ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<test_inotify_unmap_all>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <setup@plt>
  mov    $0x0,%eax
  call   <inotify_unmap_all@plt>
  mov    $0x0,%eax
  call   <inotify_map_isempty@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  call   <teardown@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void test_inotify_unmap_all() {
  setup();
  inotify_unmap_all();
  assert(inotify_map_isempty());
  teardown();
}
```
user: What is the c source code of the assembly code below:

```asm
<test_inotify_unmap_all>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <setup@plt>
  xor    %eax,%eax
  call   <inotify_unmap_all@plt>
  xor    %eax,%eax
  call   <inotify_map_isempty@plt>
  mov    %eax,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <teardown@plt>
```

assistant: ```c
void test_inotify_unmap_all() {
  setup();
  inotify_unmap_all();
  assert(inotify_map_isempty());
  teardown();
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<test_inotify_unmap_all>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <setup@plt>
  xor    %eax,%eax
  call   <inotify_unmap_all@plt>
  xor    %eax,%eax
  call   <inotify_map_isempty@plt>
  mov    %eax,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <teardown@plt>
```

assistant: ```c
void test_inotify_unmap_all() {
  setup();
  inotify_unmap_all();
  assert(inotify_map_isempty());
  teardown();
}
```
user: Translate the below assembly code into C programming language:

```asm
<convertHash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L11
L10:
  mov    -0x14(%rbp),%eax
  add    %eax,%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x16(%rbp)
  mov    -0x14(%rbp),%eax
  add    %eax,%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x15(%rbp)
  movsbl -0x16(%rbp),%eax
  mov    %eax,%edi
  call   <getHexVal@plt>
  shl    $0x4,%eax
  mov    %eax,%ebx
  movsbl -0x15(%rbp),%eax
  mov    %eax,%edi
  call   <getHexVal@plt>
  mov    %eax,%ecx
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  lea    (%rbx,%rcx,1),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x14(%rbp)
L11:
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L10
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void convertHash(const char *hexhash, unsigned char *binhash) {
  int i;
  for (i = 0; i < SHA_DIGEST_LENGTH; i++) {
    char first = *(hexhash + (2 * i));
    char second = *(hexhash + (2 * i) + 1);
    *(binhash + i) = (getHexVal(first) * 16) + getHexVal(second);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<convertHash>:
  endbr64
  mov    D13(%rip),%rax
  cmpl   $0x0,(%rax)
  jle    L7
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    $0x0,%ebp
L11:
  movzbl 0x1(%r12,%rbp,2),%r14d
  movsbl (%r12,%rbp,2),%edi
  call   <getHexVal@plt>
  mov    %eax,%ebx
  movsbl %r14b,%edi
  call   <getHexVal@plt>
  shl    $0x4,%ebx
  add    %eax,%ebx
  mov    %bl,0x0(%r13,%rbp,1)
  add    $0x1,%rbp
  mov    D13(%rip),%rax
  cmp    %ebp,(%rax)
  jg     L11
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L7:
  ret
```

assistant: ```c
void convertHash(const char *hexhash, unsigned char *binhash) {
  int i;
  for (i = 0; i < SHA_DIGEST_LENGTH; i++) {
    char first = *(hexhash + (2 * i));
    char second = *(hexhash + (2 * i) + 1);
    *(binhash + i) = (getHexVal(first) * 16) + getHexVal(second);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<convertHash>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D8(%rip),%r13
  mov    0x0(%r13),%eax
  test   %eax,%eax
  jle    L16
  mov    %rdi,%r15
  mov    %rsi,%r12
  xor    %ebp,%ebp
  nopl   0x0(%rax)
L6:
  movsbl (%r15,%rbp,2),%edi
  movzbl 0x1(%r15,%rbp,2),%r14d
  call   <getHexVal@plt>
  movsbl %r14b,%edi
  mov    %eax,%ebx
  call   <getHexVal@plt>
  shl    $0x4,%ebx
  add    %eax,%ebx
  mov    %bl,(%r12,%rbp,1)
  add    $0x1,%rbp
  cmp    %ebp,0x0(%r13)
  jg     L6
L16:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void convertHash(const char *hexhash, unsigned char *binhash) {
  int i;
  for (i = 0; i < SHA_DIGEST_LENGTH; i++) {
    char first = *(hexhash + (2 * i));
    char second = *(hexhash + (2 * i) + 1);
    *(binhash + i) = (getHexVal(first) * 16) + getHexVal(second);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<convertHash>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D7(%rip),%r13
  mov    0x0(%r13),%eax
  test   %eax,%eax
  jle    L4
  mov    %rdi,%r15
  mov    %rsi,%r12
  xor    %ebp,%ebp
  nopl   0x0(%rax)
L17:
  movsbl (%r15,%rbp,2),%edi
  movzbl 0x1(%r15,%rbp,2),%r14d
  call   <getHexVal@plt>
  movsbl %r14b,%edi
  mov    %eax,%ebx
  call   <getHexVal@plt>
  shl    $0x4,%ebx
  add    %eax,%ebx
  mov    %bl,(%r12,%rbp,1)
  add    $0x1,%rbp
  cmp    %ebp,0x0(%r13)
  jg     L17
L4:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void convertHash(const char *hexhash, unsigned char *binhash) {
  int i;
  for (i = 0; i < SHA_DIGEST_LENGTH; i++) {
    char first = *(hexhash + (2 * i));
    char second = *(hexhash + (2 * i) + 1);
    *(binhash + i) = (getHexVal(first) * 16) + getHexVal(second);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<xmalloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L8
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <exit@plt>
L8:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *xmalloc(size_t s) {
  void *p = malloc(s);
  if (!p)
    exit(ENOMEM);
  return p;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xmalloc>:
  endbr64
  push   %rbx
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L2
L12:
  mov    %rbx,%rax
  pop    %rbx
  ret
L2:
  mov    D7(%rip),%rax
  mov    (%rax),%edi
  call   <exit@plt>
  jmp    L12
```

