ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D5(%rip),%rbp
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L63:
  movsbl %bl,%edi
  call   <putchar@plt>
L20:
  cmp    %bl,0x0(%rbp)
  je     L3
L25:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  lea    -0x8(%rbx),%eax
  cmp    $0x1,%al
  jbe    L43
  cmp    $0x5c,%bl
  jne    L63
L43:
  mov    $0x5c,%edi
  call   <putchar@plt>
  cmp    $0x9,%bl
  je     L94
  cmp    $0x8,%bl
  je     L47
  cmp    $0x5c,%bl
  jne    L20
  mov    $0x5c,%edi
  call   <putchar@plt>
  cmp    %bl,0x0(%rbp)
  jne    L25
L3:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L47:
  mov    $0x62,%edi
  call   <putchar@plt>
  jmp    L20
  nopl   0x0(%rax)
L94:
  mov    $0x74,%edi
  call   <putchar@plt>
  jmp    L20
```

assistant: ```c
int main() {
  char c;
  do {
    c = getchar();
    if (c == '\t' || c == '\b' || c == '\\') {
      putchar('\\');
      if (c == '\t')
        putchar('t');
      if (c == '\b')
        putchar('b');
      if (c == '\\')
        putchar('\\');
    } else {
      putchar(c);
    }
  } while (c != EOF);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L22
L12:
  lea    -0x10(%rbp),%rcx
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%ecx
  mov    -0x18(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <count@plt>
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L22:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jl     L12
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D14
type_guessed: string
name: extract_symbolic_data
arguments: label: D25
type_guessed: string
name: extract_symbolic_data
arguments: label: D11
type_guessed: string

tool: D14: %d

tool: D25: %d %d %d

tool: D11: %d


assistant: ```c
int main() {
  int t;
  scanf("%d", &t);
  for (int a0 = 0; a0 < t; a0++) {
    int n;
    int c;
    int m;
    scanf("%d %d %d", &n, &c, &m);
    printf("%d\n", count(n, c, m));
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    D26(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x8(%rsp)
  jle    L24
  mov    $0x0,%ebx
  lea    0x14(%rsp),%r14
  lea    0x10(%rsp),%r13
  lea    D10(%rip),%r12
  lea    D19(%rip),%rbp
L28:
  lea    0xc(%rsp),%rsi
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x14(%rsp),%edx
  mov    0x10(%rsp),%esi
  mov    0xc(%rsp),%edi
  call   <count@plt>
  mov    %eax,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,0x8(%rsp)
  jg     L28
L24:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D26
type_guessed: string
name: retrieve_data
arguments: data_tag: D10
type_guessed: string
name: retrieve_data
arguments: data_tag: D19
type_guessed: string

tool: D26: %d

tool: D10: %d %d %d

tool: D19: %d


assistant: ```c
int main() {
  int t;
  scanf("%d", &t);
  for (int a0 = 0; a0 < t; a0++) {
    int n;
    int c;
    int m;
    scanf("%d %d %d", &n, &c, &m);
    printf("%d\n", count(n, c, m));
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r15
  lea    D29(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  test   %eax,%eax
  jle    L2
  xor    %r15d,%r15d
  lea    0x14(%rsp),%r14
  lea    0x10(%rsp),%r13
  lea    0xc(%rsp),%r12
  lea    D9(%rip),%rbp
  lea    D14(%rip),%rbx
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%r15d
  call   <scanf@plt>
  mov    0x14(%rsp),%edx
  mov    0x10(%rsp),%esi
  mov    0xc(%rsp),%edi
  call   <count@plt>
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  cmp    %r15d,0x8(%rsp)
  jg     L5
L2:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D29
data_type: string
name: disassemble_data
arguments: label: D9
data_type: string
name: disassemble_data
arguments: label: D14
data_type: string

tool: D29: %d

tool: D9: %d %d %d

tool: D14: %d


assistant: ```c
int main() {
  int t;
  scanf("%d", &t);
  for (int a0 = 0; a0 < t; a0++) {
    int n;
    int c;
    int m;
    scanf("%d %d %d", &n, &c, &m);
    printf("%d\n", count(n, c, m));
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  lea    D10(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  test   %eax,%eax
  jle    L5
  xor    %r15d,%r15d
  lea    0x14(%rsp),%r14
  lea    0x10(%rsp),%r13
  lea    0xc(%rsp),%r12
  lea    D3(%rip),%rbp
  lea    D13(%rip),%rbx
  nopl   0x0(%rax,%rax,1)
L4:
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%r15d
  call   <scanf@plt>
  mov    0x14(%rsp),%edx
  mov    0x10(%rsp),%esi
  mov    0xc(%rsp),%edi
  call   <count@plt>
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  cmp    %r15d,0x8(%rsp)
  jg     L4
L5:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D10
data_type: string
name: parse_data
arguments: tag: D3
data_type: string
name: parse_data
arguments: tag: D13
data_type: string

tool: D10: %d

tool: D3: %d %d %d

tool: D13: %d


assistant: ```c
int main() {
  int t;
  scanf("%d", &t);
  for (int a0 = 0; a0 < t; a0++) {
    int n;
    int c;
    int m;
    scanf("%d %d %d", &n, &c, &m);
    printf("%d\n", count(n, c, m));
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<count_newlines>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L20
L25:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L16
  addl   $0x1,-0x4(%rbp)
L16:
  addq   $0x1,-0x18(%rbp)
L20:
  mov    -0x18(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L25
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int count_newlines(const char *start, const char *end) {
  int count = 0;
  while (start < end) {
    if (*start == '\n')
      count++;
    start++;
  }
  return count;
}
```
user: What is the c source code of the assembly code below?

```asm
<count_newlines>:
  endbr64
  cmp    %rsi,%rdi
  jae    L15
  mov    $0x0,%eax
L19:
  cmpb   $0xa,(%rdi)
  sete   %dl
  movzbl %dl,%edx
  add    %edx,%eax
  add    $0x1,%rdi
  cmp    %rdi,%rsi
  jne    L19
  ret
L15:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int count_newlines(const char *start, const char *end) {
  int count = 0;
  while (start < end) {
    if (*start == '\n')
      count++;
    start++;
  }
  return count;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<count_newlines>:
  endbr64
  xor    %eax,%eax
  cmp    %rsi,%rdi
  jae    L0
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %edx,%edx
  cmpb   $0xa,(%rdi)
  sete   %dl
  add    $0x1,%rdi
  add    %edx,%eax
  cmp    %rdi,%rsi
  jne    L3
  ret
  nopl   0x0(%rax)
L0:
  ret
```

