ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<set_block_used>:
  endbr64
  test   %esi,%esi
  lea    0x7(%rsi),%eax
  mov    %esi,%ecx
  mov    $0x80,%edx
  cmovns %esi,%eax
  and    $0x7,%ecx
  sar    %cl,%edx
  sar    $0x3,%eax
  cltq
  or     %dl,(%rdi,%rax,1)
  ret
```

assistant: ```c
void set_block_used(unsigned char *block_bitmap, int blocknum) {
  unsigned char bm;
  bm = block_bitmap[blocknum / 8];
  bm |= (0x80 >> (blocknum % 8));
  block_bitmap[blocknum / 8] = bm;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getword>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L27
L30:
  mov    -0x24(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl (%rcx),%eax
  mov    %al,(%rdx)
L27:
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L38
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L30
L38:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L9
  mov    $0xffffffff,%eax
  jmp    L0
L9:
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
L0:
  leave
  ret
```

assistant: ```c
int getword(char *word, char *line, int pos) {
  int i = 0;
  while (line[pos] != '\0' && isspace(line[pos]) == 0) {
    word[i++] = line[pos++];
  }
  word[i] = '\0';
  if (line[pos] == '\0') {
    return -1;
  } else {
    return pos + 1;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<getword>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r15
  mov    %edx,%r12d
  movslq %edx,%rax
  lea    (%rsi,%rax,1),%r14
  movzbl (%r14),%edi
  mov    %r14,%rbp
  mov    $0x1,%ebx
  lea    0x1(%rsi,%rax,1),%rax
  sub    %r14,%rax
  mov    %rax,0x8(%rsp)
  mov    $0x0,%r13d
  test   %dil,%dil
  je     L12
L47:
  lea    -0x1(%rbx),%r13d
  movsbl %dil,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L12
  add    $0x1,%r12d
  mov    %ebx,%r13d
  movzbl 0x0(%rbp),%eax
  mov    %al,-0x1(%r15,%rbx,1)
  add    0x8(%rsp),%rbp
  add    $0x1,%rbx
  movzbl -0x1(%r14,%rbx,1),%edi
  test   %dil,%dil
  jne    L47
L12:
  movslq %r13d,%r13
  movb   $0x0,(%r15,%r13,1)
  cmpb   $0x0,0x0(%rbp)
  je     L9
  lea    0x1(%r12),%eax
L48:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L9:
  mov    $0xffffffff,%eax
  jmp    L48
```

assistant: ```c
int getword(char *word, char *line, int pos) {
  int i = 0;
  while (line[pos] != '\0' && isspace(line[pos]) == 0) {
    word[i++] = line[pos++];
  }
  word[i] = '\0';
  if (line[pos] == '\0') {
    return -1;
  } else {
    return pos + 1;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getword>:
  endbr64
  push   %r15
  movslq %edx,%rax
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  lea    0x1(%rsi,%rax,1),%r12
  push   %rbp
  lea    (%rsi,%rax,1),%rbp
  push   %rbx
  mov    %rbp,%r15
  mov    %rax,%rbx
  sub    %rbp,%r12
  mov    %r13,%rax
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  movsbl 0x0(%rbp),%edi
  test   %dil,%dil
  je     L68
  mov    $0x1,%r14d
  jmp    L29
  nopl   0x0(%rax,%rax,1)
L57:
  movzbl (%r15),%ecx
  movslq %r14d,%rsi
  add    $0x1,%ebx
  add    %r12,%r15
  mov    %cl,0x0(%r13)
  movsbl 0x0(%rbp,%r14,1),%edi
  lea    0x1(%rax),%r13
  add    $0x1,%r14
  test   %dil,%dil
  je     L15
L29:
  call   <isspace@plt>
  mov    %rax,%r8
  mov    %r13,%rax
  test   %r8,%r8
  je     L57
L68:
  movb   $0x0,(%rax)
  cmpb   $0x0,(%r15)
  je     L53
  lea    0x1(%rbx),%eax
L55:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  mov    0x8(%rsp),%rax
  add    %rsi,%rax
  jmp    L68
L53:
  mov    $0xffffffff,%eax
  jmp    L55
```

