ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<removeInvalidParentheses>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    %rax,(%rsp)
  movl   $0x0,(%r12)
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L52
  lea    0x1(%rbp),%rdx
  mov    $0x0,%r13d
  mov    $0x0,%ebx
  jmp    L15
L1:
  add    $0x1,%ebx
L89:
  add    $0x1,%rdx
  movzbl -0x1(%rdx),%eax
  test   %al,%al
  je     L42
L15:
  cmp    $0x28,%al
  je     L1
  cmp    $0x29,%al
  jne    L89
  test   %ebx,%ebx
  je     L8
  sub    $0x1,%ebx
  jmp    L89
L8:
  add    $0x1,%r13d
  jmp    L89
L52:
  mov    $0x0,%r13d
  mov    $0x0,%ebx
L42:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%r14d
  mov    %eax,%edi
  call   <malloc@plt>
  push   %r12
  lea    0x8(%rsp),%rdx
  push   %rdx
  push   $0xffffffffffffffff
  push   %rax
  mov    $0x0,%r9d
  mov    %r13d,%r8d
  mov    %ebx,%ecx
  mov    $0x0,%edx
  mov    %r14d,%esi
  mov    %rbp,%rdi
  call   <traverse@plt>
  mov    0x20(%rsp),%rax
  add    $0x20,%rsp
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char **removeInvalidParentheses(char *s, int *returnSize) {
  char **arr = (char **)malloc(sizeof(char *));
  *returnSize = 0;
  int left = 0, right = 0;
  for (int i = 0; s[i]; i++) {
    if (s[i] == '(')
      left++;
    else if (s[i] == ')') {
      if (left)
        left--;
      else
        right++;
    }
  }
  int len = strlen(s);
  char *stack = (char *)malloc(sizeof(char) * len);
  int top = -1;
  traverse(s, len, 0, left, right, 0, stack, top, &arr, returnSize);
  return arr;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<removeInvalidParentheses>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x8,%edi
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  movl   $0x0,(%rbx)
  mov    %rax,(%rsp)
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L18
  lea    0x1(%rbp),%rdx
  xor    %r14d,%r14d
  xor    %r12d,%r12d
  jmp    L52
  nopw   0x0(%rax,%rax,1)
L100:
  cmp    $0x29,%al
  je     L30
L105:
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L107
L52:
  cmp    $0x28,%al
  jne    L100
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  add    $0x1,%r12d
  test   %al,%al
  jne    L52
L107:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edi
  mov    %eax,%r13d
  call   <malloc@plt>
  push   %rbx
  xor    %r9d,%r9d
  mov    %r14d,%r8d
  mov    %r12d,%ecx
  mov    %r13d,%esi
  mov    %rbp,%rdi
  lea    0x8(%rsp),%rdx
  push   %rdx
  xor    %edx,%edx
  push   $0xffffffffffffffff
  push   %rax
  call   <traverse@plt>
  mov    0x20(%rsp),%rax
  add    $0x20,%rsp
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L41
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L30:
  test   %r12d,%r12d
  je     L28
  sub    $0x1,%r12d
  jmp    L105
  cs nopw 0x0(%rax,%rax,1)
L28:
  add    $0x1,%r14d
  jmp    L105
  nopl   0x0(%rax)
L18:
  xor    %r14d,%r14d
  xor    %r12d,%r12d
  jmp    L107
L41:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char **removeInvalidParentheses(char *s, int *returnSize) {
  char **arr = (char **)malloc(sizeof(char *));
  *returnSize = 0;
  int left = 0, right = 0;
  for (int i = 0; s[i]; i++) {
    if (s[i] == '(')
      left++;
    else if (s[i] == ')') {
      if (left)
        left--;
      else
        right++;
    }
  }
  int len = strlen(s);
  char *stack = (char *)malloc(sizeof(char) * len);
  int top = -1;
  traverse(s, len, 0, left, right, 0, stack, top, &arr, returnSize);
  return arr;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<removeInvalidParentheses>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x8,%edi
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  movl   $0x0,(%rbx)
  mov    %rax,(%rsp)
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L50
  lea    0x1(%rbp),%rdx