assistant: ```c
int count_newlines(const char *start, const char *end) {
  int count = 0;
  while (start < end) {
    if (*start == '\n')
      count++;
    start++;
  }
  return count;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<count_word>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L3
L32:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L5
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x5a,%al
  jle    L74
L5:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L79
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x7a,%al
  jg     L79
L74:
  movl   $0x1,-0x4(%rbp)
  jmp    L9
L79:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  je     L9
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <count_word>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  jmp    L69
L9:
  addl   $0x1,-0x8(%rbp)
L3:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L32
  mov    -0x4(%rbp),%eax
L69:
  leave
  ret
```

assistant: ```c
char count_word(char word[]) {
  int i, t;
  t = 0;
  for (i = 0; word[i]; i++)
    if ((word[i] > 64 && word[i] < 91) || (word[i] > 96 && word[i] < 123))
      t = 1;
    else if (word[i] != '\n')
      return t + count_word(word + i + 1);
  return t;
}
```
user: Translate the below assembly code into C programming language:

```asm
<count_word>:
  endbr64
  push   %rbx
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L10
  mov    $0x0,%ebx
  mov    $0x0,%ecx
  mov    $0x1,%esi
  jmp    L33
L32:
  mov    %esi,%ebx
L45:
  add    $0x1,%rcx
  movzbl (%rdi,%rcx,1),%edx
  test   %dl,%dl
  je     L4
L33:
  mov    %edx,%eax
  and    $0xffffffdf,%eax
  sub    $0x41,%eax
  cmp    $0x19,%al
  jbe    L32
  cmp    $0xa,%dl
  je     L45
  lea    0x1(%rdi,%rcx,1),%rdi
  call   <count_word>
  add    %ebx,%eax
  jmp    L19
L10:
  mov    $0x0,%ebx
L4:
  mov    %ebx,%eax
L19:
  pop    %rbx
  ret
```

