ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的：

```asm
<basename>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%rax
  cmp    %rax,%rbx
  jb     L38
  jmp    L82
  nopl   0x0(%rax,%rax,1)
L84:
  sub    $0x1,%rax
  cmp    %rax,%rbx
  je     L82
L38:
  cmpb   $0x2f,(%rax)
  je     L84
  movb   $0x0,0x1(%rax)
  cmp    %rax,%rbx
  jae    L11
  nopw   0x0(%rax,%rax,1)
L22:
  sub    $0x1,%rax
  movzbl (%rax),%edx
  cmp    %rax,%rbx
  je     L11
  cmp    $0x2f,%dl
  jne    L22
  nopl   0x0(%rax)
L67:
  add    $0x1,%rax
L11:
  movzbl (%rax),%edx
  cmp    $0x2f,%dl
  je     L67
  test   %dl,%dl
  cmove  %rbx,%rax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L82:
  movb   $0x0,0x1(%rax)
  jmp    L11
```

assistant: ```c
char *basename(char *path) {
  char *startp = path + strlen(path) - 1;
  while (startp > path && *startp == '/')
    startp--;
  startp[1] = '\0';
  while (startp > path && *startp != '/')
    startp--;
  while (*startp && *startp == '/')
    startp++;
  if (!*startp)
    return path;
  return startp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <tf_test_list@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  tf_test_list();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

