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

```asm
<wrapper_strdup>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %r13d,%edx
  mov    %r12,%rsi
  lea    0x1(%rax),%rdi
  call   <wrapper_malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *wrapper_strdup(const char *str, const char *file, const int line) {
  char *new_str = wrapper_malloc(strlen(str) + 1, file, line);
  strcpy(new_str, str);
  return new_str;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<wrapper_strdup>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %r13d,%edx
  mov    %r12,%rsi
  lea    0x1(%rax),%rdi
  call   <wrapper_malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *wrapper_strdup(const char *str, const char *file, const int line) {
  char *new_str = wrapper_malloc(strlen(str) + 1, file, line);
  strcpy(new_str, str);
  return new_str;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<binary_decimal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L49
L31:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x31,%al
  je     L33
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x30,%al
  je     L33
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
L33:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  lea    -0x30(%rax),%ebx
  mov    $0x17,%eax
  sub    -0x14(%rbp),%eax
  mov    %eax,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  imul   %ebx,%eax
  add    %eax,-0x18(%rbp)
  addl   $0x1,-0x14(%rbp)
L49:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L31
  cmpl   $0x18,-0x14(%rbp)
  je     L19
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
L19:
  mov    -0x18(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
datatype: string
name: disassemble_data
arguments: data_tag: D6
datatype: string

tool: D1: ERROR: Input must be in binary format.


tool: D6: ERROR: Input must be 21 digits in length.


assistant: ```c
int binary_decimal(char *str) {
  int decimal = 0;
  int i = 0;
  while (str[i] != '\0') {
    if (str[i] != '1' && str[i] != '0') {
      printf("ERROR: Input must be in binary format.\n");
      exit(0);
    }
    decimal += (str[i] - '0') * (int)pow(2, 23 - i);
    i++;
  }
  if (i != 24) {
    printf("ERROR: Input must be 21 digits in length.\n");
    exit(0);
  }
  return decimal;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<binary_decimal>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movzbl (%rdi),%eax
  test   %al,%al
  je     L34
  mov    %rdi,%rbp
  mov    $0x0,%r12d
  mov    $0x0,%r13d
  lea    D8(%rip),%r14
  jmp    L14
L49:
  movsbl 0x0(%rbp),%ebx
  sub    $0x30,%ebx
  mov    $0x17,%esi
  sub    %r12d,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  imul   %eax,%ebx
  add    %ebx,%r13d
  add    $0x1,%r12d
  add    $0x1,%rbp
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L69
L14:
  sub    $0x30,%eax
  cmp    $0x1,%al
  jbe    L49
  mov    %r14,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L49
L69:
  cmp    $0x18,%r12d
  jne    L29
L25:
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L34:
  mov    $0x0,%r13d
L29:
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L25
```

tool_calls: name: retrieve_data
arguments: tag: D8
datatype: string
name: retrieve_data
arguments: tag: D0
datatype: string

tool: D8: ERROR: Input must be in binary format.


tool: D0: ERROR: Input must be 21 digits in length.


assistant: ```c
int binary_decimal(char *str) {
  int decimal = 0;
  int i = 0;
  while (str[i] != '\0') {
    if (str[i] != '1' && str[i] != '0') {
      printf("ERROR: Input must be in binary format.\n");
      exit(0);
    }
    decimal += (str[i] - '0') * (int)pow(2, 23 - i);
    i++;
  }
  if (i != 24) {
    printf("ERROR: Input must be 21 digits in length.\n");
    exit(0);
  }
  return decimal;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<binary_decimal>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movzbl (%rdi),%eax
  test   %al,%al
  je     L1
  mov    %rdi,%rbp
  xor    %r14d,%r14d
  lea    D5(%rip),%r13
  xor    %r12d,%r12d
  nopl   0x0(%rax,%rax,1)
L16:
  sub    $0x30,%eax
  cmp    $0x1,%al
  jbe    L38
  mov    %r13,%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
L38:
  movsbl 0x0(%rbp),%ebx
  mov    $0x17,%esi
  mov    $0x2,%edi
  add    $0x1,%rbp
  sub    %r14d,%esi
  add    $0x1,%r14d
  call   <pow@plt>
  sub    $0x30,%ebx
  imul   %eax,%ebx
  movzbl 0x0(%rbp),%eax
  add    %ebx,%r12d
  test   %al,%al
  jne    L16
  cmp    $0x18,%r14d
  jne    L22
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L1:
  xor    %r12d,%r12d
L22:
  lea    D12(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: tag: D5
type_infered: string
name: read_data
arguments: tag: D12
type_infered: string

tool: D5: ERROR: Input must be in binary format.


tool: D12: ERROR: Input must be 21 digits in length.


assistant: ```c
int binary_decimal(char *str) {
  int decimal = 0;
  int i = 0;
  while (str[i] != '\0') {
    if (str[i] != '1' && str[i] != '0') {
      printf("ERROR: Input must be in binary format.\n");
      exit(0);
    }
    decimal += (str[i] - '0') * (int)pow(2, 23 - i);
    i++;
  }
  if (i != 24) {
    printf("ERROR: Input must be 21 digits in length.\n");
    exit(0);
  }
  return decimal;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<binary_decimal>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movsbl (%rdi),%ebx
  test   %bl,%bl
  je     L12
  mov    %rdi,%r13
  xor    %r14d,%r14d
  lea    D14(%rip),%rbp
  xor    %r12d,%r12d
  nopl   0x0(%rax,%rax,1)
L3:
  lea    -0x30(%rbx),%eax
  cmp    $0x1,%al
  jbe    L0
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  movsbl 0x0(%r13),%ebx
L0:
  mov    $0x17,%esi
  mov    $0x2,%edi
  sub    $0x30,%ebx
  add    $0x1,%r13
  sub    %r14d,%esi
  add    $0x1,%r14d
  call   <pow@plt>
  imul   %eax,%ebx
  add    %ebx,%r12d
  movsbl 0x0(%r13),%ebx
  test   %bl,%bl
  jne    L3
  cmp    $0x18,%r14d
  jne    L27
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L12:
  xor    %r12d,%r12d
L27:
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D14
type: string
name: extract_symbolic_data
arguments: data_tag: D5
type: string

tool: D14: ERROR: Input must be in binary format.


