ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x54(%rbp)
  jmp    L42
L36:
  lea    -0x50(%rbp),%rcx
  mov    -0x54(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rcx,%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x54(%rbp)
L42:
  cmpl   $0x9,-0x54(%rbp)
  jle    L36
  lea    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <sort@plt>
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x54(%rbp)
  jmp    L33
L23:
  lea    -0x50(%rbp),%rcx
  mov    -0x54(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rcx,%rax
  mov    %rax,%rdi
  call   <puts@plt>
  addl   $0x1,-0x54(%rbp)
L33:
  cmpl   $0x9,-0x54(%rbp)
  jle    L23
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L46
  call   <__stack_chk_fail@plt>
L46:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D28
datatype: string
name: unpack_data
arguments: label: D10
datatype: string
name: unpack_data
arguments: label: D11
datatype: string

tool: D28: input 10 strings with same length:

tool: D10: %s

tool: D11: new strings are:

assistant: ```c
int main() {
  void sort(char s[10][20]);
  int i;
  char str[10][6];
  printf("input 10 strings with same length:\n");
  for (i = 0; i < 10; i++)
    scanf("%s", str[i]);
  sort(str);
  printf("new strings are:\n");
  for (i = 0; i < 10; i++)
    printf("%s\n", str[i]);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    D17(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rbx
  lea    0x3c(%rsp),%r12
  mov    %rbx,%rbp
  lea    D29(%rip),%r13
L26:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x6,%rbp
  cmp    %r12,%rbp
  jne    L26
  mov    %rsp,%rdi
  call   <sort@plt>
  lea    D1(%rip),%rdi
  call   <puts@plt>
L27:
  mov    %rbx,%rdi
  call   <puts@plt>
  add    $0x6,%rbx
  cmp    %r12,%rbx
  jne    L27
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  mov    $0x0,%eax
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D17
datatype: string
name: unpack_data
arguments: tag: D29
datatype: string
name: unpack_data
arguments: tag: D1
datatype: string

tool: D17: input 10 strings with same length:

tool: D29: %s

tool: D1: new strings are:

assistant: ```c
int main() {
  void sort(char s[10][20]);
  int i;
  char str[10][6];
  printf("input 10 strings with same length:\n");
  for (i = 0; i < 10; i++)
    scanf("%s", str[i]);
  sort(str);
  printf("new strings are:\n");
  for (i = 0; i < 10; i++)
    printf("%s\n", str[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   %r14
  lea    D20(%rip),%rdi
  push   %r13
  lea    D16(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  lea    0x3c(%rsp),%r12
  call   <puts@plt>
  mov    %r14,%rbx
  mov    %r14,%rbp
  nopl   0x0(%rax)
L22:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x6,%rbp
  call   <scanf@plt>
  cmp    %r12,%rbp
  jne    L22
  mov    %r14,%rdi
  call   <sort@plt>
  lea    D17(%rip),%rdi
  call   <puts@plt>
L16:
  mov    %rbx,%rdi
  add    $0x6,%rbx
  call   <puts@plt>
  cmp    %r12,%rbx
  jne    L16
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x50,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D20
type_infered: string
name: retrieve_data
arguments: data_label: D16
type_infered: string
name: retrieve_data
arguments: data_label: D17
type_infered: string

tool: D20: input 10 strings with same length:

tool: D16: %s

tool: D17: new strings are:

assistant: ```c
int main() {
  void sort(char s[10][20]);
  int i;
  char str[10][6];
  printf("input 10 strings with same length:\n");
  for (i = 0; i < 10; i++)
    scanf("%s", str[i]);
  sort(str);
  printf("new strings are:\n");
  for (i = 0; i < 10; i++)
    printf("%s\n", str[i]);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r14
  lea    D20(%rip),%rdi
  push   %r13
  lea    D5(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  lea    0x3c(%rsp),%r12
  call   <puts@plt>
  mov    %r14,%rbx
  mov    %r14,%rbp
  nopl   0x0(%rax)
L19:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x6,%rbp
  call   <scanf@plt>
  cmp    %r12,%rbp
  jne    L19
  mov    %r14,%rdi
  call   <sort@plt>
  lea    D3(%rip),%rdi
  call   <puts@plt>
L25:
  mov    %rbx,%rdi
  add    $0x6,%rbx
  call   <puts@plt>
  cmp    %r12,%rbx
  jne    L25
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x50,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D20
datatype: string
name: read_data
arguments: label: D5
datatype: string
name: read_data
arguments: label: D3
datatype: string

