ylfeng/ReF-Decompile-dataset huggingface dataset:
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
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  lea    D31(%rip),%r13
  lea    D12(%rip),%r12
  movabs $0x6666666666666667,%rbp
  jmp    L63
L83:
  mov    %rbx,%rcx
L75:
  mov    %rcx,%rax
  imul   %rbp
  sar    $0x2,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rdx
  add    %rdx,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  movslq %eax,%rdx
  cmpl   $0x0,0x10(%rsp,%rdx,4)
  jne    L8
  movl   $0x1,0x10(%rsp,%rdx,4)
  mov    %rcx,%rax
  imul   %rbp
  sar    $0x2,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  mov    %rdx,%rbx
  sub    %rax,%rbx
  mov    %esi,%edi
  cmp    $0x9,%rcx
  jg     L83
  mov    %rbx,0x8(%rsp)
L65:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L56
L8:
  test   %dil,%dil
  je     L10
  mov    %rbx,0x8(%rsp)
L10:
  test   %rcx,%rcx
  jle    L65
  lea    D24(%rip),%rdi
  call   <printf@plt>
L56:
  lea    0x10(%rsp),%rax
  lea    0x38(%rsp),%rdx
L102:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L102
L63:
  mov    %r13,%rdi
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%rcx
  test   %rcx,%rcx
  je     L78
  jle    L65
  mov    $0x0,%edi
  mov    $0x1,%esi
  jmp    L75
L78:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L106
  mov    $0x0,%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L106:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D31
datatype: string
name: retrieve_data
arguments: data_tag: D12
datatype: string
name: retrieve_data
arguments: data_tag: D6
datatype: string
name: retrieve_data
arguments: data_tag: D24
datatype: string

tool: D31: enter a number:


tool: D12: %ld

tool: D6: no repeated digit


tool: D24: repeated digit


assistant: ```c
int main(void) {
  bool digit_repeated[10] = {false};
  int digit;
  long n;
  while (1) {
    printf("enter a number:\n");
    scanf("%ld", &n);
    if (n != 0) {
      while (n > 0) {
        digit = n % 10;
        if (digit_repeated[digit]) {
          break;
        }
        digit_repeated[digit] = true;
        n = n / 10;
      }
      if (n > 0) {
        printf("repeated digit\n");
      } else {
        printf("no repeated digit\n");
      }
      for (int i = 0; i < 10; i++) {
        digit_repeated[i] = false;
      }
    } else {
      break;
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  pxor   %xmm0,%xmm0
  movabs $0xcccccccccccccccd,%r15
  push   %r14
  push   %r13
  lea    D11(%rip),%r13
  push   %r12
  lea    D18(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D1(%rip),%rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rbp
  movaps %xmm0,0x10(%rsp)
  movq   $0x0,0x30(%rsp)
  movaps %xmm0,0x20(%rsp)
  nopl   (%rax)
L60:
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%rcx
  test   %rcx,%rcx
  je     L31
  jle    L82
  xor    %edi,%edi
  jmp    L58
  nopl   0x0(%rax,%rax,1)
L63:
  movl   $0x1,0x10(%rsp,%rax,4)
  mov    %rdx,%r14
  mov    $0x1,%edi
  cmp    $0x9,%rcx
  jle    L48
  mov    %r14,%rcx
L58:
  mov    %rcx,%rax
  mul    %r15
  mov    %rcx,%rax
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rsi
  add    %rsi,%rsi
  sub    %rsi,%rax
  cltq
  mov    0x10(%rsp,%rax,4),%esi
  test   %esi,%esi
  je     L63
  test   %dil,%dil
  je     L11
  mov    %r14,0x8(%rsp)
L11:
  lea    D9(%rip),%rdi
  call   <printf@plt>
L35:
  movq   $0x0,0x30(%rsp)
  pxor   %xmm0,%xmm0
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  jmp    L60
  nopl   0x0(%rax)
L48:
  mov    %rdx,0x8(%rsp)
L82:
  mov    %r13,%rdi
  call   <printf@plt>
  jmp    L35
  nop
L31:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D11
type_guessed: string
name: unpack_data
arguments: data_label: D18
type_guessed: string
name: unpack_data
arguments: data_label: D1
type_guessed: string
name: unpack_data
arguments: data_label: D9
type_guessed: string