  xor    %r14d,%r14d
  xor    %r12d,%r12d
  jmp    L90
  nopw   0x0(%rax,%rax,1)
L105:
  cmp    $0x29,%al
  je     L7
L59:
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L54
L90:
  cmp    $0x28,%al
  jne    L105
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  add    $0x1,%r12d
  test   %al,%al
  jne    L90
L54:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edi
  mov    %eax,%r13d
  call   <malloc@plt>
  push   %rbx
  xor    %r9d,%r9d
  mov    %r14d,%r8d
  mov    %r12d,%ecx
  mov    %r13d,%esi
  mov    %rbp,%rdi
  lea    0x8(%rsp),%rdx
  push   %rdx
  xor    %edx,%edx
  push   $0xffffffffffffffff
  push   %rax
  call   <traverse@plt>
  mov    0x20(%rsp),%rax
  add    $0x20,%rsp
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L109
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  test   %r12d,%r12d
  je     L21
  sub    $0x1,%r12d
  jmp    L59
  cs nopw 0x0(%rax,%rax,1)
L21:
  add    $0x1,%r14d
  jmp    L59
  nopl   0x0(%rax)
L50:
  xor    %r14d,%r14d
  xor    %r12d,%r12d
  jmp    L54
L109:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char **removeInvalidParentheses(char *s, int *returnSize) {
  char **arr = (char **)malloc(sizeof(char *));
  *returnSize = 0;
  int left = 0, right = 0;
  for (int i = 0; s[i]; i++) {
    if (s[i] == '(')
      left++;
    else if (s[i] == ')') {
      if (left)
        left--;
      else
        right++;
    }
  }
  int len = strlen(s);
  char *stack = (char *)malloc(sizeof(char) * len);
  int top = -1;
  traverse(s, len, 0, left, right, 0, stack, top, &arr, returnSize);
  return arr;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ofpbuf_put>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ofpbuf_put_uninit@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *ofpbuf_put(struct ofpbuf *b, const void *p, size_t size) {
  void *dst = ofpbuf_put_uninit(b, size);
  memcpy(dst, p, size);
  return dst;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ofpbuf_put>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    %rdx,%rsi
  call   <ofpbuf_put_uninit@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void *ofpbuf_put(struct ofpbuf *b, const void *p, size_t size) {
  void *dst = ofpbuf_put_uninit(b, size);
  memcpy(dst, p, size);
  return dst;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ofpbuf_put>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  mov    %rdx,%rsi
  call   <ofpbuf_put_uninit@plt>
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void *ofpbuf_put(struct ofpbuf *b, const void *p, size_t size) {
  void *dst = ofpbuf_put_uninit(b, size);
  memcpy(dst, p, size);
  return dst;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ofpbuf_put>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  mov    %rdx,%rsi
  call   <ofpbuf_put_uninit@plt>
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void *ofpbuf_put(struct ofpbuf *b, const void *p, size_t size) {
  void *dst = ofpbuf_put_uninit(b, size);
  memcpy(dst, p, size);
  return dst;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<vnum_result3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    0x8(%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x2,(%rax)
  mov    -0x18(%rbp),%rax
  movl   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void vnum_result3(int *data) {
  int k = data[2];
  data[k] = 2;
  data[0] = 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<vnum_result3>:
  endbr64
  movslq 0x8(%rdi),%rax
  movl   $0x2,(%rdi,%rax,4)
  movl   $0x0,(%rdi)
  ret
```

assistant: ```c
void vnum_result3(int *data) {
  int k = data[2];
  data[k] = 2;
  data[0] = 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
<vnum_result3>:
  endbr64
  movslq 0x8(%rdi),%rax
  movl   $0x2,(%rdi,%rax,4)
  movl   $0x0,(%rdi)
  ret
```

