ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<pvCancelCapture>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <notify_pvcall@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D9
type_guessed: string

tool: D9: f

assistant: ```c
int pvCancelCapture(char byBoardNum) {
  notify_pvcall("f");
  return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<isAnagram>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  movl   $0x1,-0x28(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x40(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,%ebx
  jne    L44
  movl   $0x0,-0x20(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L9
L21:
  movl   $0x0,-0x1c(%rbp)
  jmp    L90
L10:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L66
  movl   $0x1,-0x24(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    -0x1c(%rbp),%eax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x1(%rdx),%rcx
  mov    -0x18(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rsi
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rdx,%rcx
  mov    %eax,%edx
  mov    %rcx,%rdi
  call   <memmove@plt>
  jmp    L88
L66:
  addl   $0x1,-0x1c(%rbp)
L90:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x1c(%rbp)
  jl     L10
L88:
  cmpl   $0x0,-0x24(%rbp)
  jne    L11
  movl   $0x0,-0x28(%rbp)
  jmp    L70
L11:
  addl   $0x1,-0x20(%rbp)
L9:
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x20(%rbp)
  jl     L21
  jmp    L70
L44:
  movl   $0x0,-0x28(%rbp)
L70:
  mov    -0x28(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int isAnagram(char *a, char *b) {
  int result = 1;
  char *bCopy = NULL;
  bCopy = (char *)malloc(strlen(b) * sizeof(char));
  strcpy(bCopy, b);
  if (strlen(a) == strlen(bCopy)) {
    int flag, i, j;
    for (i = 0, flag = 0; i < strlen(a); i++) {
      for (j = 0; j < strlen(bCopy); j++) {
        if (a[i] == bCopy[j]) {
          flag = 1;
          memmove(bCopy + j, bCopy + j + 1, strlen(bCopy) - j);
          break;
        }
      }
      if (flag == 0) {
        result = 0;
        break;
      }
    }
  } else {
    result = 0;
  }
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<isAnagram>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r15
  mov    %rsi,%rbx
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %rbp,%rdi
  call   <strlen@plt>
  movl   $0x0,0xc(%rsp)
  mov    %r15,%r13
  cmp    %eax,%ebx
  je     L72
L45:
  mov    0xc(%rsp),%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L56:
  mov    %rbp,%rdi
  call   <strlen@plt>
  sub    %r14d,%eax
  mov    %eax,%edx
  lea    0x1(%rbp,%rbx,1),%rsi
  mov    %r12,%rdi
  call   <memmove@plt>
L68:
  add    $0x1,%r13
  movl   $0x1,0xc(%rsp)
L72:
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %r13d,%eax
  sub    %r15d,%eax
  cmp    %eax,%edx
  jle    L48
  mov    $0x0,%ebx
L23:
  mov    %ebx,%r14d
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L17
  lea    0x0(%rbp,%rbx,1),%r12
  lea    0x1(%rbx),%rax
  movzbl (%r12),%ecx
  cmp    %cl,0x0(%r13)
  je     L56
  mov    %rax,%rbx
  jmp    L23
L17:
  cmpl   $0x0,0xc(%rsp)
  jne    L68
  jmp    L45
L48:
  movl   $0x1,0xc(%rsp)
  jmp    L45
```

assistant: ```c
int isAnagram(char *a, char *b) {
  int result = 1;
  char *bCopy = NULL;
  bCopy = (char *)malloc(strlen(b) * sizeof(char));
  strcpy(bCopy, b);
  if (strlen(a) == strlen(bCopy)) {
    int flag, i, j;
    for (i = 0, flag = 0; i < strlen(a); i++) {
      for (j = 0; j < strlen(bCopy); j++) {
        if (a[i] == bCopy[j]) {
          flag = 1;
          memmove(bCopy + j, bCopy + j + 1, strlen(bCopy) - j);
          break;
        }
      }
      if (flag == 0) {
        result = 0;
        break;
      }
    }
  } else {
    result = 0;
  }
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<isAnagram>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  mov    %rdi,%r14
  mov    %rsi,%rdi
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <strlen@plt>
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %r12,%rsi
  mov    %r14,%r12
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <strcpy@plt>
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  cmp    %eax,%ebx
  je     L7
L62:
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L60:
  test   %r15d,%r15d
  je     L62
L5:
  add    $0x1,%r12
  mov    $0x1,%r15d
L7:
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %eax,%r8d
  mov    %r12d,%eax
  sub    %r14d,%eax
  cmp    %eax,%r8d
  jle    L44
  xor    %ebx,%ebx
  jmp    L4
  nopl   0x0(%rax)
L3:
  lea    0x0(%rbp,%rbx,1),%r8
  add    $0x1,%rbx
  movzbl (%r8),%eax
  cmp    %al,(%r12)
  je     L13
L4:
  mov    %rbp,%rdi
  mov    %ebx,%r13d
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L3
  jmp    L60
  nopl   0x0(%rax)
L13:
  mov    %rbp,%rdi
  mov    %r8,0x8(%rsp)
  call   <strlen@plt>
  mov    0x8(%rsp),%rdi
  lea    0x0(%rbp,%rbx,1),%rsi
  sub    %r13d,%eax
  mov    %eax,%edx
  call   <memmove@plt>
  jmp    L5
L44:
  add    $0x18,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int isAnagram(char *a, char *b) {
  int result = 1;
  char *bCopy = NULL;
  bCopy = (char *)malloc(strlen(b) * sizeof(char));
  strcpy(bCopy, b);
  if (strlen(a) == strlen(bCopy)) {
    int flag, i, j;
    for (i = 0, flag = 0; i < strlen(a); i++) {
      for (j = 0; j < strlen(bCopy); j++) {
        if (a[i] == bCopy[j]) {
          flag = 1;
          memmove(bCopy + j, bCopy + j + 1, strlen(bCopy) - j);
          break;
        }
      }
      if (flag == 0) {
        result = 0;
        break;
      }
    }
  } else {
    result = 0;
  }
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<isAnagram>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  mov    %rdi,%r14
  mov    %rsi,%rdi
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <strlen@plt>
