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
<caesarShift>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edi,%r12d
  push   %rbx
  movsbl %dil,%ebx
  mov    %ebx,%edi
  call   <isalpha@plt>
  test   %eax,%eax
  je     L3
  mov    %ebx,%edi
  call   <islower@plt>
  mov    $0x1a,%esi
  cmp    $0x1,%rax
  sbb    %r12d,%r12d
  and    $0xffffffe0,%r12d
  add    $0x61,%r12d
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0xffffffe0,%eax
  sub    %eax,%ebx
  lea    -0x61(%rbx,%r13,1),%edi
  call   <mod@plt>
  add    %eax,%r12d
L3:
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char caesarShift(const char toShift, const int shiftVal) {
  if (!isalpha(toShift))
    return toShift;
  int charIndex, aValue;
  if (islower(toShift))
    aValue = 97;
  else
    aValue = 65;
  charIndex = toShift - aValue;
  return (char)(mod(charIndex + shiftVal, 26) + aValue);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<caesarShift>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edi,%r12d
  push   %rbx
  movsbl %dil,%ebx
  mov    %ebx,%edi
  call   <isalpha@plt>
  test   %eax,%eax
  je     L6
  mov    %ebx,%edi
  call   <islower@plt>
  mov    $0x1a,%esi
  cmp    $0x1,%rax
  sbb    %r12d,%r12d
  and    $0xffffffe0,%r12d
  add    $0x61,%r12d
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0xffffffe0,%eax
  sub    %eax,%ebx
  lea    -0x61(%rbx,%r13,1),%edi
  call   <mod@plt>
  add    %eax,%r12d
L6:
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char caesarShift(const char toShift, const int shiftVal) {
  if (!isalpha(toShift))
    return toShift;
  int charIndex, aValue;
  if (islower(toShift))
    aValue = 97;
  else
    aValue = 65;
  charIndex = toShift - aValue;
  return (char)(mod(charIndex + shiftVal, 26) + aValue);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x24(%rbp)
  jmp    L28
L8:
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    -0x24(%rbp),%eax
  sub    $0x1,%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0x24(%rbp)
L28:
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x24(%rbp)
  jl     L8
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D32
type_guessed: string
name: disassemble_data
arguments: tag: D6
type_guessed: string
name: disassemble_data
arguments: tag: D13
type_guessed: string
name: disassemble_data
arguments: tag: D28
type_guessed: string

tool: D32: You want to say:

tool: D6: %19s

tool: D13: 
I understood: %s

tool: D28: 
I reversed: 

assistant: ```c
int main(void) {
  char string_1[20];
  printf("You want to say:");
  scanf("%19s", string_1);
  printf("\nI understood: %s", string_1);
  printf("\nI reversed: ");
  for (int index = 0; index < strlen(string_1); index++) {
    printf("%c", string_1[strlen(string_1) - index - 1]);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  lea    D32(%rip),%rdi
  call   <scanf@plt>
  mov    %rbx,%rsi
  lea    D22(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D36(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  mov    %rsp,%rbp
  jmp    L13
L27:
  mov    %rbp,%rdi
  call   <strlen@plt>
  sub    %ebx,%eax
  sub    $0x1,%eax
  cltq
  movsbl (%rsp,%rax,1),%edi
  call   <putchar@plt>
  add    $0x1,%ebx
L13:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L27
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D11
type_infered: string
name: read_data
arguments: tag: D32
type_infered: string
name: read_data
arguments: tag: D22
type_infered: string
name: read_data
arguments: tag: D36
type_infered: string

tool: D11: You want to say:

tool: D32: %19s

tool: D22: 
I understood: %s

tool: D36: 
I reversed: 