assistant: ```c
char count_word(char word[]) {
  int i, t;
  t = 0;
  for (i = 0; word[i]; i++)
    if ((word[i] > 64 && word[i] < 91) || (word[i] > 96 && word[i] < 123))
      t = 1;
    else if (word[i] != '\n')
      return t + count_word(word + i + 1);
  return t;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<count_word>:
  endbr64
  xor    %r8d,%r8d
L53:
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L55
  xor    %esi,%esi
  xor    %ecx,%ecx
  jmp    L27
  nopl   0x0(%rax)
L42:
  cmp    $0xa,%dl
  jne    L21
  add    $0x1,%rcx
  movzbl (%rdi,%rcx,1),%edx
  test   %dl,%dl
  je     L56
L27:
  mov    %edx,%eax
  and    $0xffffffdf,%eax
  sub    $0x41,%eax
  cmp    $0x19,%al
  ja     L42
  add    $0x1,%rcx
  mov    $0x1,%esi
  movzbl (%rdi,%rcx,1),%edx
  test   %dl,%dl
  jne    L27
L56:
  add    %esi,%r8d
L55:
  mov    %r8d,%eax
  ret
  nopl   (%rax)
L21:
  lea    0x1(%rdi,%rcx,1),%rdi
  add    %esi,%r8d
  jmp    L53
```

assistant: ```c
char count_word(char word[]) {
  int i, t;
  t = 0;
  for (i = 0; word[i]; i++)
    if ((word[i] > 64 && word[i] < 91) || (word[i] > 96 && word[i] < 123))
      t = 1;
    else if (word[i] != '\n')
      return t + count_word(word + i + 1);
  return t;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<count_word>:
  endbr64
  xor    %r8d,%r8d
L42:
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L56
  xor    %esi,%esi
  xor    %ecx,%ecx
  jmp    L12
  nopl   0x0(%rax)
L21:
  cmp    $0xa,%dl
  jne    L57
  add    $0x1,%rcx
  movzbl (%rdi,%rcx,1),%edx
  test   %dl,%dl
  je     L61
L12:
  mov    %edx,%eax
  and    $0xffffffdf,%eax
  sub    $0x41,%eax
  cmp    $0x19,%al
  ja     L21
  add    $0x1,%rcx
  mov    $0x1,%esi
  movzbl (%rdi,%rcx,1),%edx
  test   %dl,%dl
  jne    L12
L61:
  add    %esi,%r8d
L56:
  mov    %r8d,%eax
  ret
  nopl   (%rax)
L57:
  lea    0x1(%rdi,%rcx,1),%rdi
  add    %esi,%r8d
  jmp    L42
```

assistant: ```c
char count_word(char word[]) {
  int i, t;
  t = 0;
  for (i = 0; word[i]; i++)
    if ((word[i] > 64 && word[i] < 91) || (word[i] > 96 && word[i] < 123))
      t = 1;
    else if (word[i] != '\n')
      return t + count_word(word + i + 1);
  return t;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
L29:
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L87
  movl   $0x1,-0x10(%rbp)
  jmp    L18
L87:
  movl   $0x1,-0xc(%rbp)
L18:
  mov    -0x14(%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,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  jne    L29
  cmpl   $0x0,-0x10(%rbp)
  je     L83
  cmpl   $0x0,-0xc(%rbp)
  je     L83
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L20
L83:
  cmpl   $0x0,-0x10(%rbp)
  je     L36
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L20
L36:
  cmpl   $0x0,-0xc(%rbp)
  je     L20
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L20:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L41
  call   <__stack_chk_fail@plt>
L41:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D22
type_infered: string
name: disassemble_data
arguments: label: D43
type_infered: string
name: disassemble_data
arguments: label: D10
type_infered: string
name: disassemble_data
arguments: label: D44
type_infered: string
name: disassemble_data
arguments: label: D3
type_infered: string

tool: D22: Please input an integer:

tool: D43: %d

tool: D10: both!

tool: D44: odd number only!

tool: D3: even number only!

assistant: ```c
int main() {
  int t, n, odd = 0, even = 0;
  printf("Please input an integer:");
  scanf("%d", &n);
  do {
    if (n % 2)
      odd = 1;
    else
      even = 1;
  } while (n /= 10);
  if (odd && even)
    printf("both!");
  else if (odd)
    printf("odd number only!");
  else if (even)
    printf("even number only!");
  else
    ;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D31(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D12(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  mov    $0x0,%esi
  mov    $0x0,%ecx
  mov    $0x1,%edi
L45:
  mov    %edx,%eax
  and    $0x1,%eax
  cmove  %edi,%esi
  cmovne %edi,%ecx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,%edx
  jne    L45
  movl   $0x0,0x4(%rsp)
  test   %ecx,%ecx
  je     L43
  test   %esi,%esi
  jne    L19
L43:
  test   %ecx,%ecx
  jne    L28
  test   %esi,%esi
  je     L5
  lea    D26(%rip),%rdi
  call   <printf@plt>
  jmp    L5
L19:
  lea    D16(%rip),%rdi
  call   <printf@plt>
L5:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L28:
  lea    D11(%rip),%rdi
  call   <printf@plt>
  jmp    L5
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D31
datatype: string
name: retrieve_data
arguments: label: D12
datatype: string
name: retrieve_data
arguments: label: D26
datatype: string
name: retrieve_data
arguments: label: D16
datatype: string
name: retrieve_data
arguments: label: D11
datatype: string

tool: D31: Please input an integer:

tool: D12: %d

tool: D26: even number only!

tool: D16: both!

tool: D11: odd number only!

