ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<copy_callback>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r13
  mov    %rdx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  cltq
  cmpb   $0x2f,-0x1(%rbx,%rax,1)
  je     L24
L12:
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  cmpb   $0x2f,-0x1(%r12,%rax,1)
  je     L11
L97:
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <strstr@plt>
  cmp    %rax,%r13
  jne    L90
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %eax,%r14d
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x1(%r14,%rax,1),%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L83:
  cmp    %rdx,%rsp
  je     L30
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L83
L24:
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%rbx,%rax,1)
  jmp    L12
L11:
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%r12,%rax,1)
  jmp    L97
L30:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L80
  orq    $0x0,-0x8(%rsp,%rax,1)
L80:
  movb   $0x0,(%rsp)
  mov    %r12,%rsi
  mov    %rsp,%rdi
  call   <strcat@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  movslq %eax,%rsi
  add    %r13,%rsi
  mov    %rsp,%rdi
  call   <strcat@plt>
  mov    %rsp,%rsi
  mov    %r13,%rdi
  call   <copy_file@plt>
L4:
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L67
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L90:
  mov    $0xffffffff,%eax
  jmp    L4
L67:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int copy_callback(char *source_root, char *source, char *target_root) {
  if (source_root[strlen(source_root) - 1] == '/')
    source_root[strlen(source_root) - 1] = 0;
  if (target_root[strlen(target_root) - 1] == '/')
    target_root[strlen(target_root) - 1] = 0;
  if (strstr(source, source_root) != source)
    return -1;
  char target[strlen(source) + strlen(target_root) + 1];
  target[0] = 0;
  strcat(target, target_root);
  strcat(target, source + strlen(source_root));
  return copy_file(source, target);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<copy_callback>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  cltq
  cmpb   $0x2f,-0x1(%r12,%rax,1)
  je     L0
L63:
  mov    %r13,%rdi
  call   <strlen@plt>
  cltq
  cmpb   $0x2f,-0x1(%r13,%rax,1)
  je     L50
L18:
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <strstr@plt>
  cmp    %rax,%r14
  jne    L34
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    %rsp,%rcx
  lea    0x1(%rbx,%rax,1),%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L62
L40:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L40
L62:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L54
L77:
  mov    %r13,%rsi
  mov    %rsp,%rdi
  movb   $0x0,(%rsp)
  call   <strcat@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %rsp,%rdi
  movslq %eax,%rsi
  add    %r14,%rsi
  call   <strcat@plt>
  mov    %rsp,%rsi
  mov    %r14,%rdi
  call   <copy_file@plt>
L27:
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L82
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nop
L50:
  mov    %r13,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%r13,%rax,1)
  jmp    L18
  nopl   (%rax)
L0:
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%r12,%rax,1)
  jmp    L63
  nopl   (%rax)
L54:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L77
  nopl   0x0(%rax,%rax,1)
L34:
  mov    $0xffffffff,%eax
  jmp    L27
L82:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int copy_callback(char *source_root, char *source, char *target_root) {
  if (source_root[strlen(source_root) - 1] == '/')
    source_root[strlen(source_root) - 1] = 0;
  if (target_root[strlen(target_root) - 1] == '/')
    target_root[strlen(target_root) - 1] = 0;
  if (strstr(source, source_root) != source)
    return -1;
  char target[strlen(source) + strlen(target_root) + 1];
  target[0] = 0;
  strcat(target, target_root);
  strcat(target, source + strlen(source_root));
