ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D11(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D13(%rip),%rdi
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%rdi
  call   <primeFactors@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D11
type_infered: string
name: read_data
arguments: data_tag: D13
type_infered: string

tool: D11: Enter number to prime-factorize: 

tool: D13: %lu

assistant: ```c
int main(void) {
  unsigned long input;
  printf("Enter number to prime-factorize: ");
  scanf("%lu", &input);
  primeFactors(input);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<reverse_zero_additional_memory>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L30
  jmp    L23
L46:
  addq   $0x1,-0x8(%rbp)
L23:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L46
  subq   $0x1,-0x8(%rbp)
  jmp    L18
L3:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  xor    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  xor    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  xor    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x18(%rbp)
  subq   $0x1,-0x8(%rbp)
L18:
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L3
L30:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void reverse_zero_additional_memory(char *str) {
  char *end = str;
  if (str) {
    while (*end) {
      end++;
    }
    end--;
    while (str < end) {
      *str = *str ^ *end;
      *end = *str ^ *end;
      *str = *str ^ *end;
      str++;
      end--;
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<reverse_zero_additional_memory>:
  endbr64
  test   %rdi,%rdi
  je     L28
  cmpb   $0x0,(%rdi)
  je     L17
  mov    %rdi,%rax
L54:
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L54
L50:
  sub    $0x1,%rax
  cmp    %rax,%rdi
  jae    L28
L34:
  movzbl (%rdi),%edx
  xor    (%rax),%dl
  mov    %dl,(%rdi)
  xor    (%rax),%dl
  mov    %dl,(%rax)
  xor    %dl,(%rdi)
  add    $0x1,%rdi
  sub    $0x1,%rax
  cmp    %rax,%rdi
  jb     L34
L28:
  ret
L17:
  mov    %rdi,%rax
  jmp    L50
```

assistant: ```c
void reverse_zero_additional_memory(char *str) {
  char *end = str;
  if (str) {
    while (*end) {
      end++;
    }
    end--;
    while (str < end) {
      *str = *str ^ *end;
      *end = *str ^ *end;
      *str = *str ^ *end;
      str++;
      end--;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<reverse_zero_additional_memory>:
  endbr64
  test   %rdi,%rdi
  je     L6
  movzbl (%rdi),%eax
  test   %al,%al
  je     L69
  mov    %rdi,%rcx
  nopl   0x0(%rax,%rax,1)
L22:
  mov    %rcx,%rdx
  add    $0x1,%rcx
  cmpb   $0x0,(%rcx)
  jne    L22
L30:
  cmp    %rdx,%rdi
  jb     L68
  jmp    L6
  nopl   0x0(%rax,%rax,1)
L47:
  movzbl (%rdi),%eax
L68:
  xor    (%rdx),%al
  sub    $0x1,%rdx
  mov    %al,(%rdi)
  xor    0x1(%rdx),%al
  mov    %al,0x1(%rdx)
  xor    %al,(%rdi)
  add    $0x1,%rdi
  cmp    %rdx,%rdi
  jb     L47
L6:
  ret
  nopl   (%rax)
L69:
  lea    -0x1(%rdi),%rdx
  jmp    L30
```

