ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<irc_message_get_nick_from_host>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L72
  mov    $0x20,%esi
  movb   $0x0,D44(%rip)
  mov    %rdi,%r12
  call   <strchr@plt>
  test   %rax,%rax
  je     L34
  sub    %r12,%rax
  mov    %rax,%rbx
  cmp    $0x7f,%rax
  jg     L31
  mov    %rsp,%rbp
  mov    %r12,%rsi
  mov    %eax,%edx
  mov    %rbp,%rdi
  mov    %rbp,%r12
  call   <strncpy@plt>
  movb   $0x0,(%rsp,%rbx,1)
L34:
  xor    %eax,%eax
  cmpb   $0x3a,(%r12)
  mov    $0x21,%esi
  sete   %al
  add    %rax,%r12
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L80
  sub    %r12,%rax
  mov    %rax,%rbx
  cmp    $0x7f,%rax
  jle    L13
L80:
  lea    D44(%rip),%rbx
  mov    %r12,%rcx
  mov    $0x80,%esi
  lea    D15(%rip),%rdx
  mov    %rbx,%rdi
  call   <snprintf@plt>
  mov    %rbx,%rax
L88:
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x90,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L31:
  mov    %rsp,%rbx
  mov    %r12,%rcx
  mov    $0x80,%esi
  lea    D15(%rip),%rdx
  mov    %rbx,%rdi
  mov    %rbx,%r12
  call   <snprintf@plt>
  jmp    L34
  nopw   0x0(%rax,%rax,1)
L13:
  mov    %r12,%rsi
  lea    D44(%rip),%r12
  mov    %eax,%edx
  mov    %r12,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  jmp    L88
  cs nopw 0x0(%rax,%rax,1)
L72:
  xor    %eax,%eax
  jmp    L88
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D15
datatype: string

tool: D15: %s

assistant: ```c
const char *irc_message_get_nick_from_host(const char *host) {
  static char nick[128];
  char host2[128], *pos_space, *pos;
  const char *ptr_host;
  if (!host)
    return NULL;
  nick[0] = '\0';
  if (host) {
    ptr_host = host;
    pos_space = strchr(host, ' ');
    if (pos_space) {
      if (pos_space - host < (int)sizeof(host2)) {
        strncpy(host2, host, pos_space - host);
        host2[pos_space - host] = '\0';
      } else
        snprintf(host2, sizeof(host2), "%s", host);
      ptr_host = host2;
    }
    if (ptr_host[0] == ':')
      ptr_host++;
    pos = strchr(ptr_host, '!');
    if (pos && (pos - ptr_host < (int)sizeof(nick))) {
      strncpy(nick, ptr_host, pos - ptr_host);
      nick[pos - ptr_host] = '\0';
    } else {
      snprintf(nick, sizeof(nick), "%s", ptr_host);
    }
  }
  return nick;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<irc_message_get_nick_from_host>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L35
  mov    $0x20,%esi
  movb   $0x0,D6(%rip)
  mov    %rdi,%r12
  call   <strchr@plt>
  test   %rax,%rax
  je     L21
  sub    %r12,%rax
  mov    %rax,%rbx
  cmp    $0x7f,%rax
  jg     L76
  mov    %rsp,%rbp
  mov    %r12,%rsi
  mov    %eax,%edx
  mov    %rbp,%rdi
  mov    %rbp,%r12
  call   <strncpy@plt>
  movb   $0x0,(%rsp,%rbx,1)
L21:
  xor    %eax,%eax
  cmpb   $0x3a,(%r12)
  mov    $0x21,%esi
  sete   %al
  add    %rax,%r12
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L74
  sub    %r12,%rax
  mov    %rax,%rbx
  cmp    $0x7f,%rax
  jle    L5
L74:
  lea    D6(%rip),%rbx
  mov    %r12,%rcx
  mov    $0x80,%esi
  lea    D46(%rip),%rdx
  mov    %rbx,%rdi
  call   <snprintf@plt>
  mov    %rbx,%rax
L16:
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L20
  add    $0x90,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L76:
  mov    %rsp,%rbx
  mov    %r12,%rcx
  mov    $0x80,%esi
  lea    D46(%rip),%rdx
  mov    %rbx,%rdi
  mov    %rbx,%r12
  call   <snprintf@plt>
  jmp    L21
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %r12,%rsi
  lea    D6(%rip),%r12
  mov    %eax,%edx
  mov    %r12,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  jmp    L16
  cs nopw 0x0(%rax,%rax,1)
L35:
  xor    %eax,%eax
  jmp    L16
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D46
data_type: string

