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

```asm
<dogrep>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  lea    D8(%rip),%rsi
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  test   %rbp,%rbp
  je     L29
  mov    %rsp,%rbx
  mov    $0x401,%edx
  xor    %esi,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  jmp    L52
  nop
L51:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <OS_WordMatch@plt>
  test   %rax,%rax
  jne    L11
L52:
  mov    %rbp,%rdx
  mov    $0x400,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L51
  mov    %rbp,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
L29:
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L47
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L11:
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    $0x1,%eax
  jmp    L29
L47:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D8
type_infered: string

tool: D8: r

assistant: ```c
int dogrep(char *file, char *str) {
  char line[1024 + 1];
  FILE *fp;
  fp = fopen(file, "r");
  if (!fp) {
    return (0);
  }
  memset(line, '\0', 1024 + 1);
  while (fgets(line, 1024, fp) != NULL) {
    if (OS_WordMatch(str, line)) {
      fclose(fp);
      return (1);
    }
  }
  fclose(fp);
  return (0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dogrep>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  lea    D1(%rip),%rsi
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  test   %rbp,%rbp
  je     L42
  mov    %rsp,%rbx
  mov    $0x401,%edx
  xor    %esi,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  jmp    L46
  nop
L47:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <OS_WordMatch@plt>
  test   %rax,%rax
  jne    L11
L46:
  mov    %rbp,%rdx
  mov    $0x400,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L47
  mov    %rbp,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
L42:
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L11:
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    $0x1,%eax
  jmp    L42
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D1
type: string

tool: D1: r

assistant: ```c
int dogrep(char *file, char *str) {
  char line[1024 + 1];
  FILE *fp;
  fp = fopen(file, "r");
  if (!fp) {
    return (0);
  }
  memset(line, '\0', 1024 + 1);
  while (fgets(line, 1024, fp) != NULL) {
    if (OS_WordMatch(str, line)) {
      fclose(fp);
      return (1);
    }
  }
  fclose(fp);
  return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<dirs_equal>:
  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   <strlen@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jne    L96
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L81
L96:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x8(%rbp)
  jne    L45
  mov    -0x8(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L45
  mov    -0x4(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L45
  mov    -0x8(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L81
L45:
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jne    L32
  mov    -0x8(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L32
  mov    -0x4(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L32
  mov    -0x4(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L81
L32:
  mov    $0x0,%eax
L81:
  leave
  ret
```

assistant: ```c
int dirs_equal(const char *dir1, const char *dir2) {
  int d1len, d2len;
  d1len = strlen(dir1);
  d2len = strlen(dir2);
  ;
  if (d1len == d2len) {
    return (strcmp(dir1, dir2) == 0);
  } else if (d1len == d2len - 1 && dir1[d1len - 1] != '/' && dir2[d2len - 1] == '/') {
    return (strncmp(dir1, dir2, d1len) == 0);
  } else if (d1len - 1 == d2len && dir1[d1len - 1] == '/' && dir2[d2len - 1] != '/') {
    return (strncmp(dir1, dir2, d2len) == 0);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dirs_equal>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %eax,%ebx
