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

```asm
<str_to_str>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <strlen@plt>
  mov    %rax,%rbx
  cmp    %r13,%rax
  jbe    L2
L8:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L2:
  mov    %r13,%r14
  mov    %r12,%rdi
  mov    $0x20,%esi
  mov    %r13,%rbx
  sub    %rax,%r14
  mov    %r14,%rdx
  add    %r14,%r12
  call   <memset@plt>
  jmp    L8
```

assistant: ```c
int str_to_str(char *buf, const char *str, unsigned long count) {
  unsigned long size;
  size = strlen(str);
  if (count < size)
    count = size;
  else
    memset(buf, ' ', count - size);
  strcpy(buf + (count - size), str);
  return count;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<str_to_str>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <strlen@plt>
  mov    %rax,%rbx
  cmp    %r13,%rax
  jbe    L6
L17:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %r13,%r14
  mov    %r12,%rdi
  mov    $0x20,%esi
  mov    %r13,%rbx
  sub    %rax,%r14
  mov    %r14,%rdx
  add    %r14,%r12
  call   <memset@plt>
  jmp    L17
```

assistant: ```c
int str_to_str(char *buf, const char *str, unsigned long count) {
  unsigned long size;
  size = strlen(str);
  if (count < size)
    count = size;
  else
    memset(buf, ' ', count - size);
  strcpy(buf + (count - size), str);
  return count;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<get_command_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L2
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  jmp    L14
L2:
  mov    -0x18(%rbp),%rax
L14:
  leave
  ret
```

assistant: ```c
const char *get_command_name(const char *vektor) {
  char *last_slash;
  last_slash = strrchr(vektor, '/');
  if (last_slash)
    return last_slash + 1;
  else
    return vektor;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_command_name>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x2f,%esi
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  cmovne %rdx,%rax
  pop    %rbx
  ret
```

assistant: ```c
const char *get_command_name(const char *vektor) {
  char *last_slash;
  last_slash = strrchr(vektor, '/');
  if (last_slash)
    return last_slash + 1;
  else
    return vektor;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_command_name>:
  endbr64
  push   %rbx
  mov    $0x2f,%esi
  mov    %rdi,%rbx
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  pop    %rbx
  cmovne %rdx,%rax
  ret
```