tool: D46: %s

assistant: ```c
const char *irc_message_get_nick_from_host(const char *host) {
  static char nick[128];
  char host2[128], *pos_space, *pos;
  const char *ptr_host;
  if (!host)
    return NULL;
  nick[0] = '\0';
  if (host) {
    ptr_host = host;
    pos_space = strchr(host, ' ');
    if (pos_space) {
      if (pos_space - host < (int)sizeof(host2)) {
        strncpy(host2, host, pos_space - host);
        host2[pos_space - host] = '\0';
      } else
        snprintf(host2, sizeof(host2), "%s", host);
      ptr_host = host2;
    }
    if (ptr_host[0] == ':')
      ptr_host++;
    pos = strchr(ptr_host, '!');
    if (pos && (pos - ptr_host < (int)sizeof(nick))) {
      strncpy(nick, ptr_host, pos - ptr_host);
      nick[pos - ptr_host] = '\0';
    } else {
      snprintf(nick, sizeof(nick), "%s", ptr_host);
    }
  }
  return nick;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<LeapYear>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x7,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x190,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L22
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x5,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x64,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L35
  mov    -0x4(%rbp),%eax
  and    $0x3,%eax
  test   %eax,%eax
  jne    L35
L22:
  mov    $0x1,%eax
  jmp    L21
L35:
  mov    $0x0,%eax
L21:
  pop    %rbp
  ret
```

assistant: ```c
int LeapYear(int year) {
  if (year % 400 == 0 || (year % 100 != 0 && year % 4 == 0))
    return 1;
  else
    return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<LeapYear>:
  endbr64
  movslq %edi,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x27,%rdx
  mov    %edi,%eax
  sar    $0x1f,%eax
  sub    %eax,%edx
  imul   $0x190,%edx,%edx
  mov    $0x1,%eax
  cmp    %edx,%edi
  je     L9
  movslq %edi,%rax
  imul   $0x51eb851f,%rax,%rax
  sar    $0x25,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  imul   $0x64,%eax,%edx
  mov    %edi,%eax
  sub    %edx,%eax
  je     L9
  test   $0x3,%dil
  sete   %al
  movzbl %al,%eax
L9:
  ret
```

assistant: ```c
int LeapYear(int year) {
  if (year % 400 == 0 || (year % 100 != 0 && year % 4 == 0))
    return 1;
  else
    return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<LeapYear>:
  endbr64
  imul   $0xc28f5c29,%edi,%edx
  mov    $0x1,%eax
  add    $0x51eb850,%edx
  ror    $0x4,%edx
  cmp    $0xa3d70a,%edx
  jbe    L9
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x51eb851f,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x25,%rax
  sub    %edx,%eax
  imul   $0x64,%eax,%edx
  mov    %edi,%eax
  sub    %edx,%eax
  je     L9
  xor    %eax,%eax
  and    $0x3,%edi
  sete   %al
L9:
  ret
```

assistant: ```c
int LeapYear(int year) {
  if (year % 400 == 0 || (year % 100 != 0 && year % 4 == 0))
    return 1;
  else
    return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<LeapYear>:
  endbr64
  imul   $0xc28f5c29,%edi,%edx
  mov    $0x1,%eax
  add    $0x51eb850,%edx
  ror    $0x4,%edx
  cmp    $0xa3d70a,%edx
  jbe    L6
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x51eb851f,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x25,%rax
  sub    %edx,%eax
  imul   $0x64,%eax,%edx
  mov    %edi,%eax
  sub    %edx,%eax
  je     L6
  xor    %eax,%eax
  and    $0x3,%edi
  sete   %al
L6:
  ret
```

assistant: ```c
int LeapYear(int year) {
  if (year % 400 == 0 || (year % 100 != 0 && year % 4 == 0))
    return 1;
  else
    return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```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   $0x3,-0xc(%rbp)
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%eax
  cmp    $0x2,%eax
  je     L31
  mov    -0x14(%rbp),%eax
  cmp    $0x1,%eax
  jne    L40
L31:
  movl   $0x1,-0x10(%rbp)
  jmp    L24
L40:
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L24
L76:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L61
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jne    L24
L61:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L67
  movl   $0x1,-0x10(%rbp)
L67:
  addl   $0x2,-0xc(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  je     L76
L24:
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L84
  call   <__stack_chk_fail@plt>
L84:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D21
type: string
name: parse_data
arguments: data_tag: D4
type: string
name: parse_data
arguments: data_tag: D35
type: string
name: parse_data
arguments: data_tag: D9
type: string

tool: D21: 
Torna un 1 si el nombre es primer i 0 si no ho es.