  je     L105
  lea    -0x1(%rax),%edx
  cmp    %ebx,%edx
  je     L40
  lea    -0x1(%rbx),%ecx
  mov    $0x0,%edx
  cmp    %eax,%ecx
  je     L60
L38:
  mov    %edx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L105:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %dl
  movzbl %dl,%edx
  jmp    L38
L40:
  movslq %ebx,%rdx
  cmpb   $0x2f,-0x1(%rbp,%rdx,1)
  je     L3
  cltq
  mov    $0x0,%edx
  cmpb   $0x2f,-0x1(%r12,%rax,1)
  jne    L38
  mov    %ebx,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  sete   %dl
  movzbl %dl,%edx
  jmp    L38
L60:
  movslq %ebx,%rbx
  cmpb   $0x2f,-0x1(%rbp,%rbx,1)
  jne    L38
L117:
  movslq %eax,%rcx
  mov    $0x0,%edx
  cmpb   $0x2f,-0x1(%r12,%rcx,1)
  je     L38
  mov    %eax,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  sete   %dl
  movzbl %dl,%edx
  jmp    L38
L3:
  sub    $0x1,%ebx
  cmp    %ebx,%eax
  je     L117
  mov    $0x0,%edx
  jmp    L38
```

assistant: ```c
int dirs_equal(const char *dir1, const char *dir2) {
  int d1len, d2len;
  d1len = strlen(dir1);
  d2len = strlen(dir2);
  ;
  if (d1len == d2len) {
    return (strcmp(dir1, dir2) == 0);
  } else if (d1len == d2len - 1 && dir1[d1len - 1] != '/' && dir2[d2len - 1] == '/') {
    return (strncmp(dir1, dir2, d1len) == 0);
  } else if (d1len - 1 == d2len && dir1[d1len - 1] == '/' && dir2[d2len - 1] != '/') {
    return (strncmp(dir1, dir2, d2len) == 0);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<dirs_equal>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <strlen@plt>
  mov    %rbp,%rdi
  movslq %eax,%rbx
  call   <strlen@plt>
  cmp    %eax,%ebx
  je     L28
  lea    -0x1(%rax),%edx
  cmp    %ebx,%edx
  je     L75
  lea    -0x1(%rbx),%edx
  xor    %r8d,%r8d
  cmp    %eax,%edx
  je     L38
L4:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L75:
  movslq %edx,%rcx
  xor    %r8d,%r8d
  cmpb   $0x2f,-0x1(%r12,%rcx,1)
  je     L4
  cltq
  cmpb   $0x2f,-0x1(%rbp,%rax,1)
  jne    L4
L59:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strncmp@plt>
  xor    %r8d,%r8d
  test   %rax,%rax
  sete   %r8b
  jmp    L4
  xchg   %ax,%ax
L38:
  cmpb   $0x2f,-0x1(%r12,%rbx,1)
  jne    L4
  movslq %eax,%rdx
  cmpb   $0x2f,-0x1(%rbp,%rdx,1)
  je     L4
  mov    %eax,%edx
  jmp    L59
  cs nopw 0x0(%rax,%rax,1)
L28:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  xor    %r8d,%r8d
  pop    %rbx
  pop    %rbp
  test   %rax,%rax
  pop    %r12
  sete   %r8b
  mov    %r8d,%eax
  ret
```

assistant: ```c
int dirs_equal(const char *dir1, const char *dir2) {
  int d1len, d2len;
  d1len = strlen(dir1);
  d2len = strlen(dir2);
  ;
  if (d1len == d2len) {
    return (strcmp(dir1, dir2) == 0);
  } else if (d1len == d2len - 1 && dir1[d1len - 1] != '/' && dir2[d2len - 1] == '/') {
    return (strncmp(dir1, dir2, d1len) == 0);
  } else if (d1len - 1 == d2len && dir1[d1len - 1] == '/' && dir2[d2len - 1] != '/') {
    return (strncmp(dir1, dir2, d2len) == 0);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dirs_equal>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <strlen@plt>
  mov    %rbp,%rdi
  movslq %eax,%rbx
  call   <strlen@plt>
  cmp    %eax,%ebx
  je     L4
  lea    -0x1(%rax),%edx
  cmp    %ebx,%edx
  je     L27
  lea    -0x1(%rbx),%edx
  xor    %r8d,%r8d
  cmp    %eax,%edx
  je     L69
L18:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L27:
  movslq %edx,%rcx
  xor    %r8d,%r8d
  cmpb   $0x2f,-0x1(%r12,%rcx,1)
  je     L18
  cltq
  cmpb   $0x2f,-0x1(%rbp,%rax,1)
  jne    L18
L24:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strncmp@plt>
  xor    %r8d,%r8d
  test   %rax,%rax
  sete   %r8b
  jmp    L18
  xchg   %ax,%ax
L69:
  cmpb   $0x2f,-0x1(%r12,%rbx,1)
  jne    L18
  movslq %eax,%rdx
  cmpb   $0x2f,-0x1(%rbp,%rdx,1)
  je     L18
  mov    %eax,%edx
  jmp    L24
  cs nopw 0x0(%rax,%rax,1)
L4:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  xor    %r8d,%r8d