assistant: ```c
const char *get_command_name(const char *vektor) {
  char *last_slash;
  last_slash = strrchr(vektor, '/');
  if (last_slash)
    return last_slash + 1;
  else
    return vektor;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<get_command_name>:
  endbr64
  push   %rbx
  mov    $0x2f,%esi
  mov    %rdi,%rbx
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  pop    %rbx
  cmovne %rdx,%rax
  ret
```

assistant: ```c
const char *get_command_name(const char *vektor) {
  char *last_slash;
  last_slash = strrchr(vektor, '/');
  if (last_slash)
    return last_slash + 1;
  else
    return vektor;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<handle_carry>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rcx
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0xc(%rbp),%eax
  lea    0x2(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x14(%rbp),%eax
  lea    0x30(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %dl,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  add    $0x1,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x8(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncat@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int handle_carry(int carry, char **out_string_ptr) {
  int out_len = strlen(*out_string_ptr);
  char *tmp = (char *)calloc(out_len + 1, sizeof(char));
  strncpy(tmp, *out_string_ptr, out_len);
  *out_string_ptr = (char *)realloc(*out_string_ptr, out_len + 2);
  *(*out_string_ptr) = carry + '0';
  *(*out_string_ptr + 1) = 0;
  strncat(*out_string_ptr, tmp, strlen(tmp));
  free(tmp);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<handle_carry>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r13d
  mov    %rsi,%rbx
  mov    (%rsi),%rdi
  call   <strlen@plt>
  mov    %eax,%ebp
  lea    0x1(%rax),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,%r12
  mov    %ebp,%edx
  mov    (%rbx),%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  lea    0x2(%rbp),%esi
  mov    (%rbx),%rdi
  call   <realloc@plt>
  mov    %rax,(%rbx)
  add    $0x30,%r13d
  mov    %r13b,(%rax)
  mov    (%rbx),%rax
  movb   $0x0,0x1(%rax)
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %r12,%rsi
  mov    (%rbx),%rdi
  call   <strncat@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int handle_carry(int carry, char **out_string_ptr) {
  int out_len = strlen(*out_string_ptr);
  char *tmp = (char *)calloc(out_len + 1, sizeof(char));
  strncpy(tmp, *out_string_ptr, out_len);
  *out_string_ptr = (char *)realloc(*out_string_ptr, out_len + 2);
  *(*out_string_ptr) = carry + '0';
  *(*out_string_ptr + 1) = 0;
  strncat(*out_string_ptr, tmp, strlen(tmp));
  free(tmp);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<handle_carry>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  push   %r12
  add    $0x30,%r13d
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  mov    (%rsi),%rdi
  call   <strlen@plt>
  mov    $0x1,%esi
  lea    0x1(%rax),%edi
  mov    %eax,%ebp
  call   <calloc@plt>
  mov    (%rbx),%rsi
  mov    %ebp,%edx
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strncpy@plt>
  mov    (%rbx),%rdi
  lea    0x2(%rbp),%esi
  call   <realloc@plt>
  mov    %r12,%rdi
  mov    %rax,(%rbx)
  mov    %r13b,(%rax)
  mov    (%rbx),%rax
  movb   $0x0,0x1(%rax)
  call   <strlen@plt>
  mov    (%rbx),%rdi
  mov    %r12,%rsi
  mov    %eax,%edx
  call   <strncat@plt>
  mov    %r12,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int handle_carry(int carry, char **out_string_ptr) {
  int out_len = strlen(*out_string_ptr);
  char *tmp = (char *)calloc(out_len + 1, sizeof(char));
  strncpy(tmp, *out_string_ptr, out_len);
  *out_string_ptr = (char *)realloc(*out_string_ptr, out_len + 2);
  *(*out_string_ptr) = carry + '0';
  *(*out_string_ptr + 1) = 0;
  strncat(*out_string_ptr, tmp, strlen(tmp));
  free(tmp);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<handle_carry>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  push   %r12
  add    $0x30,%r13d
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  mov    (%rsi),%rdi
  call   <strlen@plt>
  mov    $0x1,%esi
  lea    0x1(%rax),%edi
  mov    %eax,%ebp
  call   <calloc@plt>
  mov    (%rbx),%rsi
  mov    %ebp,%edx
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strncpy@plt>
  mov    (%rbx),%rdi
  lea    0x2(%rbp),%esi
  call   <realloc@plt>
  mov    %r12,%rdi
  mov    %rax,(%rbx)
  mov    %r13b,(%rax)
  mov    (%rbx),%rax
  movb   $0x0,0x1(%rax)
  call   <strlen@plt>
  mov    (%rbx),%rdi
  mov    %r12,%rsi
  mov    %eax,%edx
  call   <strncat@plt>
  mov    %r12,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int handle_carry(int carry, char **out_string_ptr) {
  int out_len = strlen(*out_string_ptr);
  char *tmp = (char *)calloc(out_len + 1, sizeof(char));
  strncpy(tmp, *out_string_ptr, out_len);
  *out_string_ptr = (char *)realloc(*out_string_ptr, out_len + 2);
  *(*out_string_ptr) = carry + '0';
  *(*out_string_ptr + 1) = 0;
  strncat(*out_string_ptr, tmp, strlen(tmp));
  free(tmp);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<clear_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x64,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L6
L10:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0x8(%rbp)
L6:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L10
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void clear_line(char *line) {
  int len = 100;
  int i = 0;
  for (i = 0; i < len; i++) {
    *(line + i) = 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
<clear_line>:
  endbr64
  lea    0x64(%rdi),%rax
L8:
  movb   $0x0,(%rdi)
  add    $0x1,%rdi
  cmp    %rax,%rdi
  jne    L8
  ret
```

