ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<zero>:
  endbr64
  mov    %esi,%ecx
  test   %esi,%esi
  jle    L21
  lea    -0x1(%rsi),%eax
  cmp    $0x2,%eax
  jbe    L39
  shr    $0x2,%esi
  movd   %edx,%xmm1
  mov    %rdi,%rax
  sub    $0x1,%esi
  pshufd $0x0,%xmm1,%xmm0
  shl    $0x4,%rsi
  lea    0x10(%rdi,%rsi,1),%rsi
  nopl   (%rax)
L69:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rsi,%rax
  jne    L69
  mov    %ecx,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%cl
  je     L67
L68:
  movslq %eax,%rsi
  mov    %edx,(%rdi,%rsi,4)
  lea    0x0(,%rsi,4),%r8
  lea    0x1(%rax),%esi
  cmp    %esi,%ecx
  jle    L21
  add    $0x2,%eax
  mov    %edx,0x4(%rdi,%r8,1)
  cmp    %eax,%ecx
  jle    L21
  mov    %edx,0x8(%rdi,%r8,1)
L21:
  ret
  nopl   (%rax)
L67:
  ret
L39:
  xor    %eax,%eax
  jmp    L68
```

assistant: ```c
void zero(int arr[], int len, int zeroVal) {
  int i;
  for (i = 0; i < len; i++) {
    *(arr + i) = zeroVal;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<x_midi_setup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <midiin_setup@plt>
  mov    $0x0,%eax
  call   <midirealtimein_setup@plt>
  mov    $0x0,%eax
  call   <notein_setup@plt>
  mov    $0x0,%eax
  call   <ctlin_setup@plt>
  mov    $0x0,%eax
  call   <pgmin_setup@plt>
  mov    $0x0,%eax
  call   <bendin_setup@plt>
  mov    $0x0,%eax
  call   <touchin_setup@plt>
  mov    $0x0,%eax
  call   <polytouchin_setup@plt>
  mov    $0x0,%eax
  call   <midiclkin_setup@plt>
  mov    $0x0,%eax
  call   <midiout_setup@plt>
  mov    $0x0,%eax
  call   <noteout_setup@plt>
  mov    $0x0,%eax
  call   <ctlout_setup@plt>
  mov    $0x0,%eax
  call   <pgmout_setup@plt>
  mov    $0x0,%eax
  call   <bendout_setup@plt>
  mov    $0x0,%eax
  call   <touchout_setup@plt>
  mov    $0x0,%eax
  call   <polytouchout_setup@plt>
  mov    $0x0,%eax
  call   <makenote_setup@plt>
  mov    $0x0,%eax
  call   <stripnote_setup@plt>
  mov    $0x0,%eax
  call   <poly_setup@plt>
  mov    $0x0,%eax
  call   <bag_setup@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void x_midi_setup(void) {
  midiin_setup();
  midirealtimein_setup();
  notein_setup();
  ctlin_setup();
  pgmin_setup();
  bendin_setup();
  touchin_setup();
  polytouchin_setup();
  midiclkin_setup();
  midiout_setup();
  noteout_setup();
  ctlout_setup();
  pgmout_setup();
  bendout_setup();
  touchout_setup();
  polytouchout_setup();
  makenote_setup();
  stripnote_setup();
  poly_setup();
  bag_setup();
