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

```asm
<is_new_version>:
  endbr64
  push   %r12
  push   %rbp
  lea    -0x4000(%rsp),%r11
L4:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L4
  sub    $0xe38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x4e28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x2710(%rsp),%rbp
  mov    %r12,%rdi
  call   <get_cache_version@plt>
  mov    %r12,%rsi
  lea    D18(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <get_binary_version@plt>
  mov    %rbp,%rsi
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x2710,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L26
L8:
  mov    0x4e28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L21
  add    $0x4e38,%rsp
  pop    %rbp
  pop    %r12
  ret
  nop
L26:
  lea    D24(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
  jmp    L8
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D18
datatype: string
name: parse_data
arguments: tag: D14
datatype: string
name: parse_data
arguments: tag: D24
datatype: string

tool: D18: Previous MAME version is: %s


tool: D14: Current MAME version is:  %s


tool: D24: New version available

assistant: ```c
int is_new_version() {
  char buf_old[(10000)];
  char buf_new[(10000)];
  get_cache_version(buf_old);
  printf("Previous MAME version is: %s\n", buf_old);
  get_binary_version(buf_new);
  printf("Current MAME version is:  %s\n", buf_new);
  if (strncmp(buf_new, buf_old, sizeof buf_old)) {
    printf("New version available\n");
    return (1);
  }
  return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<is_new_version>:
  endbr64
  push   %r12
  push   %rbp
  lea    -0x4000(%rsp),%r11
L29:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L29
  sub    $0xe38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x4e28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x2710(%rsp),%rbp
  mov    %r12,%rdi
  call   <get_cache_version@plt>
  mov    %r12,%rsi
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <get_binary_version@plt>
  mov    %rbp,%rsi
  lea    D5(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x2710,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L6
L26:
  mov    0x4e28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x4e38,%rsp
  pop    %rbp
  pop    %r12
  ret
  nop
L6:
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
  jmp    L26
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D2
type: string
name: retrieve_data
arguments: tag: D5
type: string
name: retrieve_data
arguments: tag: D11
type: string

tool: D2: Previous MAME version is: %s


tool: D5: Current MAME version is:  %s


tool: D11: New version available

assistant: ```c
int is_new_version() {
  char buf_old[(10000)];
  char buf_new[(10000)];
  get_cache_version(buf_old);
  printf("Previous MAME version is: %s\n", buf_old);
  get_binary_version(buf_new);
  printf("Current MAME version is:  %s\n", buf_new);
  if (strncmp(buf_new, buf_old, sizeof buf_old)) {
    printf("New version available\n");
    return (1);
  }
  return (0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mdbg_timestamp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x17,-0x2c(%rbp)
  jg     L67
  mov    $0xffffffff,%eax
  jmp    L89
L67:
  lea    -0x14(%rbp),%rax
  mov    %rax,%rdi
  call   <time@plt>
  cmp    $0xffffffff,%eax
  jne    L79
  mov    $0xffffffff,%eax
  jmp    L89
L79:
  lea    -0x14(%rbp),%rax
  mov    %rax,%rdi
  call   <localtime@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L25
  mov    $0xffffffff,%eax
  jmp    L89
L25:
  mov    -0x10(%rbp),%rdx
  mov    -0x2c(%rbp),%esi
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rcx
  lea    D0(%rip),%rdx
  mov    %rax,%rdi
  call   <strftime@plt>
  cmp    $0x17,%eax
  je     L6
  mov    $0xffffffff,%eax
  jmp    L89
L6:
  mov    -0x28(%rbp),%rax
  add    $0x17,%rax
  movb   $0x0,(%rax)
  mov    $0x0,%eax
L89:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

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

tool: D0: %Y-%m-%dT%H:%M:%S.000

assistant: ```c
int mdbg_timestamp(char *out, int size) {
  time_t current_time;
  struct tm *bdtime;
