ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<find_str>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rdx,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  mov    %eax,%r14d
  mov    %r12d,%eax
  sub    %edx,%eax
  test   %eax,%eax
  jle    L34
  sub    $0x1,%eax
  lea    0x1(%rbx,%rax,1),%r13
  jmp    L4
  cs nopw 0x0(%rax,%rax,1)
L38:
  add    $0x1,%rbx
  cmp    %r13,%rbx
  je     L34
L4:
  movzbl 0x0(%rbp),%eax
  mov    %rbx,%r12
  cmp    %al,(%rbx)
  jne    L38
  mov    %r14d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L38
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L34:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *find_str(const char *haystack, int hsize, const char *needle) {
  int i = 0;
  int nsize = strlen(needle);
  hsize -= nsize - 1;
  for (i = 0; i < hsize; i++) {
    if (haystack[i] == needle[0] && strncmp(haystack + i, needle, nsize) == 0) {
      return (char *)(haystack + i);
    }
  }
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L29
L8:
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L29:
  cmpl   $0x4,-0x4(%rbp)
  jle    L8
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L34:
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L19:
  cmpl   $0x4,-0x4(%rbp)
  jle    L34
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
datatype: string

tool: D6: in for, i=%d


assistant: ```c
int main() {
  int i;
  for (i = 0; i < 5; i++)
    printf("in for, i=%d\n", i);
  for (i = 0; i < 5; ++i)
    printf("in for, i=%d\n", i);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%ebx
  lea    D17(%rip),%rbp
L7:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x5,%ebx
  jne    L7
  mov    $0x0,%ebx
  lea    D17(%rip),%rbp
L8:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x5,%ebx
  jne    L8
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D17
type_guessed: string

tool: D17: in for, i=%d


assistant: ```c
int main() {
  int i;
  for (i = 0; i < 5; i++)
    printf("in for, i=%d\n", i);
  for (i = 0; i < 5; ++i)
    printf("in for, i=%d\n", i);
  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   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
L2:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x5,%ebx
  jne    L2
  xor    %ebx,%ebx
L5:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x5,%ebx
  jne    L5
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D5
type: string

tool: D5: in for, i=%d


assistant: ```c
int main() {
  int i;
  for (i = 0; i < 5; i++)
    printf("in for, i=%d\n", i);
  for (i = 0; i < 5; ++i)
    printf("in for, i=%d\n", i);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
L18:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x5,%ebx
  jne    L18
  xor    %ebx,%ebx
L10:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x5,%ebx
  jne    L10
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
type_infered: string

tool: D2: in for, i=%d


assistant: ```c
int main() {
  int i;
  for (i = 0; i < 5; i++)
    printf("in for, i=%d\n", i);
  for (i = 0; i < 5; ++i)
    printf("in for, i=%d\n", i);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x4,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  neg    %eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0xfffffffc,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int x = 4;
  int y = -x;
  assert(y == -4);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x = 4;
  int y = -x;
  assert(y == -4);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x = 4;
  int y = -x;
  assert(y == -4);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x = 4;
  int y = -x;
  assert(y == -4);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <cpu_sig@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  cpu_sig();
  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
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <cpu_sig@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  cpu_sig();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <cpu_sig@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  cpu_sig();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <cpu_sig@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  cpu_sig();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sshbuf_putfv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x34(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <va_copy@plt>
  mov    -0x18(%rbp),%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,%ecx
  mov    %rax,%rdx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <vsnprintf@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L77
  movl   $0xfffffff6,-0x1c(%rbp)
  jmp    L50
L77:
  cmpl   $0x0,-0x14(%rbp)
  jne    L7
  movl   $0x0,-0x1c(%rbp)
  jmp    L50
L7:
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0x34(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <va_copy@plt>
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x1(%rax),%rcx
  lea    -0x10(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sshbuf_reserve@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  js     L93
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%esi
  mov    -0x10(%rbp),%rax
  mov    -0x18(%rbp),%ecx
  mov    -0x30(%rbp),%rdx
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  je     L10
  movl   $0xffffffff,-0x1c(%rbp)
  jmp    L50
L10:
  mov    -0x28(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <sshbuf_consume_end@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L55
  movl   $0x0,-0x1c(%rbp)
  jmp    L50
L93:
  nop
  jmp    L50
L55:
  nop
L50:
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0x1c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