assistant: ```c
int main(void) {
  char string_1[20];
  printf("You want to say:");
  scanf("%19s", string_1);
  printf("\nI understood: %s", string_1);
  printf("\nI reversed: ");
  for (int index = 0; index < strlen(string_1); index++) {
    printf("%c", string_1[strlen(string_1) - index - 1]);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  lea    D3(%rip),%rdi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D20(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rsi
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D34(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L3
  nopl   0x0(%rax,%rax,1)
L13:
  mov    %rbp,%rdi
  call   <strlen@plt>
  sub    %ebx,%eax
  add    $0x1,%ebx
  sub    $0x1,%eax
  cltq
  movsbl (%rsp,%rax,1),%edi
  call   <putchar@plt>
L3:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L13
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D3
datatype: string
name: read_data
arguments: data_label: D20
datatype: string
name: read_data
arguments: data_label: D27
datatype: string
name: read_data
arguments: data_label: D34
datatype: string

tool: D3: You want to say:

tool: D20: %19s

tool: D27: 
I understood: %s

tool: D34: 
I reversed: 

assistant: ```c
int main(void) {
  char string_1[20];
  printf("You want to say:");
  scanf("%19s", string_1);
  printf("\nI understood: %s", string_1);
  printf("\nI reversed: ");
  for (int index = 0; index < strlen(string_1); index++) {
    printf("%c", string_1[strlen(string_1) - index - 1]);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D6(%rip),%rdi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D4(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rsi
  lea    D32(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D30(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L27
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %rbp,%rdi
  call   <strlen@plt>
  sub    %ebx,%eax
  add    $0x1,%ebx
  sub    $0x1,%eax
  cltq
  movsbl (%rsp,%rax,1),%edi
  call   <putchar@plt>
L27:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L1
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D6
type: string
name: disassemble_data
arguments: data_label: D4
type: string
name: disassemble_data
arguments: data_label: D32
type: string
name: disassemble_data
arguments: data_label: D30
type: string

tool: D6: You want to say:

tool: D4: %19s

tool: D32: 
I understood: %s

tool: D30: 
I reversed: 

