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

```asm
<parity2>:
  endbr64
  test   %esi,%esi
  jle    L9
  lea    -0x1(%rsi),%eax
  xor    %r8d,%r8d
  lea    0x4(%rdi,%rax,4),%rsi
  nopl   0x0(%rax,%rax,1)
L14:
  mov    (%rdi),%eax
  xor    %edx,%edx
  nopl   0x0(%rax)
L26:
  mov    %eax,%ecx
  and    $0x1,%ecx
  xor    %ecx,%edx
  shr    %eax
  jne    L26
  add    $0x4,%rdi
  add    %edx,%r8d
  cmp    %rdi,%rsi
  jne    L14
  mov    %r8d,%eax
  ret
L9:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int parity2(unsigned *array, int len) {
  int i, val, res = 0;
  unsigned x;
  for (i = 0; i < len; i++) {
    x = array[i];
    val = 0;
    do {
      val ^= x & 0x1;
      x >>= 1;
    } while (x);
    res += val;
  }
  return res;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parity2>:
  endbr64
  test   %esi,%esi
  jle    L18
  lea    -0x1(%rsi),%eax
  xor    %r8d,%r8d
  lea    0x4(%rdi,%rax,4),%rsi
  nopl   0x0(%rax,%rax,1)
L17:
  mov    (%rdi),%eax
  xor    %edx,%edx
  nopl   0x0(%rax)
L12:
  mov    %eax,%ecx
  and    $0x1,%ecx
  xor    %ecx,%edx
  shr    %eax
  jne    L12
  add    $0x4,%rdi
  add    %edx,%r8d
  cmp    %rdi,%rsi
  jne    L17
  mov    %r8d,%eax
  ret
L18:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int parity2(unsigned *array, int len) {
  int i, val, res = 0;
  unsigned x;
  for (i = 0; i < len; i++) {
    x = array[i];
    val = 0;
    do {
      val ^= x & 0x1;
      x >>= 1;
    } while (x);
    res += val;
  }
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<entry_to>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <process_factory@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D8
data_type: string
name: unpack_data
arguments: tag: D0
data_type: string

tool: D8: Factory status: MANUFACTURING


tool: D0: Dispatching factory lines for order.


assistant: ```c
void entry_to() {
  printf("Factory status: MANUFACTURING\n");
  printf("Dispatching factory lines for order.\n");
  process_factory();
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<entry_to>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <process_factory@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D0
type: string
name: parse_data
arguments: data_tag: D11
type: string

tool: D0: Factory status: MANUFACTURING


tool: D11: Dispatching factory lines for order.


assistant: ```c
void entry_to() {
  printf("Factory status: MANUFACTURING\n");
  printf("Dispatching factory lines for order.\n");
  process_factory();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<entry_to>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    D10(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <process_factory@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D8
datatype: string
name: disassemble_data
arguments: tag: D10
datatype: string

tool: D8: Factory status: MANUFACTURING


tool: D10: Dispatching factory lines for order.


assistant: ```c
void entry_to() {
  printf("Factory status: MANUFACTURING\n");
  printf("Dispatching factory lines for order.\n");
  process_factory();
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<entry_to>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <printf@plt>
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <process_factory@plt>
```

tool_calls: name: read_data
arguments: data_tag: D5
type: string
name: read_data
arguments: data_tag: D8
type: string

tool: D5: Factory status: MANUFACTURING


tool: D8: Dispatching factory lines for order.


assistant: ```c
void entry_to() {
  printf("Factory status: MANUFACTURING\n");
  printf("Dispatching factory lines for order.\n");
  process_factory();
}
```
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    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D1
type_infered: string

tool: D1: Hello, Gradle!

assistant: ```c
int main(int argc, char *argv[]) {
  puts("Hello, Gradle!");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
type_guessed: string

tool: D1: Hello, Gradle!

assistant: ```c
int main(int argc, char *argv[]) {
  puts("Hello, Gradle!");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D4
datatype: string

tool: D4: Hello, Gradle!

assistant: ```c
int main(int argc, char *argv[]) {
  puts("Hello, Gradle!");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D6
type_guessed: string

tool: D6: Hello, Gradle!

