ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D12(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  lea    D0(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  mov    %r13,%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D12
type_guessed: string
name: read_data
arguments: data_label: D0
type_guessed: string

tool: D12: %3d

tool: D0: %d


assistant: ```c
int main() {
  int a;
  float b;
  scanf("%3d", &a);
  printf("%d\n", a);
  scanf("%3d", &a);
  printf("%d\n", a);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movl   $0x3e8,-0x4(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  jmp    L70
L5:
  movl   $0x1,-0xc(%rbp)
  jmp    L15
L18:
  movl   $0x1,-0x8(%rbp)
  jmp    L55
L40:
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x4(%rbp)
  jne    L84
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%ecx
  mov    -0x10(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <is_triplet@plt>
  test   %rax,%rax
  je     L84
  mov    -0x10(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    -0x8(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L23
L84:
  addl   $0x1,-0x8(%rbp)
L55:
  cmpl   $0x3e7,-0x8(%rbp)
  jle    L40
  addl   $0x1,-0xc(%rbp)
L15:
  cmpl   $0x3e7,-0xc(%rbp)
  jle    L18
  addl   $0x1,-0x10(%rbp)
L70:
  cmpl   $0x3e7,-0x10(%rbp)
  jle    L5
  nop
L23:
  mov    -0x14(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D4: The product of the triplet with sum %d is %d.


assistant: ```c
int main() {
  const int target = 1000;
  int product = 0;
  for (int a = 1; a < 1000; a++) {
    for (int b = 1; b < 1000; b++) {
      for (int c = 1; c < 1000; c++) {
        if (a + b + c == target && is_triplet(a, b, c)) {
          product = a * b * c;
          goto exit;
        }
      }
    }
  }
exit:
  printf("The product of the triplet with sum %d is %d.\n", target, product);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x1,%r12d
L67:
  mov    %r12d,%r14d
  lea    0x1(%r12),%ebp
  lea    0x3e8(%r12),%r15d
L37:
  mov    %ebp,%r13d
  sub    %r14d,%r13d
  mov    $0x1,%ebx
  jmp    L20
L36:
  add    $0x1,%ebx
  cmp    $0x3e8,%ebx
  je     L62
L20:
  lea    (%rbx,%rbp,1),%eax
  cmp    $0x3e8,%eax
  jne    L36
  mov    %ebx,%edx
  mov    %r13d,%esi
  mov    %r12d,%edi
  call   <is_triplet@plt>
  test   %rax,%rax
  je     L36
  mov    %r12d,%edx
  imul   %r13d,%edx
  imul   %ebx,%edx
L19:
  mov    $0x3e8,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L62:
  add    $0x1,%ebp
  cmp    %r15d,%ebp
  jne    L37
  add    $0x1,%r12d
  cmp    $0x3e8,%r12d
  jne    L67
  mov    $0x0,%edx
  jmp    L19
```

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

tool: D9: The product of the triplet with sum %d is %d.


assistant: ```c
int main() {
  const int target = 1000;
  int product = 0;
  for (int a = 1; a < 1000; a++) {
    for (int b = 1; b < 1000; b++) {
      for (int c = 1; c < 1000; c++) {
        if (a + b + c == target && is_triplet(a, b, c)) {
          product = a * b * c;
          goto exit;
        }
      }
    }
  }
exit:
  printf("The product of the triplet with sum %d is %d.\n", target, product);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x8,%rsp
L61:
  lea    0x1(%rbp),%r13d
  lea    0x3e8(%rbp),%r14d
  mov    %r13d,%r15d
L32:
  mov    %r15d,%r12d
  mov    $0x1,%ebx
  sub    %ebp,%r12d
  jmp    L34
  nopw   0x0(%rax,%rax,1)
L46:
  add    $0x1,%ebx
  cmp    $0x3e8,%ebx
  je     L9
L34:
  lea    (%rbx,%r15,1),%eax
  cmp    $0x3e8,%eax
  jne    L46
  mov    %ebx,%edx
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <is_triplet@plt>
  test   %rax,%rax
  je     L46
  mov    %ebp,%edx
  imul   %r12d,%edx
  imul   %ebx,%edx
L0:
  mov    $0x3e8,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L9:
  add    $0x1,%r15d
  cmp    %r14d,%r15d
  jne    L32
  mov    %r13d,%ebp
  cmp    $0x3e8,%r13d
  jne    L61
  xor    %edx,%edx
  jmp    L0
```

