ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<change_directory>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdx
  lea    D8(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <cond_printf@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <chdir@plt>
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D8
data_type: string

tool: D8: change_directory:%s


assistant: ```c
int change_directory(const char *path) {
  cond_printf(1, "change_directory:%s\n", path);
  return chdir(path);
}
```
user: What is the c source code of the assembly code below?

```asm
<change_directory>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %rdi,%rdx
  lea    D1(%rip),%rsi
  mov    $0x1,%edi
  call   <cond_printf@plt>
  mov    %rbx,%rdi
  call   <chdir@plt>
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
datatype: string

tool: D1: change_directory:%s


assistant: ```c
int change_directory(const char *path) {
  cond_printf(1, "change_directory:%s\n", path);
  return chdir(path);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<change_directory>:
  endbr64
  push   %rbp
  mov    %rdi,%rdx
  mov    %rdi,%rbp
  lea    D2(%rip),%rsi
  mov    $0x1,%edi
  call   <cond_printf@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <chdir@plt>
```

tool_calls: name: retrieve_data
arguments: label: D2
type: string

tool: D2: change_directory:%s


assistant: ```c
int change_directory(const char *path) {
  cond_printf(1, "change_directory:%s\n", path);
  return chdir(path);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<change_directory>:
  endbr64
  push   %rbp
  mov    %rdi,%rdx
  mov    %rdi,%rbp
  lea    D6(%rip),%rsi
  mov    $0x1,%edi
  call   <cond_printf@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <chdir@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D6
type_infered: string

tool: D6: change_directory:%s


assistant: ```c
int change_directory(const char *path) {
  cond_printf(1, "change_directory:%s\n", path);
  return chdir(path);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<dot_product>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L8
L6:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L8:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L6
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double dot_product(double vec1[], double vec2[], int n) {
  int i;
  double sum = 0.0;
  for (i = 0; i < n; i++) {
    sum = sum + vec1[i] * vec2[i];
  }
  return sum;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dot_product>:
  endbr64
  test   %edx,%edx
  jle    L18
  mov    %edx,%edx
  mov    $0x0,%eax
  pxor   %xmm1,%xmm1
L23:
  movsd  (%rdi,%rax,8),%xmm0
  mulsd  (%rsi,%rax,8),%xmm0
  addsd  %xmm0,%xmm1
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L23
L3:
  movapd %xmm1,%xmm0
  ret
L18:
  pxor   %xmm1,%xmm1
  jmp    L3
```

assistant: ```c
double dot_product(double vec1[], double vec2[], int n) {
  int i;
  double sum = 0.0;
  for (i = 0; i < n; i++) {
    sum = sum + vec1[i] * vec2[i];
  }
  return sum;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<dot_product>:
  endbr64
  test   %edx,%edx
  jle    L2
  movslq %edx,%rdx
  xor    %eax,%eax
  pxor   %xmm1,%xmm1
  nopl   0x0(%rax)
L13:
  movsd  (%rdi,%rax,8),%xmm0
  mulsd  (%rsi,%rax,8),%xmm0
  add    $0x1,%rax
  addsd  %xmm0,%xmm1
  cmp    %rax,%rdx
  jne    L13
  movapd %xmm1,%xmm0
  ret
  nopl   0x0(%rax)
L2:
  pxor   %xmm1,%xmm1
  movapd %xmm1,%xmm0
  ret
```

assistant: ```c
double dot_product(double vec1[], double vec2[], int n) {
  int i;
  double sum = 0.0;
  for (i = 0; i < n; i++) {
    sum = sum + vec1[i] * vec2[i];
  }
  return sum;
}
```
user: What is the c source code of the assembly code below:

```asm
<dot_product>:
  endbr64
  test   %edx,%edx
  jle    L24
  cmp    $0x1,%edx
  je     L29
  mov    %edx,%ecx
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  shr    %ecx
  shl    $0x4,%rcx
  nopl   0x0(%rax,%rax,1)
L19:
  movupd (%rdi,%rax,1),%xmm1
  movupd (%rsi,%rax,1),%xmm3
  add    $0x10,%rax
  mulpd  %xmm3,%xmm1
  addsd  %xmm1,%xmm0
  unpckhpd %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rax,%rcx
  jne    L19
  mov    %edx,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%edx
  je     L30
L18:
  cltq
  movsd  (%rsi,%rax,8),%xmm1
  mulsd  (%rdi,%rax,8),%xmm1
  addsd  %xmm1,%xmm0
  ret
  xchg   %ax,%ax
L30:
  ret
  nopl   0x0(%rax)
L24:
  pxor   %xmm0,%xmm0
  ret
L29:
  pxor   %xmm0,%xmm0
  xor    %eax,%eax
  jmp    L18
```