assistant: ```c
int main(int argc, char *argv[]) {
  puts("Hello, Gradle!");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x40(%rsp)
  movq   $0x1,0x48(%rsp)
  movq   $0x1,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movq   $0x1,0x60(%rsp)
  movq   $0x0,0x68(%rsp)
  movq   $0x0,0x70(%rsp)
  movq   $0x1,0x78(%rsp)
  movq   $0x2,0x80(%rsp)
  movq   $0x0,0x88(%rsp)
  movq   $0x0,0x90(%rsp)
  movq   $0x2,0x98(%rsp)
  movq   $0x1,0xa0(%rsp)
  movq   $0x0,0xa8(%rsp)
  movq   $0x0,0xb0(%rsp)
  movq   $0x1,0xb8(%rsp)
  mov    $0xa5f07e3c,%eax
  mov    %rax,(%rsp)
  movq   $0x7e3ca5f0,0x8(%rsp)
  movq   $0x3ca5f07e,0x10(%rsp)
  movq   $0x5a0fe7c3,0x18(%rsp)
  lea    0x5a0e81c4(%rax),%rax
  mov    %rax,0x20(%rsp)
  movq   $0xffff,0x28(%rsp)
  movq   $0xf0f0f0f,0x30(%rsp)
  lea    -0xf0e0f10(%rax),%rax
  mov    %rax,0x38(%rsp)
  mov    %rsp,%r12
  lea    0x40(%rsp),%rbp
  lea    0xc0(%rsp),%r13
  jmp    L59
L11:
  add    $0x8,%rbx
  cmp    $0x20,%rbx
  je     L46
L54:
  mov    0x20(%rsp,%rbx,1),%rsi
  mov    (%r12),%rdi
  call   <do_cntlzdm@plt>
  cmp    0x0(%rbp,%rbx,1),%rax
  je     L11
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L11
L46:
  add    $0x8,%r12
  add    $0x20,%rbp
  cmp    %r13,%rbp
  je     L18
L59:
  mov    $0x0,%ebx
  jmp    L54
L18:
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  mov    $0x0,%eax
  add    $0xd8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long long int sources[4], masks[4];
  unsigned long long int intermediates[4][4] = {
      {0x0000a5f0ll, 0x00007e3cll, 0x000050ecll, 0x0000af73ll},
      {0x00007e3cll, 0x0000a5f0ll, 0x0000ec50ll, 0x000073afll},
      {0x00003ca5ll, 0x0000f07ell, 0x0000c50ell, 0x00003af7ll},
      {0x00005a0fll, 0x0000e7c3ll, 0x0000af73ll, 0x000050ecll},
  };
  unsigned long long int results[4][4] = {
      {0, 1, 1, 0},
      {1, 0, 0, 1},
      {2, 0, 0, 2},
      {1, 0, 0, 1},
  };
  sources[0] = 0xa5f07e3cll;
  sources[1] = 0x7e3ca5f0ll;
  sources[2] = 0x3ca5f07ell;
  sources[3] = 0x5a0fe7c3ll;
  masks[0] = 0xffff0000ll;
  masks[1] = 0x0000ffffll;
  masks[2] = 0x0f0f0f0fll;
  masks[3] = 0xf0f0f0f0ll;
  unsigned long long int result;
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
      if (do_cntlzdm(sources[i], masks[j]) != results[i][j])
        abort();
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  mov    $0xffff0000,%r14d
  push   %r13
  push   %r12
  mov    $0xa5f07e3c,%r12d
  push   %rbp
  push   %rbx
  sub    $0xd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  mov    $0xf0f0f0f0,%eax
  movq   $0x0,0x40(%rsp)
  mov    %rsp,%r13
  movq   $0x1,0x48(%rsp)
  lea    0x40(%rsp),%rbp
  lea    0xc0(%rsp),%r15
  movq   $0x1,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movq   $0x1,0x60(%rsp)
  movq   $0x0,0x68(%rsp)
  movq   $0x0,0x70(%rsp)
  movq   $0x1,0x78(%rsp)
  movq   $0x2,0x80(%rsp)
  movq   $0x0,0x88(%rsp)
  movq   $0x0,0x90(%rsp)
  movq   $0x2,0x98(%rsp)
  movq   $0x1,0xa0(%rsp)
  movq   $0x0,0xa8(%rsp)
  movq   $0x0,0xb0(%rsp)
  movq   $0x1,0xb8(%rsp)
  movq   $0x7e3ca5f0,0x8(%rsp)
  movq   $0x3ca5f07e,0x10(%rsp)
  movq   $0x5a0fe7c3,0x18(%rsp)
  movq   $0xffff,0x28(%rsp)
  movq   $0xf0f0f0f,0x30(%rsp)
  mov    %rax,0x38(%rsp)
L41:
  xor    %ebx,%ebx
  mov    %r14,%rsi
  mov    %r12,%rdi
  call   <do_cntlzdm@plt>
  cmp    0x0(%rbp,%rbx,1),%rax
  je     L61
L55:
  xor    %eax,%eax
  call   <abort@plt>
L61:
  add    $0x8,%rbx
  cmp    $0x20,%rbx
  je     L68
L6:
  mov    0x20(%rsp,%rbx,1),%rsi
  mov    %r12,%rdi
  call   <do_cntlzdm@plt>
  cmp    0x0(%rbp,%rbx,1),%rax
  jne    L55
  add    $0x8,%rbx
  cmp    $0x20,%rbx
  jne    L6
L68:
  add    $0x20,%rbp
  add    $0x8,%r13
  cmp    %r15,%rbp
  je     L59
  mov    0x0(%r13),%r12
  jmp    L41
  nopl   0x0(%rax)
L59:
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0xd8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long long int sources[4], masks[4];
  unsigned long long int intermediates[4][4] = {