assistant: ```c
void reverse_zero_additional_memory(char *str) {
  char *end = str;
  if (str) {
    while (*end) {
      end++;
    }
    end--;
    while (str < end) {
      *str = *str ^ *end;
      *end = *str ^ *end;
      *str = *str ^ *end;
      str++;
      end--;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<reverse_zero_additional_memory>:
  endbr64
  test   %rdi,%rdi
  je     L42
  cmpb   $0x0,(%rdi)
  je     L9
  mov    %rdi,%rax
  nopl   0x0(%rax)
L16:
  mov    %rax,%rdx
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L16
L6:
  cmp    %rdi,%rdx
  jbe    L42
  nopl   0x0(%rax)
L54:
  movzbl (%rdi),%eax
  xor    (%rdx),%al
  sub    $0x1,%rdx
  mov    %al,(%rdi)
  xor    0x1(%rdx),%al
  mov    %al,0x1(%rdx)
  xor    %al,(%rdi)
  add    $0x1,%rdi
  cmp    %rdx,%rdi
  jb     L54
L42:
  ret
  nopl   (%rax)
L9:
  lea    -0x1(%rdi),%rdx
  jmp    L6
```

assistant: ```c
void reverse_zero_additional_memory(char *str) {
  char *end = str;
  if (str) {
    while (*end) {
      end++;
    }
    end--;
    while (str < end) {
      *str = *str ^ *end;
      *end = *str ^ *end;
      *str = *str ^ *end;
      str++;
      end--;
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<nthPathComponent>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L17
  cmpq   $0x0,-0x38(%rbp)
  jne    L66
L17:
  mov    $0x0,%eax
  jmp    L45
L66:
  movl   $0x0,-0x1c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L18
L4:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x10(%rbp)
L18:
  mov    -0x1c(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jge    L54
  mov    -0x10(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <index@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L4
L54:
  cmpq   $0x0,-0x10(%rbp)
  jne    L69
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  jmp    L45
L69:
  nop
L23:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L94
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x18(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L23
L94:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x38(%rbp),%rax
L45:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D5
type_infered: string

tool: D5: not enough path components!!!!!


assistant: ```c
char *nthPathComponent(char *source, int n, char *buf) {
  if (!source || !buf)
    return NULL;
  int i = 0;
  char *pathItem = (char *)malloc(strlen(source) + 1);
  char *os1 = pathItem;
  char *s = source;
  for (i = 0; (i < n) && (s = index(s, '/')); i++, *s++)
    ;
  if (!s) {
    printf("not enough path components!!!!!\n");
    free(pathItem);
    return NULL;
  }
  while (*s != '/' && (*os1++ = *s++))
    ;
  *os1++ = 0;
  strcpy(buf, pathItem);
  free(pathItem);
  return buf;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<nthPathComponent>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L65
  mov    %rdi,%rbx
  mov    %esi,%r12d
  mov    %rdx,%r13
  test   %rdx,%rdx
  je     L65
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r14
  mov    $0x0,%ebp
  test   %r12d,%r12d
  jle    L33
L84:
  mov    $0x2f,%esi
  mov    %rbx,%rdi
  call   <index@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L54
  add    $0x1,%ebp
  add    $0x1,%rbx
  cmp    %ebp,%r12d
  jne    L84
L33:
  mov    %r14,%rax
  sub    %r14,%rbx
  jmp    L89
L97:
  mov    %rcx,%rax
L89:
  movzbl (%rax,%rbx,1),%edx
  cmp    $0x2f,%dl
  je     L80
  lea    0x1(%rax),%rcx
  mov    %dl,(%rax)
  test   %dl,%dl
  jne    L97
  mov    %rcx,%rax
L80:
  movb   $0x0,(%rax)
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <strcpy@plt>
  mov    %r14,%rdi
  call   <free@plt>
  mov    %r13,%rbx
L98:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L65:
  mov    $0x0,%ebx
  jmp    L98
L54:
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    %r14,%rdi
  call   <free@plt>
  jmp    L98
```

tool_calls: name: disassemble_data
arguments: tag: D0
type_infered: string

tool: D0: not enough path components!!!!!


assistant: ```c
char *nthPathComponent(char *source, int n, char *buf) {
  if (!source || !buf)
    return NULL;
  int i = 0;
  char *pathItem = (char *)malloc(strlen(source) + 1);
  char *os1 = pathItem;
  char *s = source;
  for (i = 0; (i < n) && (s = index(s, '/')); i++, *s++)
    ;
  if (!s) {
    printf("not enough path components!!!!!\n");
    free(pathItem);
    return NULL;
  }
  while (*s != '/' && (*os1++ = *s++))
    ;
  *os1++ = 0;
  strcpy(buf, pathItem);
  free(pathItem);
  return buf;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<nthPathComponent>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L77
  mov    %rdx,%rbx
  test   %rdx,%rdx
  je     L77
  mov    %rdi,%r12
  mov    %esi,%r13d
  xor    %r14d,%r14d
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %r13d,%r13d
  jg     L7
  jmp    L40
  nopw   0x0(%rax,%rax,1)
L23:
  add    $0x1,%r14d
  add    $0x1,%r12
  cmp    %r14d,%r13d
  je     L40
L7:
  mov    %r12,%rdi
  mov    $0x2f,%esi
  call   <index@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L23
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  jmp    L59
L40:
  mov    %rbp,%rax
  sub    %rbp,%r12
  jmp    L82
  nopl   0x0(%rax,%rax,1)
L60:
  lea    0x1(%rax),%rcx
  mov    %dl,(%rax)
  mov    %rcx,%rax
  test   %dl,%dl
  je     L3
L82:
  movzbl (%rax,%r12,1),%edx
  cmp    $0x2f,%dl
  jne    L60
L3:
  movb   $0x0,(%rax)
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    %rbx,%r12
  call   <strcpy@plt>
  mov    %rbp,%rdi
  call   <free@plt>
L59:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L77:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: label: D6
type_guessed: string