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

assistant: ```c
int main(void) {
  tf_test_list();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <tf_test_list@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  tf_test_list();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <tf_test_list@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  tf_test_list();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <insert@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x2,%edi
  call   <insert@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x3,%edi
  call   <insert@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x4,%edi
  call   <insert@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x5,%edi
  call   <insert@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x6,%edi
  call   <insert@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x7,%edi
  call   <insert@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <forwardtraverse@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <backwardtraverse@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x63,%edi
  call   <delete@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <forwardtraverse@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <backwardtraverse@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <delete@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <forwardtraverse@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <backwardtraverse@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x4,%edi
  call   <delete@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <forwardtraverse@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <backwardtraverse@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x7,%edi
  call   <delete@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <forwardtraverse@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <backwardtraverse@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x2,%edi
  call   <delete@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <forwardtraverse@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <backwardtraverse@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
int main(void) {
  struct node *head = NULL;
  insert(1, &head);
  insert(2, &head);
  insert(3, &head);
  insert(4, &head);
  insert(5, &head);
  insert(6, &head);
  insert(7, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (99, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (1, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (4, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (7, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (2, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <insert@plt>
  mov    %rbx,%rsi
  mov    $0x2,%edi
  call   <insert@plt>
  mov    %rbx,%rsi
  mov    $0x3,%edi
  call   <insert@plt>
  mov    %rbx,%rsi
  mov    $0x4,%edi
  call   <insert@plt>
  mov    %rbx,%rsi
  mov    $0x5,%edi
  call   <insert@plt>
  mov    %rbx,%rsi
  mov    $0x6,%edi
  call   <insert@plt>
  mov    %rbx,%rsi
  mov    $0x7,%edi
  call   <insert@plt>
  mov    %rbx,%rdi
  call   <forwardtraverse@plt>
  mov    %rbx,%rdi
  call   <backwardtraverse@plt>
  mov    %rbx,%rsi
  mov    $0x63,%edi
  call   <delete@plt>
  mov    %rbx,%rdi
  call   <forwardtraverse@plt>
  mov    %rbx,%rdi
  call   <backwardtraverse@plt>
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <delete@plt>
  mov    %rbx,%rdi
  call   <forwardtraverse@plt>
  mov    %rbx,%rdi
  call   <backwardtraverse@plt>
  mov    %rbx,%rsi
  mov    $0x4,%edi
  call   <delete@plt>
  mov    %rbx,%rdi
  call   <forwardtraverse@plt>
  mov    %rbx,%rdi
  call   <backwardtraverse@plt>
  mov    %rbx,%rsi
  mov    $0x7,%edi
  call   <delete@plt>
  mov    %rbx,%rdi
  call   <forwardtraverse@plt>
  mov    %rbx,%rdi
  call   <backwardtraverse@plt>
  mov    %rbx,%rsi
  mov    $0x2,%edi
  call   <delete@plt>
  mov    %rbx,%rdi
  call   <forwardtraverse@plt>
  mov    %rbx,%rdi
  call   <backwardtraverse@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  struct node *head = NULL;
  insert(1, &head);
  insert(2, &head);
  insert(3, &head);
  insert(4, &head);
  insert(5, &head);
  insert(6, &head);
  insert(7, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (99, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (1, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (4, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (7, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (2, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x1,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movq   $0x0,(%rsp)
  mov    %rbp,%rsi
  call   <insert@plt>
  mov    %rbp,%rsi
  mov    $0x2,%edi
  call   <insert@plt>
  mov    %rbp,%rsi
  mov    $0x3,%edi