tool: D4: 
Introdueix un nombre per saber si es primer:

tool: D35: %d

tool: D9: %d


assistant: ```c
int main() {
  int n;
  int primer = 0;
  int c = 3;
  printf("\nTorna un 1 si el nombre es primer i 0 si no ho es.\n");
  printf("\nIntrodueix un nombre per saber si es primer:");
  scanf("%d", &n);
  if (n == 2 || n == 1) {
    primer = 1;
  } else if (!(n % 2 == 0)) {
    do {
      if (n % c == 0 && n != c) {
        break;
      } else if (n % c == 0) {
        primer = 1;
      }
      c += 2;
    } while (!primer);
  }
  printf("%d\n", primer);
  return primer;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D6(%rip),%rdi
  call   <puts@plt>
  lea    D39(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D23(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  lea    -0x1(%rsi),%eax
  mov    $0x1,%ebx
  cmp    $0x1,%eax
  jbe    L51
  mov    $0x0,%ebx
  test   $0x1,%sil
  je     L51
  mov    $0x3,%ecx
  mov    %esi,%eax
  cltd
  idiv   %ecx
  cmp    $0x3,%esi
  je     L47
  test   %edx,%edx
  je     L106
L47:
  test   %edx,%edx
  je     L77
  mov    $0x3,%ecx
  jmp    L72
L120:
  test   %edx,%edx
  je     L55
L72:
  add    $0x2,%ecx
  mov    %esi,%eax
  cltd
  idiv   %ecx
  cmp    %ecx,%esi
  je     L120
  test   %edx,%edx
  jne    L120
  mov    $0x0,%ebx
  jmp    L51
L106:
  mov    $0x0,%ebx
  jmp    L51
L77:
  mov    $0x1,%ebx
  jmp    L51
L55:
  mov    $0x1,%ebx
L51:
  mov    %ebx,%esi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  mov    %ebx,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D6
datatype: string
name: unpack_data
arguments: data_label: D39
datatype: string
name: unpack_data
arguments: data_label: D23
datatype: string
name: unpack_data
arguments: data_label: D1
datatype: string

tool: D6: 
Torna un 1 si el nombre es primer i 0 si no ho es.

tool: D39: 
Introdueix un nombre per saber si es primer:

tool: D23: %d

tool: D1: %d


assistant: ```c
int main() {
  int n;
  int primer = 0;
  int c = 3;
  printf("\nTorna un 1 si el nombre es primer i 0 si no ho es.\n");
  printf("\nIntrodueix un nombre per saber si es primer:");
  scanf("%d", &n);
  if (n == 2 || n == 1) {
    primer = 1;
  } else if (!(n % 2 == 0)) {
    do {
      if (n % c == 0 && n != c) {
        break;
      } else if (n % c == 0) {
        primer = 1;
      }
      c += 2;
    } while (!primer);
  }
  printf("%d\n", primer);
  return primer;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D30(%rip),%rdi
  mov    $0x1,%r12d
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  lea    -0x1(%rsi),%eax
  cmp    $0x1,%eax
  jbe    L3
  xor    %r12d,%r12d
  test   $0x1,%sil
  jne    L24
L3:
  xor    %eax,%eax
  mov    %r12d,%esi
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
L24:
  mov    %esi,%eax
  mov    $0x3,%ecx
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L32
  mov    $0x3,%ecx
  xchg   %ax,%ax
L17:
  mov    %esi,%eax
  add    $0x2,%ecx
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L17
L32:
  xor    %r12d,%r12d
  cmp    %ecx,%esi
  sete   %r12b
  jmp    L3
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D30
type_guessed: string
name: retrieve_data
arguments: label: D10
type_guessed: string
name: retrieve_data
arguments: label: D3
type_guessed: string
name: retrieve_data
arguments: label: D17
type_guessed: string

tool: D30: 
Torna un 1 si el nombre es primer i 0 si no ho es.

