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

```asm
<list_free>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L4
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L4:
  nop
  leave
  ret
```

assistant: ```c
void list_free(struct list *l) {
  if (l)
    free(l);
}
```
user: What is the c source code of the assembly code below:

```asm
<list_free>:
  endbr64
  test   %rdi,%rdi
  je     L5
  sub    $0x8,%rsp
  call   <free@plt>
  add    $0x8,%rsp
  ret
L5:
  ret
```

assistant: ```c
void list_free(struct list *l) {
  if (l)
    free(l);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<list_free>:
  endbr64
  test   %rdi,%rdi
  je     L9
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L9:
  ret
```

assistant: ```c
void list_free(struct list *l) {
  if (l)
    free(l);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<list_free>:
  endbr64
  test   %rdi,%rdi
  je     L2
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L2:
  ret
```

assistant: ```c
void list_free(struct list *l) {
  if (l)
    free(l);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <test_create@plt>
  mov    $0x0,%eax
  call   <test_size@plt>
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D9
type_infered: string

tool: D9: ----------
All tests passed
----------


assistant: ```c
int main(void) {
  test_create();
  test_size();
  printf("----------\nAll tests passed\n----------\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <test_create@plt>
  mov    $0x0,%eax
  call   <test_size@plt>
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D7
type_guessed: string

tool: D7: ----------
All tests passed
----------


assistant: ```c
int main(void) {
  test_create();
  test_size();
  printf("----------\nAll tests passed\n----------\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_create@plt>
  xor    %eax,%eax
  call   <test_size@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D1
type_infered: string

tool: D1: ----------
All tests passed
----------


assistant: ```c
int main(void) {
  test_create();
  test_size();
  printf("----------\nAll tests passed\n----------\n");
  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    $0x8,%rsp
  xor    %eax,%eax
  call   <test_create@plt>
  xor    %eax,%eax
  call   <test_size@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D9
type_guessed: string

tool: D9: ----------
All tests passed
----------


assistant: ```c
int main(void) {
  test_create();
  test_size();
  printf("----------\nAll tests passed\n----------\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %edi,-0x74(%rbp)
  mov    %rsi,-0x80(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x70(%rbp),%rdx
  mov    -0x80(%rbp),%rcx
  mov    -0x74(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <esettab@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <entab@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  char tab[100 + 1];
  esettab(argc, argv, tab);
  entab(tab);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdx
  call   <esettab@plt>
  mov    %rbx,%rdi
  call   <entab@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x70,%rsp
  pop    %rbx
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  char tab[100 + 1];
  esettab(argc, argv, tab);
  entab(tab);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdx
  call   <esettab@plt>
  mov    %rbp,%rdi
  call   <entab@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  char tab[100 + 1];
  esettab(argc, argv, tab);
  entab(tab);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdx
  call   <esettab@plt>
  mov    %rbp,%rdi
  call   <entab@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  char tab[100 + 1];
  esettab(argc, argv, tab);
  entab(tab);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getAddressFromSockAddr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    $0x100,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x100,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x80,%eax
  je     L81
  cmp    $0x81,%eax
  jne    L59
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x4(%rax),%rsi
  mov    -0x18(%rbp),%rax
  mov    $0x100,%ecx
  mov    %rax,%rdx
  mov    $0x81,%edi
  call   <inet_ntop@plt>
  test   %rax,%rax
  jne    L33
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0xffffffff,%eax
  jmp    L16
L33:
  mov    -0x30(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L47
L81:
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rsi
  mov    -0x18(%rbp),%rax
  mov    $0x100,%ecx
  mov    %rax,%rdx
  mov    $0x80,%edi
  call   <inet_ntop@plt>
  test   %rax,%rax
  jne    L50
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0xffffffff,%eax
  jmp    L16
L50:
  mov    -0x30(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L47
L59:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0xffffffff,%eax
  jmp    L16
L47:
  mov    $0x0,%eax
L16:
  leave
  ret
```

