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
<uinput_syn>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    $0x10,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    D15(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0xc(%rsp)
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x8(%rsp)
  movq   $0x0,(%rsp)
  mov    $0x10,%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <write@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void uinput_syn(int fd) {
  struct input_event ev;
  memset(&ev, 0, sizeof(ev));
  ev.type = EV_SYN;
  ev.code = SYN_REPORT;
  ev.value = 0;
  write(fd, &ev, sizeof(ev));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<uinput_syn>:
  endbr64
  push   %r12
  xor    %esi,%esi
  mov    $0x10,%edx
  push   %rbp
  mov    %edi,%ebp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <memset@plt>
  mov    $0x10,%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    D7(%rip),%rax
  movq   $0x0,(%rsp)
  mov    (%rax),%eax
  mov    %eax,0xc(%rsp)
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x8(%rsp)
  call   <write@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x28,%rsp
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void uinput_syn(int fd) {
  struct input_event ev;
  memset(&ev, 0, sizeof(ev));
  ev.type = EV_SYN;
  ev.code = SYN_REPORT;
  ev.value = 0;
  write(fd, &ev, sizeof(ev));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<uinput_syn>:
  endbr64
  push   %r12
  xor    %esi,%esi
  mov    $0x10,%edx
  push   %rbp
  mov    %edi,%ebp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <memset@plt>
  mov    D19(%rip),%rax
  mov    D13(%rip),%rdx
  mov    %ebp,%edi
  mov    %r12,%rsi
  movq   $0x0,(%rsp)
  mov    (%rax),%eax
  movd   (%rdx),%xmm0
  mov    $0x10,%edx
  movd   %eax,%xmm1
  punpckldq %xmm1,%xmm0
  movq   %xmm0,0x8(%rsp)
  call   <write@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x28,%rsp
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void uinput_syn(int fd) {
  struct input_event ev;
  memset(&ev, 0, sizeof(ev));
  ev.type = EV_SYN;
  ev.code = SYN_REPORT;
  ev.value = 0;
  write(fd, &ev, sizeof(ev));
}
```
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    $0x10,%rsp
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x1,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L15
L13:
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  shll   -0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L15:
  cmpl   $0xa,-0x8(%rbp)
  jle    L13
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D36
datatype: string
name: parse_data
arguments: label: D9
datatype: string
name: parse_data
arguments: label: D15
datatype: string
name: parse_data
arguments: label: D29
datatype: string

tool: D36: TABLE OF POWERS OF TWO


tool: D9:  n     2 to the n

tool: D15: ---    ---------------

tool: D29: %2i           %i


assistant: ```c
int main(void) {
  int n, two_to_the_n;
  printf("TABLE OF POWERS OF TWO\n\n");
  printf(" n     2 to the n\n");
  printf("---    ---------------\n");
  two_to_the_n = 1;
  for (n = 0; n <= 10; ++n) {
    printf("%2i           %i\n", n, two_to_the_n);
    two_to_the_n *= 2;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    D1(%rip),%rdi
  call   <puts@plt>
  lea    D13(%rip),%rdi
  call   <puts@plt>
  lea    D10(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%ebp
  mov    $0x0,%ebx
  lea    D2(%rip),%r12
L0:
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    %ebp,%ebp
  add    $0x1,%ebx
  cmp    $0xb,%ebx
  jne    L0
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D1
type_infered: string
name: parse_data
arguments: data_tag: D13
type_infered: string
name: parse_data
arguments: data_tag: D10
type_infered: string
name: parse_data
arguments: data_tag: D2
type_infered: string

tool: D1: TABLE OF POWERS OF TWO


tool: D13:  n     2 to the n

tool: D10: ---    ---------------

tool: D2: %2i           %i


assistant: ```c
int main(void) {
  int n, two_to_the_n;
  printf("TABLE OF POWERS OF TWO\n\n");
  printf(" n     2 to the n\n");
  printf("---    ---------------\n");
  two_to_the_n = 1;
  for (n = 0; n <= 10; ++n) {
    printf("%2i           %i\n", n, two_to_the_n);
    two_to_the_n *= 2;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r12
  lea    D15(%rip),%rdi
