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

```asm
<string>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    D11(%rip),%rbx
  test   %rbx,%rbx
  je     L17
L7:
  lea    0x8(%rbx),%rbp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L9
  mov    (%rbx),%rbx
  test   %rbx,%rbx
  jne    L7
L17:
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x10(%rax),%rdi
  mov    $0x0,%eax
  call   <alloc@plt>
  movslq %eax,%rbx
  lea    0x8(%rbx),%rbp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    D11(%rip),%rax
  mov    %rax,(%rbx)
  mov    %rbx,D18(%rip)
L9:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *string(const char *str) {
  static struct string {
    struct string *link;
    char str[1];
  } * list;
  struct string *p;
  for (p = list; p; p = p->link)
    if (strcmp(p->str, str) == 0)
      return p->str;
  p = alloc(strlen(str) + sizeof *p);
  strcpy(p->str, str);
  p->link = list;
  list = p;
  return p->str;
}
```
user: Translate the below assembly code into C programming language:

```asm
<string>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    D23(%rip),%rbx
  test   %rbx,%rbx
  jne    L7
  jmp    L5
  nopl   0x0(%rax)
L23:
  mov    (%rbx),%rbx
  test   %rbx,%rbx
  je     L5
L7:
  lea    0x8(%rbx),%r12
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L23
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x10(%rax),%rdi
  xor    %eax,%eax
  call   <alloc@plt>
  mov    %rbp,%rsi
  movslq %eax,%rbx
  lea    0x8(%rbx),%r12
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    D23(%rip),%rax
  mov    %rbx,D7(%rip)
  mov    %rax,(%rbx)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *string(const char *str) {
  static struct string {
    struct string *link;
    char str[1];
  } * list;
  struct string *p;
  for (p = list; p; p = p->link)
    if (strcmp(p->str, str) == 0)
      return p->str;
  p = alloc(strlen(str) + sizeof *p);
  strcpy(p->str, str);
  p->link = list;
  list = p;
  return p->str;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<string>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    D6(%rip),%rbx
  test   %rbx,%rbx
  jne    L29
  jmp    L12
  nopl   0x0(%rax)
L28:
  mov    (%rbx),%rbx
  test   %rbx,%rbx
  je     L12
L29:
  lea    0x8(%rbx),%r12
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L28
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L12:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x10(%rax),%rdi
  xor    %eax,%eax
  call   <alloc@plt>
  mov    %rbp,%rsi
  movslq %eax,%rbx
  lea    0x8(%rbx),%r12
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    D6(%rip),%rax
  mov    %rbx,D0(%rip)
  mov    %rax,(%rbx)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *string(const char *str) {
  static struct string {
    struct string *link;
    char str[1];
  } * list;
  struct string *p;
  for (p = list; p; p = p->link)
    if (strcmp(p->str, str) == 0)
      return p->str;
  p = alloc(strlen(str) + sizeof *p);
  strcpy(p->str, str);
  p->link = list;
  list = p;
  return p->str;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %rax,-0x8(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L2
L4:
  mov    $0x0,%eax
  call   <non_short_circuit@plt>
  addq   $0x1,-0x10(%rbp)
L2:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jl     L4
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  long i, n;
  n = atoi(argv[1]);
  for (i = 0; i < n; i++)
    non_short_circuit();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  test   %rax,%rax
  jle    L4
  mov    %rax,%rbp
  mov    $0x0,%ebx
L15:
  mov    $0x0,%eax
  call   <non_short_circuit@plt>
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L15
L4:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  long i, n;
  n = atoi(argv[1]);
  for (i = 0; i < n; i++)
    non_short_circuit();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  test   %rax,%rax
  jle    L9
  mov    %rax,%rbp
  xor    %ebx,%ebx
  nopl   (%rax)
L3:
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <non_short_circuit@plt>
  cmp    %rbx,%rbp
  jne    L3
L9:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  long i, n;
  n = atoi(argv[1]);
  for (i = 0; i < n; i++)
    non_short_circuit();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  test   %rax,%rax
  jle    L4
  mov    %rax,%rbp
  xor    %ebx,%ebx
  nopl   (%rax)
L6:
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <non_short_circuit@plt>
  cmp    %rbx,%rbp
  jne    L6
L4:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  long i, n;
  n = atoi(argv[1]);
  for (i = 0; i < n; i++)
    non_short_circuit();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<isroot>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %esi,-0x8(%rbp)
  mov    %ax,-0x4(%rbp)
  cmpw   $0x1,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
int isroot(x, y)
short x;
unsigned int y;
{ return x == 1; }
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<isroot>:
  endbr64
  cmp    $0x1,%di
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int isroot(x, y)
short x;
unsigned int y;
{ return x == 1; }
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<isroot>:
  endbr64
  xor    %eax,%eax
  cmp    $0x1,%di
  sete   %al
  ret
```

assistant: ```c
int isroot(x, y)
short x;
unsigned int y;
{ return x == 1; }
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<isroot>:
  endbr64
  xor    %eax,%eax
  cmp    $0x1,%di
  sete   %al
  ret
```

assistant: ```c
int isroot(x, y)
short x;
unsigned int y;
{ return x == 1; }
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<square1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%edx
  mov    -0xc(%rbp),%eax
  imul   %eax,%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D17
type_infered: string
name: retrieve_data
arguments: label: D6
type_infered: string
name: retrieve_data
arguments: label: D24
type_infered: string

tool: D17: Please enter an integer > 

tool: D6: %d

tool: D24: The square of %d is %d


assistant: ```c
void square1(void) {
  int x;
  printf("Please enter an integer > ");
  scanf("%d", &x);
  printf("The square of %d is %d\n", x, x * x);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<square1>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D26(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D27(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  mov    %esi,%edx
  imul   %esi,%edx
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D26
type_guessed: string
name: disassemble_data
arguments: data_tag: D27
type_guessed: string
name: disassemble_data
arguments: data_tag: D20
type_guessed: string

tool: D26: Please enter an integer > 

tool: D27: %d

tool: D20: The square of %d is %d


assistant: ```c
void square1(void) {
  int x;
  printf("Please enter an integer > ");
  scanf("%d", &x);
  printf("The square of %d is %d\n", x, x * x);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<square1>:
  endbr64
  sub    $0x18,%rsp
  lea    D4(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D20(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  xor    %eax,%eax
  lea    D23(%rip),%rdi
  mov    %esi,%edx
  imul   %esi,%edx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D4
type_guessed: string
name: unpack_data
arguments: data_label: D20
type_guessed: string
name: unpack_data
arguments: data_label: D23
type_guessed: string

tool: D4: Please enter an integer > 

tool: D20: %d

tool: D23: The square of %d is %d


assistant: ```c
void square1(void) {
  int x;
  printf("Please enter an integer > ");
  scanf("%d", &x);
  printf("The square of %d is %d\n", x, x * x);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<square1>:
  endbr64
  sub    $0x18,%rsp
  lea    D5(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D8(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  xor    %eax,%eax
  lea    D7(%rip),%rdi
  mov    %esi,%edx
  imul   %esi,%edx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D5
type_guessed: string
name: read_data
arguments: data_label: D8
type_guessed: string
name: read_data
arguments: data_label: D7
type_guessed: string

tool: D5: Please enter an integer > 

tool: D8: %d

tool: D7: The square of %d is %d


assistant: ```c
void square1(void) {
  int x;
  printf("Please enter an integer > ");
  scanf("%d", &x);
  printf("The square of %d is %d\n", x, x * x);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x18(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x14(%rbp)
  jmp    L74
L36:
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x1,-0x18(%rbp)
  movl   $0x1,-0x10(%rbp)
  jmp    L25
L79:
  mov    -0x10(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  jne    L42
  shll   -0x18(%rbp)
  jmp    L52
L42:
  mov    -0x10(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L52