assistant: ```c
int main(void) {
  char string_1[20];
  printf("You want to say:");
  scanf("%19s", string_1);
  printf("\nI understood: %s", string_1);
  printf("\nI reversed: ");
  for (int index = 0; index < strlen(string_1); index++) {
    printf("%c", string_1[strlen(string_1) - index - 1]);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<eval_6hand_fast>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xc0,%rsp
  mov    %rdi,-0xb8(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movw   $0x270f,-0xae(%rbp)
  movl   $0x0,-0x80(%rbp)
  movl   $0x1,-0x7c(%rbp)
  movl   $0x2,-0x78(%rbp)
  movl   $0x3,-0x74(%rbp)
  movl   $0x4,-0x70(%rbp)
  movl   $0x0,-0x6c(%rbp)
  movl   $0x1,-0x68(%rbp)
  movl   $0x2,-0x64(%rbp)
  movl   $0x3,-0x60(%rbp)
  movl   $0x5,-0x5c(%rbp)
  movl   $0x0,-0x58(%rbp)
  movl   $0x1,-0x54(%rbp)
  movl   $0x2,-0x50(%rbp)
  movl   $0x4,-0x4c(%rbp)
  movl   $0x5,-0x48(%rbp)
  movl   $0x0,-0x44(%rbp)
  movl   $0x1,-0x40(%rbp)
  movl   $0x3,-0x3c(%rbp)
  movl   $0x4,-0x38(%rbp)
  movl   $0x5,-0x34(%rbp)
  movl   $0x0,-0x30(%rbp)
  movl   $0x2,-0x2c(%rbp)
  movl   $0x3,-0x28(%rbp)
  movl   $0x4,-0x24(%rbp)
  movl   $0x5,-0x20(%rbp)
  movl   $0x1,-0x1c(%rbp)
  movl   $0x2,-0x18(%rbp)
  movl   $0x3,-0x14(%rbp)
  movl   $0x4,-0x10(%rbp)
  movl   $0x5,-0xc(%rbp)
  movl   $0x0,-0xac(%rbp)
  jmp    L8
L47:
  movl   $0x0,-0xa8(%rbp)
  jmp    L19
L35:
  mov    -0xa8(%rbp),%eax
  movslq %eax,%rcx
  mov    -0xac(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rcx,%rax
  mov    -0x80(%rbp,%rax,4),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0xb8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xa8(%rbp),%eax
  cltq
  mov    %edx,-0xa0(%rbp,%rax,4)
  addl   $0x1,-0xa8(%rbp)
L19:
  cmpl   $0x4,-0xa8(%rbp)
  jle    L35
  lea    -0xa0(%rbp),%rax
  mov    %rax,%rdi
  call   <eval_5hand_fast@plt>
  mov    %eax,-0xa4(%rbp)
  movswl -0xae(%rbp),%eax
  cmp    %eax,-0xa4(%rbp)
  jge    L10
  mov    -0xa4(%rbp),%eax
  mov    %ax,-0xae(%rbp)
L10:
  addl   $0x1,-0xac(%rbp)
L8:
  cmpl   $0x5,-0xac(%rbp)
  jle    L47
  movzwl -0xae(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L38
  call   <__stack_chk_fail@plt>
L38:
  leave
  ret
```

assistant: ```c
short eval_6hand_fast(int *hand) {
  int i;
  short best = 9999;
  int perm[][5] = {
      {0, 1, 2, 3, 4}, {0, 1, 2, 3, 5}, {0, 1, 2, 4, 5}, {0, 1, 3, 4, 5}, {0, 2, 3, 4, 5}, {1, 2, 3, 4, 5},
  };
  for (i = 0; i < 6; i++) {
    int subhand[5];
    int j;
    for (j = 0; j < 5; j++)
      subhand[j] = hand[perm[i][j]];
    int q = eval_5hand_fast(subhand);
    if (q < best)
      best = q;
  }
  return best;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<eval_6hand_fast>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa8,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x20(%rsp)
  movl   $0x1,0x24(%rsp)
  movl   $0x2,0x28(%rsp)
  movl   $0x3,0x2c(%rsp)
  movl   $0x4,0x30(%rsp)
  movl   $0x0,0x34(%rsp)
  movl   $0x1,0x38(%rsp)
  movl   $0x2,0x3c(%rsp)
  movl   $0x3,0x40(%rsp)
  movl   $0x5,0x44(%rsp)
  movl   $0x0,0x48(%rsp)
  movl   $0x1,0x4c(%rsp)
  movl   $0x2,0x50(%rsp)
  movl   $0x4,0x54(%rsp)
  movl   $0x5,0x58(%rsp)
  movl   $0x0,0x5c(%rsp)
  movl   $0x1,0x60(%rsp)
  movl   $0x3,0x64(%rsp)
  movl   $0x4,0x68(%rsp)
  movl   $0x5,0x6c(%rsp)
  movl   $0x0,0x70(%rsp)
  movl   $0x2,0x74(%rsp)
  movl   $0x3,0x78(%rsp)
  movl   $0x4,0x7c(%rsp)
  movl   $0x5,0x80(%rsp)
  movl   $0x1,0x84(%rsp)
  movl   $0x2,0x88(%rsp)
  movl   $0x3,0x8c(%rsp)
  movl   $0x4,0x90(%rsp)
  movl   $0x5,0x94(%rsp)
  lea    0x20(%rsp),%rbx
  lea    0x98(%rsp),%r13
  mov    $0x270f,%r12d
  jmp    L16
L49:
  mov    %rsp,%rdi
  call   <eval_5hand_fast@plt>
  movswl %r12w,%edx
  cmp    %eax,%edx
  cmovg  %eax,%r12d
  add    $0x14,%rbx
  cmp    %r13,%rbx
  je     L38
L16:
  mov    $0x0,%eax
L36:
  movslq (%rbx,%rax,1),%rdx
  mov    0x0(%rbp,%rdx,4),%edx
  mov    %edx,(%rsp,%rax,1)
  add    $0x4,%rax
  cmp    $0x14,%rax
  jne    L36
  jmp    L49
L38:
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  mov    %r12d,%eax
  add    $0xa8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