  lea    D14(%rip),%r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  xor    %ebx,%ebx
  call   <puts@plt>
  lea    D38(%rip),%rdi
  call   <puts@plt>
  lea    D7(%rip),%rdi
  call   <puts@plt>
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  add    %ebp,%ebp
  cmp    $0xb,%ebx
  jne    L4
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D15
data_type: string
name: retrieve_data
arguments: tag: D14
data_type: string
name: retrieve_data
arguments: tag: D38
data_type: string
name: retrieve_data
arguments: tag: D7
data_type: string

tool: D15: TABLE OF POWERS OF TWO


tool: D14: %2i           %i


tool: D38:  n     2 to the n

tool: D7: ---    ---------------

assistant: ```c
int main(void) {
  int n, two_to_the_n;
  printf("TABLE OF POWERS OF TWO\n\n");
  printf(" n     2 to the n\n");
  printf("---    ---------------\n");
  two_to_the_n = 1;
  for (n = 0; n <= 10; ++n) {
    printf("%2i           %i\n", n, two_to_the_n);
    two_to_the_n *= 2;
  }
  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   %r12
  lea    D33(%rip),%rdi
  lea    D37(%rip),%r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  xor    %ebx,%ebx
  call   <puts@plt>
  lea    D31(%rip),%rdi
  call   <puts@plt>
  lea    D14(%rip),%rdi
  call   <puts@plt>
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  add    %ebp,%ebp
  cmp    $0xb,%ebx
  jne    L6
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D33
type_guessed: string
name: parse_data
arguments: tag: D37
type_guessed: string
name: parse_data
arguments: tag: D31
type_guessed: string
name: parse_data
arguments: tag: D14
type_guessed: string

tool: D33: TABLE OF POWERS OF TWO


tool: D37: %2i           %i


tool: D31:  n     2 to the n

tool: D14: ---    ---------------

assistant: ```c
int main(void) {
  int n, two_to_the_n;
  printf("TABLE OF POWERS OF TWO\n\n");
  printf(" n     2 to the n\n");
  printf("---    ---------------\n");
  two_to_the_n = 1;
  for (n = 0; n <= 10; ++n) {
    printf("%2i           %i\n", n, two_to_the_n);
    two_to_the_n *= 2;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<get_cpu_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <get_cpu_str@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L2
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <atol@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L2:
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int get_cpu_int(const char *item, int cpuid) {
  char *fc = NULL;
  int ret = 0;
  fc = get_cpu_str(item, cpuid);
  if (fc) {
    ret = atol(fc);
    free(fc);
  }
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_cpu_int>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <get_cpu_str@plt>
  mov    $0x0,%ebp
  test   %rax,%rax
  je     L7
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <atol@plt>
  mov    %eax,%ebp
  mov    %rbx,%rdi
  call   <free@plt>
L7:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int get_cpu_int(const char *item, int cpuid) {
  char *fc = NULL;
  int ret = 0;
  fc = get_cpu_str(item, cpuid);
  if (fc) {
    ret = atol(fc);
    free(fc);
  }
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<get_cpu_int>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  sub    $0x8,%rsp
  call   <get_cpu_str@plt>
  test   %rax,%rax
  je     L8
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <atol@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <free@plt>
L8:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int get_cpu_int(const char *item, int cpuid) {
  char *fc = NULL;
  int ret = 0;
  fc = get_cpu_str(item, cpuid);
  if (fc) {
    ret = atol(fc);
    free(fc);
  }
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_cpu_int>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  sub    $0x8,%rsp
  call   <get_cpu_str@plt>
  test   %rax,%rax
  je     L6
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <atol@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <free@plt>
L6:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int get_cpu_int(const char *item, int cpuid) {
  char *fc = NULL;
  int ret = 0;
  fc = get_cpu_str(item, cpuid);
  if (fc) {
    ret = atol(fc);
    free(fc);
  }
  return ret;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L4:
  mov    $0x0,%eax
  call   <pthread_yield@plt>
  jmp    L4
```