  if (size < 24)
    return -1;
  if (time(&current_time) == -1)
    return -1;
  bdtime = localtime(&current_time);
  if (bdtime == NULL)
    return -1;
  if (strftime(out, size, "%Y-%m-%dT%H:%M:%S.000", bdtime) != 23)
    return -1;
  out[23] = '\0';
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<mdbg_timestamp>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x17,%esi
  jle    L54
  mov    %rdi,%rbp
  mov    %esi,%ebx
  lea    0x4(%rsp),%rdi
  call   <time@plt>
  cmp    $0xffffffff,%eax
  je     L0
  lea    0x4(%rsp),%rdi
  call   <localtime@plt>
  mov    %rax,%rcx
  test   %rax,%rax
  je     L4
  lea    D7(%rip),%rdx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <strftime@plt>
  cmp    $0x17,%eax
  jne    L70
  movb   $0x0,0x17(%rbp)
  mov    $0x0,%eax
L0:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L26
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L54:
  mov    $0xffffffff,%eax
  jmp    L0
L4:
  mov    $0xffffffff,%eax
  jmp    L0
L70:
  mov    $0xffffffff,%eax
  jmp    L0
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D7
type_guessed: string

tool: D7: %Y-%m-%dT%H:%M:%S.000

assistant: ```c
int mdbg_timestamp(char *out, int size) {
  time_t current_time;
  struct tm *bdtime;
  if (size < 24)
    return -1;
  if (time(&current_time) == -1)
    return -1;
  bdtime = localtime(&current_time);
  if (bdtime == NULL)
    return -1;
  if (strftime(out, size, "%Y-%m-%dT%H:%M:%S.000", bdtime) != 23)
    return -1;
  out[23] = '\0';
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<mdbg_timestamp>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x17,%esi
  jle    L29
  lea    0x4(%rsp),%r12
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    %r12,%rdi
  call   <time@plt>
  cmp    $0xffffffff,%eax
  je     L29
  mov    %r12,%rdi
  call   <localtime@plt>
  mov    %rax,%rcx
  test   %rax,%rax
  je     L29
  lea    D7(%rip),%rdx
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <strftime@plt>
  cmp    $0x17,%eax
  jne    L29
  movb   $0x0,0x17(%rbx)
  xor    %eax,%eax
L53:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L42
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L29:
  mov    $0xffffffff,%eax
  jmp    L53
L42:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D7
type_guessed: string

tool: D7: %Y-%m-%dT%H:%M:%S.000

assistant: ```c
int mdbg_timestamp(char *out, int size) {
  time_t current_time;
  struct tm *bdtime;
  if (size < 24)
    return -1;
  if (time(&current_time) == -1)
    return -1;
  bdtime = localtime(&current_time);
  if (bdtime == NULL)
    return -1;
  if (strftime(out, size, "%Y-%m-%dT%H:%M:%S.000", bdtime) != 23)
    return -1;
  out[23] = '\0';
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<mdbg_timestamp>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x17,%esi
  jle    L36
  lea    0x4(%rsp),%r12
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    %r12,%rdi
  call   <time@plt>
  cmp    $0xffffffff,%eax
  je     L36
  mov    %r12,%rdi
  call   <localtime@plt>
  mov    %rax,%rcx
  test   %rax,%rax
  je     L36
  lea    D9(%rip),%rdx
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <strftime@plt>
  cmp    $0x17,%eax
  jne    L36
  movb   $0x0,0x17(%rbx)
  xor    %eax,%eax
L2:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L36:
  mov    $0xffffffff,%eax
  jmp    L2
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D9
type_guessed: string

tool: D9: %Y-%m-%dT%H:%M:%S.000

assistant: ```c
int mdbg_timestamp(char *out, int size) {
  time_t current_time;
  struct tm *bdtime;
  if (size < 24)
    return -1;
  if (time(&current_time) == -1)
    return -1;
  bdtime = localtime(&current_time);
  if (bdtime == NULL)
    return -1;
  if (strftime(out, size, "%Y-%m-%dT%H:%M:%S.000", bdtime) != 23)
    return -1;
  out[23] = '\0';
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<str2nums>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %rdi,-0x68(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x60(%rbp)
  movl   $0x1,-0x5c(%rbp)
  mov    -0x68(%rbp),%rax
  mov    $0xf,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x70(%rbp),%rdx