assistant: ```c
int sshbuf_putfv(struct sshbuf *buf, const char *fmt, va_list ap) {
  va_list ap2;
  int r, len;
  u_char *p;
  va_copy(ap2, ap);
  if ((len = vsnprintf(NULL, 0, fmt, ap2)) < 0) {
    r = -10;
    goto out;
  }
  if (len == 0) {
    r = 0;
    goto out;
  }
  va_end(ap2);
  va_copy(ap2, ap);
  if ((r = sshbuf_reserve(buf, (size_t)len + 1, &p)) < 0)
    goto out;
  if ((r = vsnprintf((char *)p, len + 1, fmt, ap2)) != len) {
    r = -1;
    goto out;
  }
  if ((r = sshbuf_consume_end(buf, 1)) != 0)
    goto out;
  r = 0;
out:
  va_end(ap2);
  return r;
}
```
user: Translate the below assembly code into C programming language:

```asm
<sshbuf_putfv>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%r14
  mov    %rsi,%r13
  mov    %edx,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %edx,%esi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_copy@plt>
  mov    %ebp,%ecx
  mov    %r13,%rdx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <vsnprintf@plt>
  test   %eax,%eax
  js     L9
  mov    %eax,%ebx
  jne    L14
L37:
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L88
  mov    %ebx,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L14:
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <va_copy@plt>
  mov    %rsp,%rdx
  movslq %ebx,%rsi
  add    $0x1,%rsi
  mov    %r14,%rdi
  call   <sshbuf_reserve@plt>
  test   %eax,%eax
  js     L34
  lea    0x1(%rbx),%esi
  mov    %ebp,%ecx
  mov    %r13,%rdx
  mov    (%rsp),%rdi
  call   <vsnprintf@plt>
  cmp    %eax,%ebx
  jne    L64
  mov    $0x1,%esi
  mov    %r14,%rdi
  call   <sshbuf_consume_end@plt>
  mov    %eax,%ebx
  jmp    L37
L9:
  mov    $0xfffffff6,%ebx
  jmp    L37
L34:
  mov    %eax,%ebx
  jmp    L37
L64:
  mov    $0xffffffff,%ebx
  jmp    L37
L88:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int sshbuf_putfv(struct sshbuf *buf, const char *fmt, va_list ap) {
  va_list ap2;
  int r, len;
  u_char *p;
  va_copy(ap2, ap);
  if ((len = vsnprintf(NULL, 0, fmt, ap2)) < 0) {
    r = -10;
    goto out;
  }
  if (len == 0) {
    r = 0;
    goto out;
  }
  va_end(ap2);
  va_copy(ap2, ap);
  if ((r = sshbuf_reserve(buf, (size_t)len + 1, &p)) < 0)
    goto out;
  if ((r = vsnprintf((char *)p, len + 1, fmt, ap2)) != len) {
    r = -1;
    goto out;
  }
  if ((r = sshbuf_consume_end(buf, 1)) != 0)
    goto out;
  r = 0;
out:
  va_end(ap2);
  return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sshbuf_putfv>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  mov    %rsi,%r14
  mov    %edx,%esi
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <va_copy@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  mov    %ebp,%ecx
  mov    %r14,%rdx
  call   <vsnprintf@plt>
  test   %eax,%eax
  js     L28
  mov    %eax,%r12d
  jne    L30
L47:
  xor    %edi,%edi
  call   <va_end@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L57
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L30:
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    %r13d,%esi
  mov    %ebp,%edi
  call   <va_copy@plt>
  movslq %r12d,%rsi
  mov    %rsp,%rdx
  mov    %r15,%rdi
  add    $0x1,%rsi
  call   <sshbuf_reserve@plt>
  test   %eax,%eax
  js     L41
  mov    (%rsp),%rdi
  lea    0x1(%r12),%esi
  mov    %ebp,%ecx
  mov    %r14,%rdx
  call   <vsnprintf@plt>
  cmp    %eax,%r12d
  jne    L33
  mov    $0x1,%esi
  mov    %r15,%rdi
  call   <sshbuf_consume_end@plt>
  mov    %eax,%r12d
  jmp    L47
  nopl   0x0(%rax)
L41:
  mov    %eax,%r12d
  jmp    L47
L33:
  mov    $0xffffffff,%r12d
  jmp    L47
L28:
  mov    $0xfffffff6,%r12d
  jmp    L47
L57:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int sshbuf_putfv(struct sshbuf *buf, const char *fmt, va_list ap) {
  va_list ap2;
  int r, len;
  u_char *p;
  va_copy(ap2, ap);
  if ((len = vsnprintf(NULL, 0, fmt, ap2)) < 0) {
    r = -10;
    goto out;
  }
  if (len == 0) {
    r = 0;
    goto out;
  }
  va_end(ap2);
  va_copy(ap2, ap);
  if ((r = sshbuf_reserve(buf, (size_t)len + 1, &p)) < 0)
    goto out;
  if ((r = vsnprintf((char *)p, len + 1, fmt, ap2)) != len) {
    r = -1;
    goto out;
  }
  if ((r = sshbuf_consume_end(buf, 1)) != 0)
    goto out;
  r = 0;
out:
  va_end(ap2);
  return r;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sshbuf_putfv>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  mov    %rsi,%r14
  mov    %edx,%esi
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <va_copy@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  mov    %ebp,%ecx
  mov    %r14,%rdx
  call   <vsnprintf@plt>
  test   %eax,%eax
  js     L32
  mov    %eax,%r12d
  jne    L51
L70:
  xor    %edi,%edi
  call   <va_end@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L80
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L51:
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    %r13d,%esi
  mov    %ebp,%edi
  call   <va_copy@plt>
  movslq %r12d,%rsi
  mov    %rsp,%rdx
  mov    %r15,%rdi
  add    $0x1,%rsi
  call   <sshbuf_reserve@plt>
  test   %eax,%eax
  js     L77
  mov    (%rsp),%rdi
  lea    0x1(%r12),%esi
  mov    %ebp,%ecx
  mov    %r14,%rdx
  call   <vsnprintf@plt>
  cmp    %eax,%r12d
  jne    L2
  mov    $0x1,%esi
  mov    %r15,%rdi
  call   <sshbuf_consume_end@plt>
  mov    %eax,%r12d
  jmp    L70
  nopl   0x0(%rax)
L77:
  mov    %eax,%r12d
  jmp    L70
L2:
  mov    $0xffffffff,%r12d
  jmp    L70
L32:
  mov    $0xfffffff6,%r12d
  jmp    L70
L80:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int sshbuf_putfv(struct sshbuf *buf, const char *fmt, va_list ap) {
  va_list ap2;
  int r, len;
  u_char *p;
  va_copy(ap2, ap);
  if ((len = vsnprintf(NULL, 0, fmt, ap2)) < 0) {
    r = -10;
    goto out;
  }
  if (len == 0) {
    r = 0;
    goto out;
  }
  va_end(ap2);
  va_copy(ap2, ap);
  if ((r = sshbuf_reserve(buf, (size_t)len + 1, &p)) < 0)
    goto out;
  if ((r = vsnprintf((char *)p, len + 1, fmt, ap2)) != len) {
    r = -1;
    goto out;
  }
  if ((r = sshbuf_consume_end(buf, 1)) != 0)
    goto out;
  r = 0;
out:
  va_end(ap2);
  return r;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L4:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L4
```

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

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
L4:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L4
```