tool: D6: not enough path components!!!!!


assistant: ```c
char *nthPathComponent(char *source, int n, char *buf) {
  if (!source || !buf)
    return NULL;
  int i = 0;
  char *pathItem = (char *)malloc(strlen(source) + 1);
  char *os1 = pathItem;
  char *s = source;
  for (i = 0; (i < n) && (s = index(s, '/')); i++, *s++)
    ;
  if (!s) {
    printf("not enough path components!!!!!\n");
    free(pathItem);
    return NULL;
  }
  while (*s != '/' && (*os1++ = *s++))
    ;
  *os1++ = 0;
  strcpy(buf, pathItem);
  free(pathItem);
  return buf;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<nthPathComponent>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L69
  mov    %rdx,%rbx
  test   %rdx,%rdx
  je     L69
  mov    %rdi,%r12
  mov    %esi,%r13d
  xor    %r14d,%r14d
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %r13d,%r13d
  jg     L70
  jmp    L88
  nopw   0x0(%rax,%rax,1)
L12:
  add    $0x1,%r14d
  add    $0x1,%r12
  cmp    %r14d,%r13d
  je     L88
L70:
  mov    %r12,%rdi
  mov    $0x2f,%esi
  call   <index@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L12
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  jmp    L23
L88:
  mov    %rbp,%rax
  sub    %rbp,%r12
  jmp    L55
  nopl   0x0(%rax,%rax,1)
L49:
  lea    0x1(%rax),%rcx
  mov    %dl,(%rax)
  mov    %rcx,%rax
  test   %dl,%dl
  je     L68
L55:
  movzbl (%rax,%r12,1),%edx
  cmp    $0x2f,%dl
  jne    L49
L68:
  movb   $0x0,(%rax)
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    %rbx,%r12
  call   <strcpy@plt>
  mov    %rbp,%rdi
  call   <free@plt>
L23:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L69:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: label: D2
type_guessed: string

tool: D2: not enough path components!!!!!


