ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rax
  movsbl (%rax),%edi
  test   %edi,%edi
  jle    L12
  xor    %edx,%edx
  xor    %esi,%esi
  xor    %eax,%eax
  nopl   0x0(%rax)
L11:
  cmp    $0x1fc0,%esi
  lea    (%rsi,%rax,1),%ecx
  cmove  %edx,%ecx
  add    $0x1,%eax
  add    $0x1fc0,%edx
  mov    %ecx,%esi
  cmp    %eax,%edi
  jne    L11
L7:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L12:
  xor    %esi,%esi
  jmp    L7
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
data_type: string

tool: D6: Sum is: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  int sum = 0;
  for (i = 0; i < argv[1][0]; i++) {
    if (sum == 8128)
      sum *= i;
    else
      sum += i;
  }
  printf("Sum is: %d\n", sum);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<trim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L51
  mov    -0x18(%rbp),%rax
  jmp    L30
L21:
  addq   $0x1,-0x18(%rbp)
L51:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L21
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L21
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L2
L3:
  subq   $0x1,-0x8(%rbp)
L2:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L3
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L3
  mov    -0xc(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x8(%rbp)
  je     L28
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movb   $0x0,(%rax)
L28:
  mov    -0x18(%rbp),%rax
L30:
  leave
  ret
```

assistant: ```c
char *trim(char *_s) {
  int len;
  char *end;
  if (!_s)
    return _s;
  while ((*_s == ' ') || (*_s == '\t'))
    _s++;
  len = strlen(_s);
  end = _s + len - 1;
  while ((*end == ' ') || (*end == '\t'))
    end--;
  if (end != (_s + len - 1)) {
    *(end + 1) = '\0';
  }
  return _s;
}
```
user: Translate the below assembly code into C programming language:

```asm
<trim>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L87
  movzbl (%rdi),%eax
  cmp    $0x20,%al
  je     L82
  cmp    $0x9,%al
  jne    L12
L82:
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  cmp    $0x20,%al
  je     L82
  cmp    $0x9,%al
  je     L82
L12:
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%rcx
  movzbl (%rcx),%eax
  cmp    $0x20,%al
  je     L25
  cmp    $0x9,%al
  jne    L87
L25:
  mov    %rcx,%rax
L39:
  sub    $0x1,%rax
  movzbl (%rax),%edx
  cmp    $0x20,%dl
  je     L39
  cmp    $0x9,%dl
  je     L39
  cmp    %rax,%rcx
  je     L87
  movb   $0x0,0x1(%rax)
L87:
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *trim(char *_s) {
  int len;
  char *end;
  if (!_s)
    return _s;
  while ((*_s == ' ') || (*_s == '\t'))
    _s++;
  len = strlen(_s);
  end = _s + len - 1;
  while ((*end == ' ') || (*end == '\t'))
    end--;
  if (end != (_s + len - 1)) {
    *(end + 1) = '\0';
  }
  return _s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<trim>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  test   %rdi,%rdi
  jne    L11
  jmp    L66
L4:
  add    $0x1,%r12
L11:
  movzbl (%r12),%eax
  cmp    $0x20,%al
  je     L4
  cmp    $0x9,%al
  je     L4
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%r12,%rax,1),%rcx
  movzbl (%rcx),%eax
  cmp    $0x20,%al
  je     L73
  cmp    $0x9,%al
  jne    L66
L73:
  mov    %rcx,%rax
  xchg   %ax,%ax
L8:
  movzbl -0x1(%rax),%edx
  sub    $0x1,%rax
  cmp    $0x20,%dl
  je     L8
  cmp    $0x9,%dl
  je     L8
  cmp    %rax,%rcx
  je     L66
  movb   $0x0,0x1(%rax)
L66:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *trim(char *_s) {
  int len;
  char *end;
  if (!_s)
    return _s;
  while ((*_s == ' ') || (*_s == '\t'))
    _s++;
  len = strlen(_s);
  end = _s + len - 1;
  while ((*end == ' ') || (*end == '\t'))
    end--;
  if (end != (_s + len - 1)) {
    *(end + 1) = '\0';
  }
  return _s;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<trim>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  test   %rdi,%rdi
  jne    L20
  jmp    L77
L5:
  add    $0x1,%r12
L20:
  movzbl (%r12),%eax
  cmp    $0x20,%al
  je     L5
  cmp    $0x9,%al
  je     L5
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%r12,%rax,1),%rcx
  movzbl (%rcx),%eax
  cmp    $0x20,%al
  je     L44
  cmp    $0x9,%al
  jne    L77
L44:
  mov    %rcx,%rax
  xchg   %ax,%ax
L61:
  movzbl -0x1(%rax),%edx
  sub    $0x1,%rax
  cmp    $0x20,%dl
  je     L61
  cmp    $0x9,%dl
  je     L61
  cmp    %rax,%rcx
  je     L77
  movb   $0x0,0x1(%rax)
L77:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *trim(char *_s) {
  int len;
  char *end;
  if (!_s)
    return _s;
  while ((*_s == ' ') || (*_s == '\t'))
    _s++;
  len = strlen(_s);
  end = _s + len - 1;
  while ((*end == ' ') || (*end == '\t'))
    end--;
  if (end != (_s + len - 1)) {
    *(end + 1) = '\0';
  }
  return _s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ADCDMAEnable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    -0x8(%rbp),%rdx
  add    $0x8,%rdx
  or     $0x1,%ah
  mov    %rax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ADCDMAEnable(unsigned long ulBase) {
  ;
  (*((volatile unsigned long *)(ulBase + 0x00000008))) |= 0x00000100;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ADCDMAEnable>:
  endbr64
  mov    0x8(%rdi),%rax
  or     $0x1,%ah
  mov    %rax,0x8(%rdi)
  ret
```