      {0x0000a5f0ll, 0x00007e3cll, 0x000050ecll, 0x0000af73ll},
      {0x00007e3cll, 0x0000a5f0ll, 0x0000ec50ll, 0x000073afll},
      {0x00003ca5ll, 0x0000f07ell, 0x0000c50ell, 0x00003af7ll},
      {0x00005a0fll, 0x0000e7c3ll, 0x0000af73ll, 0x000050ecll},
  };
  unsigned long long int results[4][4] = {
      {0, 1, 1, 0},
      {1, 0, 0, 1},
      {2, 0, 0, 2},
      {1, 0, 0, 1},
  };
  sources[0] = 0xa5f07e3cll;
  sources[1] = 0x7e3ca5f0ll;
  sources[2] = 0x3ca5f07ell;
  sources[3] = 0x5a0fe7c3ll;
  masks[0] = 0xffff0000ll;
  masks[1] = 0x0000ffffll;
  masks[2] = 0x0f0f0f0fll;
  masks[3] = 0xf0f0f0f0ll;
  unsigned long long int result;
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
      if (do_cntlzdm(sources[i], masks[j]) != results[i][j])
        abort();
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<assignment_change_count>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L23
L6:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  je     L19
  addl   $0x1,-0x8(%rbp)
L19:
  addl   $0x1,-0x4(%rbp)
L23:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L6
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int assignment_change_count(int n, int a[], int b[]) {
  int change_count = 0;
  for (int ii = 0; ii < n; ii++)
    if (a[ii] != b[ii])
      change_count++;
  return change_count;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<assignment_change_count>:
  endbr64
  test   %edi,%edi
  jle    L45
  mov    %edi,%edi
  mov    $0x0,%eax
  mov    $0x0,%ecx
  jmp    L11
L35:
  add    $0x1,%rax
  cmp    %rdi,%rax
  je     L34
L11:
  mov    (%rdx,%rax,4),%r8d
  cmp    %r8d,(%rsi,%rax,4)
  je     L35
  add    $0x1,%ecx
  jmp    L35
L45:
  mov    $0x0,%ecx
L34:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int assignment_change_count(int n, int a[], int b[]) {
  int change_count = 0;
  for (int ii = 0; ii < n; ii++)
    if (a[ii] != b[ii])
      change_count++;
  return change_count;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<assignment_change_count>:
  endbr64
  test   %edi,%edi
  jle    L1
  movslq %edi,%rdi
  xor    %eax,%eax
  xor    %r8d,%r8d
L4:
  mov    (%rdx,%rax,4),%ecx
  cmp    %ecx,(%rsi,%rax,4)
  je     L23
  add    $0x1,%r8d
L23:
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L4
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L1:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int assignment_change_count(int n, int a[], int b[]) {
  int change_count = 0;
  for (int ii = 0; ii < n; ii++)
    if (a[ii] != b[ii])
      change_count++;
  return change_count;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<assignment_change_count>:
  endbr64
  mov    %rsi,%rcx
  test   %edi,%edi
  jle    L76
  lea    -0x1(%rdi),%eax
  cmp    $0x2,%eax
  jbe    L55
  mov    %edi,%esi
  pxor   %xmm1,%xmm1
  xor    %eax,%eax
  shr    $0x2,%esi
  movdqa %xmm1,%xmm2
  shl    $0x4,%rsi
  xchg   %ax,%ax
L47:
  movdqu (%rcx,%rax,1),%xmm0
  movdqu (%rdx,%rax,1),%xmm3
  add    $0x10,%rax
  pcmpeqd %xmm3,%xmm0
  pcmpeqd %xmm2,%xmm0
  psubd  %xmm0,%xmm1
  cmp    %rax,%rsi
  jne    L47
  movdqa %xmm1,%xmm0
  mov    %edi,%esi
  psrldq $0x8,%xmm0
  and    $0xfffffffc,%esi
  paddd  %xmm0,%xmm1
  movdqa %xmm1,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm1
  movd   %xmm1,%eax
  test   $0x3,%dil
  je     L0
L52:
  movslq %esi,%r9
  mov    (%rdx,%r9,4),%r10d
  lea    0x0(,%r9,4),%r8
  cmp    %r10d,(%rcx,%r9,4)
  je     L23
  add    $0x1,%eax
L23:
  lea    0x1(%rsi),%r9d
  cmp    %edi,%r9d
  jge    L68
  mov    0x4(%rcx,%r8,1),%r11d
  cmp    %r11d,0x4(%rdx,%r8,1)
  je     L65
  add    $0x1,%eax
L65:
  add    $0x2,%esi
  cmp    %esi,%edi
  jle    L68
  mov    0x8(%rdx,%r8,1),%edi
  cmp    %edi,0x8(%rcx,%r8,1)
  je     L68
  add    $0x1,%eax
  ret
  nop
L76:
  xor    %eax,%eax
L68:
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  ret
L55:
  xor    %esi,%esi
  xor    %eax,%eax
  jmp    L52
```

assistant: ```c
int assignment_change_count(int n, int a[], int b[]) {