short eval_6hand_fast(int *hand) {
  int i;
  short best = 9999;
  int perm[][5] = {
      {0, 1, 2, 3, 4}, {0, 1, 2, 3, 5}, {0, 1, 2, 4, 5}, {0, 1, 3, 4, 5}, {0, 2, 3, 4, 5}, {1, 2, 3, 4, 5},
  };
  for (i = 0; i < 6; i++) {
    int subhand[5];
    int j;
    for (j = 0; j < 5; j++)
      subhand[j] = hand[perm[i][j]];
    int q = eval_5hand_fast(subhand);
    if (q < best)
      best = q;
  }
  return best;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<eval_6hand_fast>:
  endbr64
  movabs $0x200000000,%rcx
  push   %r14
  movabs $0x400000003,%rsi
  movabs $0x300000002,%rdx
  push   %r13
  mov    $0x270f,%r14d
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  mov    %rcx,0x70(%rsp)
  lea    0x20(%rsp),%rbx
  mov    %rsp,%r12
  movabs $0x200000001,%rax
  mov    %rsi,0x78(%rsp)
  movabs $0x100000005,%rcx
  lea    0x98(%rsp),%rbp
  mov    %rax,0x38(%rsp)
  movabs $0x500000003,%rax
  mov    %rax,0x40(%rsp)
  movabs $0x100000000,%rax
  mov    %rax,0x48(%rsp)
  movabs $0x400000002,%rax
  mov    %rax,0x50(%rsp)
  movabs $0x300000001,%rax
  mov    %rax,0x60(%rsp)
  movabs $0x500000004,%rax
  movl   $0x1,0x24(%rsp)
  movq   $0x4,0x30(%rsp)
  movq   $0x5,0x58(%rsp)
  mov    %rax,0x68(%rsp)
  mov    %rcx,0x80(%rsp)
  mov    %rax,0x90(%rsp)
  mov    %rdx,0x28(%rsp)
  mov    %rdx,0x88(%rsp)
  xor    %edx,%edx
  nopl   0x0(%rax)
L0:
  xor    %eax,%eax
L20:
  mov    0x0(%r13,%rdx,4),%edx
  mov    %edx,(%r12,%rax,1)
  add    $0x4,%rax
  cmp    $0x14,%rax
  je     L37
  movslq (%rbx,%rax,1),%rdx
  jmp    L20
  nopl   0x0(%rax,%rax,1)
L37:
  mov    %r12,%rdi
  call   <eval_5hand_fast@plt>
  movswl %r14w,%edx
  cmp    %eax,%edx
  cmovg  %eax,%r14d
  add    $0x14,%rbx
  cmp    %rbp,%rbx
  je     L17
  movslq (%rbx),%rdx
  jmp    L0
  nopl   0x0(%rax,%rax,1)
L17:
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  add    $0xa0,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L32:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
short eval_6hand_fast(int *hand) {
  int i;
  short best = 9999;
  int perm[][5] = {
      {0, 1, 2, 3, 4}, {0, 1, 2, 3, 5}, {0, 1, 2, 4, 5}, {0, 1, 3, 4, 5}, {0, 2, 3, 4, 5}, {1, 2, 3, 4, 5},
  };
  for (i = 0; i < 6; i++) {
    int subhand[5];
    int j;
    for (j = 0; j < 5; j++)
      subhand[j] = hand[perm[i][j]];
    int q = eval_5hand_fast(subhand);
    if (q < best)
      best = q;
  }
  return best;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<decryptstr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  movl   $0x0,-0x1c(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  mov    $0x1,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0x0,-0x20(%rbp)
  jmp    L11
L12:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L31
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x20(%rbp),%ecx
  movslq %ecx,%rsi
  mov    -0x28(%rbp),%rcx
  lea    (%rsi,%rcx,1),%rbx
  mov    %edx,%esi
  mov    %eax,%edi
  call   <decrypt@plt>
  mov    %al,(%rbx)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  jmp    L9
L31:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x20,(%rax)
L9:
  addl   $0x1,-0x20(%rbp)
L11:
  mov    -0x20(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L12
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void decryptstr(char *plaintext, const char *keyword, const char *cyphertext) {
  int i, j = 0;
  int keylen = strlen(keyword);
  int textlen = strlen(cyphertext);
  memset(plaintext, 0, sizeof(char));
  if (sizeof(plaintext) != sizeof(cyphertext)) {
    printf("plaintext: %ld, cyphertext: %ld\n", sizeof(plaintext), sizeof(cyphertext));
    return;
  }
  for (i = 0; i < textlen; i++) {
    if (cyphertext[i] != ' ') {
      plaintext[i] = decrypt(keyword[j], cyphertext[i]);
      j = (j + 1) % keylen;
    } else {
      plaintext[i] = ' ';
    }
  }
  plaintext[textlen] = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<decryptstr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r14
  mov    %rdi,0x8(%rsp)
  mov    %rsi,%r15
  mov    %rdx,%r12
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %eax,%r13d