assistant: ```c
void ADCDMAEnable(unsigned long ulBase) {
  ;
  (*((volatile unsigned long *)(ulBase + 0x00000008))) |= 0x00000100;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ADCDMAEnable>:
  endbr64
  mov    0x8(%rdi),%rax
  or     $0x1,%ah
  mov    %rax,0x8(%rdi)
  ret
```

assistant: ```c
void ADCDMAEnable(unsigned long ulBase) {
  ;
  (*((volatile unsigned long *)(ulBase + 0x00000008))) |= 0x00000100;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ADCDMAEnable>:
  endbr64
  mov    0x8(%rdi),%rax
  or     $0x1,%ah
  mov    %rax,0x8(%rdi)
  ret
```

assistant: ```c
void ADCDMAEnable(unsigned long ulBase) {
  ;
  (*((volatile unsigned long *)(ulBase + 0x00000008))) |= 0x00000100;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<mirror_bits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x14(%rbp)
  movb   $0x0,-0x1(%rbp)
  movb   $0x0,-0x2(%rbp)
  jmp    L6
L12:
  movzbl -0x14(%rbp),%edx
  movzbl -0x2(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  and    %eax,%edx
  movzbl -0x2(%rbp),%eax
  lea    (%rax,%rax,1),%ecx
  mov    $0x7,%eax
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,%edi
  movzbl -0x14(%rbp),%edx
  movzbl -0x2(%rbp),%eax
  mov    $0x80,%esi
  mov    %eax,%ecx
  sar    %cl,%esi
  mov    %esi,%eax
  and    %eax,%edx
  movzbl -0x2(%rbp),%eax
  lea    (%rax,%rax,1),%ecx
  mov    $0x7,%eax
  sub    %ecx,%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  or     %eax,%edi
  mov    %edi,%edx
  movzbl -0x1(%rbp),%eax
  or     %edx,%eax
  mov    %al,-0x1(%rbp)
  movzbl -0x2(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x2(%rbp)
L6:
  cmpb   $0x3,-0x2(%rbp)
  jbe    L12
  movzbl -0x1(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned char mirror_bits(unsigned char bits) {
  unsigned char i, tmp = 0;
  for (i = 0; i < 4; i++) {
    tmp |= ((bits & (0x01 << i)) << (7 - i * 2)) | ((bits & (0x80 >> i)) >> (7 - i * 2));
  }
  return tmp;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<mirror_bits>:
  endbr64
  mov    $0x0,%edx
  mov    $0x0,%r9d
  movzbl %dil,%edi
  mov    $0x1,%r11d
  mov    $0x80,%r10d
L6:
  mov    %edx,%eax
  neg    %eax
  lea    0x7(%rax,%rax,1),%r8d
  mov    %r11d,%eax
  mov    %edx,%ecx
  shl    %cl,%eax
  and    %edi,%eax
  mov    %r8d,%ecx
  shl    %cl,%eax
  mov    %r10d,%esi
  mov    %edx,%ecx
  sar    %cl,%esi
  and    %edi,%esi
  mov    %r8d,%ecx
  sar    %cl,%esi
  or     %esi,%eax
  or     %eax,%r9d
  add    $0x1,%edx
  cmp    $0x4,%edx
  jne    L6
  mov    %r9d,%eax
  ret
```

assistant: ```c
unsigned char mirror_bits(unsigned char bits) {
  unsigned char i, tmp = 0;
  for (i = 0; i < 4; i++) {