  addl   $0x1,-0x18(%rbp)
L52:
  addl   $0x1,-0x10(%rbp)
L25:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L79
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L74:
  mov    -0x20(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L36
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L46
  call   <__stack_chk_fail@plt>
L46:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D0
type_guessed: string
name: read_data
arguments: tag: D5
type_guessed: string

tool: D0: %d

tool: D5: %d


assistant: ```c
int main() {
  int t, N, H = 1;
  scanf("%d", &t);
  for (int a0 = 0; a0 < t; a0++) {
    int n;
    scanf("%d", &n);
    N = n;
    H = 1;
    for (int i = 1; i <= N; i++) {
      if (i % 2 == 1)
        H = 2 * H;
      else if (i % 2 == 0)
        H = H + 1;
    }
    printf("%d\n", H);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,(%rsp)
  jle    L71
  mov    $0x0,%ebx
  lea    0x4(%rsp),%r13
  lea    D16(%rip),%r12
  lea    D12(%rip),%rbp
  jmp    L76
L2:
  mov    %edx,%eax
  and    $0x1,%eax
  cmp    $0x1,%eax
  adc    $0x0,%esi
L24:
  add    $0x1,%edx
  cmp    %edx,%edi
  je     L27
L69:
  mov    %edx,%ecx
  shr    $0x1f,%ecx
  lea    (%rdx,%rcx,1),%eax
  and    $0x1,%eax
  sub    %ecx,%eax
  cmp    $0x1,%eax
  jne    L2
  add    %esi,%esi
  jmp    L24
L37:
  mov    $0x1,%esi
L27:
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,(%rsp)
  jle    L71
L76:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  test   %edi,%edi
  jle    L37
  add    $0x1,%edi
  mov    $0x1,%edx
  mov    $0x1,%esi
  jmp    L69
L71:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D16
type_guessed: string
name: disassemble_data
arguments: data_label: D12
type_guessed: string

tool: D16: %d

tool: D12: %d


assistant: ```c
int main() {
  int t, N, H = 1;
  scanf("%d", &t);
  for (int a0 = 0; a0 < t; a0++) {
    int n;
    scanf("%d", &n);
    N = n;
    H = 1;
    for (int i = 1; i <= N; i++) {
      if (i % 2 == 1)
        H = 2 * H;
      else if (i % 2 == 0)
        H = H + 1;
    }
    printf("%d\n", H);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%eax
  test   %eax,%eax
  jle    L7
  xor    %ebx,%ebx
  lea    0x4(%rsp),%r13
  lea    D1(%rip),%r12
  nopl   (%rax)
L36:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  test   %ecx,%ecx
  jle    L54
  add    $0x1,%ecx
  mov    $0x1,%eax
  mov    $0x1,%esi
  nopl   0x0(%rax,%rax,1)
L37:
  lea    (%rsi,%rsi,1),%edx
  add    $0x1,%esi
  test   $0x1,%al
  cmovne %edx,%esi
  add    $0x1,%eax
  cmp    %eax,%ecx
  jne    L37
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,(%rsp)
  jg     L36
L7:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L41
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L54:
  mov    $0x1,%esi
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,(%rsp)
  jg     L36
  jmp    L7
L41:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D5
type: string
name: unpack_data
arguments: tag: D1
type: string

tool: D5: %d

tool: D1: %d


assistant: ```c
int main() {
  int t, N, H = 1;
  scanf("%d", &t);
  for (int a0 = 0; a0 < t; a0++) {
    int n;
    scanf("%d", &n);
    N = n;
    H = 1;
    for (int i = 1; i <= N; i++) {
      if (i % 2 == 1)
        H = 2 * H;
      else if (i % 2 == 0)
        H = H + 1;
    }
    printf("%d\n", H);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  lea    D13(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%eax
  test   %eax,%eax
  jle    L40
  xor    %ebx,%ebx
  lea    0x4(%rsp),%r13
  lea    D19(%rip),%r12
  nopl   (%rax)
L16:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  test   %ecx,%ecx
  jle    L39
  add    $0x1,%ecx
  mov    $0x1,%eax
  mov    $0x1,%esi
  nopl   0x0(%rax,%rax,1)
L60:
  lea    (%rsi,%rsi,1),%edx
  add    $0x1,%esi
  test   $0x1,%al
  cmovne %edx,%esi
  add    $0x1,%eax
  cmp    %eax,%ecx
  jne    L60
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,(%rsp)
  jg     L16
L40:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L69
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L39:
  mov    $0x1,%esi
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,(%rsp)
  jg     L16
  jmp    L40
L69:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D13
type_guessed: string
name: parse_data
arguments: label: D19
type_guessed: string

tool: D13: %d

tool: D19: %d


assistant: ```c
int main() {
  int t, N, H = 1;
  scanf("%d", &t);