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %r12,%rsi
  mov    %r14,%r12
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <strcpy@plt>
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  cmp    %eax,%ebx
  je     L14
L27:
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L73:
  test   %r15d,%r15d
  je     L27
L36:
  add    $0x1,%r12
  mov    $0x1,%r15d
L14:
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %eax,%r8d
  mov    %r12d,%eax
  sub    %r14d,%eax
  cmp    %eax,%r8d
  jle    L4
  xor    %ebx,%ebx
  jmp    L75
  nopl   0x0(%rax)
L12:
  lea    0x0(%rbp,%rbx,1),%r8
  add    $0x1,%rbx
  movzbl (%r8),%eax
  cmp    %al,(%r12)
  je     L46
L75:
  mov    %rbp,%rdi
  mov    %ebx,%r13d
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L12
  jmp    L73
  nopl   0x0(%rax)
L46:
  mov    %rbp,%rdi
  mov    %r8,0x8(%rsp)
  call   <strlen@plt>
  mov    0x8(%rsp),%rdi
  lea    0x0(%rbp,%rbx,1),%rsi
  sub    %r13d,%eax
  mov    %eax,%edx
  call   <memmove@plt>
  jmp    L36
L4:
  add    $0x18,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int isAnagram(char *a, char *b) {
  int result = 1;
  char *bCopy = NULL;
  bCopy = (char *)malloc(strlen(b) * sizeof(char));
  strcpy(bCopy, b);
  if (strlen(a) == strlen(bCopy)) {
    int flag, i, j;
    for (i = 0, flag = 0; i < strlen(a); i++) {
      for (j = 0; j < strlen(bCopy); j++) {
        if (a[i] == bCopy[j]) {
          flag = 1;
          memmove(bCopy + j, bCopy + j + 1, strlen(bCopy) - j);
          break;
        }
      }
      if (flag == 0) {
        result = 0;
        break;
      }
    }
  } else {
    result = 0;
  }
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0xe0(%rbp)
  movq   $0x0,-0xd8(%rbp)
  movq   $0x0,-0xd0(%rbp)
  movq   $0x0,-0xc8(%rbp)
  movq   $0x0,-0xc0(%rbp)
  movq   $0x0,-0xb8(%rbp)
  movq   $0x0,-0xb0(%rbp)
  movq   $0x0,-0xa8(%rbp)
  movq   $0x0,-0xa0(%rbp)
  movq   $0x0,-0x98(%rbp)
  movq   $0x0,-0x90(%rbp)
  movq   $0x0,-0x88(%rbp)
  movq   $0x0,-0x80(%rbp)
  movq   $0x0,-0x78(%rbp)
  movq   $0x0,-0x70(%rbp)
  movq   $0x0,-0x68(%rbp)
  movq   $0x0,-0x60(%rbp)
  movq   $0x0,-0x58(%rbp)
  movq   $0x0,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movb   $0x0,-0x18(%rbp)
  lea    -0xe0(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0xe0(%rbp),%rax
  mov    %rax,%rdi
  call   <hash@plt>
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D4
type_infered: string
name: read_data
arguments: tag: D8
type_infered: string

tool: D4: %s

tool: D8: %li

assistant: ```c
int main() {
  char word[201] = {0};
  scanf("%s", word);
  printf("%li", hash(word));
  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   %rbx
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  movq   $0x0,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movq   $0x0,0x60(%rsp)
  movq   $0x0,0x68(%rsp)
  movq   $0x0,0x70(%rsp)
  movq   $0x0,0x78(%rsp)
  movq   $0x0,0x80(%rsp)
  movq   $0x0,0x88(%rsp)
  movq   $0x0,0x90(%rsp)
  movq   $0x0,0x98(%rsp)
  movq   $0x0,0xa0(%rsp)
  movq   $0x0,0xa8(%rsp)
  movq   $0x0,0xb0(%rsp)
  movq   $0x0,0xb8(%rsp)
  movq   $0x0,0xc0(%rsp)
  movb   $0x0,0xc8(%rsp)
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <hash@plt>
  mov    %eax,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0xe0,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D16
datatype: string
name: extract_symbolic_data
arguments: tag: D5
datatype: string

tool: D16: %s

tool: D5: %li

assistant: ```c
int main() {
  char word[201] = {0};
  scanf("%s", word);
  printf("%li", hash(word));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  pxor   %xmm0,%xmm0
  lea    D1(%rip),%rdi
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movaps %xmm0,(%rsp)
  mov    %rbp,%rsi
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,0x60(%rsp)
  movaps %xmm0,0x70(%rsp)
  movaps %xmm0,0x80(%rsp)
  movaps %xmm0,0x90(%rsp)
  movaps %xmm0,0xa0(%rsp)
  movaps %xmm0,0xb0(%rsp)
  movq   $0x0,0xc0(%rsp)
  movb   $0x0,0xc8(%rsp)
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <hash@plt>
  lea    D12(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0xe0,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
datatype: string
name: extract_symbolic_data
arguments: data_tag: D12
datatype: string

tool: D1: %s

tool: D12: %li

assistant: ```c
int main() {
  char word[201] = {0};
  scanf("%s", word);
  printf("%li", hash(word));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  pxor   %xmm0,%xmm0
  lea    D9(%rip),%rdi
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movaps %xmm0,(%rsp)
  mov    %rbp,%rsi
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,0x60(%rsp)
  movaps %xmm0,0x70(%rsp)
  movaps %xmm0,0x80(%rsp)
  movaps %xmm0,0x90(%rsp)
  movaps %xmm0,0xa0(%rsp)
  movaps %xmm0,0xb0(%rsp)
  movq   $0x0,0xc0(%rsp)
  movb   $0x0,0xc8(%rsp)
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <hash@plt>
  lea    D4(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0xe0,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
type_infered: string
name: retrieve_data
arguments: data_tag: D4
type_infered: string

tool: D9: %s

tool: D4: %li

assistant: ```c
int main() {
  char word[201] = {0};
  scanf("%s", word);
  printf("%li", hash(word));
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<ytht_badstr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L42
L14:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L39
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x1f,%al
  jbe    L48
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xff,%al
  jne    L39
L48:
  mov    $0x1,%eax
  jmp    L40
L39:
  addl   $0x1,-0xc(%rbp)
L42:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L14
  mov    $0x0,%eax
L40:
  pop    %rbp
  ret
```