    tmp |= ((bits & (0x01 << i)) << (7 - i * 2)) | ((bits & (0x80 >> i)) >> (7 - i * 2));
  }
  return tmp;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mirror_bits>:
  endbr64
  movzbl %dil,%edi
  xor    %edx,%edx
  xor    %r9d,%r9d
  mov    $0x1,%r11d
  mov    $0x80,%r10d
L4:
  mov    %edx,%eax
  mov    %edx,%ecx
  mov    %r10d,%esi
  neg    %eax
  lea    0x7(%rax,%rax,1),%r8d
  mov    %r11d,%eax
  shl    %cl,%eax
  mov    %r8d,%ecx
  and    %edi,%eax
  shl    %cl,%eax
  mov    %edx,%ecx
  add    $0x1,%edx
  sar    %cl,%esi
  mov    %r8d,%ecx
  and    %edi,%esi
  sar    %cl,%esi
  or     %esi,%eax
  or     %eax,%r9d
  cmp    $0x4,%edx
  jne    L4
  mov    %r9d,%eax
  ret
```

assistant: ```c
unsigned char mirror_bits(unsigned char bits) {
  unsigned char i, tmp = 0;
  for (i = 0; i < 4; i++) {
    tmp |= ((bits & (0x01 << i)) << (7 - i * 2)) | ((bits & (0x80 >> i)) >> (7 - i * 2));
  }
  return tmp;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mirror_bits>:
  endbr64
  movzbl %dil,%edx
  shr    $0x7,%dil
  mov    %edx,%eax
  mov    %edx,%ecx
  mov    %edx,%esi
  shl    $0x5,%eax
  sar    $0x5,%ecx
  and    $0x2,%ecx
  and    $0x40,%eax
  sar    $0x3,%esi
  or     %ecx,%eax
  mov    %edx,%ecx
  and    $0x4,%esi
  shl    $0x7,%ecx
  or     %ecx,%edi
  lea    0x0(,%rdx,8),%ecx
  and    $0x20,%ecx
  or     %edi,%eax
  or     %esi,%ecx
  or     %ecx,%eax
  lea    (%rdx,%rdx,1),%ecx
  sar    %edx
  and    $0x10,%ecx
  and    $0x8,%edx
  or     %ecx,%edx
  or     %edx,%eax
  ret
```

assistant: ```c
unsigned char mirror_bits(unsigned char bits) {
  unsigned char i, tmp = 0;
  for (i = 0; i < 4; i++) {
    tmp |= ((bits & (0x01 << i)) << (7 - i * 2)) | ((bits & (0x80 >> i)) >> (7 - i * 2));
  }
  return tmp;
}
```
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
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <CountBits@plt>
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D9
type_infered: string
name: disassemble_data
arguments: label: D2
type_infered: string
name: disassemble_data
arguments: label: D1
type_infered: string

tool: D9: Enter the number: 

tool: D2: %d

tool: D1: Number of bits %d 


assistant: ```c
int main() {
  int num;
  printf("Enter the number: \n");
  scanf("%d", &num);
  printf("Number of bits %d \n", CountBits(num));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D13(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <CountBits@plt>
  mov    %eax,%esi
  lea    D21(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D13
type: string
name: unpack_data
arguments: tag: D18
type: string
name: unpack_data
arguments: tag: D21
type: string

tool: D13: Enter the number: 