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<x_midi_setup>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <midiin_setup@plt>
  mov    $0x0,%eax
  call   <midirealtimein_setup@plt>
  mov    $0x0,%eax
  call   <notein_setup@plt>
  mov    $0x0,%eax
  call   <ctlin_setup@plt>
  mov    $0x0,%eax
  call   <pgmin_setup@plt>
  mov    $0x0,%eax
  call   <bendin_setup@plt>
  mov    $0x0,%eax
  call   <touchin_setup@plt>
  mov    $0x0,%eax
  call   <polytouchin_setup@plt>
  mov    $0x0,%eax
  call   <midiclkin_setup@plt>
  mov    $0x0,%eax
  call   <midiout_setup@plt>
  mov    $0x0,%eax
  call   <noteout_setup@plt>
  mov    $0x0,%eax
  call   <ctlout_setup@plt>
  mov    $0x0,%eax
  call   <pgmout_setup@plt>
  mov    $0x0,%eax
  call   <bendout_setup@plt>
  mov    $0x0,%eax
  call   <touchout_setup@plt>
  mov    $0x0,%eax
  call   <polytouchout_setup@plt>
  mov    $0x0,%eax
  call   <makenote_setup@plt>
  mov    $0x0,%eax
  call   <stripnote_setup@plt>
  mov    $0x0,%eax
  call   <poly_setup@plt>
  mov    $0x0,%eax
  call   <bag_setup@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void x_midi_setup(void) {
  midiin_setup();
  midirealtimein_setup();
  notein_setup();
  ctlin_setup();
  pgmin_setup();
  bendin_setup();
  touchin_setup();
  polytouchin_setup();
  midiclkin_setup();
  midiout_setup();
  noteout_setup();
  ctlout_setup();
  pgmout_setup();
  bendout_setup();
  touchout_setup();
  polytouchout_setup();
  makenote_setup();
  stripnote_setup();
  poly_setup();
  bag_setup();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<x_midi_setup>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <midiin_setup@plt>
  xor    %eax,%eax
  call   <midirealtimein_setup@plt>
  xor    %eax,%eax
  call   <notein_setup@plt>
  xor    %eax,%eax
  call   <ctlin_setup@plt>
  xor    %eax,%eax
  call   <pgmin_setup@plt>
  xor    %eax,%eax
  call   <bendin_setup@plt>
  xor    %eax,%eax
  call   <touchin_setup@plt>
  xor    %eax,%eax
  call   <polytouchin_setup@plt>
  xor    %eax,%eax
  call   <midiclkin_setup@plt>
  xor    %eax,%eax
  call   <midiout_setup@plt>
  xor    %eax,%eax
  call   <noteout_setup@plt>
  xor    %eax,%eax
  call   <ctlout_setup@plt>
  xor    %eax,%eax
  call   <pgmout_setup@plt>
  xor    %eax,%eax
  call   <bendout_setup@plt>
  xor    %eax,%eax
  call   <touchout_setup@plt>
  xor    %eax,%eax
  call   <polytouchout_setup@plt>
  xor    %eax,%eax
  call   <makenote_setup@plt>
  xor    %eax,%eax
  call   <stripnote_setup@plt>
  xor    %eax,%eax
  call   <poly_setup@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1260 <bag_setup@plt>
```

