ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D6(%rip),%rdi
  call   <scanf@plt>
  mov    $0x2,%edi
  movsd  (%rsp),%xmm0
  call   <xround@plt>
  mov    %eax,%esi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    $0x5,%edi
  movsd  (%rsp),%xmm0
  call   <xround@plt>
  mov    %eax,%esi
  lea    D41(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  movsd  (%rsp),%xmm0
  call   <xround@plt>
  mov    %eax,%esi
  lea    D25(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  movsd  (%rsp),%xmm0
  call   <xround@plt>
  mov    %eax,%esi
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D6
type_infered: string
name: unpack_data
arguments: tag: D19
type_infered: string
name: unpack_data
arguments: tag: D41
type_infered: string
name: unpack_data
arguments: tag: D25
type_infered: string
name: unpack_data
arguments: tag: D28
type_infered: string

tool: D6: %lf

tool: D19: |%.2lf|


tool: D41: |%.5lf|


tool: D25: |      %.1lf|


tool: D28: |%.1lf      |


assistant: ```c
int main() {
  double num;
  scanf("%lf", &num);
  printf("|%.2lf|\n", xround(num, 2));
  printf("|%.5lf|\n", xround(num, 5));
  printf("|      %.1lf|\n", xround(num, 1));
  printf("|%.1lf      |\n", xround(num, 1));
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D37(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  movsd  (%rsp),%xmm0
  mov    $0x2,%edi
  call   <xround@plt>
  lea    D20(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  movsd  (%rsp),%xmm0
  mov    $0x5,%edi
  call   <xround@plt>
  lea    D19(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  movsd  (%rsp),%xmm0
  mov    $0x1,%edi
  call   <xround@plt>
  lea    D49(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  movsd  (%rsp),%xmm0
  mov    $0x1,%edi
  call   <xround@plt>
  lea    D12(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D37
type_infered: string
name: extract_symbolic_data
arguments: label: D20
type_infered: string
name: extract_symbolic_data
arguments: label: D19
type_infered: string
name: extract_symbolic_data
arguments: label: D49
type_infered: string
name: extract_symbolic_data
arguments: label: D12
type_infered: string

tool: D37: %lf

tool: D20: |%.2lf|


tool: D19: |%.5lf|


tool: D49: |      %.1lf|


tool: D12: |%.1lf      |


assistant: ```c
int main() {
  double num;
  scanf("%lf", &num);
  printf("|%.2lf|\n", xround(num, 2));
  printf("|%.5lf|\n", xround(num, 5));
  printf("|      %.1lf|\n", xround(num, 1));
  printf("|%.1lf      |\n", xround(num, 1));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D6(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  movsd  (%rsp),%xmm0
  mov    $0x2,%edi
  call   <xround@plt>
  lea    D27(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  movsd  (%rsp),%xmm0
  mov    $0x5,%edi
  call   <xround@plt>
  lea    D31(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  movsd  (%rsp),%xmm0
  mov    $0x1,%edi
  call   <xround@plt>
  lea    D42(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  movsd  (%rsp),%xmm0
  mov    $0x1,%edi
  call   <xround@plt>
  lea    D12(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D6
datatype: string
name: parse_data
arguments: label: D27
datatype: string
name: parse_data
arguments: label: D31
datatype: string
name: parse_data
arguments: label: D42
datatype: string
name: parse_data
arguments: label: D12
datatype: string

tool: D6: %lf

tool: D27: |%.2lf|


tool: D31: |%.5lf|


tool: D42: |      %.1lf|


tool: D12: |%.1lf      |


assistant: ```c
int main() {
  double num;
  scanf("%lf", &num);
  printf("|%.2lf|\n", xround(num, 2));
  printf("|%.5lf|\n", xround(num, 5));
  printf("|      %.1lf|\n", xround(num, 1));
  printf("|%.1lf      |\n", xround(num, 1));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    $0x3e8,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <stringpow@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <sumofdigits@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x1c(%rbp),%eax
  mov    -0x18(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D19
datatype: string
name: retrieve_data
arguments: data_label: D0
datatype: string
name: retrieve_data
arguments: data_label: D2
datatype: string

tool: D19: 2

tool: D0: %d

tool: D2: 
took %lu usec


assistant: ```c
int main(void) {
  struct timeval stop, start;
  gettimeofday(&start, NULL);
  char *result;
  int sum;
  result = stringpow("2", 1000);
  sum = sumofdigits(result);
  printf("%d", sum);
  gettimeofday(&stop, NULL);
  printf("\ntook %lu usec\n", stop.tv_usec - start.tv_usec);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  mov    $0x3e8,%esi
  lea    D3(%rip),%rdi
  call   <stringpow@plt>
  mov    %rax,%rdi
  call   <sumofdigits@plt>
  mov    %eax,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  mov    (%rsp),%esi
  sub    0x4(%rsp),%esi
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D3
data_type: string
name: unpack_data
arguments: data_label: D5
data_type: string
name: unpack_data
arguments: data_label: D28
data_type: string