tool: D11: no repeated digit


tool: D18: enter a number:


tool: D1: %ld

tool: D9: repeated digit


assistant: ```c
int main(void) {
  bool digit_repeated[10] = {false};
  int digit;
  long n;
  while (1) {
    printf("enter a number:\n");
    scanf("%ld", &n);
    if (n != 0) {
      while (n > 0) {
        digit = n % 10;
        if (digit_repeated[digit]) {
          break;
        }
        digit_repeated[digit] = true;
        n = n / 10;
      }
      if (n > 0) {
        printf("repeated digit\n");
      } else {
        printf("no repeated digit\n");
      }
      for (int i = 0; i < 10; i++) {
        digit_repeated[i] = false;
      }
    } else {
      break;
    }
  }
  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   %r15
  pxor   %xmm0,%xmm0
  movabs $0xcccccccccccccccd,%r15
  push   %r14
  push   %r13
  lea    D29(%rip),%r13
  push   %r12
  lea    D1(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D27(%rip),%rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rbp
  movaps %xmm0,0x10(%rsp)
  movq   $0x0,0x30(%rsp)
  movaps %xmm0,0x20(%rsp)
  nopl   (%rax)
L82:
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%rcx
  test   %rcx,%rcx
  je     L3
  jle    L38
  xor    %edi,%edi
  jmp    L59
  nopl   0x0(%rax,%rax,1)
L77:
  movl   $0x1,0x10(%rsp,%rax,4)
  mov    %rdx,%r14
  mov    $0x1,%edi
  cmp    $0x9,%rcx
  jle    L68
  mov    %r14,%rcx
L59:
  mov    %rcx,%rax
  mul    %r15
  mov    %rcx,%rax
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rsi
  add    %rsi,%rsi
  sub    %rsi,%rax
  cltq
  mov    0x10(%rsp,%rax,4),%esi
  test   %esi,%esi
  je     L77
  test   %dil,%dil
  je     L72
  mov    %r14,0x8(%rsp)
L72:
  lea    D10(%rip),%rdi
  call   <printf@plt>
L22:
  movq   $0x0,0x30(%rsp)
  pxor   %xmm0,%xmm0
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  jmp    L82
  nopl   0x0(%rax)
L68:
  mov    %rdx,0x8(%rsp)
L38:
  mov    %r13,%rdi
  call   <printf@plt>
  jmp    L22
  nop
L3:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D29
type_guessed: string
name: unpack_data
arguments: data_label: D1
type_guessed: string
name: unpack_data
arguments: data_label: D27
type_guessed: string
name: unpack_data
arguments: data_label: D10
type_guessed: string

tool: D29: no repeated digit


tool: D1: enter a number:


tool: D27: %ld

tool: D10: repeated digit


assistant: ```c
int main(void) {
  bool digit_repeated[10] = {false};
  int digit;
  long n;
  while (1) {
    printf("enter a number:\n");
    scanf("%ld", &n);
    if (n != 0) {
      while (n > 0) {
        digit = n % 10;
        if (digit_repeated[digit]) {
          break;
        }
        digit_repeated[digit] = true;
        n = n / 10;
      }
      if (n > 0) {
        printf("repeated digit\n");
      } else {
        printf("no repeated digit\n");
      }
      for (int i = 0; i < 10; i++) {
        digit_repeated[i] = false;
      }
    } else {
      break;
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<map_print>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L8
L2:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x8(%rbp),%eax
  imul   -0x18(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L24
L27:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0x4(%rbp)
L24:
  mov    -0x8(%rbp),%eax
  add    $0x1,%eax
  imul   -0x18(%rbp),%eax
  cmp    %eax,-0x4(%rbp)
  jl     L27
  addl   $0x1,-0x8(%rbp)
L8:
  mov    -0x14(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jg     L2
  mov    $0x0,%eax
  call   <getch@plt>
  nop
  leave
  ret
```

assistant: ```c
void map_print(int order_r, int order_c, char matrix[]) {
  int i, j;
  for (i = 0; i <= order_r - 1; ++i) {
    printf("\n");
    for (j = (i * order_c); j <= ((order_c) * (i + 1)) - 1; ++j) {
      printf("%c", matrix[j]);
    }
  }
  getch();
}
```
user: What is the c source code of the assembly code below:

```asm
<map_print>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %edi,%edi
  jle    L4
  mov    %edi,%r15d
  mov    %esi,%r14d
  movslq %esi,%rax
  mov    %rax,(%rsp)
  lea    -0x1(%rsi),%eax
  mov    %rax,0x8(%rsp)
  lea    0x1(%rdx,%rax,1),%rbp
  mov    $0x0,%r12d
  mov    $0x0,%r13d
L3:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r12d,%eax
  add    $0x1,%r13d
  add    %r14d,%r12d
  cmp    %r12d,%eax
  jge    L33
  mov    %rbp,%rbx
  sub    0x8(%rsp),%rbx
  sub    $0x1,%rbx
L21:
  movsbl (%rbx),%edi
  call   <putchar@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L21
L33:
  add    (%rsp),%rbp
  cmp    %r13d,%r15d
  jne    L3
L4:
  mov    $0x0,%eax
  call   <getch@plt>
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void map_print(int order_r, int order_c, char matrix[]) {
  int i, j;
  for (i = 0; i <= order_r - 1; ++i) {
    printf("\n");
    for (j = (i * order_c); j <= ((order_c) * (i + 1)) - 1; ++j) {
      printf("%c", matrix[j]);
    }
  }
  getch();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<map_print>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,0x4(%rsp)
  test   %edi,%edi
  jle    L16
  lea    -0x1(%rsi),%eax
  mov    %esi,%r13d
  movslq %esi,%r14
  xor    %ebp,%ebp
  mov    %rax,0x8(%rsp)
  lea    0x1(%rdx,%rax,1),%rbx
  xor    %r12d,%r12d
  nopw   0x0(%rax,%rax,1)
L21:
  mov    $0xa,%edi
  add    $0x1,%r12d
  call   <putchar@plt>
  mov    %ebp,%eax
  add    %r13d,%ebp
  cmp    %ebp,%eax
  jge    L27
  mov    %rbx,%rdx
  sub    0x8(%rsp),%rdx
  lea    -0x1(%rdx),%r15
  nopl   0x0(%rax,%rax,1)
L7:
  movsbl (%r15),%edi
  add    $0x1,%r15
  call   <putchar@plt>
  cmp    %rbx,%r15
  jne    L7
L27:
  add    %r14,%rbx
  cmp    %r12d,0x4(%rsp)
  jne    L21
L16:
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <getch@plt>
```

assistant: ```c
void map_print(int order_r, int order_c, char matrix[]) {
  int i, j;
  for (i = 0; i <= order_r - 1; ++i) {
    printf("\n");
    for (j = (i * order_c); j <= ((order_c) * (i + 1)) - 1; ++j) {
      printf("%c", matrix[j]);
    }
  }
  getch();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<map_print>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,0x4(%rsp)
  test   %edi,%edi
  jle    L0
  lea    -0x1(%rsi),%eax
  mov    %esi,%r13d
  movslq %esi,%r14
  xor    %ebp,%ebp
  mov    %rax,0x8(%rsp)
  lea    0x1(%rdx,%rax,1),%rbx
  xor    %r12d,%r12d
  nopw   0x0(%rax,%rax,1)
L32:
  mov    $0xa,%edi
  add    $0x1,%r12d
  call   <putchar@plt>
  mov    %ebp,%eax
  add    %r13d,%ebp
  cmp    %ebp,%eax
  jge    L26
  mov    %rbx,%rdx
  sub    0x8(%rsp),%rdx
  lea    -0x1(%rdx),%r15
  nopl   0x0(%rax,%rax,1)
L3:
  movsbl (%r15),%edi
  add    $0x1,%r15
  call   <putchar@plt>
  cmp    %rbx,%r15
  jne    L3
L26:
  add    %r14,%rbx
  cmp    %r12d,0x4(%rsp)
  jne    L32
L0:
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <getch@plt>
```

assistant: ```c
void map_print(int order_r, int order_c, char matrix[]) {
  int i, j;
  for (i = 0; i <= order_r - 1; ++i) {
    printf("\n");
    for (j = (i * order_c); j <= ((order_c) * (i + 1)) - 1; ++j) {
      printf("%c", matrix[j]);
    }
  }
  getch();
}
```
user: Translate the below assembly code into C programming language:

```asm
<WriteHostWord>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L28
L12:
  mov    -0x8(%rbp),%rax
  shl    $0x3,%eax
  mov    -0x18(%rbp),%rdx
  mov    %eax,%ecx
  shr    %cl,%rdx
  mov    %rdx,%rax
  movzbl %al,%eax
  mov    %rax,%rsi
  mov    $0x67,%edi
  call   <writeVIC@plt>
  mov    $0x0,%esi
  mov    $0x63,%edi
  call   <writeVIC@plt>
  nop
L29:
  mov    $0x63,%edi
  call   <readVIC@plt>
  test   %eax,%eax
  je     L29
  addq   $0x1,-0x8(%rbp)
L28:
  cmpq   $0x3,-0x8(%rbp)
  jbe    L12
  nop
  nop
  leave
  ret
```

