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

```asm
<bsearch>:
  endbr64
  imul   %rcx,%rdx
  push   %r15
  lea    (%rsi,%rcx,1),%rax
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %r8,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  lea    (%rsi,%rdx,1),%rbx
  sub    $0x18,%rsp
  cmp    %rax,%rbx
  jbe    L55
  lea    (%rcx,%rcx,1),%rax
  mov    %rcx,%r15
  mov    %rax,0x8(%rsp)
  nopl   0x0(%rax,%rax,1)
L54:
  mov    %rbx,%rax
  xor    %edx,%edx
  mov    %rbp,%rdi
  sub    %r14,%rax
  divq   0x8(%rsp)
  imul   %r15,%rax
  lea    (%r14,%rax,1),%r12
  mov    %r12,%rsi
  call   *%r13
  test   %eax,%eax
  js     L39
  lea    (%r12,%r15,1),%rax
  cmp    %rax,%rbx
  jbe    L2
  mov    %r12,%r14
  jmp    L54
L39:
  lea    (%r14,%r15,1),%rax
  cmp    %rax,%r12
  jbe    L55
  mov    %r12,%rbx
  jmp    L54
  xchg   %ax,%ax
L55:
  mov    %r14,%r12
L2:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   *%r13
  test   %eax,%eax
  mov    $0x0,%eax
  cmovne %rax,%r12
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void *bsearch(const void *key, const void *base, size_t nmemb, size_t size, int (*compar)(const void *, const void *)) {
  void *left = (void *)base, *right = (void *)base + size * nmemb, *mid;
  while (left + size < right) {
    mid = left + ((right - left) / (size << 1) * size);
    if (compar(key, mid) < 0) {
      right = mid;
    } else {
      left = mid;
    }
  }
  if (compar(left, key) == 0) {
    return left;
  } else {
    return NULL;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<tst_print_all_pages>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D8(%rip),%rdx
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <setup_printout@plt>
  mov    $0x0,%eax
  call   <test_printout@plt>
  mov    $0x0,%eax
  call   <mc_print_all_pages@plt>
  mov    $0x0,%eax
  call   <cleanup_printout@plt>
  mov    $0x1,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
data_type: string

tool: D8: contents of all pages

assistant: ```c
int tst_print_all_pages(const char *the_title) {
  setup_printout(the_title, (1), "contents of all pages");
  test_printout();
  mc_print_all_pages();
  cleanup_printout();
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<tst_print_all_pages>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdx
  mov    $0x1,%esi
  call   <setup_printout@plt>
  mov    $0x0,%eax
  call   <test_printout@plt>
  mov    $0x0,%eax
  call   <mc_print_all_pages@plt>
  mov    $0x0,%eax
  call   <cleanup_printout@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D2
type: string

tool: D2: contents of all pages

assistant: ```c
int tst_print_all_pages(const char *the_title) {
  setup_printout(the_title, (1), "contents of all pages");
  test_printout();
  mc_print_all_pages();
  cleanup_printout();
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<tst_print_all_pages>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdx
  mov    $0x1,%esi
  call   <setup_printout@plt>
  xor    %eax,%eax
  call   <test_printout@plt>
  xor    %eax,%eax
  call   <mc_print_all_pages@plt>
  xor    %eax,%eax
  call   <cleanup_printout@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D8
type: string

tool: D8: contents of all pages

assistant: ```c
int tst_print_all_pages(const char *the_title) {
  setup_printout(the_title, (1), "contents of all pages");
  test_printout();
  mc_print_all_pages();
  cleanup_printout();
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<tst_print_all_pages>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdx
  mov    $0x1,%esi
  call   <setup_printout@plt>
  xor    %eax,%eax
  call   <test_printout@plt>
  xor    %eax,%eax
  call   <mc_print_all_pages@plt>
  xor    %eax,%eax
  call   <cleanup_printout@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
data_type: string

tool: D3: contents of all pages

assistant: ```c
int tst_print_all_pages(const char *the_title) {
  setup_printout(the_title, (1), "contents of all pages");
  test_printout();
  mc_print_all_pages();
  cleanup_printout();
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hex2byte>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <ishexnumber@plt>
  test   %rax,%rax
  je     L56
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <ishexnumber@plt>
  test   %rax,%rax
  jne    L65
L56:
  mov    $0xffffffff,%eax
  jmp    L32
L65:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jg     L16
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  shl    $0x4,%eax
  mov    %eax,%ebx
  jmp    L64
L16:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  movsbl %al,%eax
  sub    $0x37,%eax
  shl    $0x4,%eax
  mov    %eax,%ebx
L64:
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jg     L4
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  jmp    L33
L4:
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  movsbl %al,%eax
  sub    $0x37,%eax
L33:
  add    %ebx,%eax
L32:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int hex2byte(const char c[2]) {
  if (!(ishexnumber(c[0]) && ishexnumber(c[1])))
    return -1;
  return ((c[0] <= '9' ? c[0] - '0' : toupper(c[0]) - 'A' + 10) << 4) + (c[1] <= '9' ? c[1] - '0' : toupper(c[1]) - 'A' + 10);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<hex2byte>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  movsbl (%rdi),%edi
  call   <ishexnumber@plt>
  test   %rax,%rax
  je     L1
  movsbl 0x1(%rbp),%edi
  call   <ishexnumber@plt>
  test   %rax,%rax
  je     L69
  movzbl 0x0(%rbp),%edi
  cmp    $0x39,%dil
  jg     L50
  movsbl %dil,%ebx
  sub    $0x30,%ebx
  shl    $0x4,%ebx
L12:
  movzbl 0x1(%rbp),%edi
  cmp    $0x39,%dil
  jg     L41
  movsbl %dil,%edi
  sub    $0x30,%edi
L44:
  lea    (%rbx,%rdi,1),%eax
L66:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L50:
  movsbl %dil,%edi
  call   <toupper@plt>
  movsbl %al,%ebx
  sub    $0x37,%ebx
  shl    $0x4,%ebx
  jmp    L12
L41:
  movsbl %dil,%edi
  call   <toupper@plt>
  movsbl %al,%edi
  sub    $0x37,%edi
  jmp    L44
L1:
  mov    $0xffffffff,%eax
  jmp    L66
L69:
  mov    $0xffffffff,%eax
  jmp    L66
```