assistant: ```c
void vnum_result3(int *data) {
  int k = data[2];
  data[k] = 2;
  data[0] = 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<vnum_result3>:
  endbr64
  movslq 0x8(%rdi),%rax
  movl   $0x2,(%rdi,%rax,4)
  movl   $0x0,(%rdi)
  ret
```

assistant: ```c
void vnum_result3(int *data) {
  int k = data[2];
  data[k] = 2;
  data[0] = 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<xstrcat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <_makespace@plt>
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  nop
  leave
  ret
```

assistant: ```c
void xstrcat(char **str1, char *str2) {
  _makespace(str1, strlen(str2));
  strcat(*str1, str2);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<xstrcat>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <_makespace@plt>
  mov    %rbp,%rsi
  mov    (%rbx),%rdi
  call   <strcat@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void xstrcat(char **str1, char *str2) {
  _makespace(str1, strlen(str2));
  strcat(*str1, str2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xstrcat>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <_makespace@plt>
  mov    (%rbx),%rdi
  add    $0x8,%rsp
  mov    %rbp,%rsi
  pop    %rbx
  pop    %rbp
  jmp    1090 <strcat@plt>
```

assistant: ```c
void xstrcat(char **str1, char *str2) {
  _makespace(str1, strlen(str2));
  strcat(*str1, str2);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<xstrcat>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <_makespace@plt>
  mov    (%rbx),%rdi
  add    $0x8,%rsp
  mov    %rbp,%rsi
  pop    %rbx
  pop    %rbp
  jmp    1090 <strcat@plt>
```

assistant: ```c
void xstrcat(char **str1, char *str2) {
  _makespace(str1, strlen(str2));
  strcat(*str1, str2);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ltoct>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0x10(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L8
L4:
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  mov    %eax,%esi
  mov    $0xa,%edi
  mov    $0x0,%eax
  call   <libc_pow_int@plt>
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rcx
  sar    $0x3f,%rcx
  shr    $0x3d,%rcx
  add    %rcx,%rax
  and    $0x7,%eax
  sub    %rcx,%rax
  imul   %rdx,%rax
  add    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  lea    0x7(%rax),%rdx
  test   %rax,%rax
  cmovs  %rdx,%rax
  sar    $0x3,%rax
  mov    %rax,-0x28(%rbp)
L8:
  cmpq   $0x0,-0x28(%rbp)
  jg     L4
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