  call   <insert@plt>
  mov    %rbp,%rsi
  mov    $0x4,%edi
  call   <insert@plt>
  mov    %rbp,%rsi
  mov    $0x5,%edi
  call   <insert@plt>
  mov    %rbp,%rsi
  mov    $0x6,%edi
  call   <insert@plt>
  mov    %rbp,%rsi
  mov    $0x7,%edi
  call   <insert@plt>
  mov    %rbp,%rdi
  call   <forwardtraverse@plt>
  mov    %rbp,%rdi
  call   <backwardtraverse@plt>
  mov    %rbp,%rsi
  mov    $0x63,%edi
  call   <delete@plt>
  mov    %rbp,%rdi
  call   <forwardtraverse@plt>
  mov    %rbp,%rdi
  call   <backwardtraverse@plt>
  mov    %rbp,%rsi
  mov    $0x1,%edi
  call   <delete@plt>
  mov    %rbp,%rdi
  call   <forwardtraverse@plt>
  mov    %rbp,%rdi
  call   <backwardtraverse@plt>
  mov    %rbp,%rsi
  mov    $0x4,%edi
  call   <delete@plt>
  mov    %rbp,%rdi
  call   <forwardtraverse@plt>
  mov    %rbp,%rdi
  call   <backwardtraverse@plt>
  mov    %rbp,%rsi
  mov    $0x7,%edi
  call   <delete@plt>
  mov    %rbp,%rdi
  call   <forwardtraverse@plt>
  mov    %rbp,%rdi
  call   <backwardtraverse@plt>
  mov    %rbp,%rsi
  mov    $0x2,%edi
  call   <delete@plt>
  mov    %rbp,%rdi
  call   <forwardtraverse@plt>
  mov    %rbp,%rdi
  call   <backwardtraverse@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  struct node *head = NULL;
  insert(1, &head);
  insert(2, &head);
  insert(3, &head);
  insert(4, &head);
  insert(5, &head);
  insert(6, &head);
  insert(7, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (99, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (1, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (4, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (7, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (2, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  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   %rbp
  mov    $0x1,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movq   $0x0,(%rsp)
  mov    %rbp,%rsi
  call   <insert@plt>
  mov    %rbp,%rsi
  mov    $0x2,%edi
  call   <insert@plt>
  mov    %rbp,%rsi
  mov    $0x3,%edi
  call   <insert@plt>
  mov    %rbp,%rsi
  mov    $0x4,%edi
  call   <insert@plt>
  mov    %rbp,%rsi
  mov    $0x5,%edi
  call   <insert@plt>
  mov    %rbp,%rsi
  mov    $0x6,%edi
  call   <insert@plt>
  mov    %rbp,%rsi
  mov    $0x7,%edi
  call   <insert@plt>
  mov    %rbp,%rdi
  call   <forwardtraverse@plt>
  mov    %rbp,%rdi
  call   <backwardtraverse@plt>
  mov    %rbp,%rsi
  mov    $0x63,%edi
  call   <delete@plt>
  mov    %rbp,%rdi
  call   <forwardtraverse@plt>
  mov    %rbp,%rdi
  call   <backwardtraverse@plt>
  mov    %rbp,%rsi
  mov    $0x1,%edi
  call   <delete@plt>
  mov    %rbp,%rdi
  call   <forwardtraverse@plt>
  mov    %rbp,%rdi
  call   <backwardtraverse@plt>
  mov    %rbp,%rsi
  mov    $0x4,%edi
  call   <delete@plt>
  mov    %rbp,%rdi
  call   <forwardtraverse@plt>
  mov    %rbp,%rdi
  call   <backwardtraverse@plt>
  mov    %rbp,%rsi
  mov    $0x7,%edi
  call   <delete@plt>
  mov    %rbp,%rdi
  call   <forwardtraverse@plt>
  mov    %rbp,%rdi
  call   <backwardtraverse@plt>
  mov    %rbp,%rsi
  mov    $0x2,%edi
  call   <delete@plt>
  mov    %rbp,%rdi
  call   <forwardtraverse@plt>
  mov    %rbp,%rdi
  call   <backwardtraverse@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  struct node *head = NULL;
  insert(1, &head);
  insert(2, &head);
  insert(3, &head);
  insert(4, &head);
  insert(5, &head);
  insert(6, &head);
  insert(7, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (99, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (1, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (4, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (7, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  delete (2, &head);
  forwardtraverse(&head);
  backwardtraverse(&head);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strlsave>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L14
  mov    $0x0,%eax
  jmp    L16
L14:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  cmp    %rax,%rdx
  cmovbe %rdx,%rax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strlcpy@plt>
  mov    -0x8(%rbp),%rax
L16:
  leave
  ret
```

assistant: ```c
char *strlsave(const char *s, size_t n) {
  int l;
  char *save;
  if (s == NULL)
    return NULL;
  l = strlen(s);
  l = (((n) < (l)) ? (n) : (l));
  save = malloc(l + 1);
  strlcpy(save, s, l + 1);
  return save;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strlsave>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  test   %rdi,%rdi
  je     L4
  mov    %rsi,%rbx
  call   <strlen@plt>
  cltq
  cmp    %rbx,%rax
  cmova  %rbx,%rax
  lea    0x1(%rax),%ebx
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strlcpy@plt>
L19:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  mov    %rdi,%r12
  jmp    L19
```