assistant: ```c
int getword(char *word, char *line, int pos) {
  int i = 0;
  while (line[pos] != '\0' && isspace(line[pos]) == 0) {
    word[i++] = line[pos++];
  }
  word[i] = '\0';
  if (line[pos] == '\0') {
    return -1;
  } else {
    return pos + 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
<getword>:
  endbr64
  push   %r15
  movslq %edx,%rax
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  lea    0x1(%rsi,%rax,1),%r12
  push   %rbp
  lea    (%rsi,%rax,1),%rbp
  push   %rbx
  mov    %rbp,%r15
  mov    %rax,%rbx
  sub    %rbp,%r12
  mov    %r13,%rax
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  movsbl 0x0(%rbp),%edi
  test   %dil,%dil
  je     L60
  mov    $0x1,%r14d
  jmp    L46
  nopl   0x0(%rax,%rax,1)
L42:
  movzbl (%r15),%ecx
  movslq %r14d,%rsi
  add    $0x1,%ebx
  add    %r12,%r15
  mov    %cl,0x0(%r13)
  movsbl 0x0(%rbp,%r14,1),%edi
  lea    0x1(%rax),%r13
  add    $0x1,%r14
  test   %dil,%dil
  je     L61
L46:
  call   <isspace@plt>
  mov    %rax,%r8
  mov    %r13,%rax
  test   %r8,%r8
  je     L42
L60:
  movb   $0x0,(%rax)
  cmpb   $0x0,(%r15)
  je     L18
  lea    0x1(%rbx),%eax
L1:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L61:
  mov    0x8(%rsp),%rax
  add    %rsi,%rax
  jmp    L60
L18:
  mov    $0xffffffff,%eax
  jmp    L1
```

assistant: ```c
int getword(char *word, char *line, int pos) {
  int i = 0;
  while (line[pos] != '\0' && isspace(line[pos]) == 0) {
    word[i++] = line[pos++];
  }
  word[i] = '\0';
  if (line[pos] == '\0') {
    return -1;
  } else {
    return pos + 1;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jge    L0
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <swap@plt>
  subl   $0x1,-0x10(%rbp)
  addl   $0x1,-0xc(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <reverse>
L0:
  nop
  leave
  ret
```

assistant: ```c
void reverse(char s[], int i, int j) {
  if (i < j) {
    swap(s, i, j);
    reverse(s, ++i, --j);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<reverse>:
  endbr64
  cmp    %edx,%esi
  jl     L7
  ret
L7:
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebx
  mov    %edx,%ebp
  call   <swap@plt>
  lea    -0x1(%rbp),%edx
  lea    0x1(%rbx),%esi
  mov    %r12,%rdi
  call   <reverse>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void reverse(char s[], int i, int j) {
  if (i < j) {
    swap(s, i, j);
    reverse(s, ++i, --j);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<reverse>:
  endbr64
  cmp    %edx,%esi
  jge    L11
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edx,%ebx
  nopl   0x0(%rax,%rax,1)
L10:
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  sub    $0x1,%ebx
  call   <swap@plt>
  add    $0x1,%ebp
  cmp    %ebp,%ebx
  jg     L10
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  ret
```

assistant: ```c
void reverse(char s[], int i, int j) {
  if (i < j) {
    swap(s, i, j);
    reverse(s, ++i, --j);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<reverse>:
  endbr64
  cmp    %edx,%esi
  jge    L9
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edx,%ebx
  nopl   0x0(%rax,%rax,1)
L15:
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  sub    $0x1,%ebx
  call   <swap@plt>
  add    $0x1,%ebp
  cmp    %ebp,%ebx
  jg     L15
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  ret
```