assistant: ```c
double dot_product(double vec1[], double vec2[], int n) {
  int i;
  double sum = 0.0;
  for (i = 0; i < n; i++) {
    sum = sum + vec1[i] * vec2[i];
  }
  return sum;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x3,-0x14(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x14(%rbp),%eax
  imul   %eax,%edx
  mov    -0x14(%rbp),%eax
  imul   %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  add    %edx,%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D1
type_infered: string

tool: D1: %d 


assistant: ```c
int main() {
  int i_ = 3, *j_, k_;
  j_ = &i_;
  printf("%d \n", i_ * *j_ * i_ + *j_);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1e,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D1
type_infered: string

tool: D1: %d 


assistant: ```c
int main() {
  int i_ = 3, *j_, k_;
  j_ = &i_;
  printf("%d \n", i_ * *j_ * i_ + *j_);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1e,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D5
datatype: string

tool: D5: %d 


assistant: ```c
int main() {
  int i_ = 3, *j_, k_;
  j_ = &i_;
  printf("%d \n", i_ * *j_ * i_ + *j_);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1e,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D7
type: string

tool: D7: %d 


assistant: ```c
int main() {
  int i_ = 3, *j_, k_;
  j_ = &i_;
  printf("%d \n", i_ * *j_ * i_ + *j_);
  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
<int2string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x28(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  cmpl   $0x0,-0x34(%rbp)
  jns    L40
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movb   $0x2d,(%rax)
  negl   -0x34(%rbp)
  addl   $0x1,-0x28(%rbp)
L40:
  movl   $0x0,-0x24(%rbp)
  jmp    L6
L36:
  mov    -0x34(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    -0x24(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  mov    %ecx,%edx
  cltq
  mov    %dl,-0x17(%rbp,%rax,1)
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x34(%rbp)
L6:
  cmpl   $0x0,-0x34(%rbp)
  jg     L36
  mov    -0x24(%rbp),%eax
  add    %eax,-0x28(%rbp)
  jmp    L39
L31:
  subl   $0x1,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  movzbl -0x17(%rbp,%rax,1),%eax
  lea    0x30(%rax),%ecx
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
L39:
  cmpl   $0x0,-0x24(%rbp)
  jne    L31
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x28(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L35
  call   <__stack_chk_fail@plt>
L35:
  leave
  ret
```

assistant: ```c
int int2string(int value, char *buf) {
  int count = 0;
  char *bp = buf;
  if (value < 0) {
    *bp++ = '-';
    value = -value;
    count++;
  }
  char tmp[15];
  int i = 0;
  while (value > 0) {
    tmp[i++] = value % 10;
    value /= 10;
  }
  count += i;
  while (i) {
    *bp++ = tmp[--i] + '0';
  }
  *bp = 0;
  return count;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<int2string>:
  endbr64
  sub    $0x28,%rsp
  mov    %edi,%ecx
  mov    %rsi,%r10
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  js     L54
  jle    L49
  mov    $0x0,%eax
L34:
  mov    $0x1,%esi
  lea    0x8(%rsp),%r11
L12:
  mov    %esi,%r9d
  movslq %ecx,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sub    %edi,%edx
  lea    (%rdx,%rdx,4),%edi
  add    %edi,%edi
  mov    %ecx,%r8d
  sub    %edi,%r8d
  mov    %r8b,(%r11,%rsi,1)