assistant: ```c
void clear_line(char *line) {
  int len = 100;
  int i = 0;
  for (i = 0; i < len; i++) {
    *(line + i) = 0;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<clear_line>:
  endbr64
  movq   $0x0,(%rdi)
  mov    %rdi,%rax
  lea    0x8(%rdi),%rdi
  movq   $0x0,0x54(%rdi)
  and    $0xfffffffffffffff8,%rdi
  sub    %rdi,%rax
  lea    0x64(%rax),%ecx
  xor    %eax,%eax
  shr    $0x3,%ecx
  rep stos %rax,%es:(%rdi)
  ret
```

assistant: ```c
void clear_line(char *line) {
  int len = 100;
  int i = 0;
  for (i = 0; i < len; i++) {
    *(line + i) = 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<clear_line>:
  endbr64
  movq   $0x0,(%rdi)
  mov    %rdi,%rax
  lea    0x8(%rdi),%rdi
  movq   $0x0,0x54(%rdi)
  and    $0xfffffffffffffff8,%rdi
  sub    %rdi,%rax
  lea    0x64(%rax),%ecx
  xor    %eax,%eax
  shr    $0x3,%ecx
  rep stos %rax,%es:(%rdi)
  ret
```

assistant: ```c
void clear_line(char *line) {
  int len = 100;
  int i = 0;
  for (i = 0; i < len; i++) {
    *(line + i) = 0;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<fd_nrf8001_decode_key_request_event>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpl   $0x2,-0x1c(%rbp)
  je     L9
  mov    $0x0,%eax
  call   <fd_nrf8001_error@plt>
  jmp    L4
L9:
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <fd_nrf8001_key_request_event@plt>
L4:
  leave
  ret
```

assistant: ```c
void fd_nrf8001_decode_key_request_event(uint8_t *buffer, uint32_t length) {
  if (length != 2) {
    fd_nrf8001_error();
    return;
  }
  uint8_t key_type = buffer[1];
  fd_nrf8001_key_request_event(key_type);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fd_nrf8001_decode_key_request_event>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x2,%esi
  je     L18
  mov    $0x0,%eax
  call   <fd_nrf8001_error@plt>
L15:
  add    $0x8,%rsp
  ret
L18:
  mov    0x4(%rdi),%edi
  call   <fd_nrf8001_key_request_event@plt>
  jmp    L15
```

assistant: ```c
void fd_nrf8001_decode_key_request_event(uint8_t *buffer, uint32_t length) {
  if (length != 2) {
    fd_nrf8001_error();
    return;
  }
  uint8_t key_type = buffer[1];
  fd_nrf8001_key_request_event(key_type);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fd_nrf8001_decode_key_request_event>:
  endbr64
  cmp    $0x2,%esi
  je     L4
  xor    %eax,%eax
  jmp    1060 <fd_nrf8001_error@plt>
L4:
  mov    0x4(%rdi),%edi
  jmp    1070 <fd_nrf8001_key_request_event@plt>
```

assistant: ```c
void fd_nrf8001_decode_key_request_event(uint8_t *buffer, uint32_t length) {
  if (length != 2) {
    fd_nrf8001_error();
    return;
  }
  uint8_t key_type = buffer[1];
  fd_nrf8001_key_request_event(key_type);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fd_nrf8001_decode_key_request_event>:
  endbr64
  cmp    $0x2,%esi
  je     L1
  xor    %eax,%eax
  jmp    1060 <fd_nrf8001_error@plt>
L1:
  mov    0x4(%rdi),%edi
  jmp    1070 <fd_nrf8001_key_request_event@plt>
```

assistant: ```c
void fd_nrf8001_decode_key_request_event(uint8_t *buffer, uint32_t length) {
  if (length != 2) {
    fd_nrf8001_error();
    return;
  }
  uint8_t key_type = buffer[1];
  fd_nrf8001_key_request_event(key_type);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<G729A_Get_Version>:
  endbr64
  lea    D2(%rip),%rax
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D2
type_infered: string

tool: D2: 1.1

assistant: ```c
const char *G729A_Get_Version() {
  static const char *version = "1.1";
  return version;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<G729A_Get_Version>:
  endbr64
  lea    D8(%rip),%rax
  ret
```

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

tool: D8: 1.1

assistant: ```c
const char *G729A_Get_Version() {
  static const char *version = "1.1";
  return version;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<G729A_Get_Version>:
  endbr64
  lea    D8(%rip),%rax
  ret
```