tool: D10: 
Introdueix un nombre per saber si es primer:

tool: D3: %d

tool: D17: %d


assistant: ```c
int main() {
  int n;
  int primer = 0;
  int c = 3;
  printf("\nTorna un 1 si el nombre es primer i 0 si no ho es.\n");
  printf("\nIntrodueix un nombre per saber si es primer:");
  scanf("%d", &n);
  if (n == 2 || n == 1) {
    primer = 1;
  } else if (!(n % 2 == 0)) {
    do {
      if (n % c == 0 && n != c) {
        break;
      } else if (n % c == 0) {
        primer = 1;
      }
      c += 2;
    } while (!primer);
  }
  printf("%d\n", primer);
  return primer;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r12
  lea    D23(%rip),%rdi
  mov    $0x1,%r12d
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D36(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  lea    -0x1(%rsi),%eax
  cmp    $0x1,%eax
  jbe    L47
  xor    %r12d,%r12d
  test   $0x1,%sil
  jne    L14
L47:
  xor    %eax,%eax
  mov    %r12d,%esi
  lea    D21(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
L14:
  mov    %esi,%eax
  mov    $0x3,%ecx
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L25
  mov    $0x3,%ecx
  xchg   %ax,%ax
L4:
  mov    %esi,%eax
  add    $0x2,%ecx
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L4
L25:
  xor    %r12d,%r12d
  cmp    %ecx,%esi
  sete   %r12b
  jmp    L47
L53:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D23
data_type: string
name: retrieve_data
arguments: label: D38
data_type: string
name: retrieve_data
arguments: label: D36
data_type: string
name: retrieve_data
arguments: label: D21
data_type: string

tool: D23: 
Torna un 1 si el nombre es primer i 0 si no ho es.

tool: D38: 
Introdueix un nombre per saber si es primer:

tool: D36: %d

tool: D21: %d


assistant: ```c
int main() {
  int n;
  int primer = 0;
  int c = 3;
  printf("\nTorna un 1 si el nombre es primer i 0 si no ho es.\n");
  printf("\nIntrodueix un nombre per saber si es primer:");
  scanf("%d", &n);
  if (n == 2 || n == 1) {
    primer = 1;
  } else if (!(n % 2 == 0)) {
    do {
      if (n % c == 0 && n != c) {
        break;
      } else if (n % c == 0) {
        primer = 1;
      }
      c += 2;
    } while (!primer);
  }
  printf("%d\n", primer);
  return primer;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rpl_realloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L8
  movq   $0x1,-0x10(%rbp)
L8:
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <realloc@plt>
  leave
  ret
```

assistant: ```c
void *rpl_realloc(void *ptr, size_t size) {
  if (size == 0)
    size = 1;
  return realloc(ptr, size);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rpl_realloc>:
  endbr64
  sub    $0x8,%rsp
  test   %rsi,%rsi
  mov    $0x1,%eax
  cmove  %rax,%rsi
  call   <realloc@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *rpl_realloc(void *ptr, size_t size) {
  if (size == 0)
    size = 1;
  return realloc(ptr, size);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<rpl_realloc>:
  endbr64
  test   %rsi,%rsi
  mov    $0x1,%eax
  cmove  %rax,%rsi
  jmp    1050 <realloc@plt>
```

assistant: ```c
void *rpl_realloc(void *ptr, size_t size) {
  if (size == 0)
    size = 1;
  return realloc(ptr, size);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rpl_realloc>:
  endbr64
  test   %rsi,%rsi
  mov    $0x1,%eax
  cmove  %rax,%rsi
  jmp    1050 <realloc@plt>
```

assistant: ```c
void *rpl_realloc(void *ptr, size_t size) {
  if (size == 0)
    size = 1;
  return realloc(ptr, size);
}
```
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
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <scanf_s@plt>
  movl   $0x1,-0x24(%rbp)
  jmp    L22
L2:
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  mov    $0xffffffff,%edi
  call   <pow@plt>
  movq   %xmm0,%rdx
  mov    -0x24(%rbp),%eax
  imul   %eax,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movq   %rdx,%xmm0
  divsd  %xmm1,%xmm0
  movsd  -0x20(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  addl   $0x1,-0x24(%rbp)
L22:
  mov    -0x28(%rbp),%eax
  cmp    %eax,-0x24(%rbp)
  jle    L2
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <system@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: retrieve_data
arguments: label: D33
type: string
name: retrieve_data
arguments: label: D31
type: string
name: retrieve_data
arguments: label: D2
type: string
name: retrieve_data
arguments: label: D37
type: string