assistant: ```c
int hex2byte(const char c[2]) {
  if (!(ishexnumber(c[0]) && ishexnumber(c[1])))
    return -1;
  return ((c[0] <= '9' ? c[0] - '0' : toupper(c[0]) - 'A' + 10) << 4) + (c[1] <= '9' ? c[1] - '0' : toupper(c[1]) - 'A' + 10);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<hex2byte>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  call   <ishexnumber@plt>
  test   %rax,%rax
  je     L10
  movsbl 0x1(%rbp),%edi
  call   <ishexnumber@plt>
  test   %rax,%rax
  je     L10
  movsbl 0x0(%rbp),%edi
  cmp    $0x39,%dil
  jg     L9
  lea    -0x30(%rdi),%eax
  movsbl 0x1(%rbp),%edi
  shl    $0x4,%eax
  mov    %eax,%ebx
  cmp    $0x39,%dil
  jg     L51
L34:
  sub    $0x30,%edi
L57:
  lea    (%rbx,%rdi,1),%eax
L69:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L9:
  call   <toupper@plt>
  movsbl 0x1(%rbp),%edi
  movsbl %al,%eax
  sub    $0x37,%eax
  shl    $0x4,%eax
  mov    %eax,%ebx
  cmp    $0x39,%dil
  jle    L34
L51:
  call   <toupper@plt>
  movsbl %al,%edi
  sub    $0x37,%edi
  jmp    L57
  nop
L10:
  mov    $0xffffffff,%eax
  jmp    L69
```

assistant: ```c
int hex2byte(const char c[2]) {
  if (!(ishexnumber(c[0]) && ishexnumber(c[1])))
    return -1;
  return ((c[0] <= '9' ? c[0] - '0' : toupper(c[0]) - 'A' + 10) << 4) + (c[1] <= '9' ? c[1] - '0' : toupper(c[1]) - 'A' + 10);
}
```
user: What is the c source code of the assembly code below:

```asm
<hex2byte>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  call   <ishexnumber@plt>
  test   %rax,%rax
  je     L7
  movsbl 0x1(%rbp),%edi
  call   <ishexnumber@plt>
  test   %rax,%rax
  je     L7
  movsbl 0x0(%rbp),%edi
  cmp    $0x39,%dil
  jg     L68
  lea    -0x30(%rdi),%eax
  movsbl 0x1(%rbp),%edi
  shl    $0x4,%eax
  mov    %eax,%ebx
  cmp    $0x39,%dil
  jg     L4
L69:
  sub    $0x30,%edi
L55:
  lea    (%rbx,%rdi,1),%eax
L14:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L68:
  call   <toupper@plt>
  movsbl 0x1(%rbp),%edi
  movsbl %al,%eax
  sub    $0x37,%eax
  shl    $0x4,%eax
  mov    %eax,%ebx
  cmp    $0x39,%dil
  jle    L69
L4:
  call   <toupper@plt>
  movsbl %al,%edi
  sub    $0x37,%edi
  jmp    L55
  nop
L7:
  mov    $0xffffffff,%eax
  jmp    L14
```