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

tool: D8: 1.1

assistant: ```c
const char *G729A_Get_Version() {
  static const char *version = "1.1";
  return version;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<chk1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <chk2@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x6,%eax
  and    $0x3,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  and    $0x3f,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
chk1(pkt) char *pkt;
{
  unsigned int chk;
  chk = chk2(pkt);
  chk = (((chk & 0300) >> 6) + chk) & 077;
  return (chk);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<chk1>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <chk2@plt>
  mov    %eax,%edx
  shr    $0x6,%eax
  and    $0x3,%eax
  add    %edx,%eax
  and    $0x3f,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
chk1(pkt) char *pkt;
{
  unsigned int chk;
  chk = chk2(pkt);
  chk = (((chk & 0300) >> 6) + chk) & 077;
  return (chk);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<chk1>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <chk2@plt>
  add    $0x8,%rsp
  mov    %eax,%edx
  shr    $0x6,%eax
  and    $0x3,%eax
  add    %edx,%eax
  and    $0x3f,%eax
  ret
```

assistant: ```c
chk1(pkt) char *pkt;
{
  unsigned int chk;
  chk = chk2(pkt);
  chk = (((chk & 0300) >> 6) + chk) & 077;
  return (chk);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<chk1>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <chk2@plt>
  add    $0x8,%rsp
  mov    %eax,%edx
  shr    $0x6,%eax
  and    $0x3,%eax
  add    %edx,%eax
  and    $0x3f,%eax
  ret
```

assistant: ```c
chk1(pkt) char *pkt;
{
  unsigned int chk;
  chk = chk2(pkt);
  chk = (((chk & 0300) >> 6) + chk) & 077;
  return (chk);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<delete_zeros>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L57
L71:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x31,%al
  jne    L44
  cmpl   $0x5,-0xc(%rbp)
  jle    L64
  lea    D9(%rip),%rax
  jmp    L37
L64:
  addl   $0x1,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x31,(%rax)
  addl   $0x1,-0x10(%rbp)
  jmp    L61
L44:
  cmpl   $0x5,-0xc(%rbp)
  je     L40
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x30,(%rax)
  addl   $0x1,-0x10(%rbp)
L40:
  movl   $0x0,-0xc(%rbp)
L61:
  addl   $0x1,-0x14(%rbp)
L57:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x14(%rbp)
  jl     L71
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  jmp    L38
L18:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x30,(%rax)
  addl   $0x1,-0x14(%rbp)
L38:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x14(%rbp)
  jl     L18
  mov    -0x8(%rbp),%rax
L37:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
datatype: string

tool: D9: Error: consecutive ones can't be more than 5.