assistant: ```c
void reverse(char s[], int i, int j) {
  if (i < j) {
    swap(s, i, j);
    reverse(s, ++i, --j);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<str_to_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %edi,-0x84(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x70(%rbp)
  movl   $0x0,-0x78(%rbp)
  movabs $0x3736353433323130,%rax
  mov    %rax,-0x6b(%rbp)
  movw   $0x3938,-0x63(%rbp)
  movb   $0x0,-0x61(%rbp)
  cmpl   $0x0,-0x84(%rbp)
  jns    L31
  mov    -0x70(%rbp),%eax
  cltq
  movb   $0x2d,-0x60(%rbp,%rax,1)
  addl   $0x1,-0x78(%rbp)
  negl   -0x84(%rbp)
L31:
  mov    -0x84(%rbp),%eax
  mov    %eax,-0x74(%rbp)
L32:
  addl   $0x1,-0x78(%rbp)
  mov    -0x74(%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,-0x74(%rbp)
  cmpl   $0x0,-0x74(%rbp)
  jne    L32
  mov    -0x78(%rbp),%eax
  cltq
  movb   $0x0,-0x60(%rbp,%rax,1)
L23:
  subl   $0x1,-0x78(%rbp)
  mov    -0x84(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  movslq %edx,%rax
  movzbl -0x6b(%rbp,%rax,1),%edx
  mov    -0x78(%rbp),%eax
  cltq
  mov    %dl,-0x60(%rbp,%rax,1)
  mov    -0x84(%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,-0x84(%rbp)
  cmpl   $0x0,-0x84(%rbp)
  jne    L23
  lea    -0x60(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

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

tool: D5: %s


assistant: ```c
void str_to_int(int num) {
  int i = 0;
  int length = 0;
  int sum;
  char const digits[] = "0123456789";
  char str[81];
  if (num < 0) {
    str[i] = '-';
    ++length;
    num = -num;
  }
  sum = num;
  do {
    ++length;
    sum /= 10;
  } while (sum);
  str[length] = '\0';
  do {
    --length;
    str[length] = digits[num % 10];
    num /= 10;
  } while (num);
  printf("%s\n", str);
}
```
user: Translate the below assembly code into C programming language:

```asm
<str_to_int>:
  endbr64
  sub    $0x78,%rsp
  mov    %edi,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movabs $0x3736353433323130,%rax
  mov    %rax,0x5(%rsp)
  movw   $0x3938,0xd(%rsp)
  movb   $0x0,0xf(%rsp)
  mov    $0x0,%esi
  test   %edi,%edi
  js     L21
L5:
  mov    %edx,%ecx
L36:
  add    $0x1,%esi
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  jne    L36
  movslq %esi,%rsi
  movb   $0x0,0x10(%rsp,%rsi,1)
  lea    0x10(%rsp,%rsi,1),%rsi
L38:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  movslq %edx,%rdx
  movzbl 0x5(%rsp,%rdx,1),%edx
  mov    %dl,-0x1(%rsi)
  mov    %eax,%edx
  sub    $0x1,%rsi
  test   %eax,%eax
  jne    L38
  lea    0x10(%rsp),%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x78,%rsp
  ret
L21:
  movb   $0x2d,0x10(%rsp)
  neg    %edx
  mov    $0x1,%esi
  jmp    L5
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D3
type_guessed: string

tool: D3: %s


assistant: ```c
void str_to_int(int num) {
  int i = 0;
  int length = 0;
  int sum;
  char const digits[] = "0123456789";
  char str[81];
  if (num < 0) {
    str[i] = '-';
    ++length;
    num = -num;
  }
  sum = num;
  do {
    ++length;
    sum /= 10;
  } while (sum);
  str[length] = '\0';
  do {
    --length;
    str[length] = digits[num % 10];
    num /= 10;
  } while (num);
  printf("%s\n", str);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<str_to_int>:
  endbr64
  sub    $0x78,%rsp
  mov    %edi,%eax
  xor    %ecx,%ecx
  movabs $0x3736353433323130,%rdi
  mov    %fs:0x28,%rdx
  mov    %rdx,0x68(%rsp)
  xor    %edx,%edx
  mov    $0x3938,%edx
  mov    %rdi,0x5(%rsp)
  mov    %dx,0xd(%rsp)
  movb   $0x0,0xf(%rsp)
  test   %eax,%eax
  jns    L33
  movb   $0x2d,0x10(%rsp)
  neg    %eax
  mov    $0x1,%ecx
L33:
  mov    %eax,%edx
  mov    $0xcccccccd,%esi
  nopl   0x0(%rax)
L16:
  mov    %edx,%edx
  add    $0x1,%ecx
  imul   %rsi,%rdx
  shr    $0x23,%rdx
  test   %edx,%edx
  jne    L16
  movslq %ecx,%rcx
  lea    0x10(%rsp),%r8
  mov    $0xcccccccd,%edi
  movb   $0x0,0x10(%rsp,%rcx,1)
  add    %r8,%rcx
  xchg   %ax,%ax
L18:
  mov    %eax,%edx
  sub    $0x1,%rcx
  imul   %rdi,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%esi
  add    %esi,%esi
  sub    %esi,%eax
  cltq
  movzbl 0x5(%rsp,%rax,1),%eax
  mov    %al,(%rcx)
  mov    %edx,%eax
  test   %edx,%edx
  jne    L18
  mov    %r8,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x78,%rsp
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D2
datatype: string