  return copy_file(source, target);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<copy_callback>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  cltq
  cmpb   $0x2f,-0x1(%r12,%rax,1)
  je     L12
L19:
  mov    %r13,%rdi
  call   <strlen@plt>
  cltq
  cmpb   $0x2f,-0x1(%r13,%rax,1)
  je     L50
L10:
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <strstr@plt>
  cmp    %rax,%r14
  jne    L63
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    %rsp,%rcx
  lea    0x1(%rbx,%rax,1),%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L53
L4:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L4
L53:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L18
L88:
  mov    %r13,%rsi
  mov    %rsp,%rdi
  movb   $0x0,(%rsp)
  call   <strcat@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %rsp,%rdi
  movslq %eax,%rsi
  add    %r14,%rsi
  call   <strcat@plt>
  mov    %rsp,%rsi
  mov    %r14,%rdi
  call   <copy_file@plt>
L39:
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L23
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nop
L50:
  mov    %r13,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%r13,%rax,1)
  jmp    L10
  nopl   (%rax)
L12:
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%r12,%rax,1)
  jmp    L19
  nopl   (%rax)
L18:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L88
  nopl   0x0(%rax,%rax,1)
L63:
  mov    $0xffffffff,%eax
  jmp    L39
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int copy_callback(char *source_root, char *source, char *target_root) {
  if (source_root[strlen(source_root) - 1] == '/')
    source_root[strlen(source_root) - 1] = 0;
  if (target_root[strlen(target_root) - 1] == '/')
    target_root[strlen(target_root) - 1] = 0;
  if (strstr(source, source_root) != source)
    return -1;
  char target[strlen(source) + strlen(target_root) + 1];
  target[0] = 0;
  strcat(target, target_root);
  strcat(target, source + strlen(source_root));
  return copy_file(source, target);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<eventfd_read>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x4,-0x4(%rbp)
  jne    L10
  mov    $0x0,%eax
  jmp    L0
L10:
  mov    $0xffffffff,%eax
L0:
  leave
  ret
```

assistant: ```c
int eventfd_read(int fd, eventfd_t *counter) {
  int ret = read(fd, counter, sizeof(*counter));
  if (ret == sizeof(*counter))
    return 0;
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<eventfd_read>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edx
  call   <read@plt>
  cmp    $0x4,%eax
  setne  %al
  movzbl %al,%eax
  neg    %eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int eventfd_read(int fd, eventfd_t *counter) {
  int ret = read(fd, counter, sizeof(*counter));
  if (ret == sizeof(*counter))
    return 0;
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<eventfd_read>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edx
  call   <read@plt>
  cmp    $0x4,%eax
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  neg    %eax
  ret
```

assistant: ```c
int eventfd_read(int fd, eventfd_t *counter) {
  int ret = read(fd, counter, sizeof(*counter));
  if (ret == sizeof(*counter))
    return 0;
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<eventfd_read>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edx
  call   <read@plt>
  cmp    $0x4,%eax
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  neg    %eax
  ret
```

assistant: ```c
int eventfd_read(int fd, eventfd_t *counter) {
  int ret = read(fd, counter, sizeof(*counter));
  if (ret == sizeof(*counter))
    return 0;
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  add    %eax,%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D18
type_guessed: string
name: read_data
arguments: label: D0
type_guessed: string

tool: D18: %i

tool: D0: %i minutos


assistant: ```c
int main() {
  int dist;
  scanf("%i", &dist);