assistant: ```c
int main() {
  while (1)
    pthread_yield();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
L7:
  mov    $0x0,%eax
  call   <pthread_yield@plt>
  jmp    L7
```

assistant: ```c
int main() {
  while (1)
    pthread_yield();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L1:
  xor    %eax,%eax
  call   <pthread_yield@plt>
  jmp    L1
```

assistant: ```c
int main() {
  while (1)
    pthread_yield();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L1:
  xor    %eax,%eax
  call   <pthread_yield@plt>
  jmp    L1
```

assistant: ```c
int main() {
  while (1)
    pthread_yield();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<b_stdio_putcgetc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    $0x0,%eax
  call   <tmpfile@plt>
  mov    %rax,-0x8(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L11
L15:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x78,%edi
  call   <putc@plt>
  addq   $0x1,-0x18(%rbp)
L11:
  cmpq   $0x4c4b3f,-0x18(%rbp)
  jbe    L15
  mov    D8(%rip),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <fseeko@plt>
  movq   $0x0,-0x18(%rbp)
  jmp    L24
L2:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <getc@plt>
  add    %rax,-0x10(%rbp)
  addq   $0x1,-0x18(%rbp)
L24:
  cmpq   $0x4c4b3f,-0x18(%rbp)
  jbe    L2
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
size_t b_stdio_putcgetc(void *dummy) {
  FILE *f = tmpfile();
  size_t i;
  size_t cs;
  for (i = 0; i < 5000000; i++)
    putc('x', f);
  fseeko(f, 0, SEEK_SET);
  for (i = 0; i < 5000000; i++)
    cs += getc(f);
  fclose(f);
  return cs;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<b_stdio_putcgetc>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%eax
  call   <tmpfile@plt>
  mov    %rax,%rbp
  mov    $0x4c4b40,%ebx
L7:
  mov    %rbp,%rsi
  mov    $0x78,%edi
  call   <putc@plt>
  sub    $0x1,%rbx
  jne    L7
  mov    D8(%rip),%rax
  mov    (%rax),%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <fseeko@plt>
  mov    $0x4c4b40,%ebx
L1:
  mov    %rbp,%rdi
  call   <getc@plt>
  add    %rax,%r12
  sub    $0x1,%rbx
  jne    L1
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
size_t b_stdio_putcgetc(void *dummy) {
  FILE *f = tmpfile();
  size_t i;
  size_t cs;
  for (i = 0; i < 5000000; i++)
    putc('x', f);
  fseeko(f, 0, SEEK_SET);
  for (i = 0; i < 5000000; i++)
    cs += getc(f);
  fclose(f);
  return cs;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<b_stdio_putcgetc>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  push   %rbx
  mov    $0x4c4b40,%ebx
  call   <tmpfile@plt>
  mov    %rax,%rbp
  nopw   0x0(%rax,%rax,1)
L14:
  mov    %rbp,%rsi
  mov    $0x78,%edi
  call   <putc@plt>
  sub    $0x1,%rbx
  jne    L14
  mov    D4(%rip),%rax
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    $0x4c4b40,%ebx
  mov    (%rax),%edx
  call   <fseeko@plt>
  nopl   0x0(%rax,%rax,1)
L2:
  mov    %rbp,%rdi
  call   <getc@plt>
  add    %rax,%r12
  sub    $0x1,%rbx
  jne    L2
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
size_t b_stdio_putcgetc(void *dummy) {
  FILE *f = tmpfile();
  size_t i;
  size_t cs;
  for (i = 0; i < 5000000; i++)
    putc('x', f);
  fseeko(f, 0, SEEK_SET);
  for (i = 0; i < 5000000; i++)
    cs += getc(f);
  fclose(f);
  return cs;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<b_stdio_putcgetc>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  push   %rbx
  mov    $0x4c4b40,%ebx
  call   <tmpfile@plt>
  mov    %rax,%rbp
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %rbp,%rsi
  mov    $0x78,%edi
  call   <putc@plt>
  sub    $0x1,%rbx
  jne    L7
  mov    D9(%rip),%rax
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    $0x4c4b40,%ebx
  mov    (%rax),%edx
  call   <fseeko@plt>
  nopl   0x0(%rax,%rax,1)
L2:
  mov    %rbp,%rdi
  call   <getc@plt>
  add    %rax,%r12
  sub    $0x1,%rbx
  jne    L2
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
size_t b_stdio_putcgetc(void *dummy) {
  FILE *f = tmpfile();
  size_t i;
  size_t cs;
  for (i = 0; i < 5000000; i++)
    putc('x', f);
  fseeko(f, 0, SEEK_SET);
  for (i = 0; i < 5000000; i++)
    cs += getc(f);
  fclose(f);
  return cs;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<copy02>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L4:
  addl   $0x1,-0x4(%rbp)
L12:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L4
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void copy02(char to[], char from[]) {
  int i;
  i = 0;
  while ((to[i] = from[i]) != '\0')
    ++i;
}
```
user: What is the c source code of the assembly code below:

```asm
<copy02>:
  endbr64
  movzbl (%rsi),%eax
  mov    %al,(%rdi)
  test   %al,%al
  je     L14
  mov    $0x1,%eax
L5:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L5
L14:
  ret
```