tool: D2: %s


assistant: ```c
void str_to_int(int num) {
  int i = 0;
  int length = 0;
  int sum;
  char const digits[] = "0123456789";
  char str[81];
  if (num < 0) {
    str[i] = '-';
    ++length;
    num = -num;
  }
  sum = num;
  do {
    ++length;
    sum /= 10;
  } while (sum);
  str[length] = '\0';
  do {
    --length;
    str[length] = digits[num % 10];
    num /= 10;
  } while (num);
  printf("%s\n", str);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<str_to_int>:
  endbr64
  sub    $0x78,%rsp
  mov    %edi,%eax
  xor    %ecx,%ecx
  movabs $0x3736353433323130,%rdi
  mov    %fs:0x28,%rdx
  mov    %rdx,0x68(%rsp)
  xor    %edx,%edx
  mov    $0x3938,%edx
  mov    %rdi,0x5(%rsp)
  mov    %dx,0xd(%rsp)
  movb   $0x0,0xf(%rsp)
  test   %eax,%eax
  jns    L5
  movb   $0x2d,0x10(%rsp)
  neg    %eax
  mov    $0x1,%ecx
L5:
  mov    %eax,%edx
  mov    $0xcccccccd,%esi
  nopl   0x0(%rax)
L18:
  mov    %edx,%edx
  add    $0x1,%ecx
  imul   %rsi,%rdx
  shr    $0x23,%rdx
  test   %edx,%edx
  jne    L18
  movslq %ecx,%rcx
  lea    0x10(%rsp),%r8
  mov    $0xcccccccd,%edi
  movb   $0x0,0x10(%rsp,%rcx,1)
  add    %r8,%rcx
  xchg   %ax,%ax
L14:
  mov    %eax,%edx
  sub    $0x1,%rcx
  imul   %rdi,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%esi
  add    %esi,%esi
  sub    %esi,%eax
  cltq
  movzbl 0x5(%rsp,%rax,1),%eax
  mov    %al,(%rcx)
  mov    %edx,%eax
  test   %edx,%edx
  jne    L14
  mov    %r8,%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x78,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D7
type_guessed: string

tool: D7: %s


assistant: ```c
void str_to_int(int num) {
  int i = 0;
  int length = 0;
  int sum;
  char const digits[] = "0123456789";
  char str[81];
  if (num < 0) {
    str[i] = '-';
    ++length;
    num = -num;
  }
  sum = num;
  do {
    ++length;
    sum /= 10;
  } while (sum);
  str[length] = '\0';
  do {
    --length;
    str[length] = digits[num % 10];
    num /= 10;
  } while (num);
  printf("%s\n", str);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x4,%r8d
  mov    $0x3,%ecx
  mov    $0x2,%edx
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <f@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  f(2, 1, (long long)2, 3, (long long)4);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%r8d
  mov    $0x3,%ecx
  mov    $0x2,%edx
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <f@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  f(2, 1, (long long)2, 3, (long long)4);
  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
  mov    $0x4,%r8d
  mov    $0x3,%ecx
  mov    $0x2,%edx
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <f@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  f(2, 1, (long long)2, 3, (long long)4);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%r8d
  mov    $0x3,%ecx
  mov    $0x2,%edx
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <f@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  f(2, 1, (long long)2, 3, (long long)4);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<cmx_registry_parse_filename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  lea    D5(%rip),%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  cmp    $0x2,%eax
  jne    L13
  mov    $0x1,%eax
  jmp    L3
L13:
  mov    $0x0,%eax
L3:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D5
data_type: string

tool: D5: cmx.%d.%63s

assistant: ```c
int cmx_registry_parse_filename(const char *filename_buf, int *const process_id, char *const component_name) {
  if (sscanf(filename_buf, "cmx.%d.%63s", process_id, component_name) == 2) {
    return 1;
  } else {
    return 0;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<cmx_registry_parse_filename>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdx,%rcx
  mov    %rsi,%rdx
  lea    D5(%rip),%rsi
  call   <sscanf@plt>
  cmp    $0x2,%eax
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D5
datatype: string