  printf("%i minutos\n", dist * 2);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D15(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    (%rax,%rax,1),%esi
  lea    D13(%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: label: D15
type: string
name: read_data
arguments: label: D13
type: string

tool: D15: %i

tool: D13: %i minutos


assistant: ```c
int main() {
  int dist;
  scanf("%i", &dist);
  printf("%i minutos\n", dist * 2);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D13(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    D4(%rip),%rdi
  lea    (%rax,%rax,1),%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D13
type_guessed: string
name: extract_symbolic_data
arguments: tag: D4
type_guessed: string

tool: D13: %i

tool: D4: %i minutos


assistant: ```c
int main() {
  int dist;
  scanf("%i", &dist);
  printf("%i minutos\n", dist * 2);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D5(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    D9(%rip),%rdi
  lea    (%rax,%rax,1),%esi
  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: unpack_data
arguments: data_label: D5
data_type: string
name: unpack_data
arguments: data_label: D9
data_type: string

tool: D5: %i

tool: D9: %i minutos


assistant: ```c
int main() {
  int dist;
  scanf("%i", &dist);
  printf("%i minutos\n", dist * 2);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<tmpnam>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
L62:
  lea    D31(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L56
L15:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x30,(%rax)
L56:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  je     L15
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L66
  mov    $0x0,%eax
  jmp    L48
L66:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  add    $0x1,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  mov    $0x0,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <access@plt>
  test   %rax,%rax
  je     L62
  cmpq   $0x0,-0x18(%rbp)
  je     L45
  mov    -0x18(%rbp),%rax
  lea    D2(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x18(%rbp),%rax
  jmp    L48
L45:
  lea    D2(%rip),%rax
L48:
  leave
  ret
```

assistant: ```c
char *tmpnam(char *s) {
  static char name[] = "/tmp/tn000000000000";
  char *p;
  do {
    p = name + 7;
    while (*p == '9')
      *p++ = '0';
    if (*p == '\0')
      return NULL;
    ++*p;
  } while (access(name, 0) == 0);
  if (s) {
    strcpy(s, name);
    return s;
  }
  return name;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<tmpnam>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  jmp    L27
L26:
  test   %dl,%dl
  je     L54
  add    $0x1,%edx
  mov    %dl,(%rax)
  mov    $0x0,%esi
  lea    D22(%rip),%rdi
  call   <access@plt>
  test   %rax,%rax
  jne    L60
L27:
  movzbl D28(%rip),%edx
  lea    D28(%rip),%rax
  cmp    $0x39,%dl
  jne    L26
L45:
  add    $0x1,%rax
  movb   $0x30,-0x1(%rax)
  movzbl (%rax),%edx
  cmp    $0x39,%dl
  je     L45
  jmp    L26
L60:
  lea    D22(%rip),%rax
  test   %rbx,%rbx
  je     L6
  mov    %rax,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rax
  jmp    L6
L54:
  mov    $0x0,%eax
L6:
  pop    %rbx
  ret
```

assistant: ```c
char *tmpnam(char *s) {
  static char name[] = "/tmp/tn000000000000";
  char *p;
  do {
    p = name + 7;
    while (*p == '9')
      *p++ = '0';
    if (*p == '\0')
      return NULL;
    ++*p;
  } while (access(name, 0) == 0);
  if (s) {
    strcpy(s, name);
    return s;
  }
  return name;
}
```
user: What is the c source code of the assembly code below:

```asm
<tmpnam>:
  endbr64
  push   %rbp
  lea    D38(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L4:
  movzbl D22(%rip),%edx
  lea    D22(%rip),%rax
  cmp    $0x39,%dl
  jne    L35
  nopl   0x0(%rax,%rax,1)
L8:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  movb   $0x30,-0x1(%rax)
  cmp    $0x39,%dl