assistant: ```c
void x_midi_setup(void) {
  midiin_setup();
  midirealtimein_setup();
  notein_setup();
  ctlin_setup();
  pgmin_setup();
  bendin_setup();
  touchin_setup();
  polytouchin_setup();
  midiclkin_setup();
  midiout_setup();
  noteout_setup();
  ctlout_setup();
  pgmout_setup();
  bendout_setup();
  touchout_setup();
  polytouchout_setup();
  makenote_setup();
  stripnote_setup();
  poly_setup();
  bag_setup();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x24(%rbp)
  jmp    L23
L27:
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <log10@plt>
  add    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L21
L35:
  mov    -0x20(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%eax
  lea    0x30(%rax),%edx
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x1c(%rbp),%eax
  cltq
  mov    %dl,-0x12(%rbp,%rax,1)
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x20(%rbp)
  addl   $0x1,-0x1c(%rbp)
L21:
  mov    -0x1c(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L35
  mov    -0x18(%rbp),%eax
  cltq
  movb   $0x0,-0x12(%rbp,%rax,1)
  lea    -0x12(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x24(%rbp)
L23:
  cmpl   $0x13,-0x24(%rbp)
  jle    L27
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L48
  call   <__stack_chk_fail@plt>
L48:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
type_infered: string

tool: D3: number char array = %s


assistant: ```c
int main() {
  int mid;
  for (mid = 1; mid < 20; mid++) {
    int mid_tmp = mid;
    char mid_char[10];
    int mid_size = 1 + (int)log10(mid);
    int i;
    for (i = 0; i < mid_size; i++) {
      mid_char[mid_size - 1 - i] = '0' + mid_tmp % 10;
      mid_tmp = mid_tmp / 10;
    }
    mid_char[mid_size] = '\0';
    printf("number char array = %s\n", mid_char);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    $0x1,%ebx
  lea    0xe(%rsp),%rbp
  lea    0xd(%rsp),%r13
  lea    D6(%rip),%r12
  jmp    L26
L8:
  movslq %r8d,%r8
  movb   $0x0,0xe(%rsp,%r8,1)
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x14,%ebx
  je     L13
L26:
  mov    %ebx,%edi
  call   <log10@plt>
  lea    0x1(%rax),%r8d
  test   %r8d,%r8d
  jle    L8
  movslq %eax,%rdi
  lea    0x0(%rbp,%rdi,1),%rcx
  add    %r13,%rdi
  mov    %eax,%eax
  sub    %rax,%rdi
  mov    %ebx,%edx
L57:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  add    $0x30,%edx
  mov    %dl,(%rcx)
  mov    %eax,%edx
  sub    $0x1,%rcx
  cmp    %rdi,%rcx
  jne    L57
  jmp    L8
L13:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L45:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
type: string

tool: D6: number char array = %s


assistant: ```c
int main() {
  int mid;
  for (mid = 1; mid < 20; mid++) {
    int mid_tmp = mid;
    char mid_char[10];
    int mid_size = 1 + (int)log10(mid);
    int i;
    for (i = 0; i < mid_size; i++) {
      mid_char[mid_size - 1 - i] = '0' + mid_tmp % 10;
      mid_tmp = mid_tmp / 10;
    }
    mid_char[mid_size] = '\0';
    printf("number char array = %s\n", mid_char);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  mov    $0xcccccccd,%r12d
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rbx
  cs nopw 0x0(%rax,%rax,1)
L24:
  mov    %r13d,%edi
  call   <log10@plt>
  lea    0x1(%rax),%r8d
  test   %r8d,%r8d
  jle    L6
  movslq %eax,%rdx
  mov    %eax,%eax
  lea    0xd(%rsp,%rdx,1),%rdi
  lea    (%rbx,%rdx,1),%rcx
  mov    %r13d,%edx
  sub    %rax,%rdi
  nopl   (%rax)
L10:
  mov    %edx,%eax
  sub    $0x1,%rcx
  imul   %r12,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  add    $0x30,%edx
  mov    %dl,0x1(%rcx)
  mov    %eax,%edx
  cmp    %rdi,%rcx
  jne    L10
L6:
  movslq %r8d,%r8
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  add    $0x1,%r13d
  movb   $0x0,0xe(%rsp,%r8,1)
  call   <printf@plt>
  cmp    $0x14,%r13d
  jne    L24
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
data_type: string