assistant: ```c
int hex2byte(const char c[2]) {
  if (!(ishexnumber(c[0]) && ishexnumber(c[1])))
    return -1;
  return ((c[0] <= '9' ? c[0] - '0' : toupper(c[0]) - 'A' + 10) << 4) + (c[1] <= '9' ? c[1] - '0' : toupper(c[1]) - 'A' + 10);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<os_now>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  lea    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <clock_gettime@plt>
  mov    %eax,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x18(%rbp),%rax
  imul   $0x3e8,%eax,%ecx
  mov    -0x20(%rbp),%rax
  movslq %eax,%rdx
  imul   $0x431bde83,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x12,%edx
  sar    $0x1f,%eax
  mov    %eax,%esi
  mov    %edx,%eax
  sub    %esi,%eax
  add    %ecx,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
int64_t os_now(void) {
  struct timespec ts;
  int rc = clock_gettime(CLOCK_MONOTONIC, &ts);
  assert(rc == 0);
  return ((int64_t)ts.tv_sec) * 1000 + (((int64_t)ts.tv_nsec) / 1000000);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<os_now>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    D2(%rip),%rax
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  test   %eax,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  imul   $0x3e8,0x8(%rsp),%eax
  mov    (%rsp),%rcx
  movslq %ecx,%rdx
  imul   $0x431bde83,%rdx,%rdx
  sar    $0x32,%rdx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int64_t os_now(void) {
  struct timespec ts;
  int rc = clock_gettime(CLOCK_MONOTONIC, &ts);
  assert(rc == 0);
  return ((int64_t)ts.tv_sec) * 1000 + (((int64_t)ts.tv_nsec) / 1000000);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<os_now>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D1(%rip),%rax
  mov    %rsp,%rsi
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  mov    (%rsp),%rcx
  imul   $0x3e8,0x8(%rsp),%eax
  movslq %ecx,%rdx
  sar    $0x1f,%ecx
  imul   $0x431bde83,%rdx,%rdx
  sar    $0x32,%rdx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int64_t os_now(void) {
  struct timespec ts;
  int rc = clock_gettime(CLOCK_MONOTONIC, &ts);
  assert(rc == 0);
  return ((int64_t)ts.tv_sec) * 1000 + (((int64_t)ts.tv_nsec) / 1000000);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<os_now>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D0(%rip),%rax
  mov    %rsp,%rsi
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  mov    (%rsp),%rcx
  imul   $0x3e8,0x8(%rsp),%eax
  movslq %ecx,%rdx
  sar    $0x1f,%ecx
  imul   $0x431bde83,%rdx,%rdx
  sar    $0x32,%rdx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int64_t os_now(void) {
  struct timespec ts;
  int rc = clock_gettime(CLOCK_MONOTONIC, &ts);
  assert(rc == 0);
  return ((int64_t)ts.tv_sec) * 1000 + (((int64_t)ts.tv_nsec) / 1000000);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dotest>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
L22:
  cmpl   $0x9,-0x4(%rbp)
  jle    L20
  subl   $0x1,-0x4(%rbp)
  cmpl   $0x9,-0x4(%rbp)
  jg     L22
  jmp    L1
L20:
  nop
L1:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int dotest(int x) {
  do {
    if (x < 10)
      break;
    x--;
  } while (x >= 10);
  return x;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<dotest>:
  endbr64
  mov    %edi,%eax
L12:
  cmp    $0x9,%eax
  jle    L8
  sub    $0x1,%eax
  cmp    $0x9,%eax
  jg     L12
L8:
  ret
```

assistant: ```c
int dotest(int x) {
  do {
    if (x < 10)
      break;
    x--;
  } while (x >= 10);
  return x;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<dotest>:
  endbr64
  mov    %edi,%eax
  jmp    L27
  nopl   0x0(%rax,%rax,1)
L14:
  sub    $0x1,%eax
  cmp    $0x9,%eax
  je     L10
L27:
  cmp    $0x9,%eax
  jg     L14
L10:
  ret
```