assistant: ```c
void copy02(char to[], char from[]) {
  int i;
  i = 0;
  while ((to[i] = from[i]) != '\0')
    ++i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<copy02>:
  endbr64
  movzbl (%rsi),%eax
  mov    %al,(%rdi)
  test   %al,%al
  je     L10
  mov    $0x1,%eax
  nopw   0x0(%rax,%rax,1)
L8:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L8
L10:
  ret
```

assistant: ```c
void copy02(char to[], char from[]) {
  int i;
  i = 0;
  while ((to[i] = from[i]) != '\0')
    ++i;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<copy02>:
  endbr64
  movzbl (%rsi),%eax
  mov    %al,(%rdi)
  test   %al,%al
  je     L16
  mov    $0x1,%eax
  nopw   0x0(%rax,%rax,1)
L17:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L17
L16:
  ret
```

assistant: ```c
void copy02(char to[], char from[]) {
  int i;
  i = 0;
  while ((to[i] = from[i]) != '\0')
    ++i;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<divp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <printd@plt>
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <print@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <printd@plt>
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <print@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <printd@plt>
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <print@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D25
type: string
name: retrieve_data
arguments: data_tag: D18
type: string
name: retrieve_data
arguments: data_tag: D7
type: string

tool: D25:  /= 

tool: D18:  -> 

tool: D7: 


assistant: ```c
void divp(int a, int b) {
  int *p;
  printd(a);
  print(" /= ");
  printd(b);
  print(" -> ");
  p = &a;
  *p /= b;
  printd(*p);
  print("\r\n");
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<divp>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  call   <printd@plt>
  lea    D26(%rip),%rdi
  call   <print@plt>
  mov    %ebp,%edi
  call   <printd@plt>
  lea    D16(%rip),%rdi
  call   <print@plt>
  mov    %ebx,%eax
  cltd
  idiv   %ebp
  mov    %eax,%edi
  call   <printd@plt>
  lea    D20(%rip),%rdi
  call   <print@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D26
data_type: string
name: parse_data
arguments: data_label: D16
data_type: string
name: parse_data
arguments: data_label: D20
data_type: string

tool: D26:  /= 

tool: D16:  -> 

tool: D20: 


assistant: ```c
void divp(int a, int b) {
  int *p;
  printd(a);
  print(" /= ");
  printd(b);
  print(" -> ");
  p = &a;
  *p /= b;
  printd(*p);
  print("\r\n");
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<divp>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <printd@plt>
  lea    D17(%rip),%rdi
  call   <print@plt>
  mov    %ebx,%edi
  call   <printd@plt>
  lea    D18(%rip),%rdi
  call   <print@plt>
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  mov    %eax,%edi
  call   <printd@plt>
  add    $0x8,%rsp
  lea    D22(%rip),%rdi
  pop    %rbx
  pop    %r12
  jmp    1060 <print@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D17
datatype: string
name: extract_symbolic_data
arguments: tag: D18
datatype: string
name: extract_symbolic_data
arguments: tag: D22
datatype: string