tool: D33: Enter a number : 

tool: D31: %d

tool: D2: 
 Sum is : %.2f 


tool: D37: pause

assistant: ```c
main() {
  int number, i;
  double sum = 0, A = 0, B = 0;
  printf("Enter a number : ");
  scanf_s("%d", &number);
  for (i = 1; i <= number; i++) {
    sum = sum + pow(-1, i + 1) * (double)1 / (double)(i * i);
  }
  printf("\n Sum is : %.2f \n", sum);
  printf("\n");
  system("pause");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D34(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D38(%rip),%rdi
  call   <scanf_s@plt>
  cmpl   $0x0,0x14(%rsp)
  jle    L21
  pxor   %xmm3,%xmm3
  movsd  %xmm3,0x8(%rsp)
  mov    $0x1,%ebx
L20:
  mov    %ebx,%ebp
  add    $0x1,%ebx
  mov    %ebx,%esi
  mov    $0xffffffff,%edi
  call   <pow@plt>
  imul   %ebp,%ebp
  pxor   %xmm1,%xmm1
  cvtsi2sd %ebp,%xmm1
  divsd  %xmm1,%xmm0
  addsd  0x8(%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
  cmp    0x14(%rsp),%ebx
  jle    L20
L23:
  movsd  0x8(%rsp),%xmm0
  lea    D30(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D3(%rip),%rdi
  call   <system@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L21:
  pxor   %xmm4,%xmm4
  movsd  %xmm4,0x8(%rsp)
  jmp    L23
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D34
type_infered: string
name: retrieve_data
arguments: tag: D38
type_infered: string
name: retrieve_data
arguments: tag: D30
type_infered: string
name: retrieve_data
arguments: tag: D3
type_infered: string

tool: D34: Enter a number : 

tool: D38: %d

tool: D30: 
 Sum is : %.2f 


tool: D3: pause

assistant: ```c
main() {
  int number, i;
  double sum = 0, A = 0, B = 0;
  printf("Enter a number : ");
  scanf_s("%d", &number);
  for (i = 1; i <= number; i++) {
    sum = sum + pow(-1, i + 1) * (double)1 / (double)(i * i);
  }
  printf("\n Sum is : %.2f \n", sum);
  printf("\n");
  system("pause");
}
```
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
  lea    D14(%rip),%rdi
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D11(%rip),%rdi
  call   <scanf_s@plt>
  mov    0x14(%rsp),%eax
  pxor   %xmm1,%xmm1
  test   %eax,%eax
  jle    L11
  mov    $0x1,%ebx
  nopl   0x0(%rax,%rax,1)
L25:
  mov    %ebx,%ebp
  add    $0x1,%ebx
  mov    $0xffffffff,%edi
  movsd  %xmm1,0x8(%rsp)
  imul   %ebp,%ebp
  mov    %ebx,%esi
  call   <pow@plt>
  pxor   %xmm2,%xmm2
  movsd  0x8(%rsp),%xmm1
  cvtsi2sd %ebp,%xmm2
  divsd  %xmm2,%xmm0
  addsd  %xmm0,%xmm1
  cmp    0x14(%rsp),%ebx
  jle    L25
L11:
  movapd %xmm1,%xmm0
  lea    D7(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D10(%rip),%rdi
  call   <system@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D14
type_guessed: string
name: read_data
arguments: label: D11
type_guessed: string
name: read_data
arguments: label: D7
type_guessed: string
name: read_data
arguments: label: D10
type_guessed: string

tool: D14: Enter a number : 

tool: D11: %d

tool: D7: 
 Sum is : %.2f 


tool: D10: pause

assistant: ```c
main() {
  int number, i;
  double sum = 0, A = 0, B = 0;
  printf("Enter a number : ");
  scanf_s("%d", &number);
  for (i = 1; i <= number; i++) {
    sum = sum + pow(-1, i + 1) * (double)1 / (double)(i * i);
  }
  printf("\n Sum is : %.2f \n", sum);
  printf("\n");
  system("pause");
}
```