tool: D20: input 10 strings with same length:

tool: D5: %s

tool: D3: new strings are:

assistant: ```c
int main() {
  void sort(char s[10][20]);
  int i;
  char str[10][6];
  printf("input 10 strings with same length:\n");
  for (i = 0; i < 10; i++)
    scanf("%s", str[i]);
  sort(str);
  printf("new strings are:\n");
  for (i = 0; i < 10; i++)
    printf("%s\n", str[i]);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<quoted>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  mov    %rdi,%r12
  test   %r12,%r12
  jne    L5
  lea    D7(%rip),%r12
L5:
  mov    %r12,%rdi
  call   <strlen@plt>
  shl    $0x2,%rax
  add    $0x3,%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <imalloc@plt>
  cltq
  mov    %rax,%r14
  test   %r14,%r14
  jne    L16
  mov    $0x0,%eax
  jmp    L63
L16:
  mov    %r14,%rbx
  mov    %rbx,%rax
  lea    0x1(%rax),%rbx
  movb   $0x27,(%rax)
  jmp    L1
L9:
  cmp    $0x27,%r13d
  jne    L19
  mov    %rbx,%rax
  lea    0x1(%rax),%rbx
  mov    %r13d,%edx
  mov    %dl,(%rax)
  mov    %rbx,%rax
  lea    0x1(%rax),%rbx
  movb   $0x5c,(%rax)
  mov    %rbx,%rax
  lea    0x1(%rax),%rbx
  mov    %r13d,%edx
  mov    %dl,(%rax)
  mov    %rbx,%rax
  lea    0x1(%rax),%rbx
  mov    %r13d,%edx
  mov    %dl,(%rax)
  jmp    L1
L19:
  mov    %rbx,%rax
  lea    0x1(%rax),%rbx
  mov    %r13d,%edx
  mov    %dl,(%rax)
L1:
  mov    %r12,%rax
  lea    0x1(%rax),%r12
  movzbl (%rax),%eax
  movsbl %al,%r13d
  test   %r13d,%r13d
  jne    L9
  mov    %rbx,%rax
  lea    0x1(%rax),%rbx
  movb   $0x27,(%rax)
  movb   $0x0,(%rbx)
  mov    %r14,%rax
L63:
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D7
type: string

tool: 