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %eax,0x4(%rsp)
  mov    $0x1,%edx
  mov    $0x0,%esi
  mov    %r14,%rdi
  call   <memset@plt>
  test   %ebx,%ebx
  jle    L49
  mov    %ebx,%eax
  mov    %r12,%rbx
  mov    %r14,%rbp
  lea    -0x1(%rax),%eax
  lea    0x1(%r12,%rax,1),%r14
  mov    $0x0,%r12d
  jmp    L5
L7:
  movsbl %sil,%esi
  movslq %r12d,%rax
  movsbl (%r15,%rax,1),%edi
  call   <decrypt@plt>
  mov    %al,0x0(%rbp)
  lea    0x1(%r12),%eax
  cltd
  idiv   %r13d
  mov    %edx,%r12d
L17:
  add    $0x1,%rbx
  add    $0x1,%rbp
  cmp    %r14,%rbx
  je     L49
L5:
  movzbl (%rbx),%esi
  cmp    $0x20,%sil
  jne    L7
  movb   $0x20,0x0(%rbp)
  jmp    L17
L49:
  movslq 0x4(%rsp),%rax
  mov    0x8(%rsp),%rcx
  movb   $0x0,(%rcx,%rax,1)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void decryptstr(char *plaintext, const char *keyword, const char *cyphertext) {
  int i, j = 0;
  int keylen = strlen(keyword);
  int textlen = strlen(cyphertext);
  memset(plaintext, 0, sizeof(char));
  if (sizeof(plaintext) != sizeof(cyphertext)) {
    printf("plaintext: %ld, cyphertext: %ld\n", sizeof(plaintext), sizeof(cyphertext));
    return;
  }
  for (i = 0; i < textlen; i++) {
    if (cyphertext[i] != ' ') {
      plaintext[i] = decrypt(keyword[j], cyphertext[i]);
      j = (j + 1) % keylen;
    } else {
      plaintext[i] = ' ';
    }
  }
  plaintext[textlen] = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<decryptstr>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  mov    %rdi,%r14
  mov    %rsi,%rdi
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <strlen@plt>
  mov    %r15,%rdi
  mov    %eax,0x8(%rsp)
  call   <strlen@plt>
  xor    %esi,%esi
  mov    $0x1,%edx
  mov    %r14,%rdi
  mov    %eax,%ebx
  mov    %eax,0xc(%rsp)
  call   <memset@plt>
  test   %ebx,%ebx
  jle    L16
  mov    0xc(%rsp),%eax
  mov    %r14,%rbx
  xor    %r13d,%r13d
  sub    $0x1,%eax
  lea    0x1(%r15,%rax,1),%rbp
  nopw   0x0(%rax,%rax,1)
L26:
  movsbl (%r15),%esi
  cmp    $0x20,%sil
  je     L24
  movslq %r13d,%rax
  movsbl (%r12,%rax,1),%edi
  call   <decrypt@plt>
  mov    %eax,%esi
  lea    0x1(%r13),%eax
  cltd
  idivl  0x8(%rsp)
  mov    %edx,%r13d
L24:
  add    $0x1,%r15
  mov    %sil,(%rbx)
  add    $0x1,%rbx
  cmp    %rbp,%r15
  jne    L26
L16:
  movslq 0xc(%rsp),%r13
  movb   $0x0,(%r14,%r13,1)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void decryptstr(char *plaintext, const char *keyword, const char *cyphertext) {
  int i, j = 0;
  int keylen = strlen(keyword);
  int textlen = strlen(cyphertext);
  memset(plaintext, 0, sizeof(char));
  if (sizeof(plaintext) != sizeof(cyphertext)) {
    printf("plaintext: %ld, cyphertext: %ld\n", sizeof(plaintext), sizeof(cyphertext));
    return;
  }
  for (i = 0; i < textlen; i++) {
    if (cyphertext[i] != ' ') {
      plaintext[i] = decrypt(keyword[j], cyphertext[i]);
      j = (j + 1) % keylen;
    } else {
      plaintext[i] = ' ';
    }
  }
  plaintext[textlen] = '\0';
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<decryptstr>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  mov    %rsi,%rdi
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x18,%rsp
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %eax,0x8(%rsp)
  call   <strlen@plt>
  xor    %esi,%esi
  mov    $0x1,%edx
  mov    %r15,%rdi
  mov    %eax,%r14d
  mov    %eax,0xc(%rsp)
  call   <memset@plt>
  test   %r14d,%r14d
  jle    L27
  mov    0xc(%rsp),%eax
  mov    %r15,%rbp
  xor    %r14d,%r14d
  sub    $0x1,%eax
  lea    0x1(%rbx,%rax,1),%r12
  nopl   0x0(%rax)
L1:
  movsbl (%rbx),%esi
  cmp    $0x20,%sil
  je     L36
  movslq %r14d,%rax
  add    $0x1,%rbx
  add    $0x1,%rbp
  movsbl 0x0(%r13,%rax,1),%edi
  call   <decrypt@plt>
  mov    %eax,%r8d
  lea    0x1(%r14),%eax
  cltd
  mov    %r8b,-0x1(%rbp)
  idivl  0x8(%rsp)
  mov    %edx,%r14d
  cmp    %rbx,%r12
  jne    L1
L27:
  movslq 0xc(%rsp),%r14
  movb   $0x0,(%r15,%r14,1)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L36:
  add    $0x1,%rbx
  movb   $0x20,0x0(%rbp)
  add    $0x1,%rbp
  cmp    %r12,%rbx
  jne    L1
  jmp    L27
```

