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

```asm
<rt_prio>:
  endbr64
  cmp    $0x63,%edi
  setle  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
__attribute__((always_)) int rt_prio(int prio) {
  if (__builtin_expect(!!(prio < 100), 0))
    return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<rt_prio>:
  endbr64
  xor    %eax,%eax
  cmp    $0x63,%edi
  setle  %al
  ret
```

assistant: ```c
__attribute__((always_)) int rt_prio(int prio) {
  if (__builtin_expect(!!(prio < 100), 0))
    return 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rt_prio>:
  endbr64
  xor    %eax,%eax
  cmp    $0x63,%edi
  setle  %al
  ret
```

assistant: ```c
__attribute__((always_)) int rt_prio(int prio) {
  if (__builtin_expect(!!(prio < 100), 0))
    return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<getrep>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L96
L36:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cmp    $0x9,%eax
  ja     L84
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
  addq   $0x1,-0x8(%rbp)
  addl   $0x1,-0x14(%rbp)
  cmpl   $0x32,-0xc(%rbp)
  jle    L96
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <event_error@plt>
  jmp    L7
L84:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  je     L5
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2c,%al
  jne    L58
L5:
  cmpl   $0x0,-0x14(%rbp)
  jle    L58
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cmp    $0x9,%eax
  ja     L58
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
  addq   $0x1,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  cmpl   $0x32,-0xc(%rbp)
  jle    L96
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <event_error@plt>
  jmp    L7
L58:
  movl   $0x1,-0x10(%rbp)
L96:
  cmpl   $0x0,-0x10(%rbp)
  je     L36
L7:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D0
datatype: string

tool: D0: malformed repeat

assistant: ```c
char *getrep(p, out)
char *p;
char *out;
{
  char *q;
  int digits;
  int done;
  int count;
  q = p;
  count = 0;
  done = 0;
  digits = 0;
  while (!done) {
    if (isdigit(*q)) {
      out[count] = *q;
      count = count + 1;
      q = q + 1;
      digits = digits + 1;
      if (count > 50) {
        event_error("malformed repeat");
        break;
      }
    } else {
      if (((*q == '-') || (*q == ',')) && (digits > 0) && (isdigit(*(q + 1)))) {
        out[count] = *q;
        count = count + 1;
        q = q + 1;
        digits = 0;
        if (count > 50) {
          event_error("malformed repeat");
          break;
        }
      } else {
        done = 1;
      };
    };
  };
  out[count] = '\0';
  return (q);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<getrep>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbp
  add    $0x1,%rdi
  mov    $0x1,%eax
  mov    $0x0,%esi
  jmp    L32
L46:
  lea    -0x2c(%rcx),%edx
  cmp    $0x1,%dl
  ja     L80
  test   %esi,%esi
  jg     L66
L80:
  movslq %ebx,%rbx
  movb   $0x0,0x0(%rbp,%rbx,1)
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L66:
  movsbl (%rdi),%edx
  sub    $0x30,%edx
  cmp    $0x9,%edx
  ja     L80
  mov    %cl,-0x1(%rbp,%rax,1)
  mov    %eax,%ebx
  cmp    $0x32,%eax
  jg     L82
  mov    $0x0,%esi
L15:
  add    $0x1,%rdi
  add    $0x1,%rax
L32:
  lea    -0x1(%rax),%ebx
  lea    -0x1(%rdi),%r8
  mov    %rdi,%r12
  movzbl -0x1(%rdi),%ecx
  movsbl %cl,%edx
  sub    $0x30,%edx
  cmp    $0x9,%edx
  ja     L46
  mov    %cl,-0x1(%rbp,%rax,1)
  mov    %eax,%ebx
  add    $0x1,%esi
  cmp    $0x32,%eax
  jle    L15
  lea    D8(%rip),%rdi
  mov    $0x0,%eax
  call   <event_error@plt>
  mov    %r12,%r8
  jmp    L80
L82:
  lea    D8(%rip),%rdi
  mov    $0x0,%eax
  call   <event_error@plt>
  mov    %r12,%r8
  jmp    L80
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
type_guessed: string

tool: D8: malformed repeat

assistant: ```c
char *getrep(p, out)
char *p;
char *out;
{
  char *q;
  int digits;
  int done;
  int count;
  q = p;
  count = 0;
  done = 0;
  digits = 0;
  while (!done) {
    if (isdigit(*q)) {
      out[count] = *q;
      count = count + 1;
      q = q + 1;
      digits = digits + 1;
      if (count > 50) {
        event_error("malformed repeat");
        break;
      }
    } else {
      if (((*q == '-') || (*q == ',')) && (digits > 0) && (isdigit(*(q + 1)))) {
        out[count] = *q;
        count = count + 1;
        q = q + 1;
        digits = 0;
        if (count > 50) {
          event_error("malformed repeat");
          break;
        }
      } else {
        done = 1;
      };
    };
  };
  out[count] = '\0';
  return (q);
}
```
user: Translate the below assembly code into C programming language:

```asm
<getrep>:
  endbr64
  push   %r12
  mov    $0x1,%edx
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  push   %rbx
  lea    0x1(%rdi),%rbx
  cs nopw 0x0(%rax,%rax,1)
L41:
  movsbl -0x1(%rbx),%eax
  lea    -0x1(%rdx),%ebp
  lea    -0x1(%rbx),%r8
  mov    %eax,%ecx
  sub    $0x30,%eax
  cmp    $0x9,%eax
  ja     L69
  mov    %cl,-0x1(%r12,%rdx,1)
  mov    %edx,%ebp
  add    $0x1,%esi
  cmp    $0x32,%edx
  jg     L38
L27:
  add    $0x1,%rbx
  add    $0x1,%rdx
  jmp    L41
  xchg   %ax,%ax
L69:
  lea    -0x2c(%rcx),%eax
  cmp    $0x1,%al
  ja     L16
  test   %esi,%esi
  jne    L5
L16:
  movslq %ebp,%rbp
  pop    %rbx
  mov    %r8,%rax
  movb   $0x0,(%r12,%rbp,1)
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  movsbl (%rbx),%eax
  sub    $0x30,%eax
  cmp    $0x9,%eax
  ja     L16
  mov    %cl,-0x1(%r12,%rdx,1)
  mov    %edx,%ebp
  cmp    $0x32,%edx
  jg     L38
  xor    %esi,%esi
  jmp    L27
  nopl   0x0(%rax,%rax,1)
L38:
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  movslq %ebp,%rbp
  call   <event_error@plt>
  mov    %rbx,%r8
  movb   $0x0,(%r12,%rbp,1)
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
data_type: string

tool: D2: malformed repeat

assistant: ```c
char *getrep(p, out)
char *p;
char *out;
{
  char *q;
  int digits;
  int done;
  int count;
  q = p;
  count = 0;
  done = 0;
  digits = 0;
  while (!done) {
    if (isdigit(*q)) {
      out[count] = *q;
      count = count + 1;
      q = q + 1;
      digits = digits + 1;
      if (count > 50) {
        event_error("malformed repeat");
        break;
      }
    } else {
      if (((*q == '-') || (*q == ',')) && (digits > 0) && (isdigit(*(q + 1)))) {
        out[count] = *q;
        count = count + 1;
        q = q + 1;
        digits = 0;
        if (count > 50) {
          event_error("malformed repeat");
          break;
        }
      } else {
        done = 1;
      };
    };
  };
  out[count] = '\0';
  return (q);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getrep>:
  endbr64
  push   %r12
  mov    $0x1,%edx
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  push   %rbx
  lea    0x1(%rdi),%rbx
  cs nopw 0x0(%rax,%rax,1)
L51:
  movsbl -0x1(%rbx),%eax
  lea    -0x1(%rdx),%ebp
  lea    -0x1(%rbx),%r8
  mov    %eax,%ecx
  sub    $0x30,%eax
  cmp    $0x9,%eax
  ja     L60
  mov    %cl,-0x1(%r12,%rdx,1)
  mov    %edx,%ebp
  add    $0x1,%esi
  cmp    $0x32,%edx
  jg     L73
L40:
  add    $0x1,%rbx
  add    $0x1,%rdx
  jmp    L51
  xchg   %ax,%ax
L60:
  lea    -0x2c(%rcx),%eax
  cmp    $0x1,%al
  ja     L55
  test   %esi,%esi
  jne    L64
L55:
  movslq %ebp,%rbp
  pop    %rbx
  mov    %r8,%rax
  movb   $0x0,(%r12,%rbp,1)
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L64:
  movsbl (%rbx),%eax
  sub    $0x30,%eax
  cmp    $0x9,%eax
  ja     L55
  mov    %cl,-0x1(%r12,%rdx,1)
  mov    %edx,%ebp
  cmp    $0x32,%edx
  jg     L73
  xor    %esi,%esi
  jmp    L40
  nopl   0x0(%rax,%rax,1)
L73:
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  movslq %ebp,%rbp
  call   <event_error@plt>
  mov    %rbx,%r8
  movb   $0x0,(%r12,%rbp,1)
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
type: string

tool: D2: malformed repeat

assistant: ```c
char *getrep(p, out)
char *p;
char *out;
{
  char *q;
  int digits;
  int done;
  int count;
  q = p;
  count = 0;
  done = 0;
  digits = 0;
  while (!done) {
    if (isdigit(*q)) {
      out[count] = *q;
      count = count + 1;
      q = q + 1;
      digits = digits + 1;
      if (count > 50) {
        event_error("malformed repeat");
        break;
      }
    } else {
      if (((*q == '-') || (*q == ',')) && (digits > 0) && (isdigit(*(q + 1)))) {
        out[count] = *q;
        count = count + 1;
        q = q + 1;
        digits = 0;
        if (count > 50) {
          event_error("malformed repeat");
          break;
        }
      } else {
        done = 1;
      };
    };
  };
  out[count] = '\0';
  return (q);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<linear_search>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  jmp    L75
L15:
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <compare@plt>
  cmp    $0x1,%eax
  je     L46
  cmp    $0x1,%eax
  jg     L52
  cmp    $0xffffffff,%eax
  je     L106
  test   %eax,%eax
  jne    L52
  mov    -0x10(%rbp),%rax
  jmp    L6
L106:
  mov    $0x0,%eax
  jmp    L6
L46:
  nop
  jmp    L52
L8:
  nop
L52:
  mov    -0x10(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jae    L75
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L8
L75:
  mov    -0x10(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jb     L15
  mov    $0x0,%eax
L6:
  leave
  ret
```