  pop    %rbx
  pop    %rbp
  test   %rax,%rax
  pop    %r12
  sete   %r8b
  mov    %r8d,%eax
  ret
```

assistant: ```c
int dirs_equal(const char *dir1, const char *dir2) {
  int d1len, d2len;
  d1len = strlen(dir1);
  d2len = strlen(dir2);
  ;
  if (d1len == d2len) {
    return (strcmp(dir1, dir2) == 0);
  } else if (d1len == d2len - 1 && dir1[d1len - 1] != '/' && dir2[d2len - 1] == '/') {
    return (strncmp(dir1, dir2, d1len) == 0);
  } else if (d1len - 1 == d2len && dir1[d1len - 1] == '/' && dir2[d2len - 1] != '/') {
    return (strncmp(dir1, dir2, d2len) == 0);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<exact_square>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  movd   %xmm0,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  divss  -0xc(%rbp),%xmm0
  ucomiss -0xc(%rbp),%xmm0
  jp     L32
  ucomiss -0xc(%rbp),%xmm0
  jne    L32
  movl   $0x1,-0x14(%rbp)
  jmp    L24
L32:
  movl   $0x0,-0x14(%rbp)
L24:
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D7
data_type: string

tool: D7: %d

assistant: ```c
int exact_square(int answer) {
  int num;
  float helper;
  scanf("%d", &num);
  helper = sqrt(num);
  if (num / helper == helper) {
    answer = 1;
  } else
    answer = 0;
  return answer;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<exact_square>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <sqrt@plt>
  pxor   %xmm1,%xmm1
  cvtsi2ssl 0x4(%rsp),%xmm1
  divss  %xmm0,%xmm1
  ucomiss %xmm0,%xmm1
  setnp  %al
  movzbl %al,%eax
  mov    $0x0,%edx
  cmovne %edx,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D2
type_guessed: string

tool: D2: %d

assistant: ```c
int exact_square(int answer) {
  int num;
  float helper;
  scanf("%d", &num);
  helper = sqrt(num);
  if (num / helper == helper) {
    answer = 1;
  } else
    answer = 0;
  return answer;
}
```
user: What is the c source code of the assembly code below?

```asm
<exact_square>:
  endbr64
  sub    $0x18,%rsp
  lea    D3(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <sqrt@plt>
  pxor   %xmm1,%xmm1
  xor    %eax,%eax
  mov    $0x0,%edx
  cvtsi2ssl 0x4(%rsp),%xmm1
  divss  %xmm0,%xmm1
  ucomiss %xmm0,%xmm1
  setnp  %al
  cmovne %edx,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D3
type_infered: string

tool: D3: %d

assistant: ```c
int exact_square(int answer) {
  int num;
  float helper;
  scanf("%d", &num);
  helper = sqrt(num);
  if (num / helper == helper) {
    answer = 1;
  } else
    answer = 0;
  return answer;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<exact_square>:
  endbr64
  sub    $0x18,%rsp
  lea    D0(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <sqrt@plt>
  pxor   %xmm1,%xmm1
  xor    %eax,%eax
  mov    $0x0,%edx
  cvtsi2ssl 0x4(%rsp),%xmm1
  divss  %xmm0,%xmm1
  ucomiss %xmm0,%xmm1
  setnp  %al
  cmovne %edx,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D0
data_type: string

tool: D0: %d

assistant: ```c
int exact_square(int answer) {
  int num;
  float helper;
  scanf("%d", &num);
  helper = sqrt(num);
  if (num / helper == helper) {
    answer = 1;
  } else
    answer = 0;
  return answer;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *strdup(const char *str) {
  size_t len = strlen(str);
  char *out = malloc(len + 1);
  strncpy(out, str, len);
  out[len] = '\0';
  return out;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strdup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  call   <strlen@plt>
  mov    %rax,%rbp
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%rbx,%rbp,1)
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup(const char *str) {
  size_t len = strlen(str);
  char *out = malloc(len + 1);
  strncpy(out, str, len);
  out[len] = '\0';
  return out;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup(const char *str) {
  size_t len = strlen(str);
  char *out = malloc(len + 1);
  strncpy(out, str, len);