assistant: ```c
int ytht_badstr(const char *s) {
  int i;
  const unsigned char *t = (const unsigned char *)s;
  for (i = 0; s[i]; i++)
    if (t[i] != 9 && (t[i] < 32 || t[i] == 255))
      return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<ytht_badstr>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L36
  add    $0x1,%rdi
  jmp    L24
L25:
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%eax
  test   %al,%al
  je     L18
L24:
  lea    -0x20(%rax),%edx
  cmp    $0xde,%dl
  jbe    L25
  cmp    $0x9,%al
  je     L25
  mov    $0x1,%eax
  ret
L18:
  mov    $0x0,%eax
  ret
L36:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int ytht_badstr(const char *s) {
  int i;
  const unsigned char *t = (const unsigned char *)s;
  for (i = 0; s[i]; i++)
    if (t[i] != 9 && (t[i] < 32 || t[i] == 255))
      return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ytht_badstr>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L31
  add    $0x1,%rdi
  jmp    L1
  nopl   0x0(%rax)
L37:
  movzbl (%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  je     L31
L1:
  lea    -0x20(%rax),%edx
  cmp    $0xde,%dl
  jbe    L37
  cmp    $0x9,%al
  je     L37
  mov    $0x1,%eax
  ret
  nopl   (%rax)
L31:
  xor    %eax,%eax
  ret
```

assistant: ```c
int ytht_badstr(const char *s) {
  int i;
  const unsigned char *t = (const unsigned char *)s;
  for (i = 0; s[i]; i++)
    if (t[i] != 9 && (t[i] < 32 || t[i] == 255))
      return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ytht_badstr>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L12
  add    $0x1,%rdi
  jmp    L40
  nopl   0x0(%rax)
L16:
  movzbl (%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  je     L12
L40:
  lea    -0x20(%rax),%edx
  cmp    $0xde,%dl
  jbe    L16
  cmp    $0x9,%al
  je     L16
  mov    $0x1,%eax
  ret
  nopl   (%rax)
L12:
  xor    %eax,%eax
  ret
```