assistant: ```c
char *strlsave(const char *s, size_t n) {
  int l;
  char *save;
  if (s == NULL)
    return NULL;
  l = strlen(s);
  l = (((n) < (l)) ? (n) : (l));
  save = malloc(l + 1);
  strlcpy(save, s, l + 1);
  return save;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strlsave>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  test   %rdi,%rdi
  je     L6
  mov    %rsi,%r12
  mov    %rdi,%rbp
  call   <strlen@plt>
  cltq
  cmp    %r12,%rax
  cmova  %r12,%rax
  lea    0x1(%rax),%r12d
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rax,%r13
  mov    %rax,%rdi
  call   <strlcpy@plt>
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  xor    %r13d,%r13d
  pop    %rbp
  pop    %r12
  mov    %r13,%rax
  pop    %r13
  ret
```

assistant: ```c
char *strlsave(const char *s, size_t n) {
  int l;
  char *save;
  if (s == NULL)
    return NULL;
  l = strlen(s);
  l = (((n) < (l)) ? (n) : (l));
  save = malloc(l + 1);
  strlcpy(save, s, l + 1);
  return save;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strlsave>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  test   %rdi,%rdi
  je     L4
  mov    %rsi,%r12
  mov    %rdi,%rbp
  call   <strlen@plt>
  cltq
  cmp    %r12,%rax
  cmova  %r12,%rax
  lea    0x1(%rax),%r12d
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rax,%r13
  mov    %rax,%rdi
  call   <strlcpy@plt>
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  xor    %r13d,%r13d
  pop    %rbp
  pop    %r12
  mov    %r13,%rax
  pop    %r13
  ret
```