tool: D3: 2

tool: D5: %d

tool: D28: 
took %lu usec


assistant: ```c
int main(void) {
  struct timeval stop, start;
  gettimeofday(&start, NULL);
  char *result;
  int sum;
  result = stringpow("2", 1000);
  sum = sumofdigits(result);
  printf("%d", sum);
  gettimeofday(&stop, NULL);
  printf("\ntook %lu usec\n", stop.tv_usec - start.tv_usec);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <gettimeofday@plt>
  mov    $0x3e8,%esi
  lea    D19(%rip),%rdi
  call   <stringpow@plt>
  mov    %rax,%rdi
  call   <sumofdigits@plt>
  lea    D22(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  mov    (%rsp),%esi
  lea    D24(%rip),%rdi
  sub    0x4(%rsp),%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D19
datatype: string
name: retrieve_data
arguments: data_label: D22
datatype: string
name: retrieve_data
arguments: data_label: D24
datatype: string

tool: D19: 2

tool: D22: %d

tool: D24: 
took %lu usec


assistant: ```c
int main(void) {
  struct timeval stop, start;
  gettimeofday(&start, NULL);
  char *result;
  int sum;
  result = stringpow("2", 1000);
  sum = sumofdigits(result);
  printf("%d", sum);
  gettimeofday(&stop, NULL);
  printf("\ntook %lu usec\n", stop.tv_usec - start.tv_usec);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <gettimeofday@plt>
  mov    $0x3e8,%esi
  lea    D13(%rip),%rdi
  call   <stringpow@plt>
  mov    %rax,%rdi
  call   <sumofdigits@plt>
  lea    D11(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %esi,%esi
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  mov    (%rsp),%esi
  lea    D26(%rip),%rdi
  sub    0x4(%rsp),%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D13
datatype: string
name: unpack_data
arguments: data_label: D11
datatype: string
name: unpack_data
arguments: data_label: D26
datatype: string

tool: D13: 2

tool: D11: %d

tool: D26: 
took %lu usec


assistant: ```c
int main(void) {
  struct timeval stop, start;
  gettimeofday(&start, NULL);
  char *result;
  int sum;
  result = stringpow("2", 1000);
  sum = sumofdigits(result);
  printf("%d", sum);
  gettimeofday(&stop, NULL);
  printf("\ntook %lu usec\n", stop.tv_usec - start.tv_usec);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<schreibe_menue>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  mov    -0x28(%rbp),%ebx
  movslq %ebx,%rax
  sub    $0x1,%rax
  mov    %rax,-0x18(%rbp)
  movslq %ebx,%rax
  mov    %rax,%r8
  mov    $0x0,%r9d
  movl   $0x0,-0x1c(%rbp)
  jmp    L15
L3:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  movslq %ebx,%rax
  imul   %rax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x1c(%rbp)
L15:
  mov    -0x1c(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L3
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D0
type_infered: string