  je     L8
L35:
  test   %dl,%dl
  je     L2
  add    $0x1,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    %dl,(%rax)
  call   <access@plt>
  test   %rax,%rax
  je     L4
  lea    D38(%rip),%rax
  test   %rbx,%rbx
  je     L38
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  ret
L2:
  xor    %eax,%eax
L38:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *tmpnam(char *s) {
  static char name[] = "/tmp/tn000000000000";
  char *p;
  do {
    p = name + 7;
    while (*p == '9')
      *p++ = '0';
    if (*p == '\0')
      return NULL;
    ++*p;
  } while (access(name, 0) == 0);
  if (s) {
    strcpy(s, name);
    return s;
  }
  return name;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<tmpnam>:
  endbr64
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L23:
  movzbl D24(%rip),%edx
  lea    D24(%rip),%rax
  cmp    $0x39,%dl
  jne    L43
  nopl   0x0(%rax,%rax,1)
L15:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  movb   $0x30,-0x1(%rax)
  cmp    $0x39,%dl
  je     L15
L43:
  test   %dl,%dl
  je     L29
  add    $0x1,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    %dl,(%rax)
  call   <access@plt>
  test   %rax,%rax
  je     L23
  lea    D8(%rip),%rax
  test   %rbx,%rbx
  je     L9
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  ret
L29:
  xor    %eax,%eax
L9:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *tmpnam(char *s) {
  static char name[] = "/tmp/tn000000000000";
  char *p;
  do {
    p = name + 7;
    while (*p == '9')
      *p++ = '0';
    if (*p == '\0')
      return NULL;
    ++*p;
  } while (access(name, 0) == 0);
  if (s) {
    strcpy(s, name);
    return s;
  }
  return name;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<calc_dist>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L15
L10:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  movapd %xmm0,%xmm1
  subsd  %xmm2,%xmm1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  subsd  %xmm2,%xmm0
  mulsd  %xmm1,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L15:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jne    L10
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double calc_dist(int one, double *two, double *three) {
  double answer = 0;
  int i = 0;
  while (i != one) {
    answer += (*(two + i) - *(three + i)) * (*(two + i) - *(three + i));
    i++;
  }
  return sqrt(answer);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<calc_dist>:
  endbr64
  test   %edi,%edi
  je     L1
  mov    %edi,%edi
  mov    $0x0,%eax
  pxor   %xmm0,%xmm0
L4:
  movsd  (%rsi,%rax,8),%xmm1
  subsd  (%rdx,%rax,8),%xmm1
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L4
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L31
L3:
  sqrtsd %xmm0,%xmm0
  ret
L1:
  pxor   %xmm0,%xmm0
  jmp    L3
L31:
  sub    $0x8,%rsp
  call   <sqrt@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
double calc_dist(int one, double *two, double *three) {
  double answer = 0;
  int i = 0;
  while (i != one) {
    answer += (*(two + i) - *(three + i)) * (*(two + i) - *(three + i));
    i++;
  }
  return sqrt(answer);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<calc_dist>:
  endbr64
  test   %edi,%edi
  je     L6
  pxor   %xmm2,%xmm2
  mov    %edi,%edi
  xor    %eax,%eax
  movapd %xmm2,%xmm0
  nopl   0x0(%rax)
L27:
  movsd  (%rsi,%rax,8),%xmm1
  subsd  (%rdx,%rax,8),%xmm1
  add    $0x1,%rax
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rax,%rdi
  jne    L27
  ucomisd %xmm0,%xmm2
  ja     L23
  sqrtsd %xmm0,%xmm0
  ret
  xchg   %ax,%ax
L6:
  pxor   %xmm0,%xmm0
  sqrtsd %xmm0,%xmm0
  ret
L23:
  jmp    1050 <sqrt@plt>
```

assistant: ```c
double calc_dist(int one, double *two, double *three) {
  double answer = 0;
  int i = 0;
  while (i != one) {
    answer += (*(two + i) - *(three + i)) * (*(two + i) - *(three + i));
    i++;
  }
  return sqrt(answer);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<calc_dist>:
  endbr64
  test   %edi,%edi
  je     L15
  cmp    $0x1,%edi
  je     L57
  mov    %edi,%ecx
  pxor   %xmm3,%xmm3
  xor    %eax,%eax
  shr    %ecx
  movapd %xmm3,%xmm0
  shl    $0x4,%rcx
  nop
L24:
  movupd (%rsi,%rax,1),%xmm1
  movupd (%rdx,%rax,1),%xmm4
  add    $0x10,%rax
  subpd  %xmm4,%xmm1
  mulpd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  unpckhpd %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rax,%rcx
  jne    L24
  mov    %edi,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%edi
  je     L23
L31:
  cltq
  movsd  (%rsi,%rax,8),%xmm1
  subsd  (%rdx,%rax,8),%xmm1
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
L23:
  ucomisd %xmm0,%xmm3
  ja     L44
  sqrtsd %xmm0,%xmm0
  ret
L15:
  pxor   %xmm0,%xmm0
  sqrtsd %xmm0,%xmm0
  ret
L57:
  pxor   %xmm3,%xmm3
  xor    %eax,%eax
  movapd %xmm3,%xmm0
  jmp    L31
L44:
  jmp    1050 <sqrt@plt>
```

assistant: ```c
double calc_dist(int one, double *two, double *three) {
  double answer = 0;
  int i = 0;
  while (i != one) {
    answer += (*(two + i) - *(three + i)) * (*(two + i) - *(three + i));
    i++;
  }
  return sqrt(answer);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bucketSort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %esi,-0x4c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x3c(%rbp)
  jmp    L72
L64:
  mov    -0x3c(%rbp),%eax
  cltq
  movl   $0x0,-0x30(%rbp,%rax,4)
  addl   $0x1,-0x3c(%rbp)
L72:
  cmpl   $0x9,-0x3c(%rbp)
  jle    L64
  movl   $0x0,-0x3c(%rbp)
  jmp    L76
L9:
  mov    -0x3c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x30(%rbp,%rax,4)
  addl   $0x1,-0x3c(%rbp)
L76:
  mov    -0x3c(%rbp),%eax
  cmp    -0x4c(%rbp),%eax
  jl     L9
  movl   $0x0,-0x3c(%rbp)
  movl   $0x0,-0x38(%rbp)
  jmp    L43
L67:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,-0x34(%rbp)
  jmp    L52
L42:
  mov    -0x3c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x3c(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x48(%rbp),%rax
  add    %rax,%rdx
  mov    -0x38(%rbp),%eax
  mov    %eax,(%rdx)
  subl   $0x1,-0x34(%rbp)
L52:
  cmpl   $0x0,-0x34(%rbp)
  jg     L42
  addl   $0x1,-0x38(%rbp)
L43:
  cmpl   $0x9,-0x38(%rbp)
  jle    L67
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L40
  call   <__stack_chk_fail@plt>
L40:
  leave
  ret
```

assistant: ```c
void bucketSort(int a[], int n) {
  int i, j, k, buckets[10];
  for (i = 0; i < 10; ++i)
    buckets[i] = 0;
  for (i = 0; i < n; ++i)
    ++buckets[a[i]];
  for (i = 0, j = 0; j < 10; ++j)
    for (k = buckets[j]; k > 0; --k)
      a[i++] = j;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<bucketSort>:
  endbr64
  sub    $0x38,%rsp
  mov    %rdi,%r10
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rax
  lea    0x28(%rsp),%rdx
L46:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L46
  test   %esi,%esi
  jle    L69
  mov    %r10,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%r10,%rdx,4),%rcx
L66:
  movslq (%rax),%rdx
  addl   $0x1,(%rsp,%rdx,4)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L66
L69:
  mov    $0x0,%esi
  mov    $0x0,%r8d
  lea    0x4(%r10),%r11
  jmp    L12
L51:
  add    $0x1,%rsi
  cmp    $0xa,%rsi
  je     L54
L12:
  mov    %esi,%edx
  mov    (%rsp,%rsi,4),%edi
  test   %edi,%edi
  jle    L51
  movslq %r8d,%r9
  lea    (%r10,%r9,4),%rax
  lea    -0x1(%rdi),%ecx
  add    %r9,%rcx
  lea    (%r11,%rcx,4),%rcx
L42:
  mov    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L42
  lea    (%rdi,%r8,1),%r8d
  jmp    L51
L54:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0x38,%rsp
  ret
L43:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void bucketSort(int a[], int n) {
  int i, j, k, buckets[10];
  for (i = 0; i < 10; ++i)
    buckets[i] = 0;
  for (i = 0; i < n; ++i)
    ++buckets[a[i]];
  for (i = 0, j = 0; j < 10; ++j)
    for (k = buckets[j]; k > 0; --k)
      a[i++] = j;
}
```