assistant: ```c
void WriteHostWord(unsigned long val) {
  unsigned long i;
  for (i = 0; i < 4; i++) {
    writeVIC(0x67, (val >> (8 * i)) & 0xFF);
    writeVIC(0x63, 0);
    while (!(readVIC(0x63)))
      ;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<WriteHostWord>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x0,%ebx
L1:
  mov    %rbp,%rsi
  mov    %ebx,%ecx
  shr    %cl,%rsi
  movzbl %sil,%esi
  mov    $0x67,%edi
  call   <writeVIC@plt>
  mov    $0x0,%esi
  mov    $0x63,%edi
  call   <writeVIC@plt>
L10:
  mov    $0x63,%edi
  call   <readVIC@plt>
  test   %eax,%eax
  je     L10
  add    $0x8,%ebx
  cmp    $0x20,%ebx
  jne    L1
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void WriteHostWord(unsigned long val) {
  unsigned long i;
  for (i = 0; i < 4; i++) {
    writeVIC(0x67, (val >> (8 * i)) & 0xFF);
    writeVIC(0x63, 0);
    while (!(readVIC(0x63)))
      ;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<WriteHostWord>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
L18:
  mov    %ebx,%ecx
  mov    %rbp,%rsi
  mov    $0x67,%edi
  shr    %cl,%rsi
  movzbl %sil,%esi
  call   <writeVIC@plt>
  xor    %esi,%esi
  mov    $0x63,%edi
  call   <writeVIC@plt>
  nopl   0x0(%rax)
L14:
  mov    $0x63,%edi
  call   <readVIC@plt>
  test   %eax,%eax
  je     L14
  add    $0x8,%ebx
  cmp    $0x20,%ebx
  jne    L18
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void WriteHostWord(unsigned long val) {
  unsigned long i;
  for (i = 0; i < 4; i++) {
    writeVIC(0x67, (val >> (8 * i)) & 0xFF);
    writeVIC(0x63, 0);
    while (!(readVIC(0x63)))
      ;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<WriteHostWord>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
L11:
  mov    %ebx,%ecx
  mov    %rbp,%rsi
  mov    $0x67,%edi
  shr    %cl,%rsi
  movzbl %sil,%esi
  call   <writeVIC@plt>
  xor    %esi,%esi
  mov    $0x63,%edi
  call   <writeVIC@plt>
  nopl   0x0(%rax)
L15:
  mov    $0x63,%edi
  call   <readVIC@plt>
  test   %eax,%eax
  je     L15
  add    $0x8,%ebx
  cmp    $0x20,%ebx
  jne    L11
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void WriteHostWord(unsigned long val) {
  unsigned long i;
  for (i = 0; i < 4; i++) {
    writeVIC(0x67, (val >> (8 * i)) & 0xFF);
    writeVIC(0x63, 0);
    while (!(readVIC(0x63)))
      ;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<g_us>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x190,%rsp
  mov    %rdi,-0x188(%rbp)
  mov    %rsi,-0x190(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x170(%rbp),%rax
  mov    $0x168,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0x0,-0x17c(%rbp)
  movl   $0x0,-0x178(%rbp)
  movl   $0x0,-0x174(%rbp)
  jmp    L22
L21:
  mov    -0x17c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x188(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x178(%rbp),%eax
  cltq
  movsd  %xmm0,-0x170(%rbp,%rax,8)
  addl   $0x1,-0x17c(%rbp)
  addl   $0x3,-0x178(%rbp)
  addl   $0x1,-0x174(%rbp)
L22:
  cmpl   $0xe,-0x174(%rbp)
  jle    L21
  lea    -0x170(%rbp),%rcx
  mov    -0x190(%rbp),%rax
  mov    $0x158,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