  mov    %ecx,%edi
  mov    %edx,%ecx
  add    $0x1,%rsi
  cmp    $0x9,%edi
  jg     L12
  lea    (%r9,%rax,1),%edi
  test   %r9d,%r9d
  je     L59
  movslq %r9d,%rax
  lea    0x9(%rsp,%rax,1),%rcx
  lea    -0x1(%r9),%eax
  lea    0x1(%r10,%rax,1),%rsi
  mov    %r10,%rax
L33:
  add    $0x1,%rax
  movzbl -0x1(%rcx),%edx
  add    $0x30,%edx
  mov    %dl,-0x1(%rax)
  sub    $0x1,%rcx
  cmp    %rsi,%rax
  jne    L33
  lea    -0x1(%r9),%eax
  lea    0x1(%r10,%rax,1),%r10
L59:
  movb   $0x0,(%r10)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L77
  mov    %edi,%eax
  add    $0x28,%rsp
  ret
L54:
  movb   $0x2d,(%rsi)
  neg    %ecx
  lea    0x1(%rsi),%r10
  mov    $0x1,%eax
  jmp    L34
L49:
  mov    $0x0,%edi
  jmp    L59
L77:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int int2string(int value, char *buf) {
  int count = 0;
  char *bp = buf;
  if (value < 0) {
    *bp++ = '-';
    value = -value;
    count++;
  }
  char tmp[15];
  int i = 0;
  while (value > 0) {
    tmp[i++] = value % 10;
    value /= 10;
  }
  count += i;
  while (i) {
    *bp++ = tmp[--i] + '0';
  }
  *bp = 0;
  return count;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<int2string>:
  endbr64
  sub    $0x28,%rsp
  mov    %edi,%ecx
  mov    %rsi,%r8
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  js     L14
  je     L11
  xor    %eax,%eax
L4:
  mov    $0x1,%esi
  lea    0x8(%rsp),%r11
  mov    $0xcccccccd,%r10d
  nopl   (%rax)
L76:
  mov    %ecx,%edx
  mov    %esi,%r9d
  imul   %r10,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%edi
  add    %edi,%edi
  sub    %edi,%ecx
  mov    %cl,(%r11,%rsi,1)
  mov    %ecx,%edi
  add    $0x1,%rsi
  mov    %edx,%ecx
  test   %edx,%edx
  jne    L76
  movslq %r9d,%rsi
  lea    -0x1(%r9),%ecx
  lea    (%r9,%rax,1),%r10d
  mov    %r8,%rax
  lea    0x9(%rsp,%rsi,1),%rdx
  add    %r8,%rcx
  jmp    L15
  nopl   0x0(%rax)
L9:
  movzbl -0x1(%rdx),%edi
  add    $0x1,%rax
L15:
  add    $0x30,%edi
  sub    $0x1,%rdx
  mov    %dil,(%rax)
  cmp    %rax,%rcx
  jne    L9
  add    %rsi,%r8
L24:
  movb   $0x0,(%r8)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L74
  mov    %r10d,%eax
  add    $0x28,%rsp
  ret
  cs nopw 0x0(%rax,%rax,1)
L14:
  movb   $0x2d,(%rsi)
  neg    %ecx
  add    $0x1,%r8
  mov    $0x1,%eax
  jmp    L4
  nopl   0x0(%rax,%rax,1)
L11:
  xor    %r10d,%r10d
  jmp    L24
L74:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int int2string(int value, char *buf) {
  int count = 0;
  char *bp = buf;
  if (value < 0) {
    *bp++ = '-';
    value = -value;
    count++;
  }
  char tmp[15];
  int i = 0;
  while (value > 0) {
    tmp[i++] = value % 10;
    value /= 10;
  }
  count += i;
  while (i) {
    *bp++ = tmp[--i] + '0';
  }
  *bp = 0;
  return count;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<path_has_filename>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rdx,%r12
  mov    %ecx,%r14d
  mov    %r8,%r13
  movslq %esi,%rsi
  lea    -0x1(%rdi,%rsi,1),%rbx
  movsbl (%rbx),%edi
  call   <ch_is_slash@plt>
  test   %rax,%rax
  jne    L122
L47:
  movslq %r14d,%r14
  lea    -0x1(%r12,%r14,1),%rdx
  movzbl (%rdx),%eax
  cmp    %al,(%rbx)
  jne    L49
  lea    -0x1(%rbx),%rax
  sub    $0x1,%rdx
  cmp    %rax,%rbp
  je     L162
L147:
  lea    0x1(%rax),%rsi
  movzbl (%rax),%ecx
  cmp    $0x2f,%cl
  je     L177
  cmp    %rdx,%r12
  ja     L138
  cmp    (%rdx),%cl
  jne    L64