assistant: ```c
int main() {
  int t, n, odd = 0, even = 0;
  printf("Please input an integer:");
  scanf("%d", &n);
  do {
    if (n % 2)
      odd = 1;
    else
      even = 1;
  } while (n /= 10);
  if (odd && even)
    printf("both!");
  else if (odd)
    printf("odd number only!");
  else if (even)
    printf("even number only!");
  else
    ;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D21(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D39(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  xor    %edi,%edi
  xor    %esi,%esi
  mov    $0x1,%r8d
  nopl   0x0(%rax,%rax,1)
L19:
  test   $0x1,%dl
  movslq %edx,%rax
  mov    %edx,%ecx
  cmove  %r8d,%edi
  cmovne %r8d,%esi
  sar    $0x1f,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sub    %ecx,%eax
  mov    %eax,%edx
  jne    L19
  movl   $0x0,0x4(%rsp)
  test   %edi,%esi
  jne    L9
  test   %esi,%esi
  jne    L45
  test   %edi,%edi
  jne    L69
L66:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L45:
  lea    D18(%rip),%rdi
  call   <printf@plt>
  jmp    L66
L9:
  lea    D40(%rip),%rdi
  call   <printf@plt>
  jmp    L66
L69:
  lea    D37(%rip),%rdi
  call   <printf@plt>
  jmp    L66
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D21
type_guessed: string
name: extract_symbolic_data
arguments: tag: D39
type_guessed: string
name: extract_symbolic_data
arguments: tag: D18
type_guessed: string
name: extract_symbolic_data
arguments: tag: D40
type_guessed: string
name: extract_symbolic_data
arguments: tag: D37
type_guessed: string

tool: D21: Please input an integer:

tool: D39: %d

tool: D18: odd number only!

tool: D40: both!

tool: D37: even number only!

assistant: ```c
int main() {
  int t, n, odd = 0, even = 0;
  printf("Please input an integer:");
  scanf("%d", &n);
  do {
    if (n % 2)
      odd = 1;
    else
      even = 1;
  } while (n /= 10);
  if (odd && even)
    printf("both!");
  else if (odd)
    printf("odd number only!");
  else if (even)
    printf("even number only!");
  else
    ;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D0(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  xor    %edi,%edi
  xor    %esi,%esi
  jmp    L98
  nop
L16:
  mov    %edi,%r8d
  mov    $0x1,%esi
L66:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  mov    %eax,%edx
  je     L37
L98:
  test   $0x1,%dl
  jne    L16
  mov    %esi,%r8d
  mov    $0x1,%edi
  jmp    L66
L37:
  movl   $0x0,0x4(%rsp)
  test   %r8d,%r8d
  jne    L67
  test   %esi,%esi
  jne    L64
  test   %edi,%edi
  jne    L22
L108:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L65
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L64:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  jmp    L108
L67:
  lea    D40(%rip),%rdi
  call   <printf@plt>
  jmp    L108
L22:
  lea    D45(%rip),%rdi
  call   <printf@plt>
  jmp    L108
L65:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D0
datatype: string
name: retrieve_data
arguments: data_tag: D24
datatype: string
name: retrieve_data
arguments: data_tag: D7
datatype: string
name: retrieve_data
arguments: data_tag: D40
datatype: string
name: retrieve_data
arguments: data_tag: D45
datatype: string

tool: D0: Please input an integer:

tool: D24: %d

tool: D7: odd number only!

tool: D40: both!

tool: D45: even number only!

assistant: ```c
int main() {
  int t, n, odd = 0, even = 0;
  printf("Please input an integer:");
  scanf("%d", &n);
  do {
    if (n % 2)
      odd = 1;
    else
      even = 1;
  } while (n /= 10);
  if (odd && even)
    printf("both!");
  else if (odd)
    printf("odd number only!");
  else if (even)
    printf("even number only!");
  else
    ;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  lea    D5(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    -0x20(%rbp),%rax
  lea    D12(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
type: string
name: extract_symbolic_data
arguments: data_label: D12
type: string

tool: D5: dsdsdsdso

tool: D12: .

assistant: ```c
int main() {
  char s[22];
  strcpy(s, "dsdsdsdso");
  char *str = strtok(s, ".");
  puts(str);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    D0(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  lea    D1(%rip),%rsi
  mov    %rbx,%rdi
  call   <strtok@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
data_type: string
name: extract_symbolic_data
arguments: label: D1
data_type: string

tool: D0: dsdsdsdso

tool: D1: .

assistant: ```c
int main() {
  char s[22];
  strcpy(s, "dsdsdsdso");
  char *str = strtok(s, ".");
  puts(str);
}
```
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
  lea    D13(%rip),%rsi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rdi
  lea    D3(%rip),%rsi
  call   <strtok@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D13
datatype: string
name: parse_data
arguments: data_label: D3
datatype: string

tool: D13: dsdsdsdso

tool: D3: .

assistant: ```c
int main() {
  char s[22];
  strcpy(s, "dsdsdsdso");
  char *str = strtok(s, ".");
  puts(str);
}
```