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

tool: D9: The product of the triplet with sum %d is %d.


assistant: ```c
int main() {
  const int target = 1000;
  int product = 0;
  for (int a = 1; a < 1000; a++) {
    for (int b = 1; b < 1000; b++) {
      for (int c = 1; c < 1000; c++) {
        if (a + b + c == target && is_triplet(a, b, c)) {
          product = a * b * c;
          goto exit;
        }
      }
    }
  }
exit:
  printf("The product of the triplet with sum %d is %d.\n", target, product);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x8,%rsp
L57:
  lea    0x1(%rbp),%r13d
  lea    0x3e8(%rbp),%r14d
  mov    %r13d,%r15d
L45:
  mov    %r15d,%r12d
  mov    $0x1,%ebx
  sub    %ebp,%r12d
  jmp    L29
  nopw   0x0(%rax,%rax,1)
L41:
  add    $0x1,%ebx
  cmp    $0x3e8,%ebx
  je     L4
L29:
  lea    (%rbx,%r15,1),%eax
  cmp    $0x3e8,%eax
  jne    L41
  mov    %ebx,%edx
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <is_triplet@plt>
  test   %rax,%rax
  je     L41
  mov    %ebp,%edx
  imul   %r12d,%edx
  imul   %ebx,%edx
L7:
  mov    $0x3e8,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L4:
  add    $0x1,%r15d
  cmp    %r14d,%r15d
  jne    L45
  mov    %r13d,%ebp
  cmp    $0x3e8,%r13d
  jne    L57
  xor    %edx,%edx
  jmp    L7
```

tool_calls: name: unpack_data
arguments: tag: D5
type: string

tool: D5: The product of the triplet with sum %d is %d.


assistant: ```c
int main() {
  const int target = 1000;
  int product = 0;
  for (int a = 1; a < 1000; a++) {
    for (int b = 1; b < 1000; b++) {
      for (int c = 1; c < 1000; c++) {
        if (a + b + c == target && is_triplet(a, b, c)) {
          product = a * b * c;
          goto exit;
        }
      }
    }
  }
exit:
  printf("The product of the triplet with sum %d is %d.\n", target, product);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ulz_emo_cut>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L7
  mov    $0x0,%edi
  call   <ulz_new@plt>
  mov    %rax,-0x8(%rbp)
  jmp    L5
L7:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
L5:
  call   <fork@plt>
  jmp    L5
```

assistant: ```c
void ulz_emo_cut(struct ulz *u) {
  struct ulz *uptr;
  if (!u) {
    uptr = ulz_new(NULL);
  } else {
    uptr = u;
  }
  do {
    fork();
  } while (1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ulz_emo_cut>:
  endbr64
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L27
L10:
  call   <fork@plt>
  jmp    L10
L27:
  call   <ulz_new@plt>
  jmp    L10
```

assistant: ```c
void ulz_emo_cut(struct ulz *u) {
  struct ulz *uptr;
  if (!u) {
    uptr = ulz_new(NULL);
  } else {
    uptr = u;
  }
  do {
    fork();
  } while (1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ulz_emo_cut>:
  endbr64
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L24
  nopl   (%rax)
L17:
  call   <fork@plt>
  jmp    L17
L24:
  call   <ulz_new@plt>
  jmp    L17
```