assistant: ```c
long ltoct(long dec) {
  int pow_int(int base, int exp) {
    int result = 1;
    while (exp) {
      result *= base;
      exp--;
    }
    return result;
  }
  long oct = 0;
  int n = 0;
  while (dec > 0) {
    oct += (libc_pow_int(10, n++)) * (dec % 8);
    dec /= 8;
  }
  return oct;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ltoct>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  jle    L18
  mov    %rdi,%rbx
  mov    $0x0,%ebp
  mov    $0x0,%r12d
L1:
  mov    %ebp,%esi
  add    $0x1,%ebp
  mov    $0xa,%edi
  mov    $0x0,%eax
  call   <libc_pow_int@plt>
  mov    %eax,%ecx
  mov    %rbx,%rax
  sar    $0x3f,%rax
  shr    $0x3d,%rax
  lea    (%rbx,%rax,1),%rdx
  and    $0x7,%edx
  sub    %rax,%rdx
  movslq %ecx,%rax
  imul   %rax,%rdx
  add    %rdx,%r12
  mov    %rbx,%rdx
  lea    0x7(%rbx),%rax
  test   %rbx,%rbx
  cmovns %rbx,%rax
  sar    $0x3,%rax
  mov    %rax,%rbx
  cmp    $0x7,%rdx
  jg     L1
L2:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  mov    $0x0,%r12d
  jmp    L2
```

assistant: ```c
long ltoct(long dec) {
  int pow_int(int base, int exp) {
    int result = 1;
    while (exp) {
      result *= base;
      exp--;
    }
    return result;
  }
  long oct = 0;
  int n = 0;
  while (dec > 0) {
    oct += (libc_pow_int(10, n++)) * (dec % 8);
    dec /= 8;
  }
  return oct;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ltoct>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  jle    L6
  mov    %rdi,%rbx
  xor    %ebp,%ebp
  xor    %r12d,%r12d
  nopl   (%rax)
L12:
  mov    %ebp,%esi
  xor    %eax,%eax
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <libc_pow_int@plt>
  mov    %rbx,%rdx
  and    $0x7,%edx
  cltq
  imul   %rax,%rdx
  add    %rdx,%r12
  sar    $0x3,%rbx
  jne    L12
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L6:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
long ltoct(long dec) {
  int pow_int(int base, int exp) {
    int result = 1;
    while (exp) {
      result *= base;
      exp--;
    }
    return result;
  }
  long oct = 0;
  int n = 0;
  while (dec > 0) {
    oct += (libc_pow_int(10, n++)) * (dec % 8);
    dec /= 8;
  }
  return oct;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ltoct>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  jle    L14
  mov    %rdi,%rbx
  xor    %ebp,%ebp
  xor    %r12d,%r12d
  nopl   (%rax)
L11:
  mov    %ebp,%esi
  xor    %eax,%eax
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <libc_pow_int@plt>
  mov    %rbx,%rdx
  and    $0x7,%edx
  cltq
  imul   %rax,%rdx
  add    %rdx,%r12
  sar    $0x3,%rbx
  jne    L11
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L14:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
long ltoct(long dec) {
  int pow_int(int base, int exp) {
    int result = 1;
    while (exp) {
      result *= base;
      exp--;
    }
    return result;
  }
  long oct = 0;
  int n = 0;
  while (dec > 0) {
    oct += (libc_pow_int(10, n++)) * (dec % 8);
    dec /= 8;
  }
  return oct;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_str_malloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <_str_malloc_mem@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *_str_malloc(int size) {
  char *ptr = (char *)_str_malloc_mem(size);
  return ptr;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<_str_malloc>:
  endbr64
  sub    $0x8,%rsp
  call   <_str_malloc_mem@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
char *_str_malloc(int size) {
  char *ptr = (char *)_str_malloc_mem(size);
  return ptr;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<_str_malloc>:
  endbr64
  jmp    1050 <_str_malloc_mem@plt>
```

assistant: ```c
char *_str_malloc(int size) {
  char *ptr = (char *)_str_malloc_mem(size);
  return ptr;
}
```
user: What is the c source code of the assembly code below?

```asm
<_str_malloc>:
  endbr64
  jmp    1050 <_str_malloc_mem@plt>
```

assistant: ```c
char *_str_malloc(int size) {
  char *ptr = (char *)_str_malloc_mem(size);
  return ptr;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ouchbehavior>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D19
type_infered: string
name: unpack_data
arguments: label: D13
type_infered: string

tool: D19:    OUCH! - I got signal %d , and this was my reaction.


tool: D13:    Give me a Ctrl+ \ to really stop me! 

assistant: ```c
void ouchbehavior(int sig) {
  printf("   OUCH! - I got signal %d , and this was my reaction.\n", sig);
  printf("   Give me a Ctrl+ \\ to really stop me! \n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ouchbehavior>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D9(%rip),%rdi
  call   <puts@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D14
type: string
name: retrieve_data
arguments: tag: D9
type: string

tool: D14:    OUCH! - I got signal %d , and this was my reaction.


tool: D9:    Give me a Ctrl+ \ to really stop me! 

assistant: ```c
void ouchbehavior(int sig) {
  printf("   OUCH! - I got signal %d , and this was my reaction.\n", sig);
  printf("   Give me a Ctrl+ \\ to really stop me! \n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<ouchbehavior>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D1(%rip),%rdi
  add    $0x8,%rsp
  jmp    1060 <puts@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D10
type_infered: string
name: disassemble_data
arguments: data_tag: D1
type_infered: string

tool: D10:    OUCH! - I got signal %d , and this was my reaction.


tool: D1:    Give me a Ctrl+ \ to really stop me! 

assistant: ```c
void ouchbehavior(int sig) {
  printf("   OUCH! - I got signal %d , and this was my reaction.\n", sig);
  printf("   Give me a Ctrl+ \\ to really stop me! \n");
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ouchbehavior>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D2(%rip),%rdi
  add    $0x8,%rsp
  jmp    1060 <puts@plt>
```

tool_calls: name: read_data
arguments: label: D9
type_infered: string
name: read_data
arguments: label: D2
type_infered: string

tool: D9:    OUCH! - I got signal %d , and this was my reaction.