assistant: ```c
int main() {
  while (1)
    sleep(1);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L1:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L1
```

assistant: ```c
int main() {
  while (1)
    sleep(1);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L8:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L8
```

assistant: ```c
int main() {
  while (1)
    sleep(1);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xa,-0x64(%rbp)
  movl   $0x0,-0x68(%rbp)
  jmp    L7
L29:
  lea    -0x60(%rbp),%rax
  mov    $0x14,%esi
  mov    %rax,%rdi
  call   <test@plt>
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x68(%rbp)
L7:
  mov    -0x68(%rbp),%eax
  cmp    -0x64(%rbp),%eax
  jl     L29
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D0
type_guessed: string

tool: D0: 


assistant: ```c
int main() {
  int n = 10;
  for (int i = 0; i < n; ++i) {
    int data[20];
    test(data, 20);
    printf("\n");
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    $0xa,%ebx
  lea    D1(%rip),%rbp
L6:
  mov    %rsp,%rdi
  mov    $0x14,%esi
  call   <test@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L6
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    $0x0,%eax
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D1
data_type: string

tool: D1: 


assistant: ```c
int main() {
  int n = 10;
  for (int i = 0; i < n; ++i) {
    int data[20];
    test(data, 20);
    printf("\n");
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %r12,%rdi
  mov    $0x14,%esi
  call   <test@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L8
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x60,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D6
type_infered: string

tool: D6: 