assistant: ```c
int getAddressFromSockAddr(struct sockaddr *addr, char **address, int *port) {
  struct sockaddr_in *sa4;
  struct sockaddr_in6 *sa6;
  char *addrBuf = malloc(256);
  memset(addrBuf, 0, 256);
  switch (addr->sa_family) {
  case AF_INET:
    sa4 = (struct sockaddr_in *)addr;
    if (inet_ntop(AF_INET, &sa4->sin_addr, addrBuf, 256) == NULL) {
      free(addrBuf);
      return -1;
    } else {
      *address = addrBuf;
      *port = sa4->sin_port;
    }
    break;
  case AF_INET6:
    sa6 = (struct sockaddr_in6 *)addr;
    if (inet_ntop(AF_INET6, &sa6->sin6_addr, addrBuf, 256) == NULL) {
      free(addrBuf);
      return -1;
    } else {
      *address = addrBuf;
      *port = sa6->sin6_port;
    }
    break;
  default:
    free(addrBuf);
    return -1;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getAddressFromSockAddr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%r13
  mov    $0x100,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    $0x100,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    0x0(%rbp),%eax
  cmp    $0x80,%eax
  je     L64
  cmp    $0x81,%eax
  jne    L65
  lea    0x4(%rbp),%rsi
  mov    $0x100,%ecx
  mov    %rbx,%rdx
  mov    $0x81,%edi
  call   <inet_ntop@plt>
  test   %rax,%rax
  je     L41
  mov    %rbx,(%r12)
  mov    0x0(%rbp),%eax
  mov    %eax,0x0(%r13)
  mov    $0x0,%eax
  jmp    L30
L41:
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0xffffffff,%eax
  jmp    L30
L64:
  lea    0x4(%rbp),%rsi
  mov    $0x100,%ecx
  mov    %rbx,%rdx
  mov    $0x80,%edi
  call   <inet_ntop@plt>
  test   %rax,%rax
  je     L57
  mov    %rbx,(%r12)
  mov    0x0(%rbp),%eax
  mov    %eax,0x0(%r13)
  mov    $0x0,%eax
  jmp    L30
L57:
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0xffffffff,%eax
  jmp    L30
L65:
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0xffffffff,%eax
L30:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int getAddressFromSockAddr(struct sockaddr *addr, char **address, int *port) {
  struct sockaddr_in *sa4;
  struct sockaddr_in6 *sa6;
  char *addrBuf = malloc(256);
  memset(addrBuf, 0, 256);
  switch (addr->sa_family) {
  case AF_INET:
    sa4 = (struct sockaddr_in *)addr;
    if (inet_ntop(AF_INET, &sa4->sin_addr, addrBuf, 256) == NULL) {
      free(addrBuf);
      return -1;
    } else {
      *address = addrBuf;
      *port = sa4->sin_port;
    }
    break;
  case AF_INET6:
    sa6 = (struct sockaddr_in6 *)addr;
    if (inet_ntop(AF_INET6, &sa6->sin6_addr, addrBuf, 256) == NULL) {
      free(addrBuf);
      return -1;
    } else {
      *address = addrBuf;
      *port = sa6->sin6_port;
    }
    break;
  default:
    free(addrBuf);
    return -1;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<getAddressFromSockAddr>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x100,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    $0x100,%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <memset@plt>
  mov    (%rbx),%eax
  cmp    $0x80,%eax
  je     L11
  cmp    $0x81,%eax
  jne    L9
  lea    0x4(%rbx),%rsi
  mov    $0x100,%ecx
  mov    %rbp,%rdx
  mov    $0x81,%edi
  call   <inet_ntop@plt>
  test   %rax,%rax
  je     L9
L2:
  mov    (%rbx),%eax
  mov    %rbp,(%r12)
  mov    %eax,0x0(%r13)
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L11:
  lea    0x4(%rbx),%rsi
  mov    $0x100,%ecx
  mov    %rbp,%rdx
  mov    $0x80,%edi
  call   <inet_ntop@plt>
  test   %rax,%rax
  jne    L2
L9:
  mov    %rbp,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int getAddressFromSockAddr(struct sockaddr *addr, char **address, int *port) {
  struct sockaddr_in *sa4;
  struct sockaddr_in6 *sa6;
  char *addrBuf = malloc(256);
  memset(addrBuf, 0, 256);
  switch (addr->sa_family) {
  case AF_INET:
    sa4 = (struct sockaddr_in *)addr;
    if (inet_ntop(AF_INET, &sa4->sin_addr, addrBuf, 256) == NULL) {
      free(addrBuf);
      return -1;
    } else {
      *address = addrBuf;
      *port = sa4->sin_port;
    }
    break;
  case AF_INET6:
    sa6 = (struct sockaddr_in6 *)addr;