tool: D0: 
(%d)	  %s

assistant: ```c
void schreibe_menue(int menuelaenge, int textlaenge, char menue[menuelaenge][textlaenge]) {
  for (int i = 0; i < menuelaenge; i++)
    printf("\n(%d)\t  %s", i + 1, menue[i]);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<schreibe_menue>:
  endbr64
  test   %edi,%edi
  jle    L4
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  movslq %esi,%r13
  mov    %rdx,%rbp
  mov    $0x0,%ebx
  lea    D8(%rip),%r14
L1:
  add    $0x1,%ebx
  mov    %rbp,%rdx
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <printf@plt>
  add    %r13,%rbp
  cmp    %r12d,%ebx
  jne    L1
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L4:
  ret
```

tool_calls: name: read_data
arguments: tag: D8
type_guessed: string

tool: D8: 
(%d)	  %s

assistant: ```c
void schreibe_menue(int menuelaenge, int textlaenge, char menue[menuelaenge][textlaenge]) {
  for (int i = 0; i < menuelaenge; i++)
    printf("\n(%d)\t  %s", i + 1, menue[i]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<schreibe_menue>:
  endbr64
  test   %edi,%edi
  jle    L11
  push   %r14
  lea    D0(%rip),%r14
  push   %r13
  movslq %esi,%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L5:
  add    $0x1,%ebx
  mov    %rbp,%rdx
  mov    %r14,%rdi
  add    %r13,%rbp
  mov    %ebx,%esi
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L5
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L11:
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D0
type_infered: string

tool: D0: 
(%d)	  %s

assistant: ```c
void schreibe_menue(int menuelaenge, int textlaenge, char menue[menuelaenge][textlaenge]) {
  for (int i = 0; i < menuelaenge; i++)
    printf("\n(%d)\t  %s", i + 1, menue[i]);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<schreibe_menue>:
  endbr64
  test   %edi,%edi
  jle    L15
  push   %r14
  lea    D4(%rip),%r14
  push   %r13
  movslq %esi,%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L1:
  add    $0x1,%ebx
  mov    %rbp,%rdx
  mov    %r14,%rdi
  add    %r13,%rbp
  mov    %ebx,%esi
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L1
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L15:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D4
type_infered: string

tool: D4: 
(%d)	  %s

assistant: ```c
void schreibe_menue(int menuelaenge, int textlaenge, char menue[menuelaenge][textlaenge]) {
  for (int i = 0; i < menuelaenge; i++)
    printf("\n(%d)\t  %s", i + 1, menue[i]);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  movl   $0x2a,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x18(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  cltq
  add    $0x4,%rax
  mov    %rax,%rcx
  mov    $0x0,%ebx
  mov    %rcx,%rax
  mov    %rbx,%rdx
  shld   $0x3,%rax,%rdx
  shl    $0x3,%rax
  add    $0x1f,%rax
  adc    $0x0,%rdx
  mov    %rax,%rbx
  and    $0xffffffffffffffe0,%rbx
  mov    %rbx,%rsi
  mov    %rdx,%rax
  and    $0xff,%ah
  mov    %rax,%rdi
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int main() {
  int size = 42;
  struct S {
    int size;
    unsigned char data[size];
  };
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  int size = 42;
  struct S {
    int size;
    unsigned char data[size];
  };
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int size = 42;
  struct S {
    int size;
    unsigned char data[size];
  };
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int size = 42;
  struct S {
    int size;
    unsigned char data[size];
  };
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L48
L29:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L0
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L32
L23:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x2(%rdx),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  subl   $0x1,-0x8(%rbp)
L32:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jg     L23
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x2(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x25,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x32,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x2(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x30,(%rax)
  addl   $0x2,-0x1c(%rbp)
L0:
  addl   $0x1,-0x4(%rbp)
L48:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L29
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *rs(char str[], int len) {
  int j, k;
  for (k = 0; k < len; k++)
    if (str[k] == ' ') {
      for (j = len - 1; j > k; j--)
        str[j + 2] = str[j];
      str[len + 2] = '\0';
      str[k] = '%';
      str[k + 1] = '2';
      str[k + 2] = '0';
      len = len + 2;
    }
  return str;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<rs>:
  endbr64
  test   %esi,%esi
  jle    L62
  mov    $0x0,%edx
  lea    -0x2(%rdi),%r9
  jmp    L21
L66:
  movslq %esi,%rax
  movb   $0x0,0x2(%rdi,%rax,1)
  movb   $0x25,(%rdi,%rdx,1)
  movb   $0x32,0x1(%rdi,%rdx,1)
  movb   $0x30,0x2(%rdi,%rdx,1)
  add    $0x2,%esi
L56:
  add    $0x1,%rdx
  cmp    %edx,%esi
  jle    L62