tool: D18: %d

tool: D21: Number of bits %d 


assistant: ```c
int main() {
  int num;
  printf("Enter the number: \n");
  scanf("%d", &num);
  printf("Number of bits %d \n", CountBits(num));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D28(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <CountBits@plt>
  lea    D22(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D28
type: string
name: unpack_data
arguments: data_tag: D1
type: string
name: unpack_data
arguments: data_tag: D22
type: string

tool: D28: Enter the number: 

tool: D1: %d

tool: D22: Number of bits %d 


assistant: ```c
int main() {
  int num;
  printf("Enter the number: \n");
  scanf("%d", &num);
  printf("Number of bits %d \n", CountBits(num));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D16(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D20(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <CountBits@plt>
  lea    D26(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D16
type_guessed: string
name: read_data
arguments: data_label: D20
type_guessed: string
name: read_data
arguments: data_label: D26
type_guessed: string

tool: D16: Enter the number: 

tool: D20: %d

tool: D26: Number of bits %d 


assistant: ```c
int main() {
  int num;
  printf("Enter the number: \n");
  scanf("%d", &num);
  printf("Number of bits %d \n", CountBits(num));
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <f@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L17
L29:
  movl   $0x2,-0x18(%rbp)
  addl   $0x1,-0x14(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
L17:
  cmpl   $0x9,-0x14(%rbp)
  jle    L29
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

assistant: ```c
int main(void) {
  int *p = 0;
  {
    int x;
    p = &x;
  }
  p = f();
  int i = 0;
  while (i < 10) {
    int x = 2;
    i++;
    p = &x;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <f@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int *p = 0;
  {
    int x;
    p = &x;
  }
  p = f();
  int i = 0;
  while (i < 10) {
    int x = 2;
    i++;
    p = &x;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <f@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int *p = 0;
  {
    int x;
    p = &x;
  }
  p = f();
  int i = 0;
  while (i < 10) {
    int x = 2;
    i++;
    p = &x;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sys$cretva>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x30,%rax
  int    $0x81
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int sys$cretva(struct _va_range *inadr, struct _va_range *retadr, unsigned int acmode) {
  return ({
    unsigned long resultvar;
    register long int _a3 asm("rdx") = (long)(acmode);
    register long int _a2 asm("rsi") = (long)(retadr);
    register long int _a1 asm("rdi") = (long)(inadr);
    asm volatile("movq %1, %%rax\n\t"
                 "int $0x81\n\t"
                 : "=a"(resultvar)
                 : "i"(48), "r"(_a1), "r"(_a2), "r"(_a3)
                 : "memory", "cc", "r11", "cx");
    (long)resultvar;
  });
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sys$cretva>:
  endbr64
  mov    %edx,%edx
  mov    $0x30,%rax
  int    $0x81
  ret
```

assistant: ```c
int sys$cretva(struct _va_range *inadr, struct _va_range *retadr, unsigned int acmode) {
  return ({
    unsigned long resultvar;
    register long int _a3 asm("rdx") = (long)(acmode);
    register long int _a2 asm("rsi") = (long)(retadr);
    register long int _a1 asm("rdi") = (long)(inadr);
    asm volatile("movq %1, %%rax\n\t"
                 "int $0x81\n\t"
                 : "=a"(resultvar)
                 : "i"(48), "r"(_a1), "r"(_a2), "r"(_a3)
                 : "memory", "cc", "r11", "cx");
    (long)resultvar;
  });
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sys$cretva>:
  endbr64
  mov    %edx,%edx
  mov    $0x30,%rax
  int    $0x81
  ret
```