assistant: ```c
int dotest(int x) {
  do {
    if (x < 10)
      break;
    x--;
  } while (x >= 10);
  return x;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dotest>:
  endbr64
  mov    %edi,%eax
  jmp    L26
  nopl   0x0(%rax,%rax,1)
L13:
  sub    $0x1,%eax
  cmp    $0x9,%eax
  je     L5
L26:
  cmp    $0x9,%eax
  jg     L13
L5:
  ret
```

assistant: ```c
int dotest(int x) {
  do {
    if (x < 10)
      break;
    x--;
  } while (x >= 10);
  return x;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fn12>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x1,%edi
  call   <fn11@plt>
  mov    $0x1,%edi
  call   <fn11@plt>
  mov    $0x1,%edi
  call   <fn11@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void fn12(void) {
  fn11(1);
  fn11(1);
  fn11(1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<fn12>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <fn11@plt>
  mov    $0x1,%edi
  call   <fn11@plt>
  mov    $0x1,%edi
  call   <fn11@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void fn12(void) {
  fn11(1);
  fn11(1);
  fn11(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fn12>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <fn11@plt>
  mov    $0x1,%edi
  call   <fn11@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1050 <fn11@plt>
```

assistant: ```c
void fn12(void) {
  fn11(1);
  fn11(1);
  fn11(1);
}
```
user: What is the c source code of the assembly code below?

```asm
<fn12>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <fn11@plt>
  mov    $0x1,%edi
  call   <fn11@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1050 <fn11@plt>
```

assistant: ```c
void fn12(void) {
  fn11(1);
  fn11(1);
  fn11(1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
int main() {
  int x, *p;
  if (sizeof(0) < 2)
    return 1;
  if (sizeof 0 < 2)
    return 1;
  if (sizeof(char) < 1)
    return 1;
  if (sizeof(int) - 2 < 0)
    return 1;
  if (sizeof(&x) != sizeof p)
    return 1;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

assistant: ```c
int main() {
  int x, *p;
  if (sizeof(0) < 2)
    return 1;
  if (sizeof 0 < 2)
    return 1;
  if (sizeof(char) < 1)
    return 1;
  if (sizeof(int) - 2 < 0)
    return 1;
  if (sizeof(&x) != sizeof p)
    return 1;
  return 0;
}
```
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
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int x, *p;
  if (sizeof(0) < 2)
    return 1;
  if (sizeof 0 < 2)
    return 1;
  if (sizeof(char) < 1)
    return 1;
  if (sizeof(int) - 2 < 0)
    return 1;
  if (sizeof(&x) != sizeof p)
    return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