    if (inet_ntop(AF_INET6, &sa6->sin6_addr, addrBuf, 256) == NULL) {
      free(addrBuf);
      return -1;
    } else {
      *address = addrBuf;
      *port = sa6->sin6_port;
    }
    break;
  default:
    free(addrBuf);
    return -1;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<getAddressFromSockAddr>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x100,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    $0x100,%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <memset@plt>
  mov    (%rbx),%eax
  cmp    $0x80,%eax
  je     L15
  cmp    $0x81,%eax
  jne    L10
  lea    0x4(%rbx),%rsi
  mov    $0x100,%ecx
  mov    %rbp,%rdx
  mov    $0x81,%edi
  call   <inet_ntop@plt>
  test   %rax,%rax
  je     L10
L14:
  mov    (%rbx),%eax
  mov    %rbp,(%r12)
  mov    %eax,0x0(%r13)
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L15:
  lea    0x4(%rbx),%rsi
  mov    $0x100,%ecx
  mov    %rbp,%rdx
  mov    $0x80,%edi
  call   <inet_ntop@plt>
  test   %rax,%rax
  jne    L14
L10:
  mov    %rbp,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int getAddressFromSockAddr(struct sockaddr *addr, char **address, int *port) {
  struct sockaddr_in *sa4;
  struct sockaddr_in6 *sa6;
  char *addrBuf = malloc(256);
  memset(addrBuf, 0, 256);
  switch (addr->sa_family) {
  case AF_INET:
    sa4 = (struct sockaddr_in *)addr;
    if (inet_ntop(AF_INET, &sa4->sin_addr, addrBuf, 256) == NULL) {
      free(addrBuf);
      return -1;
    } else {
      *address = addrBuf;
      *port = sa4->sin_port;
    }
    break;
  case AF_INET6:
    sa6 = (struct sockaddr_in6 *)addr;
    if (inet_ntop(AF_INET6, &sa6->sin6_addr, addrBuf, 256) == NULL) {
      free(addrBuf);
      return -1;
    } else {
      *address = addrBuf;
      *port = sa6->sin6_port;
    }
    break;
  default:
    free(addrBuf);
    return -1;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_least_squares_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  jmp    L27
L18:
  mov    -0x18(%rbp),%rax
  shr    $0x2,%rax
  mov    %rax,-0x18(%rbp)
L27:
  mov    -0x18(%rbp),%rax
  and    $0x3,%eax
  test   %rax,%rax
  je     L18
  mov    -0x18(%rbp),%rax
  and    $0x7,%eax
  cmp    $0x7,%rax
  jne    L71
  mov    $0x4,%eax
  jmp    L7
L71:
  movl   $0x0,-0x8(%rbp)
  jmp    L77
L66:
  mov    -0x8(%rbp),%eax
  imul   %eax,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  sub    %rdx,%rax
  mov    %rax,%rdi
  call   <isqrt@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  imul   %eax,%eax
  add    %edx,%eax
  cltq
  cmp    %rax,-0x18(%rbp)
  jne    L21
  cmpl   $0x0,-0x8(%rbp)
  setne  %al
  movzbl %al,%edx
  cmpl   $0x0,-0x4(%rbp)
  setne  %al
  movzbl %al,%eax
  add    %edx,%eax
  jmp    L7
L21:
  addl   $0x1,-0x8(%rbp)
L77:
  mov    -0x8(%rbp),%eax
  imul   %eax,%eax
  cltq
  cmp    %rax,-0x18(%rbp)
  jae    L66
  mov    $0x3,%eax
L7:
  leave
  ret
```

assistant: ```c
int get_least_squares_0(size_t num) {
  int a, b;
  while (num % 4 == 0)
    num /= 4;
  if (num % 8 == 7)
    return 4;
  for (a = 0; a * a <= num; ++a) {
    b = isqrt(num - a * a);
    if (a * a + b * b == num) {
      return !!a + !!b;
    }
  }
  return 3;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_least_squares_0>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  test   $0x3,%dil
  jne    L47
L44:
  mov    %rbx,%rax
  shr    $0x2,%rax
  mov    %rax,%rbx
  test   $0x3,%al
  je     L44
L47:
  mov    %rbx,%rax
  and    $0x7,%eax
  cmp    $0x7,%rax
  je     L52
  mov    $0x0,%ebp
  mov    $0x0,%eax
  mov    $0x0,%r12d
L33:
  mov    %rbx,%rdi
  sub    %rax,%rdi
  call   <isqrt@plt>
  mov    %eax,%edx
  imul   %eax,%edx
  add    %r12d,%edx
  movslq %edx,%rdx
  cmp    %rbx,%rdx
  je     L20
  add    $0x1,%ebp
  mov    %ebp,%r12d
  imul   %ebp,%r12d
  movslq %r12d,%rax
  cmp    %rbx,%rax
  jbe    L33
  mov    $0x3,%eax
  jmp    L8
L20:
  test   %ebp,%ebp
  setne  %dl
  movzbl %dl,%edx
  cmp    $0x1,%eax
  mov    %edx,%eax
  sbb    $0xffffffff,%eax
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L52:
  mov    $0x4,%eax
  jmp    L8
```