  out[len] = '\0';
  return out;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup(const char *str) {
  size_t len = strlen(str);
  char *out = malloc(len + 1);
  strncpy(out, str, len);
  out[len] = '\0';
  return out;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<check_linux_version_newer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %esi,-0x38(%rbp)
  mov    %edx,-0x3c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x30(%rbp)
  movl   $0x0,-0x2c(%rbp)
  movl   $0x0,-0x28(%rbp)
  movl   $0x0,-0x24(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <uname@plt>
  mov    -0x18(%rbp),%rax
  lea    D22(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L3
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x2c(%rbp)
L3:
  lea    D22(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L43
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x28(%rbp)
L43:
  lea    D22(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L17
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x24(%rbp)
L17:
  mov    -0x34(%rbp),%eax
  shl    $0x18,%eax
  mov    %eax,%edx
  mov    -0x38(%rbp),%eax
  shl    $0x10,%eax
  and    $0xff0000,%eax
  or     %eax,%edx
  mov    -0x3c(%rbp),%eax
  shl    $0x8,%eax
  movzwl %ax,%eax
  or     %edx,%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x2c(%rbp),%eax
  shl    $0x18,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%eax
  shl    $0x10,%eax
  and    $0xff0000,%eax
  or     %eax,%edx
  mov    -0x24(%rbp),%eax
  shl    $0x8,%eax
  movzwl %ax,%eax
  or     %edx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L49
  movl   $0x1,-0x30(%rbp)
L49:
  mov    -0x30(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D22
type_infered: string

tool: D22: .

assistant: ```c
int check_linux_version_newer(int major, int minor, int subminor) {
  int newer = 0;
  int check_version, current_version;
  struct utsname uname_buffer;
  char *ptr;
  int current_major = 0, current_minor = 0, current_sub = 0;
  uname(&uname_buffer);
  ptr = strtok(uname_buffer.release, ".");
  if (ptr != NULL)
    current_major = atoi(ptr);
  ptr = strtok(NULL, ".");
  if (ptr != NULL)
    current_minor = atoi(ptr);
  ptr = strtok(NULL, ".");
  if (ptr != NULL)
    current_sub = atoi(ptr);
  check_version = (((major & 0xff) << 24) | ((minor & 0xff) << 16) | ((subminor & 0xff) << 8));
  current_version = (((current_major & 0xff) << 24) | ((current_minor & 0xff) << 16) | ((current_sub & 0xff) << 8));
  if (current_version >= check_version)
    newer = 1;
  return newer;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<check_linux_version_newer>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%r14d
  mov    %esi,%ebx
  mov    %edx,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <uname@plt>
  lea    D27(%rip),%rsi
  mov    (%rsp),%rdi
  call   <strtok@plt>
  mov    $0x0,%ebp
  test   %rax,%rax
  je     L18
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,%ebp
L18:
  lea    D27(%rip),%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    $0x0,%r12d
  test   %rax,%rax
  je     L10
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,%r12d
L10:
  lea    D27(%rip),%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    $0x0,%edx
  test   %rax,%rax
  je     L3
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,%edx
L3:
  shl    $0x10,%ebx
  and    $0xff0000,%ebx
  shl    $0x18,%r14d
  or     %r14d,%ebx
  shl    $0x8,%r13d
  movzwl %r13w,%r13d
  or     %r13d,%ebx
  mov    %r12d,%eax
  shl    $0x10,%eax
  and    $0xff0000,%eax
  shl    $0x18,%ebp
  or     %ebp,%eax
  shl    $0x8,%edx
  movzwl %dx,%edx
  or     %edx,%eax
  cmp    %eax,%ebx
  setle  %al
  movzbl %al,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L37
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L37:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D27
datatype: string

tool: D27: .