assistant: ```c
char *quoted(name)
register const char *name;
{
  register char *result;
  register char *cp;
  register int c;
  if (name == ((void *)0))
    name = "";
  result = imalloc(4 * strlen(name) + 3);
  if (result == ((void *)0))
    return ((void *)0);
  cp = result;
  *cp++ = '\'';
  while ((c = *name++) != '\0')
    if (c == '\'') {
      *cp++ = c;
      *cp++ = '\\';
      *cp++ = c;
      *cp++ = c;
    } else
      *cp++ = c;
  *cp++ = '\'';
  *cp = '\0';
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<quoted>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L74
  call   <strlen@plt>
  lea    0x3(,%rax,4),%rdi
  mov    $0x0,%eax
  call   <imalloc@plt>
  movslq %eax,%rdi
  test   %eax,%eax
  je     L60
  lea    0x1(%rdi),%rax
  movb   $0x27,(%rdi)
  lea    0x1(%rbx),%rsi
  movzbl (%rbx),%edx
  movsbl %dl,%ecx
  test   %ecx,%ecx
  jne    L79
  jmp    L80
L25:
  mov    %dl,(%rax)
  lea    0x1(%rax),%rax
L38:
  add    $0x1,%rsi
  movzbl -0x1(%rsi),%edx
  movsbl %dl,%ecx
  test   %ecx,%ecx
  je     L80
L79:
  cmp    $0x27,%ecx
  jne    L25
  mov    %dl,(%rax)
  movb   $0x5c,0x1(%rax)
  mov    %dl,0x2(%rax)
  mov    %dl,0x3(%rax)
  lea    0x4(%rax),%rax
  jmp    L38
L60:
  mov    $0x0,%edi
  jmp    L71
L74:
  mov    $0x3,%edi
  mov    $0x0,%eax
  call   <imalloc@plt>
  mov    %rbx,%rdi
  test   %eax,%eax
  je     L71
  movslq %eax,%rdi
  lea    0x1(%rdi),%rax
  movb   $0x27,(%rdi)
L80:
  movb   $0x27,(%rax)
  movb   $0x0,0x1(%rax)
L71:
  mov    %rdi,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *quoted(name)
register const char *name;
{
  register char *result;
  register char *cp;
  register int c;
  if (name == ((void *)0))
    name = "";
  result = imalloc(4 * strlen(name) + 3);
  if (result == ((void *)0))
    return ((void *)0);
  cp = result;
  *cp++ = '\'';
  while ((c = *name++) != '\0')
    if (c == '\'') {
      *cp++ = c;
      *cp++ = '\\';
      *cp++ = c;
      *cp++ = c;
    } else
      *cp++ = c;
  *cp++ = '\'';
  *cp = '\0';
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<quoted>:
  endbr64
  push   %rbx
  test   %rdi,%rdi
  je     L57
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x3(,%rax,4),%rdi
  xor    %eax,%eax
  call   <imalloc@plt>
  movslq %eax,%r8
  test   %eax,%eax
  je     L20
  movb   $0x27,(%r8)
  movsbl (%rbx),%edx
  lea    0x1(%r8),%rax
  lea    0x1(%rbx),%rcx
  mov    %edx,%esi
  test   %edx,%edx
  jne    L31
  jmp    L49
  nop
L78:
  add    $0x1,%rcx
  mov    %sil,(%rax)
  add    $0x1,%rax
  movsbl -0x1(%rcx),%edx
  mov    %edx,%esi
  test   %edx,%edx
  je     L49
L31:
  cmp    $0x27,%edx
  jne    L78
  add    $0x1,%rcx
  movl   $0x27275c27,(%rax)
  add    $0x4,%rax
  movsbl -0x1(%rcx),%edx
  mov    %edx,%esi
  test   %edx,%edx
  jne    L31
L49:
  mov    $0x27,%edx
  mov    %dx,(%rax)
  mov    %r8,%rax
  pop    %rbx
  ret
  nop
L57:
  xor    %eax,%eax
  mov    $0x3,%edi
  call   <imalloc@plt>
  test   %eax,%eax
  jne    L76
L20:
  xor    %r8d,%r8d
  pop    %rbx
  mov    %r8,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L76:
  movslq %eax,%r8
  movb   $0x27,(%r8)
  lea    0x1(%r8),%rax
  jmp    L49
```

assistant: ```c
char *quoted(name)
register const char *name;
{
  register char *result;
  register char *cp;
  register int c;
  if (name == ((void *)0))
    name = "";
  result = imalloc(4 * strlen(name) + 3);
  if (result == ((void *)0))
    return ((void *)0);
  cp = result;
  *cp++ = '\'';
  while ((c = *name++) != '\0')
    if (c == '\'') {
      *cp++ = c;
      *cp++ = '\\';
      *cp++ = c;
      *cp++ = c;
    } else
      *cp++ = c;
  *cp++ = '\'';
  *cp = '\0';
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<quoted>:
  endbr64
  push   %rbx
  test   %rdi,%rdi
  je     L71
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x3(,%rax,4),%rdi
  xor    %eax,%eax
  call   <imalloc@plt>
  movslq %eax,%r8
  test   %eax,%eax
  je     L57
  movb   $0x27,(%r8)
  movsbl (%rbx),%eax
  lea    0x1(%r8),%rcx
  lea    0x1(%rbx),%rdx
  mov    %eax,%esi
  test   %eax,%eax
  jne    L39
  jmp    L63
  nop
L87:
  add    $0x1,%rdx
  mov    %sil,(%rcx)
  lea    0x1(%rcx),%rdi
  movsbl -0x1(%rdx),%eax
  mov    %eax,%esi
  test   %eax,%eax
  je     L4
L59:
  mov    %rdi,%rcx
L39:
  cmp    $0x27,%eax
  jne    L87
  add    $0x1,%rdx
  movl   $0x27275c27,(%rcx)
  lea    0x4(%rcx),%rdi
  movsbl -0x1(%rdx),%eax
  mov    %eax,%esi
  test   %eax,%eax
  jne    L59
L4:
  mov    $0x27,%eax
  mov    %ax,(%rdi)
  mov    %r8,%rax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L71:
  xor    %eax,%eax
  mov    $0x3,%edi
  call   <imalloc@plt>
  test   %eax,%eax
  jne    L18
L57:
  xor    %r8d,%r8d
  pop    %rbx
  mov    %r8,%rax
  ret
L18:
  movslq %eax,%r8
  movb   $0x27,(%r8)
  lea    0x1(%r8),%rdi
  jmp    L4
L63:
  mov    %rcx,%rdi
  jmp    L4
```