tool: D8: number char array = %s


assistant: ```c
int main() {
  int mid;
  for (mid = 1; mid < 20; mid++) {
    int mid_tmp = mid;
    char mid_char[10];
    int mid_size = 1 + (int)log10(mid);
    int i;
    for (i = 0; i < mid_size; i++) {
      mid_char[mid_size - 1 - i] = '0' + mid_tmp % 10;
      mid_tmp = mid_tmp / 10;
    }
    mid_char[mid_size] = '\0';
    printf("number char array = %s\n", mid_char);
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  mov    $0xcccccccd,%r12d
  push   %rbp
  push   %rbx
  lea    D2(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rbp
  cs nopw 0x0(%rax,%rax,1)
L87:
  mov    %r13d,%edi
  call   <log10@plt>
  lea    0x1(%rax),%ecx
  test   %ecx,%ecx
  jle    L0
  mov    %r13d,%edx
  mov    %r13d,%esi
  movslq %eax,%r8
  imul   %r12,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%edi
  add    %edi,%edi
  sub    %edi,%esi
  add    $0x30,%esi
  mov    %sil,0xe(%rsp,%r8,1)
  test   %eax,%eax
  jle    L0
  lea    -0x1(%rax),%esi
  add    $0x30,%edx
  movslq %esi,%rsi
  mov    %dl,0xe(%rsp,%rsi,1)
  cmp    $0x1,%eax
  je     L0
  lea    -0x2(%rax),%edx
  movslq %edx,%rdx
  movb   $0x30,0xe(%rsp,%rdx,1)
  cmp    $0x2,%eax
  je     L0
  lea    -0x3(%rax),%edx
  movslq %edx,%rdx
  movb   $0x30,0xe(%rsp,%rdx,1)
  cmp    $0x3,%eax
  je     L0
  lea    -0x4(%rax),%edx
  movslq %edx,%rdx
  movb   $0x30,0xe(%rsp,%rdx,1)
  cmp    $0x4,%eax
  je     L0
  lea    -0x5(%rax),%edx
  movslq %edx,%rdx
  movb   $0x30,0xe(%rsp,%rdx,1)
  cmp    $0x5,%eax
  je     L0
  lea    -0x6(%rax),%edx
  movslq %edx,%rdx
  movb   $0x30,0xe(%rsp,%rdx,1)
  cmp    $0x6,%eax
  je     L0
  lea    -0x7(%rax),%edx
  movslq %edx,%rdx
  movb   $0x30,0xe(%rsp,%rdx,1)
  cmp    $0x7,%eax
  je     L0
  lea    -0x8(%rax),%edx
  movslq %edx,%rdx
  movb   $0x30,0xe(%rsp,%rdx,1)
  cmp    $0x8,%eax
  je     L0
  sub    $0x9,%eax
  cltq
  movb   $0x30,0xe(%rsp,%rax,1)
L0:
  movslq %ecx,%rcx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  add    $0x1,%r13d
  movb   $0x0,0xe(%rsp,%rcx,1)
  call   <printf@plt>
  cmp    $0x14,%r13d
  jne    L87
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L46:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D2
type: string

tool: D2: number char array = %s


assistant: ```c
int main() {
  int mid;
  for (mid = 1; mid < 20; mid++) {
    int mid_tmp = mid;
    char mid_char[10];
    int mid_size = 1 + (int)log10(mid);
    int i;
    for (i = 0; i < mid_size; i++) {
      mid_char[mid_size - 1 - i] = '0' + mid_tmp % 10;
      mid_tmp = mid_tmp / 10;
    }
    mid_char[mid_size] = '\0';
    printf("number char array = %s\n", mid_char);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```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),%rax
  lea    D16(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x30(%rbp),%rax
  add    $0x2,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x28(%rbp),%rax
  lea    D4(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D16
type_guessed: string
name: read_data
arguments: tag: D4
type_guessed: string

tool: D16: Hello World!


tool: D4: y you guys!


assistant: ```c
int main() {
  char str_a[20];
  char *pointer;
  char *pointer2;
  strcpy(str_a, "Hello World!\n");
  pointer = str_a;
  printf(pointer);
  pointer2 = pointer + 2;
  printf(pointer2);
  strcpy(pointer2, "y you guys!\n");
  printf(pointer);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    D18(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    0x2(%rsp),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    D5(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D18
data_type: string
name: retrieve_data
arguments: data_label: D5
data_type: string

tool: D18: Hello World!


tool: D5: y you guys!


assistant: ```c
int main() {
  char str_a[20];
  char *pointer;
  char *pointer2;
  strcpy(str_a, "Hello World!\n");
  pointer = str_a;
  printf(pointer);
  pointer2 = pointer + 2;
  printf(pointer2);
  strcpy(pointer2, "y you guys!\n");
  printf(pointer);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r12
  lea    D15(%rip),%rsi
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x2(%rsp),%r12
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  lea    D2(%rip),%rsi
  call   <strcpy@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D15
data_type: string
name: disassemble_data
arguments: data_tag: D2
data_type: string

tool: D15: Hello World!