assistant: ```c
void g_us(const double o[15], double u[43]) {
  double b_u[45];
  int ix;
  int iy;
  int k;
  memset(&b_u[0], 0, 45U * sizeof(double));
  ix = 0;
  iy = 0;
  for (k = 0; k < 15; k++) {
    b_u[iy] = o[ix];
    ix++;
    iy += 3;
  }
  memcpy(&u[0], &b_u[0], 43U * sizeof(double));
}
```
user: What is the c source code of the assembly code below?

```asm
<g_us>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x170,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x168(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    $0x168,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    %rbp,%rdi
  mov    %rbx,%rax
  lea    0x168(%rsp),%rdx
L9:
  movsd  (%rdi),%xmm0
  movsd  %xmm0,(%rax)
  add    $0x8,%rdi
  add    $0x18,%rax
  cmp    %rdx,%rax
  jne    L9
  mov    %rsp,%rsi
  mov    $0x158,%edx
  mov    %r12,%rdi
  call   <memcpy@plt>
  mov    0x168(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x170,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void g_us(const double o[15], double u[43]) {
  double b_u[45];
  int ix;
  int iy;
  int k;
  memset(&b_u[0], 0, 45U * sizeof(double));
  ix = 0;
  iy = 0;
  for (k = 0; k < 15; k++) {
    b_u[iy] = o[ix];
    ix++;
    iy += 3;
  }
  memcpy(&u[0], &b_u[0], 43U * sizeof(double));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<g_us>:
  endbr64
  push   %r12
  mov    $0x168,%edx
  push   %rbp
  mov    %rsi,%rbp
  xor    %esi,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x170,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x168(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <memset@plt>
  mov    %rbx,%rdi
  mov    %r12,%rax
  lea    0x168(%rsp),%rdx
  nopl   0x0(%rax,%rax,1)
L4:
  movsd  (%rdi),%xmm0
  add    $0x18,%rax
  add    $0x8,%rdi
  movsd  %xmm0,-0x18(%rax)
  cmp    %rdx,%rax
  jne    L4
  mov    $0x158,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  mov    0x168(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x170,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void g_us(const double o[15], double u[43]) {
  double b_u[45];
  int ix;
  int iy;
  int k;
  memset(&b_u[0], 0, 45U * sizeof(double));
  ix = 0;
  iy = 0;
  for (k = 0; k < 15; k++) {
    b_u[iy] = o[ix];
    ix++;
    iy += 3;
  }
  memcpy(&u[0], &b_u[0], 43U * sizeof(double));
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<g_us>:
  endbr64
  push   %r12
  mov    $0x168,%edx
  push   %rbp
  mov    %rsi,%rbp
  xor    %esi,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x170,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x168(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <memset@plt>
  movupd (%rbx),%xmm0
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    $0x158,%edx
  movlpd %xmm0,(%rsp)
  movhpd %xmm0,0x18(%rsp)
  movupd 0x10(%rbx),%xmm0
  movlpd %xmm0,0x30(%rsp)
  movhpd %xmm0,0x48(%rsp)
  movupd 0x20(%rbx),%xmm0
  movlpd %xmm0,0x60(%rsp)
  movhpd %xmm0,0x78(%rsp)
  movupd 0x30(%rbx),%xmm0
  movlpd %xmm0,0x90(%rsp)
  movhpd %xmm0,0xa8(%rsp)
  movupd 0x40(%rbx),%xmm0
  movlpd %xmm0,0xc0(%rsp)
  movhpd %xmm0,0xd8(%rsp)
  movupd 0x50(%rbx),%xmm0
  movlpd %xmm0,0xf0(%rsp)
  movhpd %xmm0,0x108(%rsp)
  movupd 0x60(%rbx),%xmm0
  movlpd %xmm0,0x120(%rsp)
  movhpd %xmm0,0x138(%rsp)
  movsd  0x70(%rbx),%xmm0
  movsd  %xmm0,0x150(%rsp)
  call   <memcpy@plt>
  mov    0x168(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x170,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void g_us(const double o[15], double u[43]) {
  double b_u[45];
  int ix;
  int iy;
  int k;
  memset(&b_u[0], 0, 45U * sizeof(double));
  ix = 0;
  iy = 0;
  for (k = 0; k < 15; k++) {
    b_u[iy] = o[ix];
    ix++;
    iy += 3;
  }
  memcpy(&u[0], &b_u[0], 43U * sizeof(double));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<trashInput>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    $0x0,%edi
  call   <timeout@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L9
L11:
  mov    $0x0,%eax
  call   <getch@plt>
  addl   $0x1,-0x4(%rbp)
L9:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L11
  mov    $0xffffffff,%edi
  call   <timeout@plt>
  nop
  leave
  ret
```

assistant: ```c
void trashInput(int quantity) {
  int j;
  timeout(0);
  for (j = 0; j < quantity; j++)
    getch();
  timeout(-1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<trashInput>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    $0x0,%edi
  call   <timeout@plt>
  test   %ebp,%ebp
  jle    L13
  mov    $0x0,%ebx
L7:
  mov    $0x0,%eax
  call   <getch@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L7
L13:
  mov    $0xffffffff,%edi
  call   <timeout@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void trashInput(int quantity) {
  int j;
  timeout(0);
  for (j = 0; j < quantity; j++)
    getch();
  timeout(-1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<trashInput>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  xor    %edi,%edi
  push   %rbx
  sub    $0x8,%rsp
  call   <timeout@plt>
  test   %ebp,%ebp
  jle    L6
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L7:
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <getch@plt>
  cmp    %ebx,%ebp
  jne    L7
L6:
  add    $0x8,%rsp
  mov    $0xffffffff,%edi
  pop    %rbx
  pop    %rbp
  jmp    1070 <timeout@plt>
```