assistant: ```c
char *strlsave(const char *s, size_t n) {
  int l;
  char *save;
  if (s == NULL)
    return NULL;
  l = strlen(s);
  l = (((n) < (l)) ? (n) : (l));
  save = malloc(l + 1);
  strlcpy(save, s, l + 1);
  return save;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ConvertCase>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L4:
  sarl   -0x14(%rbp)
  addl   $0x1,-0x4(%rbp)
L12:
  cmpl   $0x0,-0x14(%rbp)
  jg     L4
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  pop    %rbp
  ret
```

assistant: ```c
ushort ConvertCase(ushort caseij) {
  ushort inc = 0;
  while (caseij > 0) {
    caseij = caseij >> 1;
    inc++;
  }
  return inc - 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<ConvertCase>:
  endbr64
  test   %edi,%edi
  jle    L15
  mov    $0x0,%eax
L24:
  sar    %edi
  add    $0x1,%eax
  test   %edi,%edi
  jg     L24
L12:
  sub    $0x1,%eax
  ret
L15:
  mov    $0x0,%eax
  jmp    L12
```

assistant: ```c
ushort ConvertCase(ushort caseij) {
  ushort inc = 0;
  while (caseij > 0) {
    caseij = caseij >> 1;
    inc++;
  }
  return inc - 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ConvertCase>:
  endbr64
  test   %edi,%edi
  jle    L2
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L14:
  mov    %eax,%r8d
  add    $0x1,%eax
  sar    %edi
  jne    L14
  mov    %r8d,%eax
  ret
  xchg   %ax,%ax
L2:
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
ushort ConvertCase(ushort caseij) {
  ushort inc = 0;
  while (caseij > 0) {
    caseij = caseij >> 1;
    inc++;
  }
  return inc - 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ConvertCase>:
  endbr64
  test   %edi,%edi
  jle    L2
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L10:
  mov    %eax,%r8d
  add    $0x1,%eax
  sar    %edi
  jne    L10
  mov    %r8d,%eax
  ret
  xchg   %ax,%ax
L2:
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
ushort ConvertCase(ushort caseij) {
  ushort inc = 0;
  while (caseij > 0) {
    caseij = caseij >> 1;
    inc++;
  }
  return inc - 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<nand_smatrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %rcx,-0x50(%rbp)
  mov    %r8,-0x58(%rbp)
  mov    %r9,-0x60(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  jmp    L25
L29:
  mov    -0x28(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x60(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x50(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  jmp    L13
L3:
  mov    -0x20(%rbp),%rax
  lea    (%rax,%rax,1),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  mov    %ax,-0x2e(%rbp)
  mov    -0x20(%rbp),%rax
  lea    (%rax,%rax,1),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  mov    %ax,-0x2c(%rbp)
  movzwl -0x2e(%rbp),%eax
  and    -0x2c(%rbp),%ax
  test   %ax,%ax
  sete   %al
  movzbl %al,%eax
  mov    %ax,-0x2a(%rbp)
  mov    -0x20(%rbp),%rax
  lea    (%rax,%rax,1),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzwl -0x2a(%rbp),%eax
  mov    %ax,(%rdx)
  addq   $0x1,-0x20(%rbp)
L13:
  mov    -0x20(%rbp),%rax
  cmp    -0x58(%rbp),%rax
  jle    L3
  addq   $0x1,-0x28(%rbp)
L25:
  mov    -0x28(%rbp),%rax
  cmp    -0x48(%rbp),%rax
  jle    L29
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void nand_smatrix(short **m1, long nrl, long nrh, long ncl, long nch, short **m2, short **m3) {
  long i, j;
  short x, y, z, *Xi, *Yi, *Zi;
  for (i = nrl; i <= nrh; i++) {
    Xi = m1[i];
    Yi = m2[i];
    Zi = m3[i];
    for (j = ncl; j <= nch; j++) {
      x = Xi[j];
      y = Yi[j];
      z = !(x & y);
      Zi[j] = z;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<nand_smatrix>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x28(%rsp),%r13
  cmp    %rdx,%rsi
  jg     L28
  mov    %rdi,%rbx
  mov    %rsi,%r11
  mov    %rcx,%rbp
  mov    %r8,%r12
  lea    0x1(%rdx),%r8
  lea    0x1(%r12),%r10
L34:
  mov    (%rbx,%r11,8),%rcx
  mov    (%r9,%r11,8),%rsi
  mov    0x0(%r13,%r11,8),%rdi
  cmp    %r12,%rbp
  jg     L22
  mov    %rbp,%rax
L27:
  movzwl (%rcx,%rax,2),%edx
  and    (%rsi,%rax,2),%dx
  sete   %dl
  movzbl %dl,%edx
  mov    %dx,(%rdi,%rax,2)
  add    $0x1,%rax
  cmp    %r10,%rax
  jne    L27
L22:
  add    $0x1,%r11
  cmp    %r8,%r11
  jne    L34
L28:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void nand_smatrix(short **m1, long nrl, long nrh, long ncl, long nch, short **m2, short **m3) {
  long i, j;
  short x, y, z, *Xi, *Yi, *Zi;
  for (i = nrl; i <= nrh; i++) {
    Xi = m1[i];
    Yi = m2[i];
    Zi = m3[i];
    for (j = ncl; j <= nch; j++) {
      x = Xi[j];
      y = Yi[j];
      z = !(x & y);
      Zi[j] = z;
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<nand_smatrix>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x28(%rsp),%r13
  cmp    %rdx,%rsi
  jg     L14
  mov    %r8,%r12
  mov    %rdi,%rbx
  mov    %rsi,%r11
  mov    %rcx,%rbp
  lea    0x1(%rdx),%r8
  lea    0x1(%r12),%r10
  nopl   0x0(%rax)
L21:
  mov    (%rbx,%r11,8),%rcx
  mov    (%r9,%r11,8),%rsi
  mov    0x0(%r13,%r11,8),%rdi
  cmp    %r12,%rbp
  jg     L4
  mov    %rbp,%rax
  nopl   (%rax)
L10:
  movzwl (%rcx,%rax,2),%edx
  and    (%rsi,%rax,2),%dx
  sete   %dl
  movzbl %dl,%edx
  mov    %dx,(%rdi,%rax,2)
  add    $0x1,%rax
  cmp    %r10,%rax
  jne    L10
L4:
  add    $0x1,%r11
  cmp    %r8,%r11
  jne    L21
L14:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void nand_smatrix(short **m1, long nrl, long nrh, long ncl, long nch, short **m2, short **m3) {
  long i, j;
  short x, y, z, *Xi, *Yi, *Zi;
  for (i = nrl; i <= nrh; i++) {
    Xi = m1[i];
    Yi = m2[i];
    Zi = m3[i];
    for (j = ncl; j <= nch; j++) {
      x = Xi[j];
      y = Yi[j];
      z = !(x & y);
      Zi[j] = z;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<monteCarlo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L24
L31:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x3e8,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  movslq %edx,%rax
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x3e8,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  movslq %edx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  imul   %rax,%rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  imul   %rax,%rax
  add    %rdx,%rax
  cmp    $0xf4240,%rax
  jg     L21
  addl   $0x1,-0x18(%rbp)
L21:
  addl   $0x1,-0x14(%rbp)
L24:
  mov    -0x14(%rbp),%eax
  cltq
  cmp    %rax,-0x28(%rbp)
  ja     L31
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x18(%rbp),%xmm1
  movsd  D15(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  mov    -0x28(%rbp),%rax
  test   %rax,%rax
  js     L17
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  jmp    L20
L17:
  mov    %rax,%rdx
  shr    %rdx
  and    $0x1,%eax
  or     %rax,%rdx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdx,%xmm0
  addsd  %xmm0,%xmm0
L20:
  divsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D15
type_guessed: byte[8]
name: extract_symbolic_data
arguments: data_tag: D12
type_guessed: string

tool: {"D15": [0x00,0x00,0x00,0x00,0x00,0x00,0x10,0x40]}

tool: D12: %e


assistant: ```c
void monteCarlo(unsigned long long times) {
  long long x, y;
  int in = 0;
  int a;
  for (a = 0; a < times; a++) {
    x = rand() % 1000;
    y = rand() % 1000;
    if ((x * x) + (y * y) <= 1000 * 1000)
      in++;
  }
  printf("%e\n", ((double)(in)*4) / times);
}
```