  for (int a0 = 0; a0 < t; a0++) {
    int n;
    scanf("%d", &n);
    N = n;
    H = 1;
    for (int i = 1; i <= N; i++) {
      if (i % 2 == 1)
        H = 2 * H;
      else if (i % 2 == 0)
        H = H + 1;
    }
    printf("%d\n", H);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<str_copy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <malloc_a@plt>
  mov    %rax,-0x8(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L11
L8:
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addq   $0x1,-0x18(%rbp)
L11:
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  cmp    %rax,-0x18(%rbp)
  jb     L8
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *str_copy(const char *from) {
  size_t length = strlen(from);
  char *new_string = malloc_a(length + 1, sizeof(*from));
  for (size_t i = 0; i < length + 1; i++) {
    new_string[i] = from[i];
  }
  return new_string;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<str_copy>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %rax,%rbx
  lea    0x1(%rax),%rdi
  mov    $0x1,%esi
  call   <malloc_a@plt>
  mov    $0x0,%edx
L5:
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  mov    %rdx,%rcx
  add    $0x1,%rdx
  cmp    %rcx,%rbx
  jne    L5
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *str_copy(const char *from) {
  size_t length = strlen(from);
  char *new_string = malloc_a(length + 1, sizeof(*from));
  for (size_t i = 0; i < length + 1; i++) {
    new_string[i] = from[i];
  }
  return new_string;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<str_copy>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    $0x1,%esi
  lea    0x1(%rax),%rdi
  mov    %rax,%rbx
  call   <malloc_a@plt>
  xor    %edx,%edx
  nopl   (%rax)
L4:
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  mov    %rdx,%rcx
  add    $0x1,%rdx
  cmp    %rcx,%rbx
  jne    L4
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *str_copy(const char *from) {
  size_t length = strlen(from);
  char *new_string = malloc_a(length + 1, sizeof(*from));
  for (size_t i = 0; i < length + 1; i++) {
    new_string[i] = from[i];
  }
  return new_string;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<str_copy>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    $0x1,%esi
  lea    0x1(%rax),%r12
  mov    %rax,%rbp
  mov    %r12,%rdi
  call   <malloc_a@plt>
  lea    0x1(%rbx),%rdx
  mov    %rax,%r8
  sub    %rdx,%rax
  cmp    $0xe,%rax
  jbe    L104
  cmp    $0x6,%rbp
  jbe    L104
  cmp    $0xe,%rbp
  jbe    L92
  mov    %r12,%rdx
  xor    %eax,%eax
  and    $0xfffffffffffffff0,%rdx
  nopl   (%rax)
L73:
  movdqu (%rbx,%rax,1),%xmm0
  movups %xmm0,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L73
  mov    %r12,%rax
  and    $0xfffffffffffffff0,%rax
  test   $0xf,%r12b
  je     L129
  sub    %rax,%rbp
  lea    0x1(%rbp),%rcx
  cmp    $0x6,%rbp
  jbe    L84
L23:
  mov    (%rbx,%rax,1),%rdx
  mov    %rdx,(%r8,%rax,1)
  mov    %rcx,%rdx
  and    $0xfffffffffffffff8,%rdx
  add    %rdx,%rax
  cmp    %rdx,%rcx
  je     L129
L84:
  movzbl (%rbx,%rax,1),%edx
  mov    %dl,(%r8,%rax,1)
  lea    0x1(%rax),%rdx
  cmp    %rdx,%r12
  jbe    L129
  movzbl 0x1(%rbx,%rax,1),%edx
  mov    %dl,0x1(%r8,%rax,1)
  lea    0x2(%rax),%rdx
  cmp    %rdx,%r12
  jbe    L129
  movzbl 0x2(%rbx,%rax,1),%edx
  mov    %dl,0x2(%r8,%rax,1)
  lea    0x3(%rax),%rdx
  cmp    %rdx,%r12
  jbe    L129
  movzbl 0x3(%rbx,%rax,1),%edx
  mov    %dl,0x3(%r8,%rax,1)
  lea    0x4(%rax),%rdx
  cmp    %rdx,%r12
  jbe    L129
  movzbl 0x4(%rbx,%rax,1),%edx
  mov    %dl,0x4(%r8,%rax,1)
  lea    0x5(%rax),%rdx
  cmp    %rdx,%r12
  jbe    L129
  movzbl 0x5(%rbx,%rax,1),%edx
  mov    %dl,0x5(%r8,%rax,1)
  lea    0x6(%rax),%rdx
  cmp    %rdx,%r12
  jbe    L129
  movzbl 0x6(%rbx,%rax,1),%edx
  mov    %dl,0x6(%r8,%rax,1)
L129:
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L104:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L91:
  movzbl (%rbx,%rax,1),%edx
  mov    %dl,(%r8,%rax,1)
  mov    %rax,%rdx
  add    $0x1,%rax
  cmp    %rdx,%rbp
  jne    L91
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
L92:
  mov    %r12,%rcx
  xor    %eax,%eax
  jmp    L23
```

assistant: ```c
char *str_copy(const char *from) {
  size_t length = strlen(from);
  char *new_string = malloc_a(length + 1, sizeof(*from));
  for (size_t i = 0; i < length + 1; i++) {
    new_string[i] = from[i];
  }
  return new_string;
}
```