  lea    -0x50(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    -0x50(%rbp),%rax
  lea    D6(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x58(%rbp)
  cmpq   $0x0,-0x58(%rbp)
  jne    L13
  mov    $0x0,%eax
  jmp    L7
L13:
  mov    -0x58(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    -0x68(%rbp),%rdx
  mov    %rax,(%rdx)
  jmp    L19
L40:
  addl   $0x1,-0x60(%rbp)
  mov    -0x60(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x68(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    -0x58(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %rax,(%rbx)
  addl   $0x1,-0x5c(%rbp)
L19:
  lea    D6(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x58(%rbp)
  cmpq   $0x0,-0x58(%rbp)
  jne    L40
  mov    -0x5c(%rbp),%eax
L7:
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D6
data_type: string

tool: D6: ,

assistant: ```c
int str2nums(int *nums[], const char *ns) {
  char *tk, *ps;
  char cns[3 * 15];
  int pos = 0, qtd = 1;
  memset(nums, 0, 15);
  strcpy(cns, ns);
  if (!(tk = strtok(cns, ",")))
    return 0;
  nums[0] = atoi(tk);
  while ((tk = strtok(NULL, ","))) {
    nums[++pos] = atoi(tk);
    qtd++;
  }
  return qtd;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<str2nums>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    $0xf,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  mov    %rsp,%r12
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  lea    D10(%rip),%rsi
  mov    %r12,%rdi
  call   <strtok@plt>
  mov    $0x0,%r13d
  test   %rax,%rax
  je     L11
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %rax,0x0(%rbp)
  mov    $0x1,%ebx
  lea    D10(%rip),%r12
  jmp    L6
L14:
  call   <atoi@plt>
  mov    %rax,0x0(%rbp,%rbx,8)
  add    $0x1,%rbx
L6:
  mov    %ebx,%r13d
  mov    %r12,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L14
L11:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  mov    %r13d,%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D10
datatype: string

tool: D10: ,

assistant: ```c
int str2nums(int *nums[], const char *ns) {
  char *tk, *ps;
  char cns[3 * 15];
  int pos = 0, qtd = 1;
  memset(nums, 0, 15);
  strcpy(cns, ns);
  if (!(tk = strtok(cns, ",")))
    return 0;
  nums[0] = atoi(tk);
  while ((tk = strtok(NULL, ","))) {
    nums[++pos] = atoi(tk);
    qtd++;
  }
  return qtd;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<str2nums>:
  endbr64
  push   %r13
  mov    $0xf,%edx
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  call   <memset@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  lea    D0(%rip),%r12
  call   <strcpy@plt>
  mov    %r13,%rdi
  mov    %r12,%rsi
  xor    %r13d,%r13d
  call   <strtok@plt>
  test   %rax,%rax
  je     L4
  mov    %rax,%rdi
  mov    $0x1,%ebx
  call   <atoi@plt>
  mov    %rax,0x0(%rbp)
  jmp    L5
  nopl   0x0(%rax,%rax,1)
L36:
  call   <atoi@plt>
  mov    %rax,0x0(%rbp,%rbx,8)
  add    $0x1,%rbx
L5:
  xor    %edi,%edi
  mov    %r12,%rsi
  mov    %ebx,%r13d
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L36
L4:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x48,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D0
type: string

tool: D0: ,

assistant: ```c
int str2nums(int *nums[], const char *ns) {
  char *tk, *ps;
  char cns[3 * 15];
  int pos = 0, qtd = 1;
  memset(nums, 0, 15);
  strcpy(cns, ns);
  if (!(tk = strtok(cns, ",")))
    return 0;
  nums[0] = atoi(tk);
  while ((tk = strtok(NULL, ","))) {
    nums[++pos] = atoi(tk);
    qtd++;
  }
  return qtd;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<str2nums>:
  endbr64
  push   %r13
  mov    $0xf,%edx
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  call   <memset@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  lea    D6(%rip),%r12
  call   <strcpy@plt>
  mov    %r13,%rdi
  mov    %r12,%rsi
  xor    %r13d,%r13d
  call   <strtok@plt>
  test   %rax,%rax
  je     L11
  mov    %rax,%rdi
  mov    $0x1,%ebx
  call   <atoi@plt>
  mov    %rax,0x0(%rbp)
  jmp    L36
  nopl   0x0(%rax,%rax,1)
L12:
  call   <atoi@plt>
  mov    %rax,0x0(%rbp,%rbx,8)
  add    $0x1,%rbx
L36:
  xor    %edi,%edi
  mov    %r12,%rsi
  mov    %ebx,%r13d