assistant: ```c
char *quoted(name)
register const char *name;
{
  register char *result;
  register char *cp;
  register int c;
  if (name == ((void *)0))
    name = "";
  result = imalloc(4 * strlen(name) + 3);
  if (result == ((void *)0))
    return ((void *)0);
  cp = result;
  *cp++ = '\'';
  while ((c = *name++) != '\0')
    if (c == '\'') {
      *cp++ = c;
      *cp++ = '\\';
      *cp++ = c;
      *cp++ = c;
    } else
      *cp++ = c;
  *cp++ = '\'';
  *cp = '\0';
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L16
L10:
  mov    $0x0,%eax
  call   <sched_yield@plt>
  addl   $0x1,-0x4(%rbp)
L16:
  cmpl   $0x0,-0x4(%rbp)
  jle    L10
  nop
  nop
  leave
  ret
```

assistant: ```c
void foo(void) {
#pragma omp target
#pragma omp parallel for
  for (int i = 0; i < 1; i++)
    sched_yield();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<foo>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <sched_yield@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void foo(void) {
#pragma omp target
#pragma omp parallel for
  for (int i = 0; i < 1; i++)
    sched_yield();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<foo>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <sched_yield@plt>
```

assistant: ```c
void foo(void) {
#pragma omp target
#pragma omp parallel for
  for (int i = 0; i < 1; i++)
    sched_yield();
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<foo>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <sched_yield@plt>
```

assistant: ```c
void foo(void) {
#pragma omp target
#pragma omp parallel for
  for (int i = 0; i < 1; i++)
    sched_yield();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L8
L18:
  mov    -0x4(%rbp),%eax
  sub    -0x8(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L8:
  mov    -0x8(%rbp),%eax
  sub    -0x4(%rbp),%eax
  test   %eax,%eax
  jg     L18
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int x = __VERIFIER_nondet_int();
  int y = __VERIFIER_nondet_int();
  while (x - y > 0) {
    x = y - x;
    y++;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  cmp    %eax,%ebx
  jle    L10
L0:
  mov    %eax,%edx
  sub    %ebx,%edx
  mov    %edx,%ebx
  add    $0x1,%eax
  cmp    %eax,%edx
  jg     L0
L10:
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int x = __VERIFIER_nondet_int();
  int y = __VERIFIER_nondet_int();
  while (x - y > 0) {
    x = y - x;
    y++;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x = __VERIFIER_nondet_int();
  int y = __VERIFIER_nondet_int();
  while (x - y > 0) {
    x = y - x;
    y++;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x = __VERIFIER_nondet_int();
  int y = __VERIFIER_nondet_int();
  while (x - y > 0) {
    x = y - x;
    y++;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<VARZCurrentDoubleTime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x18(%rbp),%rax
  pxor   %xmm1,%xmm1
  cvtsi2sd %rax,%xmm1
  mov    -0x20(%rbp),%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movsd  D4(%rip),%xmm2
  divsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D4
type: f64

tool: D4: 1000000

assistant: ```c
double VARZCurrentDoubleTime() {
  struct timeval tp;
  gettimeofday(&tp, NULL);
  return ((double)tp.tv_sec) + ((double)tp.tv_usec / 1.0e6);
}
```
user: Translate the below assembly code into C programming language:

```asm
<VARZCurrentDoubleTime>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sdq (%rsp),%xmm0
  divsd  D7(%rip),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sdq 0x8(%rsp),%xmm1
  addsd  %xmm1,%xmm0
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D7
type_infered: f64

tool: D7: 1000000

assistant: ```c
double VARZCurrentDoubleTime() {
  struct timeval tp;
  gettimeofday(&tp, NULL);