assistant: ```c
char *linear_search(char *string, char *front, char *back) {
  while (front < back) {
    switch (compare(string, front, back)) {
    case 0:
      return (front);
      break;
    case (-1):
      return (NULL);
      break;
    case 1:
      break;
    }
    while (front < back && *front++ != '\n')
      continue;
    ;
  }
  return (NULL);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<linear_search>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %rdx,%rbp
  cmp    %rdx,%rsi
  jae    L27
L34:
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <compare@plt>
  cmp    $0xffffffff,%eax
  je     L15
  test   %eax,%eax
  je     L35
L6:
  cmp    %rbp,%rbx
  jae    L70
  add    $0x1,%rbx
  cmpb   $0xa,-0x1(%rbx)
  jne    L6
  cmp    %rbx,%rbp
  ja     L34
  mov    $0x0,%ebx
  jmp    L35
L27:
  mov    $0x0,%ebx
  jmp    L35
L15:
  mov    $0x0,%ebx
L35:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L70:
  mov    $0x0,%ebx
  jmp    L35
```

assistant: ```c
char *linear_search(char *string, char *front, char *back) {
  while (front < back) {
    switch (compare(string, front, back)) {
    case 0:
      return (front);
      break;
    case (-1):
      return (NULL);
      break;
    case 1:
      break;
    }
    while (front < back && *front++ != '\n')
      continue;
    ;
  }
  return (NULL);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<linear_search>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    %rdx,%rsi
  jae    L54
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%rbx
L4:
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <compare@plt>
  cmp    $0xffffffff,%eax
  je     L54
  test   %eax,%eax
  je     L63
L33:
  add    $0x1,%r12
  cmpb   $0xa,-0x1(%r12)
  je     L69
  cmp    %r12,%rbx
  ja     L33
L54:
  xor    %r12d,%r12d
L63:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L69:
  cmp    %r12,%rbx
  ja     L4
  jmp    L54
```

assistant: ```c
char *linear_search(char *string, char *front, char *back) {
  while (front < back) {
    switch (compare(string, front, back)) {
    case 0:
      return (front);
      break;
    case (-1):
      return (NULL);
      break;
    case 1:
      break;
    }
    while (front < back && *front++ != '\n')
      continue;
    ;
  }
  return (NULL);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<linear_search>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    %rdx,%rsi
  jae    L20
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%rbx
L62:
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <compare@plt>
  cmp    $0xffffffff,%eax
  je     L20
  test   %eax,%eax
  je     L52
L55:
  add    $0x1,%r12
  cmpb   $0xa,-0x1(%r12)
  je     L49
  cmp    %r12,%rbx
  ja     L55
L20:
  xor    %r12d,%r12d
L52:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L49:
  cmp    %r12,%rbx
  ja     L62
  jmp    L20
```

assistant: ```c
char *linear_search(char *string, char *front, char *back) {
  while (front < back) {
    switch (compare(string, front, back)) {
    case 0:
      return (front);
      break;
    case (-1):
      return (NULL);
      break;
    case 1:
      break;
    }
    while (front < back && *front++ != '\n')
      continue;
    ;
  }
  return (NULL);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L0:
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L0
```