tool: D2:    Give me a Ctrl+ \ to really stop me! 

assistant: ```c
void ouchbehavior(int sig) {
  printf("   OUCH! - I got signal %d , and this was my reaction.\n", sig);
  printf("   Give me a Ctrl+ \\ to really stop me! \n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  add    $0x3686696a,%rax
  xor    $0xffffffffce5d000b,%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L39
L9:
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp,%rax,8),%rax
  shl    $0x4,%rax
  and    $0x1f0,%eax
  or     %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,-0x10(%rbp,%rax,8)
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  shl    $0x2,%rax
  and    $0x3c,%eax
  or     %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,-0x10(%rbp,%rax,8)
  addq   $0x1,-0x18(%rbp)
L39:
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  cmp    %rax,-0x18(%rbp)
  jb     L9
  movq   $0x0,-0x18(%rbp)
  jmp    L31
L18:
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  shl    $0x2,%rax
  and    $0x1c,%eax
  or     %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,-0x10(%rbp,%rax,8)
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp,%rax,8),%rax
  shl    $0x2,%rax
  and    $0x1c,%eax
  or     %rdx,%rax
  mov    %rax,-0x10(%rbp)
  addq   $0x1,-0x18(%rbp)
L31:
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  cmp    %rax,-0x18(%rbp)
  jb     L18
  mov    -0x10(%rbp),%rax
  imul   $0x20bbe199,%rax,%rax
  lea    0x1fd7f33e(%rax),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ 0xffffffffce5d000bUL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[local1] |= (state[local1] & 31UL) << 4UL;
      state[local1] |= (state[0UL] & 15UL) << 2UL;
      local1++;
    }
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[local1] |= (state[0UL] & 7UL) << 2UL;
      state[0UL] |= (state[local1] & 7UL) << 2UL;
      local1++;
    }
    output[0UL] = state[0UL] * 549183897UL + 534246206UL;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<RandomFunc>:
  endbr64
  mov    %rsi,%rcx
  mov    (%rdi),%rax
  add    $0x3686696a,%rax
  xor    $0xffffffffce5d000b,%rax
  mov    0x8(%rdi),%rdi
  test   %rdi,%rdi
  je     L4
  mov    $0x0,%esi
L2:
  mov    %rax,%rdx
  shl    $0x4,%rdx
  and    $0x1f0,%edx
  or     %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  and    $0x3c,%edx
  or     %rdx,%rax
  add    $0x1,%rsi
  cmp    %rdi,%rsi
  jne    L2
  mov    $0x0,%esi
L27:
  lea    0x0(,%rax,4),%rdx
  and    $0x1c,%edx
  or     %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  and    $0x1c,%edx
  or     %rdx,%rax
  add    $0x1,%rsi
  cmp    %rdi,%rsi
  jne    L27
L4:
  imul   $0x20bbe199,%rax,%rax
  add    $0x1fd7f33e,%rax
  mov    %rax,(%rcx)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  {
    state[0UL] = (input[0UL] + 914778474UL) ^ 0xffffffffce5d000bUL;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[local1] |= (state[local1] & 31UL) << 4UL;
      state[local1] |= (state[0UL] & 15UL) << 2UL;
      local1++;
    }
    local1 = 0UL;
    while (local1 < input[1UL]) {
      state[local1] |= (state[0UL] & 7UL) << 2UL;
      state[0UL] |= (state[local1] & 7UL) << 2UL;
      local1++;
    }
    output[0UL] = state[0UL] * 549183897UL + 534246206UL;
  }
}
```