tool: D2: y you guys!


assistant: ```c
int main() {
  char str_a[20];
  char *pointer;
  char *pointer2;
  strcpy(str_a, "Hello World!\n");
  pointer = str_a;
  printf(pointer);
  pointer2 = pointer + 2;
  printf(pointer2);
  strcpy(pointer2, "y you guys!\n");
  printf(pointer);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D0(%rip),%rsi
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x2(%rsp),%r12
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  lea    D5(%rip),%rsi
  call   <strcpy@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

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

tool: D0: Hello World!


tool: D5: y you guys!


assistant: ```c
int main() {
  char str_a[20];
  char *pointer;
  char *pointer2;
  strcpy(str_a, "Hello World!\n");
  pointer = str_a;
  printf(pointer);
  pointer2 = pointer + 2;
  printf(pointer2);
  strcpy(pointer2, "y you guys!\n");
  printf(pointer);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<transpose_3x3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movsd  0x8(%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  movsd  0x18(%rax),%xmm0
  movsd  %xmm0,(%rdx)
  mov    -0x18(%rbp),%rax
  add    $0x18,%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  movsd  0x10(%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x10(%rax),%rdx
  mov    -0x18(%rbp),%rax
  movsd  0x30(%rax),%xmm0
  movsd  %xmm0,(%rdx)
  mov    -0x18(%rbp),%rax
  add    $0x30,%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  movsd  0x28(%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x28(%rax),%rdx
  mov    -0x18(%rbp),%rax
  movsd  0x38(%rax),%xmm0
  movsd  %xmm0,(%rdx)
  mov    -0x18(%rbp),%rax
  add    $0x38,%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void transpose_3x3(double *mat) {
  double a;
  a = mat[1];
  mat[1] = mat[3];
  mat[3] = a;
  a = mat[2];
  mat[2] = mat[6];
  mat[6] = a;
  a = mat[5];
  mat[5] = mat[7];
  mat[7] = a;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<transpose_3x3>:
  endbr64
  movsd  0x8(%rdi),%xmm0
  movsd  0x18(%rdi),%xmm1
  movsd  %xmm1,0x8(%rdi)
  movsd  %xmm0,0x18(%rdi)
  movsd  0x10(%rdi),%xmm0
  movsd  0x30(%rdi),%xmm1
  movsd  %xmm1,0x10(%rdi)
  movsd  %xmm0,0x30(%rdi)
  movsd  0x28(%rdi),%xmm0
  movsd  0x38(%rdi),%xmm1
  movsd  %xmm1,0x28(%rdi)
  movsd  %xmm0,0x38(%rdi)
  ret
```

assistant: ```c
void transpose_3x3(double *mat) {
  double a;
  a = mat[1];
  mat[1] = mat[3];
  mat[3] = a;
  a = mat[2];
  mat[2] = mat[6];
  mat[6] = a;
  a = mat[5];
  mat[5] = mat[7];
  mat[7] = a;
}
```
user: Translate the below assembly code into C programming language:

```asm
<transpose_3x3>:
  endbr64
  movsd  0x8(%rdi),%xmm0
  movsd  0x18(%rdi),%xmm1
  movsd  %xmm0,0x18(%rdi)
  movsd  0x10(%rdi),%xmm0
  movsd  %xmm1,0x8(%rdi)
  movsd  0x30(%rdi),%xmm1
  movsd  %xmm0,0x30(%rdi)
  movsd  0x28(%rdi),%xmm0
  movsd  %xmm1,0x10(%rdi)
  movsd  0x38(%rdi),%xmm1
  movsd  %xmm0,0x38(%rdi)
  movsd  %xmm1,0x28(%rdi)
  ret
```