tool_calls: name: retrieve_data
arguments: label: D9
data_type: string

tool: D9: wait for login


assistant: ```c
int main(void) {
  while (1) {
    printf("wait for login\n");
    sleep(1);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D0(%rip),%rbx
L7:
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L7
```

tool_calls: name: unpack_data
arguments: data_tag: D0
type_infered: string

tool: D0: wait for login


assistant: ```c
int main(void) {
  while (1) {
    printf("wait for login\n");
    sleep(1);
  }
  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   %rbx
  lea    D0(%rip),%rbx
  nopl   0x0(%rax)
L3:
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L3
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D0
type_infered: string

tool: D0: wait for login


assistant: ```c
int main(void) {
  while (1) {
    printf("wait for login\n");
    sleep(1);
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D6(%rip),%rbx
  nopl   0x0(%rax)
L6:
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L6
```

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

tool: D6: wait for login


assistant: ```c
int main(void) {
  while (1) {
    printf("wait for login\n");
    sleep(1);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x19(%rbp),%rax
  sub    $0x2,%rax
  mov    %rax,-0x18(%rbp)
  lea    -0x19(%rbp),%rax
  add    $0x2,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  lea    -0x19(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D8
type_infered: string

tool: D8: &a = 	 %p 
&a - 2 = %p 
&a + 2 = %p


assistant: ```c
int main() {
  char a, *b, *c;
  b = &a - 2;
  c = &a + 2;
  printf("&a = \t %p \n&a - 2 = %p \n&a + 2 = %p\n", &a, b, c);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rsi
  lea    0x9(%rsp),%rcx
  lea    0x5(%rsp),%rdx
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D8
data_type: string

tool: D8: &a = 	 %p 
&a - 2 = %p 
&a + 2 = %p


assistant: ```c
int main() {
  char a, *b, *c;
  b = &a - 2;
  c = &a + 2;
  printf("&a = \t %p \n&a - 2 = %p \n&a + 2 = %p\n", &a, b, c);
  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
  lea    D2(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rsi
  lea    0x9(%rsp),%rcx
  lea    0x5(%rsp),%rdx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

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

tool: D2: &a = 	 %p 
&a - 2 = %p 
&a + 2 = %p


assistant: ```c
int main() {
  char a, *b, *c;
  b = &a - 2;
  c = &a + 2;
  printf("&a = \t %p \n&a - 2 = %p \n&a + 2 = %p\n", &a, b, c);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D4(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rsi
  lea    0x9(%rsp),%rcx
  lea    0x5(%rsp),%rdx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D4
type: string

tool: D4: &a = 	 %p 
&a - 2 = %p 
&a + 2 = %p


assistant: ```c
int main() {
  char a, *b, *c;
  b = &a - 2;
  c = &a + 2;
  printf("&a = \t %p \n&a - 2 = %p \n&a + 2 = %p\n", &a, b, c);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<norm2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  fldt   0x10(%rbp)
  fldt   0x30(%rbp)
  fsubrp %st,%st(1)
  fstpt  -0x30(%rbp)
  push   0x28(%rbp)
  push   0x20(%rbp)
  call   <cos@plt>
  add    $0x10,%rsp
  fldt   -0x30(%rbp)
  fmulp  %st,%st(1)
  fstpt  -0x30(%rbp)
  fldt   0x10(%rbp)
  fldt   0x30(%rbp)
  fsubrp %st,%st(1)
  fstpt  -0x40(%rbp)
  push   0x28(%rbp)
  push   0x20(%rbp)
  call   <cos@plt>
  add    $0x10,%rsp
  fldt   -0x40(%rbp)
  fmulp  %st,%st(1)
  fldt   -0x30(%rbp)
  fmulp  %st,%st(1)
  fstpt  -0x20(%rbp)
  fldt   0x20(%rbp)
  fldt   0x40(%rbp)
  fsubrp %st,%st(1)
  fldt   0x20(%rbp)
  fldt   0x40(%rbp)
  fsubrp %st,%st(1)
  fmulp  %st,%st(1)
  fstpt  -0x10(%rbp)
  fldt   -0x20(%rbp)
  fldt   -0x10(%rbp)
  faddp  %st,%st(1)
  leave
  ret
```

assistant: ```c
long double norm2(long double x1, long double y1, long double x2, long double y2) {
  long double a = ((x1 - x2) * cos(y1)) * ((x1 - x2) * cos(y1));
  long double b = (y1 - y2) * (y1 - y2);
  return a + b;
}
```
user: Translate the below assembly code into C programming language:

```asm
<norm2>:
  endbr64
  sub    $0x28,%rsp
  fldt   0x30(%rsp)
  fldt   0x50(%rsp)
  fsubrp %st,%st(1)
  fstpt  (%rsp)
  push   0x48(%rsp)
  push   0x48(%rsp)
  call   <cos@plt>
  fldt   0x10(%rsp)
  fmulp  %st,%st(1)
  fstpt  0x20(%rsp)
  push   0x58(%rsp)
  push   0x58(%rsp)
  call   <cos@plt>
  fldt   0x80(%rsp)
  fldt   0x60(%rsp)
  fsubp  %st,%st(1)
  fldt   0x20(%rsp)
  fmulp  %st,%st(2)
  fldt   0x30(%rsp)
  fmulp  %st,%st(2)
  fmul   %st(0),%st
  faddp  %st,%st(1)
  add    $0x48,%rsp
  ret
```