assistant: ```c
int main() {
  int n = 10;
  for (int i = 0; i < n; ++i) {
    int data[20];
    test(data, 20);
    printf("\n");
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  nopl   0x0(%rax,%rax,1)
L16:
  mov    %r12,%rdi
  mov    $0x14,%esi
  call   <test@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L16
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x60,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D2
datatype: string

tool: D2: 


assistant: ```c
int main() {
  int n = 10;
  for (int i = 0; i < n; ++i) {
    int data[20];
    test(data, 20);
    printf("\n");
  }
  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
<incrementXbyY>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    $0x1,%edi
  call   <pushCells@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <codeGoTo@plt>
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <codeStr@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <codeGoTo@plt>
  lea    D74(%rip),%rax
  mov    %rax,%rdi
  call   <codeStr@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <codeGoTo@plt>
  lea    D77(%rip),%rax
  mov    %rax,%rdi
  call   <codeStr@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <codeGoTo@plt>
  lea    D77(%rip),%rax
  mov    %rax,%rdi
  call   <codeStr@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <codeGoTo@plt>
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  call   <codeStr@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <codeGoTo@plt>
  lea    D74(%rip),%rax
  mov    %rax,%rdi
  call   <codeStr@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <codeGoTo@plt>
  lea    D77(%rip),%rax
  mov    %rax,%rdi
  call   <codeStr@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <codeGoTo@plt>
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  call   <codeStr@plt>
  mov    $0x1,%edi
  call   <popCells@plt>
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D45
data_type: string
name: parse_data
arguments: data_tag: D74
data_type: string
name: parse_data
arguments: data_tag: D77
data_type: string
name: parse_data
arguments: data_tag: D53
data_type: string

tool: D45: [-]

tool: D74: [

tool: D77: +

tool: D53: -]

assistant: ```c
void incrementXbyY(int x, int y) {
  int temp0 = pushCells(1);
  codeGoTo(temp0);
  codeStr("[-]");
  codeGoTo(y);
  codeStr("[");
  codeGoTo(x);
  codeStr("+");
  codeGoTo(temp0);
  codeStr("+");
  codeGoTo(y);
  codeStr("-]");
  codeGoTo(temp0);
  codeStr("[");
  codeGoTo(y);
  codeStr("+");
  codeGoTo(temp0);
  codeStr("-]");
  popCells(1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<incrementXbyY>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %esi,%ebp
  mov    $0x1,%edi
  call   <pushCells@plt>
  mov    %eax,%ebx
  mov    %eax,%edi
  call   <codeGoTo@plt>
  lea    D21(%rip),%rdi
  call   <codeStr@plt>
  mov    %ebp,%edi
  call   <codeGoTo@plt>
  lea    D26(%rip),%r14
  mov    %r14,%rdi
  call   <codeStr@plt>
  mov    %r12d,%edi
  call   <codeGoTo@plt>
  lea    D11(%rip),%r12
  mov    %r12,%rdi
  call   <codeStr@plt>
  mov    %ebx,%edi
  call   <codeGoTo@plt>
  mov    %r12,%rdi
  call   <codeStr@plt>
  mov    %ebp,%edi
  call   <codeGoTo@plt>
  lea    D20(%rip),%r13
  mov    %r13,%rdi
  call   <codeStr@plt>
  mov    %ebx,%edi
  call   <codeGoTo@plt>
  mov    %r14,%rdi
  call   <codeStr@plt>
  mov    %ebp,%edi
  call   <codeGoTo@plt>
  mov    %r12,%rdi
  call   <codeStr@plt>
  mov    %ebx,%edi
  call   <codeGoTo@plt>
  mov    %r13,%rdi
  call   <codeStr@plt>
  mov    $0x1,%edi
  call   <popCells@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: label: D21
type_infered: string
name: parse_data
arguments: label: D26
type_infered: string
name: parse_data
arguments: label: D11
type_infered: string
name: parse_data
arguments: label: D20
type_infered: string

tool: D21: [-]

tool: D26: [

tool: D11: +

tool: D20: -]

assistant: ```c
void incrementXbyY(int x, int y) {
  int temp0 = pushCells(1);
  codeGoTo(temp0);
  codeStr("[-]");
  codeGoTo(y);
  codeStr("[");
  codeGoTo(x);
  codeStr("+");
  codeGoTo(temp0);
  codeStr("+");
  codeGoTo(y);
  codeStr("-]");
  codeGoTo(temp0);
  codeStr("[");
  codeGoTo(y);
  codeStr("+");
  codeGoTo(temp0);
  codeStr("-]");
  popCells(1);
}
```