  int change_count = 0;
  for (int ii = 0; ii < n; ii++)
    if (a[ii] != b[ii])
      change_count++;
  return change_count;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<QRinput_checkModeNum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L9
L28:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L24
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jle    L45
L24:
  mov    $0xffffffff,%eax
  jmp    L39
L45:
  addl   $0x1,-0x4(%rbp)
L9:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L28
  mov    $0x0,%eax
L39:
  pop    %rbp
  ret
```

assistant: ```c
int QRinput_checkModeNum(int size, const char *data) {
  int i;
  for (i = 0; i < size; i++) {
    if (data[i] < '0' || data[i] > '9')
      return -1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<QRinput_checkModeNum>:
  endbr64
  test   %edi,%edi
  jle    L7
  mov    %rsi,%rdx
  lea    -0x1(%rdi),%eax
  lea    0x1(%rsi,%rax,1),%rcx
L1:
  movzbl (%rdx),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L11
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L1
  mov    $0x0,%eax
  ret
L7:
  mov    $0x0,%eax
  ret
L11:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int QRinput_checkModeNum(int size, const char *data) {
  int i;
  for (i = 0; i < size; i++) {
    if (data[i] < '0' || data[i] > '9')
      return -1;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<QRinput_checkModeNum>:
  endbr64
  test   %edi,%edi
  jle    L37
  lea    -0x1(%rdi),%eax
  lea    0x1(%rsi,%rax,1),%rdx
  jmp    L9
  nopw   0x0(%rax,%rax,1)
L35:
  add    $0x1,%rsi
  cmp    %rdx,%rsi
  je     L37
L9:
  movzbl (%rsi),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L35
  mov    $0xffffffff,%eax
  ret
  nopl   0x0(%rax)
L37:
  xor    %eax,%eax
  ret
```

assistant: ```c
int QRinput_checkModeNum(int size, const char *data) {
  int i;
  for (i = 0; i < size; i++) {
    if (data[i] < '0' || data[i] > '9')
      return -1;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<QRinput_checkModeNum>:
  endbr64
  test   %edi,%edi
  jle    L17
  lea    -0x1(%rdi),%eax
  lea    0x1(%rsi,%rax,1),%rdx
  jmp    L23
  nopw   0x0(%rax,%rax,1)
L36:
  add    $0x1,%rsi
  cmp    %rdx,%rsi
  je     L17
L23:
  movzbl (%rsi),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L36
  mov    $0xffffffff,%eax
  ret
  nopl   0x0(%rax)
L17:
  xor    %eax,%eax
  ret
```

assistant: ```c
int QRinput_checkModeNum(int size, const char *data) {
  int i;
  for (i = 0; i < size; i++) {
    if (data[i] < '0' || data[i] > '9')
      return -1;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x170,%rsp
  mov    %edi,-0x164(%rbp)
  mov    %rsi,-0x170(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x150(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x28,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  mov    %rdi,%rdx
  mov    %eax,(%rdx)
  add    $0x4,%rdx
  mov    $0x0,%eax
  call   <randomize@plt>
  movl   $0x1,-0x15c(%rbp)
  jmp    L36
L23:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x38e38e39,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0x154(%rbp)
  mov    -0x154(%rbp),%ecx
  mov    %ecx,%edx
  shl    $0x3,%edx
  add    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0x154(%rbp)
  mov    -0x154(%rbp),%eax
  cltq
  mov    -0x150(%rbp,%rax,4),%eax
  test   %eax,%eax
  jne    L23
  mov    -0x154(%rbp),%eax
  cltq
  mov    -0x15c(%rbp),%edx
  mov    %edx,-0x150(%rbp,%rax,4)
  addl   $0x1,-0x15c(%rbp)
L36:
  cmpl   $0x9,-0x15c(%rbp)
  jle    L23
  lea    -0x150(%rbp),%rax
  mov    %rax,%rdi
  call   <print_board@plt>
  lea    -0x150(%rbp),%rax
  mov    %rax,%rdi
  call   <recursive_iterate@plt>
  mov    %eax,-0x158(%rbp)
  lea    -0x150(%rbp),%rax
  mov    %rax,%rdi
  call   <print_board@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int board[81] = {0};
  randomize();
  for (int i = 1; i < 10; i++) {
    int val;
    do {
      val = (rand() % 9);
    } while (board[val] != 0);
    board[val] = i;
  }