assistant: ```c
int get_least_squares_0(size_t num) {
  int a, b;
  while (num % 4 == 0)
    num /= 4;
  if (num % 8 == 7)
    return 4;
  for (a = 0; a * a <= num; ++a) {
    b = isqrt(num - a * a);
    if (a * a + b * b == num) {
      return !!a + !!b;
    }
  }
  return 3;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_least_squares_0>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  and    $0x3,%edi
  jne    L27
L6:
  shr    $0x2,%rbx
  test   $0x3,%bl
  je     L6
L27:
  mov    %rbx,%rax
  and    $0x7,%eax
  cmp    $0x7,%rax
  je     L40
  xor    %ebp,%ebp
  xor    %eax,%eax
  xor    %r12d,%r12d
  jmp    L53
  xchg   %ax,%ax
L20:
  add    $0x1,%ebp
  mov    %ebp,%r12d
  imul   %ebp,%r12d
  movslq %r12d,%rax
  cmp    %rbx,%rax
  ja     L46
L53:
  mov    %rbx,%rdi
  sub    %rax,%rdi
  call   <isqrt@plt>
  mov    %eax,%edx
  imul   %eax,%edx
  add    %r12d,%edx
  movslq %edx,%rdx
  cmp    %rbx,%rdx
  jne    L20
  xor    %edx,%edx
  test   %ebp,%ebp
  pop    %rbx
  pop    %rbp
  setne  %dl
  cmp    $0x1,%eax
  pop    %r12
  mov    %edx,%eax
  sbb    $0xffffffff,%eax
  ret
  nopl   0x0(%rax)
L46:
  pop    %rbx
  mov    $0x3,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L40:
  pop    %rbx
  mov    $0x4,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int get_least_squares_0(size_t num) {
  int a, b;
  while (num % 4 == 0)
    num /= 4;
  if (num % 8 == 7)
    return 4;
  for (a = 0; a * a <= num; ++a) {
    b = isqrt(num - a * a);
    if (a * a + b * b == num) {
      return !!a + !!b;
    }
  }
  return 3;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_least_squares_0>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  and    $0x3,%edi
  jne    L12
L57:
  shr    $0x2,%rbx
  test   $0x3,%bl
  je     L57
L12:
  mov    %rbx,%rax
  and    $0x7,%eax
  cmp    $0x7,%rax
  je     L23
  xor    %ebp,%ebp
  xor    %eax,%eax
  xor    %r12d,%r12d
  jmp    L40
  xchg   %ax,%ax
L10:
  add    $0x1,%ebp
  mov    %ebp,%r12d
  imul   %ebp,%r12d
  movslq %r12d,%rax
  cmp    %rbx,%rax
  ja     L16
L40:
  mov    %rbx,%rdi
  sub    %rax,%rdi
  call   <isqrt@plt>
  mov    %eax,%edx
  imul   %eax,%edx
  add    %r12d,%edx
  movslq %edx,%rdx
  cmp    %rbx,%rdx
  jne    L10
  xor    %edx,%edx
  test   %ebp,%ebp
  pop    %rbx
  pop    %rbp
  setne  %dl
  cmp    $0x1,%eax
  pop    %r12
  mov    %edx,%eax
  sbb    $0xffffffff,%eax
  ret
  nopl   0x0(%rax)
L16:
  pop    %rbx
  mov    $0x3,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L23:
  pop    %rbx
  mov    $0x4,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int get_least_squares_0(size_t num) {
  int a, b;
  while (num % 4 == 0)
    num /= 4;
  if (num % 8 == 7)
    return 4;
  for (a = 0; a * a <= num; ++a) {
    b = isqrt(num - a * a);
    if (a * a + b * b == num) {
      return !!a + !!b;
    }
  }
  return 3;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x340,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x334(%rbp)
L69:
  lea    -0x340(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x340(%rbp),%eax
  test   %eax,%eax
  js     L69
  mov    -0x340(%rbp),%eax
  cmp    $0x7,%eax
  jg     L69
L68:
  lea    -0x33c(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x340(%rbp),%eax
  test   %eax,%eax
  js     L68
  mov    -0x340(%rbp),%eax
  cmp    $0xc8,%eax
  jg     L68
  jmp    L100
L13:
  mov    -0x33c(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  mov    -0x334(%rbp),%eax
  cltq
  mov    %edx,-0x330(%rbp,%rax,4)
  mov    -0x33c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x33c(%rbp)
  addl   $0x1,-0x334(%rbp)
L100:
  mov    -0x33c(%rbp),%eax
  test   %eax,%eax
  jne    L13
  mov    -0x334(%rbp),%eax
  mov    %eax,-0x338(%rbp)
  jmp    L102
L98:
  mov    -0x338(%rbp),%eax
  cltq
  movl   $0x0,-0x330(%rbp,%rax,4)
  addl   $0x1,-0x338(%rbp)
L102:
  cmpl   $0x8,-0x338(%rbp)
  jle    L98
  movl   $0x0,-0x334(%rbp)
  jmp    L48
L85:
  addl   $0x1,-0x334(%rbp)
L48:
  mov    -0x340(%rbp),%eax
  cmp    %eax,-0x334(%rbp)
  jne    L85
  mov    -0x334(%rbp),%eax
  cltq
  mov    -0x330(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D18
type_guessed: string
name: parse_data
arguments: label: D15
type_guessed: string