assistant: ```c
long double norm2(long double x1, long double y1, long double x2, long double y2) {
  long double a = ((x1 - x2) * cos(y1)) * ((x1 - x2) * cos(y1));
  long double b = (y1 - y2) * (y1 - y2);
  return a + b;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<norm2>:
  endbr64
  sub    $0x28,%rsp
  fldt   0x30(%rsp)
  fldt   0x50(%rsp)
  fsubrp %st,%st(1)
  fstpt  (%rsp)
  push   0x48(%rsp)
  push   0x48(%rsp)
  call   <cos@plt>
  fldt   0x10(%rsp)
  fmulp  %st,%st(1)
  fstpt  0x20(%rsp)
  push   0x58(%rsp)
  push   0x58(%rsp)
  call   <cos@plt>
  fldt   0x80(%rsp)
  fldt   0x60(%rsp)
  fsubp  %st,%st(1)
  fldt   0x20(%rsp)
  fmulp  %st,%st(2)
  fldt   0x30(%rsp)
  add    $0x48,%rsp
  fmulp  %st,%st(2)
  fmul   %st(0),%st
  faddp  %st,%st(1)
  ret
```

assistant: ```c
long double norm2(long double x1, long double y1, long double x2, long double y2) {
  long double a = ((x1 - x2) * cos(y1)) * ((x1 - x2) * cos(y1));
  long double b = (y1 - y2) * (y1 - y2);
  return a + b;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<norm2>:
  endbr64
  sub    $0x28,%rsp
  fldt   0x30(%rsp)
  fldt   0x50(%rsp)
  fsubrp %st,%st(1)
  fstpt  (%rsp)
  push   0x48(%rsp)
  push   0x48(%rsp)
  call   <cos@plt>
  fldt   0x10(%rsp)
  fmulp  %st,%st(1)
  fstpt  0x20(%rsp)
  push   0x58(%rsp)
  push   0x58(%rsp)
  call   <cos@plt>
  fldt   0x80(%rsp)
  fldt   0x60(%rsp)
  fsubp  %st,%st(1)
  fldt   0x20(%rsp)
  fmulp  %st,%st(2)
  fldt   0x30(%rsp)
  add    $0x48,%rsp
  fmulp  %st,%st(2)
  fmul   %st(0),%st
  faddp  %st,%st(1)
  ret
```

assistant: ```c
long double norm2(long double x1, long double y1, long double x2, long double y2) {
  long double a = ((x1 - x2) * cos(y1)) * ((x1 - x2) * cos(y1));
  long double b = (y1 - y2) * (y1 - y2);
  return a + b;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<dump_stack_data>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x18(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movl   $0xfffffff0,-0x14(%rbp)
  jmp    L1
L0:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%eax
  add    $0x1,%eax
  and    $0x3,%eax
  test   %eax,%eax
  jne    L19
  mov    $0xa,%edi
  call   <putchar@plt>
L19:
  mov    -0x18(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  addl   $0x1,-0x14(%rbp)
L1:
  cmpl   $0x0,-0x14(%rbp)
  jle    L0
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L39
  call   <__stack_chk_fail@plt>
L39:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D8
datatype: string

tool: D8: 0x%-12x 

assistant: ```c
void dump_stack_data() {
  int i;
  int *pi;
  int len;
  i = 0;
  pi = &i;
  for (len = -0x10; len <= 0; len++) {
    printf("0x%-12x ", *(pi + len));
    if ((i + 1) % 4 == 0)
      printf("\n");
    ++i;
  }
}
```