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L12
L11:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x48,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D6
type: string

tool: D6: ,

assistant: ```c
int str2nums(int *nums[], const char *ns) {
  char *tk, *ps;
  char cns[3 * 15];
  int pos = 0, qtd = 1;
  memset(nums, 0, 15);
  strcpy(cns, ns);
  if (!(tk = strtok(cns, ",")))
    return 0;
  nums[0] = atoi(tk);
  while ((tk = strtok(NULL, ","))) {
    nums[++pos] = atoi(tk);
    qtd++;
  }
  return qtd;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<handler>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x30(%rbp),%rax
  mov    0x4(%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    0x8(%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x30(%rbp),%rax
  mov    0x10(%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%edi
  call   <_exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void handler(int sig, siginfo_t *info, void *context) {
  int ssi_errno = info->si_errno;
  int ssi_signo = info->si_signo;
  int ssi_code = info->si_code;
  void *ssi_addr = info->si_addr;
  _exit(0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<handler>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <_exit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void handler(int sig, siginfo_t *info, void *context) {
  int ssi_errno = info->si_errno;
  int ssi_signo = info->si_signo;
  int ssi_code = info->si_code;
  void *ssi_addr = info->si_addr;
  _exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<handler>:
  endbr64
  xor    %edi,%edi
  jmp    1050 <_exit@plt>
```

assistant: ```c
void handler(int sig, siginfo_t *info, void *context) {
  int ssi_errno = info->si_errno;
  int ssi_signo = info->si_signo;
  int ssi_code = info->si_code;
  void *ssi_addr = info->si_addr;
  _exit(0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<handler>:
  endbr64
  xor    %edi,%edi
  jmp    1050 <_exit@plt>
```

assistant: ```c
void handler(int sig, siginfo_t *info, void *context) {
  int ssi_errno = info->si_errno;
  int ssi_signo = info->si_signo;
  int ssi_code = info->si_code;
  void *ssi_addr = info->si_addr;
  _exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<tst>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x7,%eax
  jle    L9
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    -0x10(%rbp),%rax
  add    $0x5,%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x8(%rbp),%rax
  lea    D1(%rip),%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
  jmp    L17
L9:
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
L17:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D1
type_guessed: string
name: parse_data
arguments: data_label: D13
type_guessed: string

tool: D1: 8%c%c0 

tool: D13: Unable to resolve code written at line %d


assistant: ```c
void tst(FILE *fo, char input[], int counter) {
  if (strlen(input) >= 8)
    fprintf(fo, "8%c%c0 ", input[5], input[8]);
  else
    printf("Unable to resolve code written at line %d\n", counter);
  return;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<tst>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %edx,%ebp
  mov    %rsi,%rdi
  call   <strlen@plt>
  cmp    $0x7,%eax
  jle    L3
  movsbl 0x8(%rbx),%ecx
  movsbl 0x5(%rbx),%edx
  lea    D10(%rip),%rsi
  mov    %r12,%rdi
  call   <fprintf@plt>
L10:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  mov    %ebp,%esi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  jmp    L10
```

tool_calls: name: read_data
arguments: label: D10
type: string
name: read_data
arguments: label: D18
type: string

tool: D10: 8%c%c0 

tool: D18: Unable to resolve code written at line %d


assistant: ```c
void tst(FILE *fo, char input[], int counter) {
  if (strlen(input) >= 8)
    fprintf(fo, "8%c%c0 ", input[5], input[8]);
  else
    printf("Unable to resolve code written at line %d\n", counter);
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<tst>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rsi,%rbx
  call   <strlen@plt>
  cmp    $0x7,%eax
  jle    L5
  movsbl 0x8(%rbx),%ecx
  movsbl 0x5(%rbx),%edx