assistant: ```c
void decryptstr(char *plaintext, const char *keyword, const char *cyphertext) {
  int i, j = 0;
  int keylen = strlen(keyword);
  int textlen = strlen(cyphertext);
  memset(plaintext, 0, sizeof(char));
  if (sizeof(plaintext) != sizeof(cyphertext)) {
    printf("plaintext: %ld, cyphertext: %ld\n", sizeof(plaintext), sizeof(cyphertext));
    return;
  }
  for (i = 0; i < textlen; i++) {
    if (cyphertext[i] != ' ') {
      plaintext[i] = decrypt(keyword[j], cyphertext[i]);
      j = (j + 1) % keylen;
    } else {
      plaintext[i] = ' ';
    }
  }
  plaintext[textlen] = '\0';
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<PPM_scan_ws>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x23,%al
  jne    L6
  jmp    L65
L8:
  nop
L65:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <getc@plt>
  mov    -0x8(%rbp),%rdx
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L8
  jmp    L6
L0:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <getc@plt>
  mov    -0x8(%rbp),%rdx
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x23,%al
  jne    L6
  jmp    L5
L73:
  nop
L5:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <getc@plt>
  mov    -0x8(%rbp),%rdx
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L73
L6:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L0
  nop
  nop
  leave
  ret
```

assistant: ```c
void PPM_scan_ws(unsigned char *c, FILE *fp) {
  if (*c == '#')
    while ((*c = getc(fp)) != '\n')
      continue;
  while (isspace(*c)) {
    *c = getc(fp);
    if (*c == '#')
      while ((*c = getc(fp)) != '\n')
        continue;
  }
}
```