assistant: ```c
char *nthPathComponent(char *source, int n, char *buf) {
  if (!source || !buf)
    return NULL;
  int i = 0;
  char *pathItem = (char *)malloc(strlen(source) + 1);
  char *os1 = pathItem;
  char *s = source;
  for (i = 0; (i < n) && (s = index(s, '/')); i++, *s++)
    ;
  if (!s) {
    printf("not enough path components!!!!!\n");
    free(pathItem);
    return NULL;
  }
  while (*s != '/' && (*os1++ = *s++))
    ;
  *os1++ = 0;
  strcpy(buf, pathItem);
  free(pathItem);
  return buf;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<string_nocase_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  movq   $0x1505,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L8
L17:
  mov    -0x28(%rbp),%rax
  shl    $0x5,%rax
  xor    -0x28(%rbp),%rax
  mov    %rax,%rbx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  xor    %rbx,%rax
  mov    %rax,-0x28(%rbp)
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
L8:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L4
  mov    -0x20(%rbp),%rax
  cmp    -0x40(%rbp),%rax
  jb     L17
L4:
  mov    -0x28(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
unsigned long string_nocase_hash(void *string, unsigned long length) {
  unsigned long result = 5381;
  unsigned long acc = 0;
  unsigned char *p;
  p = (unsigned char *)string;
  while (*p != '\0' && acc < length) {
    result = ((result << 5) ^ result) ^ tolower(*p);
    ++p;
    ++acc;
  }
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<string_nocase_hash>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  movzbl (%rdi),%edi
  test   %rsi,%rsi
  je     L41
  mov    %rsi,%r13
  test   %dil,%dil
  je     L41
  mov    $0x0,%ebx
  mov    $0x1505,%ebp
L37:
  movzbl %dil,%edi
  call   <tolower@plt>
  mov    %rax,%rdx
  mov    %rbp,%rax
  shl    $0x5,%rax
  xor    %rdx,%rax
  xor    %rax,%rbp
  add    $0x1,%rbx
  movzbl (%r12,%rbx,1),%edi
  cmp    %rbx,%r13
  jbe    L25
  test   %dil,%dil
  jne    L37
L25:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L41:
  mov    $0x1505,%ebp
  jmp    L25
```

assistant: ```c
unsigned long string_nocase_hash(void *string, unsigned long length) {
  unsigned long result = 5381;
  unsigned long acc = 0;
  unsigned char *p;
  p = (unsigned char *)string;
  while (*p != '\0' && acc < length) {
    result = ((result << 5) ^ result) ^ tolower(*p);
    ++p;
    ++acc;
  }
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<string_nocase_hash>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%edi
  test   %rsi,%rsi
  je     L46
  test   %dil,%dil
  je     L46
  mov    %rsi,%r12
  xor    %ebx,%ebx
  mov    $0x1505,%r13d
  jmp    L14
  nopl   0x0(%rax,%rax,1)
L5:
  test   %dil,%dil
  je     L15
L14:
  call   <tolower@plt>
  add    $0x1,%rbx
  mov    %rax,%r8
  mov    %r13,%rax
  movzbl 0x0(%rbp,%rbx,1),%edi
  shl    $0x5,%rax
  xor    %r8,%rax
  xor    %rax,%r13
  cmp    %rbx,%r12
  ja     L5
L15:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L46:
  add    $0x8,%rsp
  mov    $0x1505,%r13d
  pop    %rbx
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
unsigned long string_nocase_hash(void *string, unsigned long length) {
  unsigned long result = 5381;
  unsigned long acc = 0;
  unsigned char *p;
  p = (unsigned char *)string;
  while (*p != '\0' && acc < length) {
    result = ((result << 5) ^ result) ^ tolower(*p);
    ++p;
    ++acc;
  }
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<string_nocase_hash>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%edi
  test   %rsi,%rsi
  je     L8
  test   %dil,%dil
  je     L8
  mov    %rsi,%r12
  xor    %ebx,%ebx
  mov    $0x1505,%r13d
  jmp    L39
  nopl   0x0(%rax,%rax,1)
L22:
  test   %dil,%dil
  je     L44
L39:
  call   <tolower@plt>
  add    $0x1,%rbx
  mov    %rax,%r8
  mov    %r13,%rax
  movzbl 0x0(%rbp,%rbx,1),%edi
  shl    $0x5,%rax
  xor    %r8,%rax
  xor    %rax,%r13
  cmp    %rbx,%r12
  ja     L22
L44:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L8:
  add    $0x8,%rsp
  mov    $0x1505,%r13d
  pop    %rbx
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
unsigned long string_nocase_hash(void *string, unsigned long length) {
  unsigned long result = 5381;
  unsigned long acc = 0;
  unsigned char *p;
  p = (unsigned char *)string;
  while (*p != '\0' && acc < length) {
    result = ((result << 5) ^ result) ^ tolower(*p);
    ++p;
    ++acc;
  }
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的？

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

assistant: ```c
int main() {
  _list_test();
  _map_test();
  _config_test();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

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

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

assistant: ```c
int main() {
  _list_test();
  _map_test();
  _config_test();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

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

assistant: ```c
int main() {
  _list_test();
  _map_test();
  _config_test();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<a>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <b@plt>
  nop
  leave
  ret
```

assistant: ```c
void a() {
  int a_local = 0;
  b(a_local);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<a>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <b@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void a() {
  int a_local = 0;
  b(a_local);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<a>:
  endbr64
  xor    %edi,%edi
  jmp    1050 <b@plt>
```

assistant: ```c
void a() {
  int a_local = 0;
  b(a_local);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<a>:
  endbr64
  xor    %edi,%edi
  jmp    1050 <b@plt>
```