assistant: ```c
void trashInput(int quantity) {
  int j;
  timeout(0);
  for (j = 0; j < quantity; j++)
    getch();
  timeout(-1);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<trashInput>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  xor    %edi,%edi
  push   %rbx
  sub    $0x8,%rsp
  call   <timeout@plt>
  test   %ebp,%ebp
  jle    L2
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L19:
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <getch@plt>
  cmp    %ebx,%ebp
  jne    L19
L2:
  add    $0x8,%rsp
  mov    $0xffffffff,%edi
  pop    %rbx
  pop    %rbp
  jmp    1070 <timeout@plt>
```

assistant: ```c
void trashInput(int quantity) {
  int j;
  timeout(0);
  for (j = 0; j < quantity; j++)
    getch();
  timeout(-1);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<parse_Letter>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movb   $0x0,-0x9(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <getCharacter@plt>
  mov    %al,-0x9(%rbp)
  mov    -0x20(%rbp),%rax
  movzbl -0x9(%rbp),%edx
  mov    %dl,(%rax)
  cmpb   $0x40,-0x9(%rbp)
  jle    L0
  cmpb   $0x5a,-0x9(%rbp)
  jle    L37
L0:
  cmpb   $0x60,-0x9(%rbp)
  jle    L16
  cmpb   $0x7a,-0x9(%rbp)
  jle    L37
L16:
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    $0x1,%eax
  jmp    L45
L37:
  mov    $0x0,%eax
L45:
  leave
  ret
```

assistant: ```c
int parse_Letter(char **xml, char *letter) {
  char *pos1;
  char c = '\0';
  pos1 = *xml;
  c = getCharacter(xml);
  *letter = c;
  if (!((c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z'))) {
    *xml = pos1;
    return 1;
  }
  return 0;
}
```