assistant: ```c
void ulz_emo_cut(struct ulz *u) {
  struct ulz *uptr;
  if (!u) {
    uptr = ulz_new(NULL);
  } else {
    uptr = u;
  }
  do {
    fork();
  } while (1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ulz_emo_cut>:
  endbr64
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L26
  nopl   (%rax)
L14:
  call   <fork@plt>
  jmp    L14
L26:
  call   <ulz_new@plt>
  jmp    L14
```

assistant: ```c
void ulz_emo_cut(struct ulz *u) {
  struct ulz *uptr;
  if (!u) {
    uptr = ulz_new(NULL);
  } else {
    uptr = u;
  }
  do {
    fork();
  } while (1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xstrdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <assert@plt>
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *xstrdup(const char *s) {
  void *ret = malloc(strlen(s) + 1);
  assert(ret);
  strcpy(ret, s);
  return ret;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<xstrdup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <assert@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void *xstrdup(const char *s) {
  void *ret = malloc(strlen(s) + 1);
  assert(ret);
  strcpy(ret, s);
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xstrdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <assert@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void *xstrdup(const char *s) {
  void *ret = malloc(strlen(s) + 1);
  assert(ret);
  strcpy(ret, s);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<xstrdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <assert@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void *xstrdup(const char *s) {
  void *ret = malloc(strlen(s) + 1);
  assert(ret);
  strcpy(ret, s);
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<log2pad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L8
L27:
  shll   -0x4(%rbp)
L8:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jb     L27
  cmpl   $0x0,-0x18(%rbp)
  jne    L12
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jbe    L12
  shrl   -0x4(%rbp)
L12:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int log2pad(unsigned int v, int roundup) {
  unsigned int x = 1;
  while (x < v)
    x <<= 1;
  if (roundup == 0) {
    if (x > v) {
      x >>= 1;
    }
  }
  return x;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<log2pad>:
  endbr64
  cmp    $0x1,%edi
  jbe    L27
  mov    $0x1,%eax
L36:
  add    %eax,%eax
  cmp    %eax,%edi
  ja     L36
L18:
  cmp    %eax,%edi
  jae    L29
  test   %esi,%esi
  sete   %cl
  mov    %eax,%edx
  shr    %edx
  test   %cl,%cl
  cmovne %edx,%eax
L29:
  ret
L27:
  mov    $0x1,%eax
  jmp    L18
```

assistant: ```c
unsigned int log2pad(unsigned int v, int roundup) {
  unsigned int x = 1;
  while (x < v)
    x <<= 1;
  if (roundup == 0) {
    if (x > v) {
      x >>= 1;
    }
  }
  return x;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<log2pad>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jbe    L20
  xchg   %ax,%ax
L21:
  add    %eax,%eax
  cmp    %eax,%edi
  ja     L21
L20:
  cmp    %eax,%edi
  jae    L1
  test   %esi,%esi
  mov    %eax,%edx
  sete   %cl
  shr    %edx
  test   %cl,%cl
  cmovne %edx,%eax
L1:
  ret
```

assistant: ```c
unsigned int log2pad(unsigned int v, int roundup) {
  unsigned int x = 1;
  while (x < v)
    x <<= 1;
  if (roundup == 0) {
    if (x > v) {
      x >>= 1;
    }
  }
  return x;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<log2pad>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jbe    L2
  xchg   %ax,%ax
L8:
  add    %eax,%eax
  cmp    %eax,%edi
  ja     L8
L2:
  cmp    %eax,%edi
  jae    L3
  test   %esi,%esi
  mov    %eax,%edx
  sete   %cl
  shr    %edx
  test   %cl,%cl
  cmovne %edx,%eax
L3:
  ret
```