assistant: ```c
void *xmalloc(size_t s) {
  void *p = malloc(s);
  if (!p)
    exit(ENOMEM);
  return p;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xmalloc>:
  endbr64
  push   %r12
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L2
L4:
  mov    %r12,%rax
  pop    %r12
  ret
L2:
  mov    D3(%rip),%rax
  mov    (%rax),%edi
  call   <exit@plt>
  jmp    L4
```

assistant: ```c
void *xmalloc(size_t s) {
  void *p = malloc(s);
  if (!p)
    exit(ENOMEM);
  return p;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<xmalloc>:
  endbr64
  push   %r12
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L9
L13:
  mov    %r12,%rax
  pop    %r12
  ret
L9:
  mov    D7(%rip),%rax
  mov    (%rax),%edi
  call   <exit@plt>
  jmp    L13
```

assistant: ```c
void *xmalloc(size_t s) {
  void *p = malloc(s);
  if (!p)
    exit(ENOMEM);
  return p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<decode>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rax,%rbx
  lea    (%rax,%rax,1),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  movb   $0x0,(%rax)
  movb   $0x0,0x10(%rsp)
  test   %rbx,%rbx
  je     L78
  lea    (%rbx,%rbx,1),%r15
  mov    %r13,%r12
  lea    0x0(%r13,%rbx,1),%rax
  mov    %rax,0x8(%rsp)
  mov    $0x0,%ebx
  jmp    L100
L52:
  lea    0x10(%rsp),%rdi
  mov    $0x1,%edx
  mov    %r12,%rsi
  call   <strncat@plt>
  jmp    L69
L135:
  lea    0x10(%rsp),%rdi
  call   <atoi@plt>
  test   %rax,%rax
  je     L103
  add    %rbx,%rax
  mov    %rax,(%rsp)
  jmp    L74
L68:
  cmp    %rbx,(%rsp)
  je     L38
L74:
  mov    $0x1,%edx
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <strncat@plt>
  add    $0x1,%rbx
  lea    (%r15,%r15,1),%r13
  movabs $0xaaaaaaaaaaaaaaab,%rax
  mul    %r13
  shr    %rdx
  cmp    %rbx,%rdx
  jae    L68
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <realloc@plt>
  mov    %rax,%rbp
  mov    %r13,%r15
  jmp    L68
L38:
  mov    (%rsp),%rbx
L103:
  movb   $0x0,0x10(%rsp)
L69:
  add    $0x1,%r12
  cmp    %r12,0x8(%rsp)
  je     L78
L100:
  mov    %r12,%r14
  movsbl (%r12),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L52
  lea    0x10(%rsp),%rdi
  call   <strlen@plt>
  test   %rax,%rax
  jg     L135
  mov    $0x1,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strncat@plt>
  add    $0x1,%rbx
  lea    (%r15,%r15,1),%r13
  movabs $0xaaaaaaaaaaaaaaab,%rax
  mul    %r13
  shr    %rdx
  cmp    %rbx,%rdx
  jae    L103
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <realloc@plt>
  mov    %rax,%rbp
  mov    %r13,%r15
  jmp    L103
L78:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L119