  print_board(board);
  int result = recursive_iterate(board);
  print_board(board);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x150,%rsp
  mov    $0x28,%ecx
  mov    %fs:(%rcx),%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x0,(%rdi)
  call   <randomize@plt>
  mov    $0x1,%ebx
L13:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x38e38e39,%rdx,%rdx
  sar    $0x21,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,8),%ecx
  sub    %ecx,%eax
  cltq
  cmpl   $0x0,(%rsp,%rax,4)
  jne    L13
  mov    %ebx,(%rsp,%rax,4)
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  jne    L13
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <print_board@plt>
  mov    %rbx,%rdi
  call   <recursive_iterate@plt>
  mov    %rbx,%rdi
  call   <print_board@plt>
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  mov    $0x0,%eax
  add    $0x150,%rsp
  pop    %rbx
  ret
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int board[81] = {0};
  randomize();
  for (int i = 1; i < 10; i++) {
    int val;
    do {
      val = (rand() % 9);
    } while (board[val] != 0);
    board[val] = i;
  }
  print_board(board);
  int result = recursive_iterate(board);
  print_board(board);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x28,%ecx
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x158,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x0,(%rdi)
  call   <randomize@plt>
  xchg   %ax,%ax
L28:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x38e38e39,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x21,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,8),%ecx
  sub    %ecx,%eax
  movslq %eax,%rdx
  mov    (%rsp,%rdx,4),%eax
  test   %eax,%eax
  jne    L28
  mov    %ebx,(%rsp,%rdx,4)
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  jne    L28
  mov    %rbp,%rdi
  call   <print_board@plt>
  mov    %rbp,%rdi
  call   <recursive_iterate@plt>
  mov    %rbp,%rdi
  call   <print_board@plt>
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x158,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int board[81] = {0};
  randomize();
  for (int i = 1; i < 10; i++) {
    int val;
    do {
      val = (rand() % 9);
    } while (board[val] != 0);
    board[val] = i;
  }
  print_board(board);
  int result = recursive_iterate(board);
  print_board(board);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x28,%ecx
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x158,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x0,(%rdi)
  call   <randomize@plt>
  xchg   %ax,%ax
L19:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x38e38e39,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x21,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,8),%ecx
  sub    %ecx,%eax
  movslq %eax,%rdx
  mov    (%rsp,%rdx,4),%eax
  test   %eax,%eax
  jne    L19
  mov    %ebx,(%rsp,%rdx,4)
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  jne    L19
  mov    %rbp,%rdi
  call   <print_board@plt>
  mov    %rbp,%rdi
  call   <recursive_iterate@plt>
  mov    %rbp,%rdi
  call   <print_board@plt>
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x158,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int board[81] = {0};
  randomize();
  for (int i = 1; i < 10; i++) {
    int val;
    do {
      val = (rand() % 9);
    } while (board[val] != 0);
    board[val] = i;
  }
  print_board(board);
  int result = recursive_iterate(board);
  print_board(board);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<fsize>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    D15(%rip),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <fseek@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <ftell@plt>
  mov    %eax,-0x4(%rbp)
  mov    D3(%rip),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <fseek@plt>
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int fsize(FILE *f) {
  int end;
  fseek(f, 0, SEEK_END);
  end = ftell(f);
  fseek(f, 0, SEEK_SET);
  return end;
}
```