  sub    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %rax,%rbp
  jne    L147
L162:
  mov    $0x0,%ecx
  cmp    %rdx,%r12
  jbe    L175
  mov    $0x1,%ecx
  test   %r13,%r13
  je     L175
  mov    %rax,0x0(%r13)
  jmp    L175
L122:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <get_front_of_slashes@plt>
  cmp    %rax,%rbp
  je     L39
  lea    -0x1(%rax),%rbx
  jmp    L47
L177:
  mov    $0x0,%ecx
  cmp    %rdx,%r12
  jbe    L175
  mov    $0x1,%ecx
  test   %r13,%r13
  je     L175
  mov    %rsi,0x0(%r13)
  jmp    L175
L64:
  mov    $0x0,%ecx
  jmp    L175
L39:
  mov    $0x0,%ecx
  jmp    L175
L49:
  mov    $0x0,%ecx
  jmp    L175
L138:
  mov    $0x0,%ecx
L175:
  mov    %ecx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
bool path_has_filename(const char *__attribute__((nonnull)) const path, const uint64_t path_length, const char *__attribute__((nonnull)) const filename, const uint64_t filename_length, const char **const filename_out) {
  const char *const path_back = path + (path_length - 1);
  const char *path_iter = path_back;
  if (ch_is_slash(*path_back)) {
    const char *const row_front = get_front_of_slashes(path, path_back);
    if (row_front == path) {
      return false;
    }
    path_iter = row_front - 1;
  }
  const char *filename_iter = filename + (filename_length - 1);
  char path_ch = *path_iter;
  char filename_ch = *filename_iter;
  do {
    if (path_ch != filename_ch) {
      return false;
    }
    --path_iter;
    --filename_iter;
    if (path_iter == path) {
      if (filename_iter < filename) {
        if (filename_out != NULL) {
          *filename_out = path_iter;
        }
        return true;
      }
      return false;
    }
    path_ch = *path_iter;
    if (path_ch == '/') {
      if (filename_iter < filename) {
        if (filename_out != NULL) {
          *filename_out = path_iter + 1;
        }
        return true;
      }
      return false;
    }
    if (filename_iter < filename) {
      return false;
    }
    filename_ch = *filename_iter;
  } while (true);
  return true;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<path_has_filename>:
  endbr64
  movslq %esi,%rsi
  push   %r14
  movslq %ecx,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  lea    -0x1(%rdi,%rsi,1),%rbp
  push   %rbx
  movsbl 0x0(%rbp),%edi
  mov    %r8,%rbx
  call   <ch_is_slash@plt>
  test   %rax,%rax
  jne    L158
L52:
  lea    -0x1(%r13,%r14,1),%rax
  movzbl (%rax),%edi
  cmp    %dil,0x0(%rbp)
  jne    L62
  lea    -0x1(%rbp),%rsi
  sub    $0x1,%rax
  cmp    %rsi,%r12
  jne    L90
  jmp    L42
  nopl   0x0(%rax,%rax,1)
L41:
  cmp    %rax,%r13
  ja     L62
  cmp    (%rax),%dl
  jne    L62
  sub    $0x1,%rsi
  sub    $0x1,%rax
  cmp    %rsi,%r12
  je     L42
L90:
  movzbl (%rsi),%edx
  lea    0x1(%rsi),%rcx
  cmp    $0x2f,%dl
  jne    L41
  cmp    %rax,%r13
  jbe    L62
  mov    $0x1,%eax
  test   %rbx,%rbx
  je     L61
  mov    %rcx,(%rbx)
  jmp    L61
  cs nopw 0x0(%rax,%rax,1)
L42:
  cmp    %rax,%r13
  ja     L70
  nopl   (%rax)
L62:
  xor    %eax,%eax
L61:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L158:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <get_front_of_slashes@plt>
  lea    -0x1(%rax),%rbp
  cmp    %rax,%r12
  jne    L52
  jmp    L62
  nopw   0x0(%rax,%rax,1)
L70:
  mov    $0x1,%eax
  test   %rbx,%rbx
  je     L61
  mov    %rsi,(%rbx)
  jmp    L61
```