assistant: ```c
unsigned int log2pad(unsigned int v, int roundup) {
  unsigned int x = 1;
  while (x < v)
    x <<= 1;
  if (roundup == 0) {
    if (x > v) {
      x >>= 1;
    }
  }
  return x;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<factorial>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L18:
  addl   $0x1,-0x4(%rbp)
L13:
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jle    L18
  nop
  pop    %rbp
  ret
```

assistant: ```c
int factorial(int n) {
  int i;
  for (i = 0; i <= INT_MAX; i++)
    ;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<factorial>:
  endbr64
  mov    D6(%rip),%rax
  mov    (%rax),%edx
  test   %edx,%edx
  js     L14
  add    $0x1,%edx
  mov    $0x0,%eax
L7:
  add    $0x1,%eax
  cmp    %edx,%eax
  jne    L7
L14:
  ret
```

assistant: ```c
int factorial(int n) {
  int i;
  for (i = 0; i <= INT_MAX; i++)
    ;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<factorial>:
  endbr64
  ret
```

assistant: ```c
int factorial(int n) {
  int i;
  for (i = 0; i <= INT_MAX; i++)
    ;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<factorial>:
  endbr64
  ret
```

assistant: ```c
int factorial(int n) {
  int i;
  for (i = 0; i <= INT_MAX; i++)
    ;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<getlogin_r>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x0,%eax
  call   <getlogin@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L8
  mov    D19(%rip),%rax
  mov    (%rax),%eax
  jmp    L18
L8:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %rax,-0x20(%rbp)
  ja     L23
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  jmp    L18
L23:
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    $0x0,%eax
L18:
  leave
  ret
```

assistant: ```c
int getlogin_r(char *name, size_t size) {
  char *logname = getlogin();
  if (!logname)
    return ENXIO;
  if (strlen(name) >= size)
    return ERANGE;
  strcpy(name, logname);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getlogin_r>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    $0x0,%eax
  call   <getlogin@plt>
  test   %rax,%rax
  je     L21
  mov    %rax,%rbx
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %r12,%rax
  jb     L18
  mov    D10(%rip),%rax
  mov    (%rax),%eax
L4:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L21:
  mov    D17(%rip),%rax
  mov    (%rax),%eax
  jmp    L4
L18:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    $0x0,%eax
  jmp    L4
```

assistant: ```c
int getlogin_r(char *name, size_t size) {
  char *logname = getlogin();
  if (!logname)
    return ENXIO;
  if (strlen(name) >= size)
    return ERANGE;
  strcpy(name, logname);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getlogin_r>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <getlogin@plt>
  test   %rax,%rax
  je     L7
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <strlen@plt>
  cmp    %rbx,%rax
  jb     L0
  mov    D1(%rip),%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  mov    (%rax),%eax
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  mov    D18(%rip),%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  mov    (%rax),%eax
  ret
```

assistant: ```c
int getlogin_r(char *name, size_t size) {
  char *logname = getlogin();
  if (!logname)
    return ENXIO;
  if (strlen(name) >= size)
    return ERANGE;
  strcpy(name, logname);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<getlogin_r>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <getlogin@plt>
  test   %rax,%rax
  je     L5
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <strlen@plt>
  cmp    %rbx,%rax
  jb     L14
  mov    D2(%rip),%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  mov    (%rax),%eax
  ret
  nopl   0x0(%rax,%rax,1)
L14:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  mov    D5(%rip),%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  mov    (%rax),%eax
  ret
```