assistant: ```c
char *delete_zeros(char *input) {
  int i, current, one_count;
  char *output;
  output = (char *)malloc(sizeof(input));
  current = 0;
  one_count = 0;
  for (i = 0; i < strlen(input); i++) {
    if (input[i] == '1') {
      if (one_count > 5) {
        return "Error: consecutive ones can't be more than 5.";
      } else {
        one_count++;
        output[current] = '1';
        current++;
      }
    } else {
      if (one_count != 5) {
        output[current] = '0';
        current++;
      }
      one_count = 0;
    }
  }
  for (i = strlen(output); i < strlen(input); i++) {
    output[i] = '0';
  }
  return output;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<delete_zeros>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    %rax,%r15
  mov    %rax,%r14
  mov    $0x0,%ebx
  mov    $0x0,%r12d
  mov    $0x0,%r13d
  jmp    L76
L51:
  cmp    $0x5,%r12d
  jg     L58
  add    $0x1,%r12d
  movslq %r13d,%rax
  movb   $0x31,(%r14,%rax,1)
  add    $0x1,%r13d
L16:
  add    $0x1,%rbx
L76:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L18
  cmpb   $0x31,0x0(%rbp,%rbx,1)
  je     L51
  cmp    $0x5,%r12d
  je     L59
  movslq %r13d,%rax
  movb   $0x30,(%r14,%rax,1)
  add    $0x1,%r13d
  mov    $0x0,%r12d
  jmp    L16
L59:
  mov    $0x0,%r12d
  jmp    L16
L18:
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %eax,%r12d
  movslq %eax,%rdx
  add    %r15,%rdx
  mov    %rdx,%rbx
  sub    %edx,%r12d
L72:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    (%r12,%rbx,1),%edx
  cmp    %edx,%eax
  jle    L21
  movb   $0x30,(%rbx)
  add    $0x1,%rbx
  jmp    L72
L58:
  lea    D5(%rip),%r14
L21:
  mov    %r14,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D5
datatype: string

tool: D5: Error: consecutive ones can't be more than 5.

assistant: ```c
char *delete_zeros(char *input) {
  int i, current, one_count;
  char *output;
  output = (char *)malloc(sizeof(input));
  current = 0;
  one_count = 0;
  for (i = 0; i < strlen(input); i++) {
    if (input[i] == '1') {
      if (one_count > 5) {
        return "Error: consecutive ones can't be more than 5.";
      } else {
        one_count++;
        output[current] = '1';
        current++;
      }
    } else {
      if (one_count != 5) {
        output[current] = '0';
        current++;
      }
      one_count = 0;
    }
  }
  for (i = strlen(output); i < strlen(input); i++) {
    output[i] = '0';
  }
  return output;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<delete_zeros>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x8,%edi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r15
  mov    %rax,%r14
  jmp    L7
  nop
L2:
  cmp    $0x5,%r12d
  je     L20
  movslq %r13d,%rax
  xor    %r12d,%r12d
  add    $0x1,%r13d
  movb   $0x30,(%r15,%rax,1)
L3:
  add    $0x1,%rbx
L7:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L26
  cmpb   $0x31,0x0(%rbp,%rbx,1)
  jne    L2
  cmp    $0x6,%r12d
  je     L10
  movslq %r13d,%rax
  add    $0x1,%r12d
  add    $0x1,%r13d
  movb   $0x31,(%r15,%rax,1)
  jmp    L3
  nopl   0x0(%rax)
L20:
  xor    %r12d,%r12d
  jmp    L3
  nopl   (%rax)
L26:
  mov    %r15,%rdi
  call   <strlen@plt>
  movslq %eax,%rbx
  mov    %rbx,%r12
  add    %r15,%rbx
  sub    %ebx,%r12d
  jmp    L49
  cs nopw 0x0(%rax,%rax,1)
L72:
  movb   $0x30,(%rbx)
  add    $0x1,%rbx
L49:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    (%r12,%rbx,1),%edx
  cmp    %edx,%eax
  jg     L72
L13:
  add    $0x8,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L10:
  lea    D7(%rip),%r14
  jmp    L13
```

tool_calls: name: read_data
arguments: label: D7
datatype: string

tool: D7: Error: consecutive ones can't be more than 5.

assistant: ```c
char *delete_zeros(char *input) {
  int i, current, one_count;
  char *output;
  output = (char *)malloc(sizeof(input));
  current = 0;
  one_count = 0;
  for (i = 0; i < strlen(input); i++) {
    if (input[i] == '1') {
      if (one_count > 5) {
        return "Error: consecutive ones can't be more than 5.";
      } else {
        one_count++;
        output[current] = '1';
        current++;
      }
    } else {
      if (one_count != 5) {
        output[current] = '0';
        current++;
      }
      one_count = 0;
    }
  }
  for (i = strlen(output); i < strlen(input); i++) {
    output[i] = '0';
  }
  return output;
}
```