tool: D17:  /= 

tool: D18:  -> 

tool: D22: 


assistant: ```c
void divp(int a, int b) {
  int *p;
  printd(a);
  print(" /= ");
  printd(b);
  print(" -> ");
  p = &a;
  *p /= b;
  printd(*p);
  print("\r\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<divp>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <printd@plt>
  lea    D21(%rip),%rdi
  call   <print@plt>
  mov    %ebx,%edi
  call   <printd@plt>
  lea    D25(%rip),%rdi
  call   <print@plt>
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  mov    %eax,%edi
  call   <printd@plt>
  add    $0x8,%rsp
  lea    D18(%rip),%rdi
  pop    %rbx
  pop    %r12
  jmp    1060 <print@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D21
data_type: string
name: extract_symbolic_data
arguments: label: D25
data_type: string
name: extract_symbolic_data
arguments: label: D18
data_type: string

tool: D21:  /= 

tool: D25:  -> 

tool: D18: 


assistant: ```c
void divp(int a, int b) {
  int *p;
  printd(a);
  print(" /= ");
  printd(b);
  print(" -> ");
  p = &a;
  *p /= b;
  printd(*p);
  print("\r\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bitvector_resize>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x1f,%rax
  shr    $0x5,%rax
  shl    $0x2,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L1
  mov    $0x0,%eax
  jmp    L8
L1:
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x8(%rbp),%rax
L8:
  leave
  ret
```

assistant: ```c
u_int32_t *bitvector_resize(u_int32_t *b, size_t n) {
  u_int32_t *p;
  size_t sz = ((n + 31) >> 5) * 4;
  p = realloc(b, sz);
  if (p == NULL)
    return NULL;
  memset(p, 0, sz);
  return p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bitvector_resize>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    0x1f(%rsi),%rbx
  shr    $0x5,%rbx
  shl    $0x2,%rbx
  mov    %rbx,%rsi
  call   <realloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L1
  mov    %rbx,%rdx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
L1:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
u_int32_t *bitvector_resize(u_int32_t *b, size_t n) {
  u_int32_t *p;
  size_t sz = ((n + 31) >> 5) * 4;
  p = realloc(b, sz);
  if (p == NULL)
    return NULL;
  memset(p, 0, sz);
  return p;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<bitvector_resize>:
  endbr64
  push   %r13
  push   %r12
  lea    0x1f(%rsi),%r12
  shr    $0x5,%r12
  shl    $0x2,%r12
  sub    $0x8,%rsp
  mov    %r12,%rsi
  call   <realloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L9
  mov    %r12,%rdx
  xor    %esi,%esi
  mov    %rax,%rdi
  call   <memset@plt>
L9:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
u_int32_t *bitvector_resize(u_int32_t *b, size_t n) {
  u_int32_t *p;
  size_t sz = ((n + 31) >> 5) * 4;
  p = realloc(b, sz);
  if (p == NULL)
    return NULL;
  memset(p, 0, sz);
  return p;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<bitvector_resize>:
  endbr64
  push   %r13
  push   %r12
  lea    0x1f(%rsi),%r12
  shr    $0x5,%r12
  shl    $0x2,%r12
  sub    $0x8,%rsp
  mov    %r12,%rsi
  call   <realloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L4
  mov    %r12,%rdx
  xor    %esi,%esi
  mov    %rax,%rdi
  call   <memset@plt>
L4:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
u_int32_t *bitvector_resize(u_int32_t *b, size_t n) {
  u_int32_t *p;
  size_t sz = ((n + 31) >> 5) * 4;
  p = realloc(b, sz);
  if (p == NULL)
    return NULL;
  memset(p, 0, sz);
  return p;
}
```