdx,8),%rsi
  mov    %r12,%rdx
  call   <memcpy@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
uint8_t data_fragmenter(uint8_t fragment[], uint8_t fragSize, uint8_t offset, uint8_t data[], uint8_t dataSize) {
  if (offset < dataSize) {
    if (dataSize - offset < fragSize) {
      fragSize = dataSize - offset;
    }
    memcpy(fragment, &data[offset], fragSize);
    return fragSize;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<offset>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  subl   $0x1,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L6
L10:
  mov    -0x4(%rbp),%eax
  and    -0x1c(%rbp),%eax
  test   %eax,%eax
  je     L21
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  add    %eax,-0x8(%rbp)
L21:
  subl   $0x1,-0x14(%rbp)
  sarl   -0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L6:
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jg     L10
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int offset(int bit, int current_bit, int id) {
  int offset;
  int mask;
  for (offset = 0, bit--, mask = 1 << bit; bit > current_bit; bit--, mask >>= 1, offset++)
    if (mask & id)
      offset += mask - 1;
  return offset;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<offset>:
  endbr64
  lea    -0x1(%rdi),%ecx
  mov    $0x1,%r8d
  shl    %cl,%r8d
  cmp    %esi,%ecx
  jle    L6
  mov    $0x0,%eax
L7:
  lea    -0x1(%r8,%rax,1),%edi
  test   %r8d,%edx
  cmovne %edi,%eax
  sub    $0x1,%ecx
  sar    %r8d
  add    $0x1,%eax
  cmp    %ecx,%esi
  jne    L7
  ret
L6:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int offset(int bit, int current_bit, int id) {
  int offset;
  int mask;
  for (offset = 0, bit--, mask = 1 << bit; bit > current_bit; bit--, mask >>= 1, offset++)
    if (mask & id)
      offset += mask - 1;
  return offset;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<offset>:
  endbr64
  lea    -0x1(%rdi),%ecx
  mov    $0x1,%r8d
  xor    %eax,%eax
  shl    %cl,%r8d
  cmp    %esi,%ecx
  jle    L15
  cs nopw 0x0(%rax,%rax,1)
L14:
  test   %r8d,%edx
  lea    -0x1(%r8,%rax,1),%edi
  cmovne %edi,%eax
  sub    $0x1,%ecx
  sar    %r8d
  add    $0x1,%eax
  cmp    %ecx,%esi
  jne    L14
  ret
  nopl   0x0(%rax)
L15:
  ret
```

assistant: ```c
int offset(int bit, int current_bit, int id) {
  int offset;
  int mask;
  for (offset = 0, bit--, mask = 1 << bit; bit > current_bit; bit--, mask >>= 1, offset++)
    if (mask & id)
      offset += mask - 1;
  return offset;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<offset>:
  endbr64
  lea    -0x1(%rdi),%ecx
  mov    $0x1,%r8d
  xor    %eax,%eax
  shl    %cl,%r8d
  cmp    %esi,%ecx
  jle    L26
  cs nopw 0x0(%rax,%rax,1)
L36:
  test   %edx,%r8d
  je     L27
  lea    -0x1(%r8,%rax,1),%eax
  sub    $0x1,%ecx
  sar    %r8d
  add    $0x1,%eax
  cmp    %esi,%ecx
  jne    L36
  ret
  nopl   0x0(%rax,%rax,1)
L27:
  sub    $0x1,%ecx
  sar    %r8d
  add    $0x1,%eax
  cmp    %ecx,%esi
  jne    L36
  ret
  xchg   %ax,%ax
L26:
  ret
```

assistant: ```c
int offset(int bit, int current_bit, int id) {
  int offset;
  int mask;
  for (offset = 0, bit--, mask = 1 << bit; bit > current_bit; bit--, mask >>= 1, offset++)
    if (mask & id)
      offset += mask - 1;
  return offset;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<adjust_pre>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L22
L29:
  subl   $0x1,-0x8(%rbp)
  subl   $0x1,-0x24(%rbp)
  addl   $0x1,-0x4(%rbp)
L22:
  cmpl   $0x0,-0x8(%rbp)
  je     L46
  cmpl   $0x0,-0x24(%rbp)
  je     L46
  mov    -0x24(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <csame@plt>
  test   %rax,%rax
  jne    L29
L46:
  cmpl   $0x0,-0x8(%rbp)
  je     L5
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rax,%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memmove@plt>
L5:
  nop
  leave
  ret
```

assistant: ```c
void adjust_pre(char *common, const char *new, int len) {
  int l = strlen(common);
  int newlen = 0;
  while (l && len && csame(common[l - 1], new[len - 1])) {
    l -= 1;
    len -= 1;
    newlen += 1;
  }
  if (l)
    memmove(common, common + l, newlen + 1);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<adjust_pre>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r15
  mov    %edx,%r12d
  call   <strlen@plt>
  test   %eax,%eax
  je     L33
  test   %r12d,%r12d
  je     L33
  lea    -0x1(%r12),%ebx
  movslq %ebx,%rbx
  mov    %eax,%ebp
  mov    $0x0,%r14d
  cltq
  movslq %r12d,%r12
  sub    %r12,%rax
  lea    0x0(%r13,%rax,1),%r12
L7:
  movsbl (%r15,%rbx,1),%esi
  movsbl (%r12,%rbx,1),%edi
  call   <csame@plt>
  test   %rax,%rax
  je     L64
  add    $0x1,%r14d
  sub    $0x1,%ebp
  setne  %dl
  test   %ebx,%ebx
  setne  %al
  sub    $0x1,%rbx
  test   %al,%dl
  jne    L7
L64:
  test   %ebp,%ebp
  jne    L29
L0:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L33:
  mov    %eax,%ebp
  mov    $0x0,%r14d
  jmp    L64
L29:
  lea    0x1(%r14),%edx
  movslq %ebp,%rsi
  add    %r13,%rsi
  mov    %r13,%rdi
  call   <memmove@plt>
  jmp    L0
```