  return ((double)tp.tv_sec) + ((double)tp.tv_usec / 1.0e6);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<VARZCurrentDoubleTime>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  pxor   %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sdq 0x8(%rsp),%xmm1
  cvtsi2sdq (%rsp),%xmm0
  divsd  D4(%rip),%xmm0
  addsd  %xmm1,%xmm0
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x28,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D4
data_type: f64

tool: D4: 1000000

assistant: ```c
double VARZCurrentDoubleTime() {
  struct timeval tp;
  gettimeofday(&tp, NULL);
  return ((double)tp.tv_sec) + ((double)tp.tv_usec / 1.0e6);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<VARZCurrentDoubleTime>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  pxor   %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sdq 0x8(%rsp),%xmm1
  cvtsi2sdq (%rsp),%xmm0
  divsd  D5(%rip),%xmm0
  addsd  %xmm1,%xmm0
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D5
datatype: f64

tool: D5: 1000000

assistant: ```c
double VARZCurrentDoubleTime() {
  struct timeval tp;
  gettimeofday(&tp, NULL);
  return ((double)tp.tv_sec) + ((double)tp.tv_usec / 1.0e6);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<sstrdup>:
  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    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *sstrdup(char *s) {
  char *copy = malloc(strlen(s) + 1);
  strcpy(copy, s);
  return copy;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sstrdup>:
  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    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

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

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

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

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

assistant: ```c
char *sstrdup(char *s) {
  char *copy = malloc(strlen(s) + 1);
  strcpy(copy, s);
  return copy;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<destroy_image>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L5
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L5:
  nop
  leave
  ret
```

assistant: ```c
void destroy_image(unsigned char *pimg) {
  if (pimg != NULL) {
    free(pimg);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<destroy_image>:
  endbr64
  test   %rdi,%rdi
  je     L1
  sub    $0x8,%rsp
  call   <free@plt>
  add    $0x8,%rsp
  ret
L1:
  ret
```

assistant: ```c
void destroy_image(unsigned char *pimg) {
  if (pimg != NULL) {
    free(pimg);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

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

assistant: ```c
void destroy_image(unsigned char *pimg) {
  if (pimg != NULL) {
    free(pimg);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<destroy_image>:
  endbr64
  test   %rdi,%rdi
  je     L8
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L8:
  ret
```

assistant: ```c
void destroy_image(unsigned char *pimg) {
  if (pimg != NULL) {
    free(pimg);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_scan_pattern>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
L23:
  lea    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <_copy_string@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L26
  mov    $0x0,%eax
  jmp    L10
L26:
  addl   $0x1,-0x4(%rbp)
  lea    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <_copy_string@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L23
  mov    -0x4(%rbp),%eax
L10:
  leave
  ret
```

assistant: ```c
int _scan_pattern(const char *format, char *temp) {
  int n = 0;
  for (;;) {
    format = _copy_string(format, &temp);
    if (format[0] == '\0')
      return 0;
    ++n;
    format = _copy_string(format, &temp);
    if (format[0] == '\0')
      return n;
  }
}
```