assistant: ```c
int sys$cretva(struct _va_range *inadr, struct _va_range *retadr, unsigned int acmode) {
  return ({
    unsigned long resultvar;
    register long int _a3 asm("rdx") = (long)(acmode);
    register long int _a2 asm("rsi") = (long)(retadr);
    register long int _a1 asm("rdi") = (long)(inadr);
    asm volatile("movq %1, %%rax\n\t"
                 "int $0x81\n\t"
                 : "=a"(resultvar)
                 : "i"(48), "r"(_a1), "r"(_a2), "r"(_a3)
                 : "memory", "cc", "r11", "cx");
    (long)resultvar;
  });
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<aloca_matriz>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L7
L15:
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L7:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L15
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char **aloca_matriz(int m, int n) {
  char **M;
  int i;
  M = (char **)malloc(m * sizeof(char *));
  for (i = 0; i < m; i++) {
    M[i] = (char *)malloc(n * sizeof(char));
  }
  return M;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<aloca_matriz>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r12d
  mov    %esi,%ebp
  lea    0x0(,%rdi,8),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  jle    L7
  mov    %rax,%rbx
  mov    %r12d,%r12d
  lea    (%rax,%r12,8),%r12
L2:
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L2
L7:
  mov    %r13,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **aloca_matriz(int m, int n) {
  char **M;
  int i;
  M = (char **)malloc(m * sizeof(char *));
  for (i = 0; i < m; i++) {
    M[i] = (char *)malloc(n * sizeof(char));
  }
  return M;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<aloca_matriz>:
  endbr64
  push   %r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  lea    0x0(,%r12,8),%edi
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  jle    L8
  mov    %rax,%rbx
  lea    (%rax,%r12,8),%r12
  nop
L17:
  mov    %ebp,%edi
  add    $0x8,%rbx
  call   <malloc@plt>
  mov    %rax,-0x8(%rbx)
  cmp    %r12,%rbx
  jne    L17
L8:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **aloca_matriz(int m, int n) {
  char **M;
  int i;
  M = (char **)malloc(m * sizeof(char *));
  for (i = 0; i < m; i++) {
    M[i] = (char *)malloc(n * sizeof(char));
  }
  return M;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<aloca_matriz>:
  endbr64
  push   %r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  lea    0x0(,%r12,8),%edi
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  jle    L0
  mov    %rax,%rbx
  lea    (%rax,%r12,8),%r12
  nop
L9:
  mov    %ebp,%edi
  add    $0x8,%rbx
  call   <malloc@plt>
  mov    %rax,-0x8(%rbx)
  cmp    %r12,%rbx
  jne    L9
L0:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **aloca_matriz(int m, int n) {
  char **M;
  int i;
  M = (char **)malloc(m * sizeof(char *));
  for (i = 0; i < m; i++) {
    M[i] = (char *)malloc(n * sizeof(char));
  }
  return M;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<clickplc_exe_EndOfSubProgramm_enc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x4,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  movb   $0x81,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x2,%rax
  movb   $0xff,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x3,%rax
  movb   $0xff,(%rax)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int clickplc_exe_EndOfSubProgramm_enc(void *obj, unsigned char *stream) {
  int encsize = 4;
  stream[0] = 0x00;
  stream[1] = 0x81;
  stream[2] = 0xff;
  stream[3] = 0xff;
  return (encsize);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<clickplc_exe_EndOfSubProgramm_enc>:
  endbr64
  movb   $0x0,(%rsi)
  movb   $0x81,0x1(%rsi)
  movb   $0xff,0x2(%rsi)
  movb   $0xff,0x3(%rsi)
  mov    $0x4,%eax
  ret
```

assistant: ```c
int clickplc_exe_EndOfSubProgramm_enc(void *obj, unsigned char *stream) {
  int encsize = 4;
  stream[0] = 0x00;
  stream[1] = 0x81;
  stream[2] = 0xff;
  stream[3] = 0xff;
  return (encsize);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<clickplc_exe_EndOfSubProgramm_enc>:
  endbr64
  movl   $0xffff8100,(%rsi)
  mov    $0x4,%eax
  ret
```

assistant: ```c
int clickplc_exe_EndOfSubProgramm_enc(void *obj, unsigned char *stream) {
  int encsize = 4;
  stream[0] = 0x00;
  stream[1] = 0x81;
  stream[2] = 0xff;
  stream[3] = 0xff;
  return (encsize);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<clickplc_exe_EndOfSubProgramm_enc>:
  endbr64
  movl   $0xffff8100,(%rsi)
  mov    $0x4,%eax
  ret
```

assistant: ```c
int clickplc_exe_EndOfSubProgramm_enc(void *obj, unsigned char *stream) {
  int encsize = 4;
  stream[0] = 0x00;
  stream[1] = 0x81;
  stream[2] = 0xff;
  stream[3] = 0xff;
  return (encsize);
}
```