assistant: ```c
void adjust_pre(char *common, const char *new, int len) {
  int l = strlen(common);
  int newlen = 0;
  while (l && len && csame(common[l - 1], new[len - 1])) {
    l -= 1;
    len -= 1;
    newlen += 1;
  }
  if (l)
    memmove(common, common + l, newlen + 1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<adjust_pre>:
  endbr64
  push   %r15
  mov    %edx,%r15d
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  test   %eax,%eax
  je     L21
  test   %r15d,%r15d
  je     L21
  movslq %r15d,%rdx
  lea    -0x1(%rax),%ebx
  movslq %eax,%r12
  xor    %ebp,%ebp
  mov    %rdx,%rax
  movslq %ebx,%rbx
  sub    %r12,%rax
  add    %rax,%r13
  jmp    L19
  nopw   0x0(%rax,%rax,1)
L68:
  test   %ebx,%ebx
  movslq %ebx,%r12
  setne  %dl
  cmp    %ebp,%r15d
  setne  %al
  sub    $0x1,%rbx
  test   %al,%dl
  je     L43
L19:
  movsbl 0x0(%r13,%rbx,1),%esi
  movsbl (%r14,%rbx,1),%edi
  lea    0x1(%rbx),%r12
  call   <csame@plt>
  mov    %ebp,%edx
  add    $0x1,%ebp
  test   %rax,%rax
  jne    L68
L15:
  add    $0x8,%rsp
  lea    (%r14,%r12,1),%rsi
  mov    %r14,%rdi
  add    $0x1,%edx
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1090 <memmove@plt>
L21:
  movslq %eax,%r12
  xor    %ebp,%ebp
  xchg   %ax,%ax
L43:
  test   %r12d,%r12d
  jne    L13
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  mov    %ebp,%edx
  jmp    L15
```

assistant: ```c
void adjust_pre(char *common, const char *new, int len) {
  int l = strlen(common);
  int newlen = 0;
  while (l && len && csame(common[l - 1], new[len - 1])) {
    l -= 1;
    len -= 1;
    newlen += 1;
  }
  if (l)
    memmove(common, common + l, newlen + 1);
}
```
user: What is the c source code of the assembly code below:

```asm
<adjust_pre>:
  endbr64
  push   %r15
  mov    %edx,%r15d
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  test   %eax,%eax
  je     L28
  test   %r15d,%r15d
  je     L28
  movslq %r15d,%rdx
  lea    -0x1(%rax),%ebx
  movslq %eax,%r12
  xor    %ebp,%ebp
  mov    %rdx,%rax
  movslq %ebx,%rbx
  sub    %r12,%rax
  add    %rax,%r13
  jmp    L47
  nopw   0x0(%rax,%rax,1)
L36:
  test   %ebx,%ebx
  movslq %ebx,%r12
  setne  %dl
  cmp    %ebp,%r15d
  setne  %al
  sub    $0x1,%rbx
  test   %al,%dl
  je     L40
L47:
  movsbl 0x0(%r13,%rbx,1),%esi
  movsbl (%r14,%rbx,1),%edi
  lea    0x1(%rbx),%r12
  call   <csame@plt>
  mov    %ebp,%edx
  add    $0x1,%ebp
  test   %rax,%rax
  jne    L36
L65:
  add    $0x8,%rsp
  lea    (%r14,%r12,1),%rsi
  mov    %r14,%rdi
  add    $0x1,%edx
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1090 <memmove@plt>
L28:
  movslq %eax,%r12
  xor    %ebp,%ebp
  xchg   %ax,%ax
L40:
  test   %r12d,%r12d
  jne    L4
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L4:
  mov    %ebp,%edx
  jmp    L65
```

assistant: ```c
void adjust_pre(char *common, const char *new, int len) {
  int l = strlen(common);
  int newlen = 0;
  while (l && len && csame(common[l - 1], new[len - 1])) {
    l -= 1;
    len -= 1;
    newlen += 1;
  }
  if (l)
    memmove(common, common + l, newlen + 1);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ez_net_close_socket>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jle    L4
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
L4:
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int ez_net_close_socket(int s) {
  if (s > 0)
    close(s);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ez_net_close_socket>:
  endbr64
  test   %edi,%edi
  jg     L1
  mov    $0x0,%eax
  ret
L1:
  sub    $0x8,%rsp
  call   <close@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int ez_net_close_socket(int s) {
  if (s > 0)
    close(s);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ez_net_close_socket>:
  endbr64
  test   %edi,%edi
  jg     L4
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  sub    $0x8,%rsp
  call   <close@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int ez_net_close_socket(int s) {
  if (s > 0)
    close(s);
  return 0;
}
```