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

assistant: ```c
int main() {
  int x, *p;
  if (sizeof(0) < 2)
    return 1;
  if (sizeof 0 < 2)
    return 1;
  if (sizeof(char) < 1)
    return 1;
  if (sizeof(int) - 2 < 0)
    return 1;
  if (sizeof(&x) != sizeof p)
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cpuid_width_fixed_perf_counters>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xa,-0xc(%rbp)
  lea    -0x10(%rbp),%rdi
  lea    -0x14(%rbp),%rcx
  lea    -0x18(%rbp),%rdx
  lea    -0x1c(%rbp),%rsi
  mov    -0xc(%rbp),%eax
  mov    %rdi,%r8
  mov    %eax,%edi
  call   <cpuid@plt>
  mov    -0x10(%rbp),%eax
  sar    $0x5,%eax
  movzbl %al,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
int cpuid_width_fixed_perf_counters() {
  uint64_t rax, rbx, rcx, rdx;
  int leaf = 10;
  cpuid(leaf, &rax, &rbx, &rcx, &rdx);
  return (((uint64_t)(rdx) & ((((uint64_t)1 << (((12)) - ((5)) + 1)) - 1) << ((5)))) >> (5));
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<cpuid_width_fixed_perf_counters>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  mov    $0xa,%edi
  call   <cpuid@plt>
  mov    0x14(%rsp),%eax
  sar    $0x5,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  movzbl %al,%eax
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int cpuid_width_fixed_perf_counters() {
  uint64_t rax, rbx, rcx, rdx;
  int leaf = 10;
  cpuid(leaf, &rax, &rbx, &rcx, &rdx);
  return (((uint64_t)(rdx) & ((((uint64_t)1 << (((12)) - ((5)) + 1)) - 1) << ((5)))) >> (5));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cpuid_width_fixed_perf_counters>:
  endbr64
  sub    $0x28,%rsp
  mov    $0xa,%edi
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  call   <cpuid@plt>
  mov    0x14(%rsp),%eax
  sar    $0x5,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  movzbl %al,%eax
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int cpuid_width_fixed_perf_counters() {
  uint64_t rax, rbx, rcx, rdx;
  int leaf = 10;
  cpuid(leaf, &rax, &rbx, &rcx, &rdx);
  return (((uint64_t)(rdx) & ((((uint64_t)1 << (((12)) - ((5)) + 1)) - 1) << ((5)))) >> (5));
}
```
user: Translate the below assembly code into C programming language:

```asm
<cpuid_width_fixed_perf_counters>:
  endbr64
  sub    $0x28,%rsp
  mov    $0xa,%edi
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  call   <cpuid@plt>
  mov    0x14(%rsp),%eax
  sar    $0x5,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  movzbl %al,%eax
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int cpuid_width_fixed_perf_counters() {
  uint64_t rax, rbx, rcx, rdx;
  int leaf = 10;
  cpuid(leaf, &rax, &rbx, &rcx, &rdx);
  return (((uint64_t)(rdx) & ((((uint64_t)1 << (((12)) - ((5)) + 1)) - 1) << ((5)))) >> (5));
}
```
user: Translate the below assembly code into C programming language:

```asm
<rmstr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L25:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L31
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L1:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L25
  jmp    L36
L31:
  nop
L36:
  nop
  pop    %rbp
  ret
```

assistant: ```c
int rmstr(char *str, int len) {
  int i;
  for (i = 0; i < len; i++) {
    if (str[i] == '\0')
      break;
    str[i] = str[i + 1];
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<rmstr>:
  endbr64
  test   %esi,%esi
  jle    L23
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x1(%rdi,%rdx,1),%rcx
L29:
  cmpb   $0x0,(%rax)
  je     L23
  movzbl 0x1(%rax),%edx
  mov    %dl,(%rax)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L29
L23:
  ret
```

assistant: ```c
int rmstr(char *str, int len) {
  int i;
  for (i = 0; i < len; i++) {
    if (str[i] == '\0')
      break;
    str[i] = str[i + 1];
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<rmstr>:
  endbr64
  test   %esi,%esi
  jle    L30
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%rdx
  jmp    L23
  nopw   0x0(%rax,%rax,1)
L39:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  mov    %al,-0x1(%rdi)
  cmp    %rdx,%rdi
  je     L30
L23:
  cmpb   $0x0,(%rdi)
  jne    L39
L30:
  ret
```

assistant: ```c
int rmstr(char *str, int len) {
  int i;
  for (i = 0; i < len; i++) {
    if (str[i] == '\0')
      break;
    str[i] = str[i + 1];
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<rmstr>:
  endbr64
  test   %esi,%esi
  jle    L39
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%rdx
  jmp    L26
  nopw   0x0(%rax,%rax,1)
L2:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  mov    %al,-0x1(%rdi)
  cmp    %rdx,%rdi
  je     L39
L26:
  cmpb   $0x0,(%rdi)
  jne    L2
L39:
  ret
```

assistant: ```c
int rmstr(char *str, int len) {
  int i;
  for (i = 0; i < len; i++) {
    if (str[i] == '\0')
      break;
    str[i] = str[i + 1];
  }
}
```