assistant: ```c
void a() {
  int a_local = 0;
  b(a_local);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<__nv_ffsll>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L48
L49:
  mov    -0x18(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  shr    %cl,%rdx
  mov    %rdx,%rax
  and    $0x1,%eax
  test   %rax,%rax
  jne    L58
  addl   $0x1,-0x4(%rbp)
L48:
  cmpl   $0x3f,-0x4(%rbp)
  jle    L49
  jmp    L26
L58:
  nop
L26:
  cmpl   $0x40,-0x4(%rbp)
  jne    L32
  mov    $0x0,%eax
  jmp    L6
L32:
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
L6:
  pop    %rbp
  ret
```

assistant: ```c
int __nv_ffsll(long long a) {
  int i = 0;
  while (i < 64) {
    if (a & (1ull << i))
      break;
    i++;
  }
  if (i == 64)
    return 0;
  return i + 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<__nv_ffsll>:
  endbr64
  test   $0x1,%dil
  jne    L11
  mov    $0x1,%eax
L21:
  bt     %rax,%rdi
  jb     L6
  add    $0x1,%eax
  cmp    $0x40,%eax
  jne    L21
  mov    $0x0,%eax
  ret
L11:
  mov    $0x0,%eax
L6:
  add    $0x1,%eax
  ret
```

assistant: ```c
int __nv_ffsll(long long a) {
  int i = 0;
  while (i < 64) {
    if (a & (1ull << i))
      break;
    i++;
  }
  if (i == 64)
    return 0;
  return i + 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<__nv_ffsll>:
  endbr64
  xor    %ecx,%ecx
  jmp    L23
  nopl   0x0(%rax,%rax,1)
L8:
  cmp    $0x40,%ecx
  je     L29
L23:
  mov    %rdi,%rax
  shr    %cl,%rax
  add    $0x1,%ecx
  test   $0x1,%al
  je     L8
  mov    %ecx,%eax
  ret
  nopl   (%rax)
L29:
  xor    %ecx,%ecx
  mov    %ecx,%eax
  ret
```

assistant: ```c
int __nv_ffsll(long long a) {
  int i = 0;
  while (i < 64) {
    if (a & (1ull << i))
      break;
    i++;
  }
  if (i == 64)
    return 0;
  return i + 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<__nv_ffsll>:
  endbr64
  xor    %ecx,%ecx
  jmp    L1
  nopl   0x0(%rax,%rax,1)
L15:
  cmp    $0x40,%ecx
  je     L27
L1:
  mov    %rdi,%rax
  shr    %cl,%rax
  add    $0x1,%ecx
  test   $0x1,%al
  je     L15
  mov    %ecx,%eax
  ret
  nopl   (%rax)
L27:
  xor    %ecx,%ecx
  mov    %ecx,%eax
  ret
```

assistant: ```c
int __nv_ffsll(long long a) {
  int i = 0;
  while (i < 64) {
    if (a & (1ull << i))
      break;
    i++;
  }
  if (i == 64)
    return 0;
  return i + 1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x30(%rbp),%rax
  add    $0x14,%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x38(%rbp),%rax
  movl   $0x2,(%rax)
  mov    -0x1c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
main() {
  int ar[10];
  int *p;
  p = &ar[5];
  *p = 2;
  return ar[5];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  mov    $0x2,%eax
  ret
```

assistant: ```c
main() {
  int ar[10];
  int *p;
  p = &ar[5];
  *p = 2;
  return ar[5];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  mov    $0x2,%eax
  ret
```

assistant: ```c
main() {
  int ar[10];
  int *p;
  p = &ar[5];
  *p = 2;
  return ar[5];
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  mov    $0x2,%eax
  ret
```

assistant: ```c
main() {
  int ar[10];
  int *p;
  p = &ar[5];
  *p = 2;
  return ar[5];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<add_test_newline_delimiter>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <add@plt>
  cmp    $0x19,%eax
  jne    L12
  mov    $0x1,%eax
  jmp    L2
L12:
  mov    $0x0,%eax
L2:
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
type_guessed: string

tool: D4: 7,9
5,1,3

assistant: ```c
bool add_test_newline_delimiter() {
  if (add("7,9\n5,1,3") == 25)
    return true;
  return false;
}
```