  mov    %rbp,%rax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L119:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *decode(const char *data) {
  size_t length = strlen(data);
  char *result = malloc(sizeof(char) * length * 2);
  size_t result_size = length * 2;
  size_t result_len = 0;
  result[0] = '\0';
  char temp[20];
  temp[0] = '\0';
  if (length == 0)
    return result;
  for (size_t i = 0; i < length; ++i) {
    if (isdigit(data[i])) {
      strncat(temp, &data[i], 1);
    } else {
      if (strlen(temp) > 0) {
        size_t number = atoi(temp);
        for (size_t j = 0; j < number; ++j) {
          strncat(result, &data[i], 1);
          result_len++;
          if (result_len > result_size * 2 / 3) {
            result = realloc(result, result_size * 2);
            result_size *= 2;
          }
        }
      } else {
        strncat(result, &data[i], 1);
        result_len++;
        if (result_len > result_size * 2 / 3) {
          result = realloc(result, result_size * 2);
          result_size *= 2;
        }
      }
      temp[0] = '\0';
    }
  }
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1,-0x10(%rbp)
  movl   $0x1,-0xc(%rbp)
  movl   $0x1,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L21:
  mov    -0x8(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L6
  mov    -0x8(%rbp),%eax
  add    %eax,-0x4(%rbp)
L6:
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L2:
  cmpl   $0x3d0900,-0x8(%rbp)
  jle    L21
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D6
data_type: string

tool: D6: %d


assistant: ```c
int main() {
  int a = 1, b = 1, c = 1, total = 0;
  while (c <= 4000000) {
    if (c % 2 == 0) {
      total += c;
    }
    c = a + b;
    a = b;
    b = c;
  }
  printf("%d\n", total);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edx
  mov    $0x1,%ecx
  mov    $0x0,%esi
L19:
  lea    (%rcx,%rdx,1),%eax
  cmp    $0x3d0900,%eax
  jg     L8
  lea    (%rsi,%rax,1),%ecx
  test   $0x1,%al
  cmove  %ecx,%esi
  mov    %edx,%ecx
  mov    %eax,%edx
  jmp    L19
L8:
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D0
datatype: string

tool: D0: %d


assistant: ```c
int main() {
  int a = 1, b = 1, c = 1, total = 0;
  while (c <= 4000000) {
    if (c % 2 == 0) {
      total += c;
    }
    c = a + b;
    a = b;
    b = c;
  }
  printf("%d\n", total);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edx
  mov    $0x1,%ecx
  xor    %esi,%esi
  jmp    L17
  cs nopw 0x0(%rax,%rax,1)
L6:
  lea    (%rsi,%rax,1),%ecx
  test   $0x1,%al
  cmove  %ecx,%esi
  mov    %edx,%ecx
  mov    %eax,%edx
L17:
  lea    (%rdx,%rcx,1),%eax
  cmp    $0x3d0900,%eax
  jle    L6
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D0
datatype: string