assistant: ```c
int getlogin_r(char *name, size_t size) {
  char *logname = getlogin();
  if (!logname)
    return ENXIO;
  if (strlen(name) >= size)
    return ERANGE;
  strcpy(name, logname);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<delay>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  imul   -0x24(%rbp),%eax
  cltq
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <clock@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L15
L12:
  mov    $0x0,%eax
  call   <clock@plt>
  mov    %rax,-0x18(%rbp)
L15:
  mov    -0x18(%rbp),%rax
  sub    -0x8(%rbp),%rax
  cmp    %rax,-0x10(%rbp)
  jg     L12
  nop
  leave
  ret
```

assistant: ```c
void delay(int milliseconds) {
  long pause;
  clock_t now, then;
  pause = milliseconds * (CLOCKS_PER_SEC / 1000);
  now = then = clock();
  while ((now - then) < pause)
    now = clock();
  return;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<delay>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  movslq %eax,%rbx
  imul   $0x10624dd3,%rbx,%rbx
  sar    $0x26,%rbx
  sar    $0x1f,%eax
  sub    %eax,%ebx
  imul   %edi,%ebx
  movslq %ebx,%rbx
  mov    $0x0,%eax
  call   <clock@plt>
  mov    %rax,%rbp
  test   %rbx,%rbx
  jle    L17
L15:
  mov    $0x0,%eax
  call   <clock@plt>
  sub    %rbp,%rax
  cmp    %rbx,%rax
  jl     L15
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void delay(int milliseconds) {
  long pause;
  clock_t now, then;
  pause = milliseconds * (CLOCKS_PER_SEC / 1000);
  now = then = clock();
  while ((now - then) < pause)
    now = clock();
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<delay>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D0(%rip),%rax
  movslq (%rax),%rbx
  mov    %rbx,%rax
  imul   $0x10624dd3,%rbx,%rbx
  sar    $0x1f,%eax
  sar    $0x26,%rbx
  sub    %eax,%ebx
  xor    %eax,%eax
  imul   %edi,%ebx
  movslq %ebx,%rbx
  call   <clock@plt>
  test   %rbx,%rbx
  jle    L8
  mov    %rax,%rbp
  nopl   0x0(%rax)
L9:
  xor    %eax,%eax
  call   <clock@plt>
  sub    %rbp,%rax
  cmp    %rbx,%rax
  jl     L9
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void delay(int milliseconds) {
  long pause;
  clock_t now, then;
  pause = milliseconds * (CLOCKS_PER_SEC / 1000);
  now = then = clock();
  while ((now - then) < pause)
    now = clock();
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<delay>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D9(%rip),%rax
  movslq (%rax),%rbx
  mov    %rbx,%rax
  imul   $0x10624dd3,%rbx,%rbx
  sar    $0x1f,%eax
  sar    $0x26,%rbx
  sub    %eax,%ebx
  xor    %eax,%eax
  imul   %edi,%ebx
  movslq %ebx,%rbx
  call   <clock@plt>
  test   %rbx,%rbx
  jle    L19
  mov    %rax,%rbp
  nopl   0x0(%rax)
L9:
  xor    %eax,%eax
  call   <clock@plt>
  sub    %rbp,%rax
  cmp    %rbx,%rax
  jl     L9
L19:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void delay(int milliseconds) {
  long pause;
  clock_t now, then;
  pause = milliseconds * (CLOCKS_PER_SEC / 1000);
  now = then = clock();
  while ((now - then) < pause)
    now = clock();
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<malloc_out_parameter_local_mutation_bad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  movq   $0x0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void malloc_out_parameter_local_mutation_bad(int **x) {
  *x = (int *)malloc(sizeof(int));
  *x = NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<malloc_out_parameter_local_mutation_bad>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x4,%edi
  call   <malloc@plt>
  movq   $0x0,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void malloc_out_parameter_local_mutation_bad(int **x) {
  *x = (int *)malloc(sizeof(int));
  *x = NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<malloc_out_parameter_local_mutation_bad>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x4,%edi
  call   <malloc@plt>
  movq   $0x0,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void malloc_out_parameter_local_mutation_bad(int **x) {
  *x = (int *)malloc(sizeof(int));
  *x = NULL;
}
```
user: Translate the below assembly code into C programming language:

```asm
<malloc_out_parameter_local_mutation_bad>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x4,%edi
  call   <malloc@plt>
  movq   $0x0,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void malloc_out_parameter_local_mutation_bad(int **x) {
  *x = (int *)malloc(sizeof(int));
  *x = NULL;
}
```