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

```asm
<pthread_rwlock_wrlock>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nop
L36:
  mov    %rbp,%rdi
  call   <pthread_rwlock_trywrlock@plt>
  test   %eax,%eax
  je     L3
L8:
  cmp    $0x63,%ebx
  ja     L20
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <pthread_rwlock_trywrlock@plt>
  test   %eax,%eax
  jne    L8
L3:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L20:
  xor    %eax,%eax
  call   <sched_yield@plt>
  jmp    L36
```

assistant: ```c
int pthread_rwlock_wrlock(pthread_rwlock_t *rwlock) {
  unsigned int n;
  for (n = 0;;) {
    int r = pthread_rwlock_trywrlock(rwlock);
    if (r == 0)
      return r;
    if (n < 100) {
      ++n;
      continue;
    }
    sched_yield();
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<pthread_rwlock_wrlock>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nop
L32:
  mov    %rbp,%rdi
  call   <pthread_rwlock_trywrlock@plt>
  test   %eax,%eax
  je     L1
L30:
  cmp    $0x63,%ebx
  ja     L21
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <pthread_rwlock_trywrlock@plt>
  test   %eax,%eax
  jne    L30
L1:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L21:
  xor    %eax,%eax
  call   <sched_yield@plt>
  jmp    L32
```

assistant: ```c
int pthread_rwlock_wrlock(pthread_rwlock_t *rwlock) {
  unsigned int n;
  for (n = 0;;) {
    int r = pthread_rwlock_trywrlock(rwlock);
    if (r == 0)
      return r;
    if (n < 100) {
      ++n;
      continue;
    }
    sched_yield();
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<subString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x18(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *subString(int offset, int len, const char *input) {
  char *sub = malloc(sizeof(char) * len);
  int input_len = strlen(input);
  strncpy(sub, input + offset, len);
  return sub;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<subString>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%r13d
  mov    %rdx,%rbx
  mov    %esi,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rbx,%rdi
  call   <strlen@plt>
  movslq %ebp,%rbp
  lea    (%rbx,%rbp,1),%rsi
  mov    %r13d,%edx
  mov    %r12,%rdi
  call   <strncpy@plt>
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *subString(int offset, int len, const char *input) {
  char *sub = malloc(sizeof(char) * len);
  int input_len = strlen(input);
  strncpy(sub, input + offset, len);
  return sub;
}
```
user: Translate the below assembly code into C programming language:

```asm
<subString>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  mov    %esi,%edi
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rbx,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  lea    (%rbx,%rbp,1),%rsi
  mov    %r13d,%edx
  mov    %r12,%rdi
  call   <strncpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *subString(int offset, int len, const char *input) {
  char *sub = malloc(sizeof(char) * len);
  int input_len = strlen(input);
  strncpy(sub, input + offset, len);
  return sub;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<subString>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  mov    %esi,%edi
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rbx,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  lea    (%rbx,%rbp,1),%rsi
  mov    %r13d,%edx
  mov    %r12,%rdi
  call   <strncpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *subString(int offset, int len, const char *input) {
  char *sub = malloc(sizeof(char) * len);
  int input_len = strlen(input);
  strncpy(sub, input + offset, len);
  return sub;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<readlines>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L18
L39:
  cmpl   $0xa,-0x4(%rbp)
  jne    L19
  mov    -0x8(%rbp),%eax
  imul   -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0x8(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  je     L48
  movl   $0x0,-0xc(%rbp)
  jmp    L18
L19:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jge    L18
  mov    -0x8(%rbp),%eax
  imul   -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0xc(%rbp)
L18:
  mov    -0x8(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L62
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L39
  jmp    L62
L48:
  nop
L62:
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int readlines(char *lptr, int max_length, int max_hight) {
  int c;
  int x = 0;
  int y = 0;
  while (y < max_hight && (c = getchar()) != EOF) {
    if (c == '\n') {
      *(lptr + y * max_length + x) = 0;
      y++;
      if (x == 0)
        break;
      x = 0;
    } else if (x < max_length - 1) {
      *(lptr + y * max_length + x) = c;
      x++;
    }
  }
  return y;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<readlines>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r14
  mov    %esi,%r13d
  mov    %edx,%r12d
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  lea    -0x1(%rsi),%r15d
  test   %edx,%edx
  jg     L35
L53:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L65:
  mov    %r13d,%eax
  imul   %ebp,%eax
  cltq
  movslq %ebx,%rdx
  add    %r14,%rax
  movb   $0x0,(%rax,%rdx,1)
  add    $0x1,%ebp
  test   %ebx,%ebx
  je     L53
  mov    $0x0,%ebx
L43:
  cmp    %r12d,%ebp
  jge    L53
L35:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D8(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L53
  cmp    $0xa,%eax
  je     L65
  cmp    %ebx,%r15d
  jle    L43
  mov    %r13d,%edx
  imul   %ebp,%edx
  movslq %edx,%rdx
  movslq %ebx,%rcx
  add    %r14,%rdx
  mov    %al,(%rdx,%rcx,1)
  add    $0x1,%ebx
  jmp    L43
```