tool: D0: %d


assistant: ```c
int main() {
  int a = 1, b = 1, c = 1, total = 0;
  while (c <= 4000000) {
    if (c % 2 == 0) {
      total += c;
    }
    c = a + b;
    a = b;
    b = c;
  }
  printf("%d\n", total);
}
```
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    $0x8,%rsp
  mov    $0x1,%ecx
  mov    $0x2,%eax
  xor    %esi,%esi
  jmp    L12
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    %edx,%eax
L12:
  lea    (%rsi,%rax,1),%edx
  test   $0x1,%al
  cmove  %edx,%esi
  lea    (%rcx,%rax,1),%edx
  mov    %eax,%ecx
  cmp    $0x3d0900,%edx
  jle    L14
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D8: %d


assistant: ```c
int main() {
  int a = 1, b = 1, c = 1, total = 0;
  while (c <= 4000000) {
    if (c % 2 == 0) {
      total += c;
    }
    c = a + b;
    a = b;
    b = c;
  }
  printf("%d\n", total);
}
```
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
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  lea    -0x10(%rbp),%rax
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <push@plt>
  lea    -0x10(%rbp),%rax
  mov    $0x7,%esi
  mov    %rax,%rdi
  call   <push@plt>
  lea    -0x10(%rbp),%rax
  mov    $0x9,%esi
  mov    %rax,%rdi
  call   <push@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <peek@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <pop@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <peek@plt>
  lea    -0x10(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <push@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <pop@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <pop@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <pop@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <peek@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int main() {
  struct Stack *stack = NULL;
  push(&stack, 4);
  push(&stack, 7);
  push(&stack, 9);
  peek(stack);
  pop(&stack);
  peek(stack);
  push(&stack, 1);
  pop(&stack);
  pop(&stack);
  pop(&stack);
  peek(stack);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%rbx
  mov    $0x4,%esi
  mov    %rbx,%rdi
  call   <push@plt>
  mov    $0x7,%esi
  mov    %rbx,%rdi
  call   <push@plt>
  mov    $0x9,%esi
  mov    %rbx,%rdi
  call   <push@plt>
  mov    (%rsp),%rdi
  call   <peek@plt>
  mov    %rbx,%rdi
  call   <pop@plt>
  mov    (%rsp),%rdi
  call   <peek@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <push@plt>
  mov    %rbx,%rdi
  call   <pop@plt>
  mov    %rbx,%rdi
  call   <pop@plt>
  mov    %rbx,%rdi
  call   <pop@plt>
  mov    (%rsp),%rdi
  call   <peek@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  struct Stack *stack = NULL;
  push(&stack, 4);
  push(&stack, 7);
  push(&stack, 9);
  peek(stack);
  pop(&stack);
  peek(stack);
  push(&stack, 1);
  pop(&stack);
  pop(&stack);
  pop(&stack);
  peek(stack);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x4,%esi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movq   $0x0,(%rsp)
  mov    %rbp,%rdi
  call   <push@plt>
  mov    $0x7,%esi
  mov    %rbp,%rdi
  call   <push@plt>
  mov    $0x9,%esi
  mov    %rbp,%rdi
  call   <push@plt>
  mov    (%rsp),%rdi
  call   <peek@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  mov    (%rsp),%rdi
  call   <peek@plt>
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <push@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  mov    (%rsp),%rdi
  call   <peek@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  struct Stack *stack = NULL;
  push(&stack, 4);
  push(&stack, 7);
  push(&stack, 9);
  peek(stack);
  pop(&stack);
  peek(stack);
  push(&stack, 1);
  pop(&stack);
  pop(&stack);
  pop(&stack);
  peek(stack);
}
```
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    $0x4,%esi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movq   $0x0,(%rsp)
  mov    %rbp,%rdi
  call   <push@plt>
  mov    $0x7,%esi
  mov    %rbp,%rdi
  call   <push@plt>
  mov    $0x9,%esi
  mov    %rbp,%rdi
  call   <push@plt>
  mov    (%rsp),%rdi
  call   <peek@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  mov    (%rsp),%rdi
  call   <peek@plt>
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <push@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  mov    %rbp,%rdi
  call   <pop@plt>
  mov    (%rsp),%rdi
  call   <peek@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  struct Stack *stack = NULL;
  push(&stack, 4);
  push(&stack, 7);
  push(&stack, 9);
  peek(stack);
  pop(&stack);
  peek(stack);
  push(&stack, 1);
  pop(&stack);
  pop(&stack);
  pop(&stack);
  peek(stack);
}
```
user: What is the c source code of the assembly code below?

```asm
<librlc_rs_encode_cleanup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int librlc_rs_encode_cleanup(char *encoded_data, char *encoded_parity) {
  free(encoded_data);
  free(encoded_parity);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<librlc_rs_encode_cleanup>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  call   <free@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int librlc_rs_encode_cleanup(char *encoded_data, char *encoded_parity) {
  free(encoded_data);
  free(encoded_parity);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<librlc_rs_encode_cleanup>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int librlc_rs_encode_cleanup(char *encoded_data, char *encoded_parity) {
  free(encoded_data);
  free(encoded_parity);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<librlc_rs_encode_cleanup>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int librlc_rs_encode_cleanup(char *encoded_data, char *encoded_parity) {
  free(encoded_data);
  free(encoded_parity);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<MacIsValid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xff,%al
  jne    L10
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L10
  mov    -0x8(%rbp),%eax
  jmp    L42
L10:
  movl   $0x0,-0x4(%rbp)
  jmp    L21
L53:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L25
  movl   $0x1,-0x8(%rbp)
  jmp    L61
L25:
  addl   $0x1,-0x4(%rbp)
L21:
  cmpl   $0x5,-0x4(%rbp)
  jle    L53
L61:
  mov    -0x8(%rbp),%eax
L42:
  pop    %rbp
  ret
```