assistant: ```c
bool path_has_filename(const char *__attribute__((nonnull)) const path, const uint64_t path_length, const char *__attribute__((nonnull)) const filename, const uint64_t filename_length, const char **const filename_out) {
  const char *const path_back = path + (path_length - 1);
  const char *path_iter = path_back;
  if (ch_is_slash(*path_back)) {
    const char *const row_front = get_front_of_slashes(path, path_back);
    if (row_front == path) {
      return false;
    }
    path_iter = row_front - 1;
  }
  const char *filename_iter = filename + (filename_length - 1);
  char path_ch = *path_iter;
  char filename_ch = *filename_iter;
  do {
    if (path_ch != filename_ch) {
      return false;
    }
    --path_iter;
    --filename_iter;
    if (path_iter == path) {
      if (filename_iter < filename) {
        if (filename_out != NULL) {
          *filename_out = path_iter;
        }
        return true;
      }
      return false;
    }
    path_ch = *path_iter;
    if (path_ch == '/') {
      if (filename_iter < filename) {
        if (filename_out != NULL) {
          *filename_out = path_iter + 1;
        }
        return true;
      }
      return false;
    }
    if (filename_iter < filename) {
      return false;
    }
    filename_ch = *filename_iter;
  } while (true);
  return true;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<path_has_filename>:
  endbr64
  movslq %esi,%rsi
  push   %r14
  movslq %ecx,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  lea    -0x1(%rdi,%rsi,1),%rbp
  push   %rbx
  movsbl 0x0(%rbp),%edi
  mov    %r8,%rbx
  call   <ch_is_slash@plt>
  test   %rax,%rax
  jne    L118
L115:
  lea    -0x1(%r13,%r14,1),%rax
  movzbl (%rax),%edi
  cmp    %dil,0x0(%rbp)
  jne    L104
  lea    -0x1(%rbp),%rsi
  sub    $0x1,%rax
  cmp    %rsi,%r12
  jne    L105
  jmp    L103
  nopl   0x0(%rax,%rax,1)
L7:
  cmp    %rax,%r13
  ja     L104
  cmp    (%rax),%dl
  jne    L104
  sub    $0x1,%rsi
  sub    $0x1,%rax
  cmp    %rsi,%r12
  je     L103
L105:
  movzbl (%rsi),%edx
  lea    0x1(%rsi),%rcx
  cmp    $0x2f,%dl
  jne    L7
  cmp    %rax,%r13
  jbe    L104
  mov    $0x1,%eax
  test   %rbx,%rbx
  je     L29
  mov    %rcx,(%rbx)
  jmp    L29
  cs nopw 0x0(%rax,%rax,1)
L103:
  cmp    %rax,%r13
  ja     L20
  nopl   (%rax)
L104:
  xor    %eax,%eax
L29:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L118:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <get_front_of_slashes@plt>
  lea    -0x1(%rax),%rbp
  cmp    %rax,%r12
  jne    L115
  jmp    L104
  nopw   0x0(%rax,%rax,1)
L20:
  mov    $0x1,%eax
  test   %rbx,%rbx
  je     L29
  mov    %rsi,(%rbx)
  jmp    L29
```

assistant: ```c
bool path_has_filename(const char *__attribute__((nonnull)) const path, const uint64_t path_length, const char *__attribute__((nonnull)) const filename, const uint64_t filename_length, const char **const filename_out) {
  const char *const path_back = path + (path_length - 1);
  const char *path_iter = path_back;
  if (ch_is_slash(*path_back)) {
    const char *const row_front = get_front_of_slashes(path, path_back);
    if (row_front == path) {
      return false;
    }
    path_iter = row_front - 1;
  }
  const char *filename_iter = filename + (filename_length - 1);
  char path_ch = *path_iter;
  char filename_ch = *filename_iter;
  do {
    if (path_ch != filename_ch) {
      return false;
    }
    --path_iter;
    --filename_iter;
    if (path_iter == path) {
      if (filename_iter < filename) {
        if (filename_out != NULL) {
          *filename_out = path_iter;
        }
        return true;
      }
      return false;
    }
    path_ch = *path_iter;
    if (path_ch == '/') {
      if (filename_iter < filename) {
        if (filename_out != NULL) {
          *filename_out = path_iter + 1;
        }
        return true;
      }
      return false;
    }
    if (filename_iter < filename) {
      return false;
    }
    filename_ch = *filename_iter;
  } while (true);
  return true;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_zipcode_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x30,%al
  jle    L1
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x34,%al
  jg     L1
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x21,%al
  jne    L1
  mov    $0x1,%eax
  jmp    L34
L1:
  mov    $0x0,%eax
L34:
  pop    %rbp
  ret
```