L21:
  cmpb   $0x20,(%rdi,%rdx,1)
  jne    L56
  lea    -0x1(%rsi),%eax
  cmp    %edx,%eax
  jle    L66
  cltq
  add    %rdi,%rax
  movslq %esi,%r8
  add    %r9,%r8
  lea    -0x2(%rsi),%ecx
  sub    %edx,%ecx
  sub    %rcx,%r8
L31:
  movzbl (%rax),%ecx
  mov    %cl,0x2(%rax)
  sub    $0x1,%rax
  cmp    %r8,%rax
  jne    L31
  jmp    L66
L62:
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *rs(char str[], int len) {
  int j, k;
  for (k = 0; k < len; k++)
    if (str[k] == ' ') {
      for (j = len - 1; j > k; j--)
        str[j + 2] = str[j];
      str[len + 2] = '\0';
      str[k] = '%';
      str[k + 1] = '2';
      str[k + 2] = '0';
      len = len + 2;
    }
  return str;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rs>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L42
  mov    %esi,%ebp
  xor    %ebx,%ebx
  jmp    L50
  nopl   (%rax)
L18:
  add    $0x1,%rbx
  cmp    %ebx,%ebp
  jle    L42
L50:
  cmpb   $0x20,(%r12,%rbx,1)
  jne    L18
  lea    -0x1(%rbp),%eax
  cmp    %ebx,%eax
  jle    L36
  mov    %ebp,%edx
  cltq
  sub    %ebx,%edx
  mov    %rax,%rsi
  add    $0x2,%rax
  sub    $0x2,%edx
  sub    %rdx,%rsi
  sub    %rdx,%rax
  lea    0x1(%rdx),%r8
  add    %r12,%rsi
  lea    (%r12,%rax,1),%rdi
  mov    %r8,%rdx
  call   <memmove@plt>
L36:
  movslq %ebp,%rax
  add    $0x2,%ebp
  movb   $0x0,0x2(%r12,%rax,1)
  movb   $0x25,(%r12,%rbx,1)
  movb   $0x32,0x1(%r12,%rbx,1)
  movb   $0x30,0x2(%r12,%rbx,1)
  add    $0x1,%rbx
  cmp    %ebx,%ebp
  jg     L50
L42:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *rs(char str[], int len) {
  int j, k;
  for (k = 0; k < len; k++)
    if (str[k] == ' ') {
      for (j = len - 1; j > k; j--)
        str[j + 2] = str[j];
      str[len + 2] = '\0';
      str[k] = '%';
      str[k + 1] = '2';
      str[k + 2] = '0';
      len = len + 2;
    }
  return str;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<rs>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L55
  mov    %esi,%ebp
  xor    %ebx,%ebx
  jmp    L29
  nopl   (%rax)
L20:
  add    $0x1,%rbx
  cmp    %ebx,%ebp
  jle    L55
L29:
  cmpb   $0x20,(%r12,%rbx,1)
  jne    L20
  lea    -0x1(%rbp),%eax
  cmp    %ebx,%eax
  jle    L9
  mov    %ebp,%edx
  cltq
  sub    %ebx,%edx
  mov    %rax,%rsi
  add    $0x2,%rax
  sub    $0x2,%edx
  sub    %rdx,%rsi
  sub    %rdx,%rax
  lea    0x1(%rdx),%r8
  add    %r12,%rsi
  lea    (%r12,%rax,1),%rdi
  mov    %r8,%rdx
  call   <memmove@plt>
L9:
  movslq %ebp,%rax
  add    $0x2,%ebp
  movb   $0x0,0x2(%r12,%rax,1)
  movb   $0x25,(%r12,%rbx,1)
  movb   $0x32,0x1(%r12,%rbx,1)
  movb   $0x30,0x2(%r12,%rbx,1)
  add    $0x1,%rbx
  cmp    %ebx,%ebp
  jg     L29
L55:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *rs(char str[], int len) {
  int j, k;
  for (k = 0; k < len; k++)
    if (str[k] == ' ') {
      for (j = len - 1; j > k; j--)
        str[j + 2] = str[j];
      str[len + 2] = '\0';
      str[k] = '%';
      str[k + 1] = '2';
      str[k + 2] = '0';
      len = len + 2;
    }
  return str;
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_window_size>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D4(%rip),%rax
  mov    (%rax),%eax
  lea    -0x10(%rbp),%rdx
  mov    %eax,%esi
  mov    $0x0,%edi
  call   <ioctl@plt>
  mov    -0x10(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0xc(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
void get_window_size(int *width, int *height) {
  struct winsize w;
  ioctl(0, TIOCGWINSZ, &w);
  *width = w.ws_col;
  *height = w.ws_row;
}
```
user: What is the c source code of the assembly code below:

```asm
<get_window_size>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  mov    D6(%rip),%rax
  mov    (%rax),%esi
  mov    $0x0,%edi
  call   <ioctl@plt>
  mov    (%rsp),%eax
  mov    %eax,0x0(%rbp)
  mov    0x4(%rsp),%eax
  mov    %eax,(%rbx)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void get_window_size(int *width, int *height) {
  struct winsize w;
  ioctl(0, TIOCGWINSZ, &w);
  *width = w.ws_col;
  *height = w.ws_row;
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_window_size>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D3(%rip),%rax
  mov    %rsp,%rdx
  mov    (%rax),%esi
  call   <ioctl@plt>
  mov    (%rsp),%eax
  mov    %eax,0x0(%rbp)
  mov    0x4(%rsp),%eax
  mov    %eax,(%rbx)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void get_window_size(int *width, int *height) {
  struct winsize w;
  ioctl(0, TIOCGWINSZ, &w);
  *width = w.ws_col;
  *height = w.ws_row;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<get_window_size>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D0(%rip),%rax
  mov    %rsp,%rdx
  mov    (%rax),%esi
  call   <ioctl@plt>
  mov    (%rsp),%eax
  mov    %eax,0x0(%rbp)
  mov    0x4(%rsp),%eax
  mov    %eax,(%rbx)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void get_window_size(int *width, int *height) {
  struct winsize w;
  ioctl(0, TIOCGWINSZ, &w);
  *width = w.ws_col;
  *height = w.ws_row;
}
```