assistant: ```c
int MacIsValid(unsigned char *mac) {
  int fvalid = 0;
  int i;
  if (mac[0] == 0xFF && mac[1] == 0x00)
    return (fvalid);
  for (i = 0; i < 6; i++)
    if (mac[i] != 0) {
      fvalid = 1;
      break;
    }
  return (fvalid);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<MacIsValid>:
  endbr64
  cmpb   $0xff,(%rdi)
  je     L0
L19:
  mov    %rdi,%rax
  add    $0x6,%rdi
L5:
  cmpb   $0x0,(%rax)
  jne    L13
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L5
  mov    $0x0,%eax
  ret
L0:
  mov    $0x0,%eax
  cmpb   $0x0,0x1(%rdi)
  jne    L19
  ret
L13:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int MacIsValid(unsigned char *mac) {
  int fvalid = 0;
  int i;
  if (mac[0] == 0xFF && mac[1] == 0x00)
    return (fvalid);
  for (i = 0; i < 6; i++)
    if (mac[i] != 0) {
      fvalid = 1;
      break;
    }
  return (fvalid);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<MacIsValid>:
  endbr64
  movzbl (%rdi),%edx
  cmp    $0xff,%dl
  je     L15
L19:
  lea    0x1(%rdi),%rax
  add    $0x6,%rdi
  test   %dl,%dl
  jne    L36
L2:
  cmp    %rdi,%rax
  je     L45
  movzbl (%rax),%edx
  add    $0x1,%rax
  test   %dl,%dl
  je     L2
L36:
  mov    $0x1,%eax
  ret
  xchg   %ax,%ax
L45:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  xor    %eax,%eax
  cmpb   $0x0,0x1(%rdi)
  jne    L19
  ret
```

assistant: ```c
int MacIsValid(unsigned char *mac) {
  int fvalid = 0;
  int i;
  if (mac[0] == 0xFF && mac[1] == 0x00)
    return (fvalid);
  for (i = 0; i < 6; i++)
    if (mac[i] != 0) {
      fvalid = 1;
      break;
    }
  return (fvalid);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<MacIsValid>:
  endbr64
  movzbl (%rdi),%eax
  cmp    $0xff,%al
  je     L15
  test   %al,%al
  jne    L55
  cmpb   $0x0,0x1(%rdi)
  jne    L55
  cmpb   $0x0,0x2(%rdi)
  jne    L55
  cmpb   $0x0,0x3(%rdi)
  jne    L55
  cmpb   $0x0,0x4(%rdi)
  jne    L55
  xor    %eax,%eax
  cmpb   $0x0,0x5(%rdi)
  setne  %al
  ret
  nopl   0x0(%rax)
L15:
  xor    %eax,%eax
  cmpb   $0x0,0x1(%rdi)
  setne  %al
  ret
  nopw   0x0(%rax,%rax,1)
L55:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int MacIsValid(unsigned char *mac) {
  int fvalid = 0;
  int i;
  if (mac[0] == 0xFF && mac[1] == 0x00)
    return (fvalid);
  for (i = 0; i < 6; i++)
    if (mac[i] != 0) {
      fvalid = 1;
      break;
    }
  return (fvalid);
}
```