assistant: ```c
int is_zipcode_name(char *name) {
  if (name[0] >= '1' && name[0] <= '4' && name[1] == '!')
    return 1;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_zipcode_name>:
  endbr64
  movzbl (%rdi),%eax
  sub    $0x31,%eax
  mov    $0x0,%edx
  cmp    $0x3,%al
  ja     L6
  cmpb   $0x21,0x1(%rdi)
  sete   %dl
  movzbl %dl,%edx
L6:
  mov    %edx,%eax
  ret
```

assistant: ```c
int is_zipcode_name(char *name) {
  if (name[0] >= '1' && name[0] <= '4' && name[1] == '!')
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_zipcode_name>:
  endbr64
  movzbl (%rdi),%eax
  xor    %r8d,%r8d
  sub    $0x31,%eax
  cmp    $0x3,%al
  ja     L4
  xor    %r8d,%r8d
  cmpb   $0x21,0x1(%rdi)
  sete   %r8b
L4:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int is_zipcode_name(char *name) {
  if (name[0] >= '1' && name[0] <= '4' && name[1] == '!')
    return 1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<is_zipcode_name>:
  endbr64
  movzbl (%rdi),%eax
  xor    %r8d,%r8d
  sub    $0x31,%eax
  cmp    $0x3,%al
  ja     L0
  xor    %r8d,%r8d
  cmpb   $0x21,0x1(%rdi)
  sete   %r8b
L0:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int is_zipcode_name(char *name) {
  if (name[0] >= '1' && name[0] <= '4' && name[1] == '!')
    return 1;
  return 0;
}
```