assistant: ```c
int check_linux_version_newer(int major, int minor, int subminor) {
  int newer = 0;
  int check_version, current_version;
  struct utsname uname_buffer;
  char *ptr;
  int current_major = 0, current_minor = 0, current_sub = 0;
  uname(&uname_buffer);
  ptr = strtok(uname_buffer.release, ".");
  if (ptr != NULL)
    current_major = atoi(ptr);
  ptr = strtok(NULL, ".");
  if (ptr != NULL)
    current_minor = atoi(ptr);
  ptr = strtok(NULL, ".");
  if (ptr != NULL)
    current_sub = atoi(ptr);
  check_version = (((major & 0xff) << 24) | ((minor & 0xff) << 16) | ((subminor & 0xff) << 8));
  current_version = (((current_major & 0xff) << 24) | ((current_minor & 0xff) << 16) | ((current_sub & 0xff) << 8));
  if (current_version >= check_version)
    newer = 1;
  return newer;
}
```
user: What is the c source code of the assembly code below?

```asm
<check_linux_version_newer>:
  endbr64
  push   %r14
  lea    D8(%rip),%r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <uname@plt>
  mov    (%rsp),%rdi
  mov    %r14,%rsi
  call   <strtok@plt>
  test   %rax,%rax
  je     L10
  mov    %rax,%rdi
  call   <atoi@plt>
  shl    $0x18,%eax
  mov    %eax,%ebp
L10:
  xor    %edi,%edi
  mov    %r14,%rsi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  je     L4
  call   <atoi@plt>
  shl    $0x10,%eax
  and    $0xff0000,%eax
  or     %eax,%ebp
L4:
  xor    %edi,%edi
  mov    %r14,%rsi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  je     L30
  call   <atoi@plt>
  shl    $0x8,%eax
  movzwl %ax,%eax
  or     %eax,%ebp
L30:
  shl    $0x10,%ebx
  shl    $0x18,%r13d
  xor    %eax,%eax
  and    $0xff0000,%ebx
  shl    $0x8,%r12d
  or     %r13d,%ebx
  movzwl %r12w,%r12d
  or     %r12d,%ebx
  cmp    %ebp,%ebx
  setle  %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L23
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D8
type_guessed: string

tool: D8: .

assistant: ```c
int check_linux_version_newer(int major, int minor, int subminor) {
  int newer = 0;
  int check_version, current_version;
  struct utsname uname_buffer;
  char *ptr;
  int current_major = 0, current_minor = 0, current_sub = 0;
  uname(&uname_buffer);
  ptr = strtok(uname_buffer.release, ".");
  if (ptr != NULL)
    current_major = atoi(ptr);
  ptr = strtok(NULL, ".");
  if (ptr != NULL)
    current_minor = atoi(ptr);
  ptr = strtok(NULL, ".");
  if (ptr != NULL)
    current_sub = atoi(ptr);
  check_version = (((major & 0xff) << 24) | ((minor & 0xff) << 16) | ((subminor & 0xff) << 8));
  current_version = (((current_major & 0xff) << 24) | ((current_minor & 0xff) << 16) | ((current_sub & 0xff) << 8));
  if (current_version >= check_version)
    newer = 1;
  return newer;
}
```