tool: D5: cmx.%d.%63s

assistant: ```c
int cmx_registry_parse_filename(const char *filename_buf, int *const process_id, char *const component_name) {
  if (sscanf(filename_buf, "cmx.%d.%63s", process_id, component_name) == 2) {
    return 1;
  } else {
    return 0;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cmx_registry_parse_filename>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdx,%rcx
  mov    %rsi,%rdx
  lea    D6(%rip),%rsi
  call   <sscanf@plt>
  cmp    $0x2,%eax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

tool_calls: name: read_data
arguments: label: D6
type: string

tool: D6: cmx.%d.%63s

assistant: ```c
int cmx_registry_parse_filename(const char *filename_buf, int *const process_id, char *const component_name) {
  if (sscanf(filename_buf, "cmx.%d.%63s", process_id, component_name) == 2) {
    return 1;
  } else {
    return 0;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cmx_registry_parse_filename>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdx,%rcx
  mov    %rsi,%rdx
  lea    D7(%rip),%rsi
  call   <sscanf@plt>
  cmp    $0x2,%eax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
type_infered: string

tool: D7: cmx.%d.%63s

assistant: ```c
int cmx_registry_parse_filename(const char *filename_buf, int *const process_id, char *const component_name) {
  if (sscanf(filename_buf, "cmx.%d.%63s", process_id, component_name) == 2) {
    return 1;
  } else {
    return 0;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<get_ind_count>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %rdi,-0x88(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x70(%rbp)
  movq   $0x0,-0x68(%rbp)
  movq   $0x0,-0x60(%rbp)
  movq   $0x0,-0x58(%rbp)
  movq   $0x0,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0x88(%rbp),%rdx
  lea    -0x70(%rbp),%rax
  lea    D11(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    -0x70(%rbp),%rax
  lea    D14(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <popen@plt>
  mov    %rax,-0x78(%rbp)
  movl   $0x0,-0x7c(%rbp)
  lea    -0x7c(%rbp),%rdx
  mov    -0x78(%rbp),%rax
  lea    D5(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fscanf@plt>
  mov    -0x7c(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x78(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    -0x7c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D11
type_guessed: string
name: read_data
arguments: data_tag: D14
type_guessed: string
name: read_data
arguments: data_tag: D5
type_guessed: string
name: read_data
arguments: data_tag: D9
type_guessed: string

tool: D11: wc -l %s

tool: D14: r

tool: D5: %d

tool: D9: # samples = %d


assistant: ```c
int get_ind_count(char *fname) {
  char command[100] = "";
  sprintf(command, "wc -l %s", fname);
  FILE *fp = popen(command, "r");
  int n = 0;
  fscanf(fp, "%d", &n);
  printf("# samples = %d\n", n);
  fclose(fp);
  return n;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<get_ind_count>:
  endbr64
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %rdi,%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  movq   $0x0,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movq   $0x0,0x60(%rsp)
  movq   $0x0,0x68(%rsp)
  movl   $0x0,0x70(%rsp)
  lea    0x10(%rsp),%rbx
  lea    D25(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  lea    D16(%rip),%rsi
  mov    %rbx,%rdi
  call   <popen@plt>
  mov    %rax,%rbx
  movl   $0x0,0xc(%rsp)
  lea    0xc(%rsp),%rdx
  lea    D5(%rip),%rsi
  mov    %rax,%rdi
  call   <fscanf@plt>
  mov    0xc(%rsp),%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <fclose@plt>
  mov    0xc(%rsp),%eax
  mov    0x78(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D25
data_type: string
name: extract_symbolic_data
arguments: data_tag: D16
data_type: string
name: extract_symbolic_data
arguments: data_tag: D5
data_type: string
name: extract_symbolic_data
arguments: data_tag: D7
data_type: string

tool: D25: wc -l %s

tool: D16: r

tool: D5: %d

tool: D7: # samples = %d


assistant: ```c
int get_ind_count(char *fname) {
  char command[100] = "";
  sprintf(command, "wc -l %s", fname);
  FILE *fp = popen(command, "r");
  int n = 0;
  fscanf(fp, "%d", &n);
  printf("# samples = %d\n", n);
  fclose(fp);
  return n;
}
```