tool: D18: %d

tool: D15: 
%d

assistant: ```c
int main() {
  int hp;
  int helper;
  int i = 0;
  int base[200];
  int box;
  do {
    scanf("%d", &helper);
  } while (helper < 0 || helper > 7);
  do {
    scanf("%d", &box);
  } while (helper < 0 || helper > 200);
  while (box != 0) {
    base[i] = box % 2;
    box = box / 2;
    i++;
  }
  hp = i;
  while (hp <= 8) {
    base[hp] = 0;
    hp++;
  }
  i = 0;
  while (i != helper) {
    i++;
  }
  printf("\n%d", base[i]);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x340,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x338(%rsp)
  xor    %eax,%eax
  lea    D13(%rip),%rbx
L80:
  lea    0x8(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmpl   $0x7,0x8(%rsp)
  ja     L80
  lea    D13(%rip),%rbx
L106:
  lea    0xc(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%edi
  cmp    $0xc8,%edi
  ja     L106
  mov    0xc(%rsp),%ecx
  test   %ecx,%ecx
  je     L24
  mov    $0x1,%esi
L5:
  mov    %ecx,%edx
  shr    $0x1f,%edx
  lea    (%rcx,%rdx,1),%eax
  and    $0x1,%eax
  sub    %edx,%eax
  mov    %eax,0xc(%rsp,%rsi,4)
  mov    %edx,%eax
  add    %ecx,%eax
  sar    %eax
  mov    %eax,%ecx
  mov    %esi,%edx
  add    $0x1,%rsi
  test   %eax,%eax
  jne    L5
  movl   $0x0,0xc(%rsp)
  cmp    $0x8,%edx
  jg     L63
L47:
  movslq %edx,%rax
L56:
  movl   $0x0,0x10(%rsp,%rax,4)
  add    $0x1,%rax
  cmp    $0x8,%eax
  jle    L56
L63:
  test   %edi,%edi
  je     L11
  mov    $0x0,%eax
L48:
  add    $0x1,%eax
  cmp    %eax,%edi
  jne    L48
L20:
  cltq
  mov    0x10(%rsp,%rax,4),%esi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    0x338(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  mov    $0x0,%eax
  add    $0x340,%rsp
  pop    %rbx
  ret
L24:
  mov    %ecx,%edx
  jmp    L47
L11:
  mov    %edi,%eax
  jmp    L20
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D13
datatype: string
name: read_data
arguments: tag: D16
datatype: string

tool: D13: %d

tool: D16: 
%d

assistant: ```c
int main() {
  int hp;
  int helper;
  int i = 0;
  int base[200];
  int box;
  do {
    scanf("%d", &helper);
  } while (helper < 0 || helper > 7);
  do {
    scanf("%d", &box);
  } while (helper < 0 || helper > 200);
  while (box != 0) {
    base[i] = box % 2;
    box = box / 2;
    i++;
  }
  hp = i;
  while (hp <= 8) {
    base[hp] = 0;
    hp++;
  }
  i = 0;
  while (i != helper) {
    i++;
  }
  printf("\n%d", base[i]);
  return 0;
}
```