assistant: ```c
int readlines(char *lptr, int max_length, int max_hight) {
  int c;
  int x = 0;
  int y = 0;
  while (y < max_hight && (c = getchar()) != EOF) {
    if (c == '\n') {
      *(lptr + y * max_length + x) = 0;
      y++;
      if (x == 0)
        break;
      x = 0;
    } else if (x < max_length - 1) {
      *(lptr + y * max_length + x) = c;
      x++;
    }
  }
  return y;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<readlines>:
  endbr64
  push   %r15
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %edx,%edx
  jle    L10
  lea    -0x1(%rsi),%eax
  mov    D8(%rip),%rbp
  mov    %rdi,%r13
  mov    %esi,%r12d
  mov    %eax,0xc(%rsp)
  mov    %edx,%ebx
  xor    %r14d,%r14d
  xor    %r15d,%r15d
  jmp    L75
  nopw   0x0(%rax,%rax,1)
L37:
  cmp    %r15d,0xc(%rsp)
  jle    L36
  mov    %r12d,%edx
  movslq %r15d,%rcx
  add    $0x1,%r15d
  imul   %r14d,%edx
  movslq %edx,%rdx
  add    %r13,%rdx
  mov    %al,(%rdx,%rcx,1)
L36:
  cmp    %ebx,%r14d
  jge    L10
L75:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %eax,0x0(%rbp)
  je     L10
  cmp    $0xa,%eax
  jne    L37
  mov    %r12d,%eax
  movslq %r15d,%rdx
  imul   %r14d,%eax
  add    $0x1,%r14d
  cltq
  add    %r13,%rax
  movb   $0x0,(%rax,%rdx,1)
  test   %r15d,%r15d
  je     L10
  xor    %r15d,%r15d
  cmp    %ebx,%r14d
  jl     L75
L10:
  add    $0x18,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int readlines(char *lptr, int max_length, int max_hight) {
  int c;
  int x = 0;
  int y = 0;
  while (y < max_hight && (c = getchar()) != EOF) {
    if (c == '\n') {
      *(lptr + y * max_length + x) = 0;
      y++;
      if (x == 0)
        break;
      x = 0;
    } else if (x < max_length - 1) {
      *(lptr + y * max_length + x) = c;
      x++;
    }
  }
  return y;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<readlines>:
  endbr64
  push   %r15
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %edx,%edx
  jle    L48
  lea    -0x1(%rsi),%eax
  mov    D1(%rip),%rbp
  mov    %rdi,%r13
  mov    %esi,%r12d
  mov    %eax,0xc(%rsp)
  mov    %edx,%ebx
  xor    %r14d,%r14d
  xor    %r15d,%r15d
  jmp    L73
  nopw   0x0(%rax,%rax,1)
L47:
  cmp    %r15d,0xc(%rsp)
  jle    L11
  mov    %r12d,%edx
  movslq %r15d,%rcx
  add    $0x1,%r15d
  imul   %r14d,%edx
  movslq %edx,%rdx
  add    %r13,%rdx
  mov    %al,(%rdx,%rcx,1)
L11:
  cmp    %ebx,%r14d
  jge    L48
L73:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %eax,0x0(%rbp)
  je     L48
  cmp    $0xa,%eax
  jne    L47
  mov    %r12d,%eax
  movslq %r15d,%rdx
  imul   %r14d,%eax
  add    $0x1,%r14d
  cltq
  add    %r13,%rax
  movb   $0x0,(%rax,%rdx,1)
  test   %r15d,%r15d
  je     L48
  xor    %r15d,%r15d
  cmp    %ebx,%r14d
  jl     L73
L48:
  add    $0x18,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int readlines(char *lptr, int max_length, int max_hight) {
  int c;
  int x = 0;
  int y = 0;
  while (y < max_hight && (c = getchar()) != EOF) {
    if (c == '\n') {
      *(lptr + y * max_length + x) = 0;
      y++;
      if (x == 0)
        break;
      x = 0;
    } else if (x < max_length - 1) {
      *(lptr + y * max_length + x) = c;
      x++;
    }
  }
  return y;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<test_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D16(%rip),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ecx
  mov    -0x10(%rbp),%rax
  lea    -0x10(%rbp),%rdx
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D16
datatype: string
name: retrieve_data
arguments: label: D17
datatype: string

tool: D16: hellow world


tool: D17: %s
%p ,len = %zu


assistant: ```c
void test_string() {
  typedef char *sds;
  const sds s = "hellow world\n";
  printf("%s\n%p ,len = %zu\n", s, &s, strlen(s));
  free(s);
}
```
user: What is the c source code of the assembly code below?