assistant: ```c
int ytht_badstr(const char *s) {
  int i;
  const unsigned char *t = (const unsigned char *)s;
  for (i = 0; s[i]; i++)
    if (t[i] != 9 && (t[i] < 32 || t[i] == 255))
      return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<maxnum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L15
  mov    -0x4(%rbp),%eax
  jmp    L4
L15:
  mov    -0x8(%rbp),%eax
L4:
  pop    %rbp
  ret
```

assistant: ```c
int maxnum(int a, int b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<maxnum>:
  endbr64
  cmp    %esi,%edi
  mov    %esi,%eax
  cmovge %edi,%eax
  ret
```

assistant: ```c
int maxnum(int a, int b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<maxnum>:
  endbr64
  cmp    %esi,%edi
  mov    %esi,%eax
  cmovge %edi,%eax
  ret
```

assistant: ```c
int maxnum(int a, int b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<maxnum>:
  endbr64
  cmp    %esi,%edi
  mov    %esi,%eax
  cmovge %edi,%eax
  ret
```

assistant: ```c
int maxnum(int a, int b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<startswith>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <assert@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <assert@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L7
  mov    $0x1,%eax
  jmp    L23
L7:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jae    L39
  mov    $0x0,%eax
  jmp    L23
L39:
  mov    -0x8(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
L23:
  leave
  ret
```

assistant: ```c
bool startswith(const char *s, const char *prefix) {
  size_t sl, pl;
  assert(s);
  assert(prefix);
  sl = strlen(s);
  pl = strlen(prefix);
  if (pl == 0)
    return true;
  if (sl < pl)
    return false;
  return memcmp(s, prefix, pl) == 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<startswith>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %rax,%r12
  mov    $0x1,%eax
  test   %r12,%r12
  je     L6
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  cmp    %r12,%rdx
  jae    L19
L6:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  mov    %r12,%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L6
```

assistant: ```c
bool startswith(const char *s, const char *prefix) {
  size_t sl, pl;
  assert(s);
  assert(prefix);
  sl = strlen(s);
  pl = strlen(prefix);
  if (pl == 0)
    return true;
  if (sl < pl)
    return false;
  return memcmp(s, prefix, pl) == 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<startswith>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,%r13
  mov    $0x1,%eax
  test   %r13,%r13
  je     L5
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  cmp    %r13,%r8
  jae    L8
L5:
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L8:
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <memcmp@plt>
  pop    %rbp
  pop    %r12
  test   %rax,%rax
  pop    %r13
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
bool startswith(const char *s, const char *prefix) {
  size_t sl, pl;
  assert(s);
  assert(prefix);
  sl = strlen(s);
  pl = strlen(prefix);
  if (pl == 0)
    return true;
  if (sl < pl)
    return false;
  return memcmp(s, prefix, pl) == 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<startswith>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,%r13
  mov    $0x1,%eax
  test   %r13,%r13
  je     L11
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  cmp    %r13,%r8
  jae    L0
L11:
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L0:
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <memcmp@plt>
  pop    %rbp
  pop    %r12
  test   %rax,%rax
  pop    %r13
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
bool startswith(const char *s, const char *prefix) {
  size_t sl, pl;
  assert(s);
  assert(prefix);
  sl = strlen(s);
  pl = strlen(prefix);
  if (pl == 0)
    return true;
  if (sl < pl)
    return false;
  return memcmp(s, prefix, pl) == 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<init_palavra>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *init_palavra(char *texto_palavra) {
  char *palavra = (char *)malloc(sizeof(char) * (strlen(texto_palavra) + 1));
  strcpy(palavra, texto_palavra);
  return palavra;
}
```
user: Translate the below assembly code into C programming language:

```asm
<init_palavra>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *init_palavra(char *texto_palavra) {
  char *palavra = (char *)malloc(sizeof(char) * (strlen(texto_palavra) + 1));
  strcpy(palavra, texto_palavra);
  return palavra;
}
```