assistant: ```c
void transpose_3x3(double *mat) {
  double a;
  a = mat[1];
  mat[1] = mat[3];
  mat[3] = a;
  a = mat[2];
  mat[2] = mat[6];
  mat[6] = a;
  a = mat[5];
  mat[5] = mat[7];
  mat[7] = a;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<transpose_3x3>:
  endbr64
  movsd  0x8(%rdi),%xmm1
  movupd 0x28(%rdi),%xmm0
  movsd  0x10(%rdi),%xmm2
  movlpd 0x18(%rdi),%xmm0
  movsd  %xmm1,0x18(%rdi)
  movsd  0x38(%rdi),%xmm1
  movups %xmm0,0x8(%rdi)
  movsd  0x28(%rdi),%xmm0
  unpcklpd %xmm2,%xmm1
  movups %xmm1,0x28(%rdi)
  movsd  %xmm0,0x38(%rdi)
  ret
```

assistant: ```c
void transpose_3x3(double *mat) {
  double a;
  a = mat[1];
  mat[1] = mat[3];
  mat[3] = a;
  a = mat[2];
  mat[2] = mat[6];
  mat[6] = a;
  a = mat[5];
  mat[5] = mat[7];
  mat[7] = a;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<__uio_num_from_filename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0xffffffff,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  jmp    L115
L206:
  movsbl -0xd(%rbp),%eax
  cmp    $0x75,%eax
  je     L118
  cmp    $0x75,%eax
  jg     L128
  cmp    $0x69,%eax
  je     L184
  cmp    $0x6f,%eax
  je     L163
  jmp    L128
L118:
  cmpl   $0x0,-0xc(%rbp)
  jne    L129
  movl   $0x1,-0xc(%rbp)
  jmp    L124
L129:
  movl   $0x4,-0xc(%rbp)
  jmp    L124
L184:
  cmpl   $0x1,-0xc(%rbp)
  jne    L200
  movl   $0x2,-0xc(%rbp)
  jmp    L124
L200:
  movl   $0x4,-0xc(%rbp)
  jmp    L124
L163:
  cmpl   $0x2,-0xc(%rbp)
  jne    L36
  movl   $0x3,-0xc(%rbp)
  jmp    L124
L36:
  movl   $0x4,-0xc(%rbp)
  jmp    L124
L128:
  cmpb   $0x2f,-0xd(%rbp)
  jle    L119
  cmpb   $0x39,-0xd(%rbp)
  jg     L119
  cmpl   $0x3,-0xc(%rbp)
  jne    L119
  cmpl   $0x0,-0x4(%rbp)
  jns    L80
  movsbl -0xd(%rbp),%eax
  sub    $0x30,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L124
L80:
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  movsbl -0xd(%rbp),%eax
  sub    $0x30,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L124
L119:
  movl   $0x4,-0xc(%rbp)
L124:
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
L115:
  cmpb   $0x0,-0xd(%rbp)
  je     L112
  cmpl   $0x4,-0xc(%rbp)
  jne    L206
L112:
  cmpl   $0x4,-0xc(%rbp)
  jne    L3
  movl   $0xffffffff,-0x4(%rbp)
L3:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int __uio_num_from_filename(char *name) {
  enum scan_states { ss_u, ss_i, ss_o, ss_num, ss_err };
  enum scan_states state = ss_u;
  int i = 0, num = -1;
  char ch = name[0];
  while (ch && (state != ss_err)) {
    switch (ch) {
    case 'u':
      if (state == ss_u)
        state = ss_i;
      else
        state = ss_err;
      break;
    case 'i':
      if (state == ss_i)
        state = ss_o;
      else
        state = ss_err;
      break;
    case 'o':
      if (state == ss_o)
        state = ss_num;
      else
        state = ss_err;
      break;
    default:
      if ((ch >= '0') && (ch <= '9') && (state == ss_num)) {
        if (num < 0)
          num = (ch - '0');
        else
          num = (num * 10) + (ch - '0');
      } else
        state = ss_err;
    }
    i++;
    ch = name[i];
  }
  if (state == ss_err)
    num = -1;
  return num;
}
```