```asm
<test_string>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D13(%rip),%rbx
  mov    %rbx,(%rsp)
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%ecx
  mov    %rsp,%rdx
  mov    %rbx,%rsi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x10,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D13
type: string
name: unpack_data
arguments: data_label: D12
type: string

tool: D13: hellow world


tool: D12: %s
%p ,len = %zu


assistant: ```c
void test_string() {
  typedef char *sds;
  const sds s = "hellow world\n";
  printf("%s\n%p ,len = %zu\n", s, &s, strlen(s));
  free(s);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<test_string>:
  endbr64
  push   %rbp
  lea    D17(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rbp,(%rsp)
  call   <strlen@plt>
  lea    D15(%rip),%rdi
  mov    %rsp,%rdx
  mov    %rbp,%rsi
  mov    %eax,%ecx
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D17
data_type: string
name: retrieve_data
arguments: data_label: D15
data_type: string

tool: D17: hellow world


tool: D15: %s
%p ,len = %zu


assistant: ```c
void test_string() {
  typedef char *sds;
  const sds s = "hellow world\n";
  printf("%s\n%p ,len = %zu\n", s, &s, strlen(s));
  free(s);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<test_string>:
  endbr64
  push   %rbp
  lea    D2(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rbp,(%rsp)
  call   <strlen@plt>
  lea    D5(%rip),%rdi
  mov    %rsp,%rdx
  mov    %rbp,%rsi
  mov    %eax,%ecx
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x10,%rsp
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D2
datatype: string
name: disassemble_data
arguments: label: D5
datatype: string