  mov    %r12,%rdi
  lea    D2(%rip),%rsi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1090 <fprintf@plt>
  nopl   0x0(%rax,%rax,1)
L5:
  pop    %rbx
  mov    %ebp,%esi
  lea    D12(%rip),%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D2
type: string
name: disassemble_data
arguments: data_label: D12
type: string

tool: D2: 8%c%c0 

tool: D12: Unable to resolve code written at line %d


assistant: ```c
void tst(FILE *fo, char input[], int counter) {
  if (strlen(input) >= 8)
    fprintf(fo, "8%c%c0 ", input[5], input[8]);
  else
    printf("Unable to resolve code written at line %d\n", counter);
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<tst>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rsi,%rbx
  call   <strlen@plt>
  cmp    $0x7,%eax
  jle    L6
  movsbl 0x8(%rbx),%ecx
  movsbl 0x5(%rbx),%edx
  mov    %r12,%rdi
  lea    D17(%rip),%rsi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1090 <fprintf@plt>
  nopl   0x0(%rax,%rax,1)
L6:
  pop    %rbx
  mov    %ebp,%esi
  lea    D12(%rip),%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <printf@plt>
```

tool_calls: name: read_data
arguments: data_tag: D17
type_infered: string
name: read_data
arguments: data_tag: D12
type_infered: string

tool: D17: 8%c%c0 

tool: D12: Unable to resolve code written at line %d


assistant: ```c
void tst(FILE *fo, char input[], int counter) {
  if (strlen(input) >= 8)
    fprintf(fo, "8%c%c0 ", input[5], input[8]);
  else
    printf("Unable to resolve code written at line %d\n", counter);
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<get_container_work_directory>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    -0x18(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  sub    $0x8,%rsp
  push   -0x20(%rbp)
  mov    %rcx,%r9
  mov    %rdx,%r8
  mov    %rax,%rcx
  mov    $0x4,%edx
  lea    D7(%rip),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <concatenate@plt>
  add    $0x10,%rsp
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D7
data_type: string
name: unpack_data
arguments: label: D17
data_type: string

tool: D7: container_dir_path

tool: D17: %s/usercache/%s/appcache/%s/%s

assistant: ```c
char *get_container_work_directory(const char *nm_root, const char *user, const char *app_id, const char *container_id) {
  return concatenate("%s/usercache/%s"
                     "/appcache/%s"
                     "/%s",
                     "container_dir_path", 4, nm_root, user, app_id, container_id);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<get_container_work_directory>:
  endbr64
  sub    $0x10,%rsp
  mov    %rsi,%r8
  mov    %rdx,%r9
  push   %rcx
  mov    %rdi,%rcx
  mov    $0x4,%edx
  lea    D9(%rip),%rsi
  lea    D10(%rip),%rdi
  call   <concatenate@plt>
  add    $0x18,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D9
type_guessed: string
name: retrieve_data
arguments: label: D10
type_guessed: string

tool: D9: container_dir_path

tool: D10: %s/usercache/%s/appcache/%s/%s

assistant: ```c
char *get_container_work_directory(const char *nm_root, const char *user, const char *app_id, const char *container_id) {
  return concatenate("%s/usercache/%s"
                     "/appcache/%s"
                     "/%s",
                     "container_dir_path", 4, nm_root, user, app_id, container_id);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_container_work_directory>:
  endbr64
  sub    $0x10,%rsp
  mov    %rsi,%r8
  mov    %rdx,%r9
  mov    $0x4,%edx
  push   %rcx
  lea    D8(%rip),%rsi
  mov    %rdi,%rcx
  lea    D9(%rip),%rdi
  call   <concatenate@plt>
  add    $0x18,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D8
type_guessed: string
name: unpack_data
arguments: label: D9
type_guessed: string

tool: D8: container_dir_path

tool: D9: %s/usercache/%s/appcache/%s/%s

assistant: ```c
char *get_container_work_directory(const char *nm_root, const char *user, const char *app_id, const char *container_id) {
  return concatenate("%s/usercache/%s"
                     "/appcache/%s"
                     "/%s",
                     "container_dir_path", 4, nm_root, user, app_id, container_id);
}
```