tool: D2: hellow world


tool: D5: %s
%p ,len = %zu


assistant: ```c
void test_string() {
  typedef char *sds;
  const sds s = "hellow world\n";
  printf("%s\n%p ,len = %zu\n", s, &s, strlen(s));
  free(s);
}
```
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    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x24(%rbp)
  mov    -0x30(%rbp),%eax
  mov    -0x24(%rbp),%edx
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x1,-0x2c(%rbp)
  mov    -0x30(%rbp),%eax
  mov    %eax,-0x28(%rbp)
  jmp    L20
L2:
  mov    -0x2c(%rbp),%eax
  imul   -0x28(%rbp),%eax
  mov    %eax,-0x2c(%rbp)
  mov    -0x24(%rbp),%eax
  sub    %eax,-0x28(%rbp)
L20:
  cmpl   $0x0,-0x28(%rbp)
  jg     L2
  mov    -0x2c(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D29
datatype: string
name: read_data
arguments: data_label: D20
datatype: string
name: read_data
arguments: data_label: D7
datatype: string

tool: D29: %d %s

tool: D20: n is %d, k is %d


tool: D7: %d


assistant: ```c
int main() {
  int n;
  char str[21];
  scanf("%d %s", &n, str);
  int k = strlen(str);
  printf("n is %d, k is %d\n", n, k);
  int result = 1;
  for (int i = n; i > 0; i -= k) {
    result *= i;
  }
  printf("%d\n", result);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  lea    0xc(%rsp),%rsi
  mov    %rbx,%rdx
  lea    D0(%rip),%rdi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %eax,%edx
  mov    0xc(%rsp),%esi
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L18
  mov    $0x1,%esi
L17:
  imul   %eax,%esi
  sub    %ebx,%eax
  test   %eax,%eax
  jg     L17
L36:
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  ret
L18:
  mov    $0x1,%esi
  jmp    L36
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D0
data_type: string
name: disassemble_data
arguments: data_label: D14
data_type: string
name: disassemble_data
arguments: data_label: D3
data_type: string

tool: D0: %d %s

tool: D14: n is %d, k is %d


tool: D3: %d


assistant: ```c
int main() {
  int n;
  char str[21];
  scanf("%d %s", &n, str);
  int k = strlen(str);
  printf("n is %d, k is %d\n", n, k);
  int result = 1;
  for (int i = n; i > 0; i -= k) {
    result *= i;
  }
  printf("%d\n", result);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D16(%rip),%rdi
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdx
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    0xc(%rsp),%esi
  lea    D12(%rip),%rdi
  mov    %eax,%edx
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  mov    $0x1,%esi
  test   %eax,%eax
  jle    L9
  xchg   %ax,%ax
L7:
  imul   %eax,%esi
  sub    %ebx,%eax
  test   %eax,%eax
  jg     L7
L9:
  xor    %eax,%eax
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D16
data_type: string
name: extract_symbolic_data
arguments: tag: D12
data_type: string
name: extract_symbolic_data
arguments: tag: D19
data_type: string

tool: D16: %d %s

tool: D12: n is %d, k is %d


tool: D19: %d


assistant: ```c
int main() {
  int n;
  char str[21];
  scanf("%d %s", &n, str);
  int k = strlen(str);
  printf("n is %d, k is %d\n", n, k);
  int result = 1;
  for (int i = n; i > 0; i -= k) {
    result *= i;
  }
  printf("%d\n", result);
  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
  lea    D19(%rip),%rdi
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdx
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    0xc(%rsp),%esi
  lea    D1(%rip),%rdi
  mov    %eax,%edx
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  mov    $0x1,%esi
  test   %eax,%eax
  jle    L4
  xchg   %ax,%ax
L17:
  imul   %eax,%esi
  sub    %ebx,%eax
  test   %eax,%eax
  jg     L17
L4:
  xor    %eax,%eax
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D19
type: string
name: unpack_data
arguments: label: D1
type: string
name: unpack_data
arguments: label: D7
type: string

tool: D19: %d %s

tool: D1: n is %d, k is %d


tool: D7: %d


assistant: ```c
int main() {
  int n;
  char str[21];
  scanf("%d %s", &n, str);
  int k = strlen(str);
  printf("n is %d, k is %d\n", n, k);
  int result = 1;
  for (int i = n; i > 0; i -= k) {
    result *= i;
  }
  printf("%d\n", result);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<findDuplicates>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  movl   $0x0,(%rax)
  movl   $0x0,-0xc(%rbp)
  jmp    L32
L16:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jns    L11
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x28(%rbp),%rdx
  mov    (%rdx),%edx
  lea    0x1(%rdx),%esi
  mov    -0x28(%rbp),%rcx
  mov    %esi,(%rcx)
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %eax,%ecx
  neg    %ecx
  cmovns %ecx,%eax
  mov    %eax,(%rdx)
  jmp    L35
L11:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  mov    %eax,%ecx
  neg    %ecx
  cmovns %ecx,%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  neg    %edx
  mov    %edx,(%rax)
L35:
  addl   $0x1,-0xc(%rbp)
L32:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L16
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *findDuplicates(int *nums, int numsSize, int *returnSize) {
  int *numsReturn;
  int i;
  numsReturn = (int *)malloc(sizeof(int) * numsSize);
  *returnSize = 0;
  for (i = 0; i < numsSize; i++) {
    if (nums[abs(nums[i]) - 1] < 0) {
      numsReturn[(*returnSize)++] = abs(nums[i]);
    } else {
      nums[abs(nums[i]) - 1] *= -1;
    }
  }
  return numsReturn;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<findDuplicates>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%r12d
  mov    %rdx,%rbp
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  mov    %rax,%r8
  movl   $0x0,0x0(%rbp)
  test   %r12d,%r12d
  jle    L25
  mov    %rbx,%rdx
  lea    -0x1(%r12),%eax
  lea    0x4(%rbx,%rax,4),%rdi
  jmp    L39
L3:
  neg    %ecx
  mov    %ecx,(%rsi)
L35:
  add    $0x4,%rdx
  cmp    %rdi,%rdx
  je     L25
L39:
  mov    (%rdx),%ecx
  mov    %ecx,%eax
  neg    %eax
  cmovs  %ecx,%eax
  movslq %eax,%rcx
  lea    -0x4(%rbx,%rcx,4),%rsi
  mov    (%rsi),%ecx
  test   %ecx,%ecx
  jns    L3
  mov    0x0(%rbp),%ecx
  lea    0x1(%rcx),%esi
  mov    %esi,0x0(%rbp)
  movslq %ecx,%rcx
  mov    %eax,(%r8,%rcx,4)
  jmp    L35
L25:
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int *findDuplicates(int *nums, int numsSize, int *returnSize) {
  int *numsReturn;
  int i;
  numsReturn = (int *)malloc(sizeof(int) * numsSize);
  *returnSize = 0;
  for (i = 0; i < numsSize; i++) {
    if (nums[abs(nums[i]) - 1] < 0) {
      numsReturn[(*returnSize)++] = abs(nums[i]);
    } else {
      nums[abs(nums[i]) - 1] *= -1;
    }
  }
  return numsReturn;
}
```