tool: D5: ERROR: Input must be 21 digits in length.


assistant: ```c
int binary_decimal(char *str) {
  int decimal = 0;
  int i = 0;
  while (str[i] != '\0') {
    if (str[i] != '1' && str[i] != '0') {
      printf("ERROR: Input must be in binary format.\n");
      exit(0);
    }
    decimal += (str[i] - '0') * (int)pow(2, 23 - i);
    i++;
  }
  if (i != 24) {
    printf("ERROR: Input must be 21 digits in length.\n");
    exit(0);
  }
  return decimal;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<_removelast>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    -0x18(%rbp),%rdx
  lea    (%rax,%rdx,1),%rbx
  jmp    L19
L16:
  movb   $0x0,(%rbx)
L19:
  sub    $0x1,%rbx
  movzbl (%rbx),%eax
  cmp    $0x2f,%al
  je     L15
  cmp    -0x18(%rbp),%rbx
  jae    L16
L15:
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void _removelast(path) char *path;
{
  register char *p;
  for (p = path + strlen(path); *--p != '/' && p >= path;)
    *p = '\0';
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<_removelast>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  add    %rbx,%rax
  cmpb   $0x2f,-0x1(%rax)
  je     L17
  lea    -0x1(%rax),%rdx
  cmp    %rdx,%rbx
  ja     L17
L23:
  movb   $0x0,(%rdx)
  sub    $0x1,%rdx
  cmpb   $0x2f,(%rdx)
  je     L17
  cmp    %rdx,%rbx
  jbe    L23
L17:
  pop    %rbx
  ret
```

assistant: ```c
void _removelast(path) char *path;
{
  register char *p;
  for (p = path + strlen(path); *--p != '/' && p >= path;)
    *p = '\0';
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<_removelast>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  add    %rbx,%rax
  cmpb   $0x2f,-0x1(%rax)
  je     L11
  lea    -0x1(%rax),%rdx
  jmp    L4
  nopl   0x0(%rax)
L35:
  movb   $0x0,(%rdx)
  sub    $0x1,%rdx
  cmpb   $0x2f,(%rdx)
  je     L11
L4:
  cmp    %rdx,%rbx
  jbe    L35
L11:
  pop    %rbx
  ret
```

assistant: ```c
void _removelast(path) char *path;
{
  register char *p;
  for (p = path + strlen(path); *--p != '/' && p >= path;)
    *p = '\0';
}
```
user: Translate the below assembly code into C programming language:

```asm
<_removelast>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  add    %rbx,%rax
  cmpb   $0x2f,-0x1(%rax)
  je     L1
  lea    -0x1(%rax),%rdx
  jmp    L12
  nopl   0x0(%rax)
L7:
  movb   $0x0,(%rdx)
  sub    $0x1,%rdx
  cmpb   $0x2f,(%rdx)
  je     L1
L12:
  cmp    %rdx,%rbx
  jbe    L7
L1:
  pop    %rbx
  ret
```

assistant: ```c
void _removelast(path) char *path;
{
  register char *p;
  for (p = path + strlen(path); *--p != '/' && p >= path;)
    *p = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mk_result_date>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %edx,-0x4c(%rbp)
  mov    %rcx,-0x58(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x76c,-0x44(%rbp)
  jbe    L16
  subl   $0x76c,-0x44(%rbp)
L16:
  mov    -0x44(%rbp),%eax
  mov    %eax,-0x40(%rbp)
  mov    -0x48(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x3c(%rbp)
  mov    -0x4c(%rbp),%eax
  mov    %eax,-0x38(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  movq   $0x0,-0x30(%rbp)
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <mktime@plt>
  mov    -0x58(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x58(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0xffffffff,%eax
  jne    L13
  mov    $0x0,%eax
  jmp    L30
L13:
  mov    $0x1,%eax
L30:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L37
  call   <__stack_chk_fail@plt>
L37:
  leave
  ret
```

assistant: ```c
int mk_result_date(unsigned year, unsigned month, unsigned day, time_t *tp) {
  struct tm tms;
  if (year > 1900)
    year -= 1900;
  tms.tm_year = year;
  tms.tm_mon = month - 1;
  tms.tm_mday = day;
  tms.tm_hour = 0;
  tms.tm_min = tms.tm_sec = 0;
  tms.tm_isdst = 0;
  *tp = mktime(&tms);
  if (*tp == (time_t)(-1))
    return 0;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mk_result_date>:
  endbr64
  push   %rbx
  sub    $0x40,%rsp
  mov    %rcx,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    -0x76c(%rdi),%eax
  cmp    $0x76c,%edi
  cmova  %eax,%edi
  mov    %edi,(%rsp)
  sub    $0x1,%esi
  mov    %esi,0x4(%rsp)
  mov    %edx,0x8(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x10(%rsp)
  mov    %rsp,%rdi
  call   <mktime@plt>
  mov    %eax,(%rbx)
  cmp    $0xffffffff,%eax
  setne  %al
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  movzbl %al,%eax
  add    $0x40,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int mk_result_date(unsigned year, unsigned month, unsigned day, time_t *tp) {
  struct tm tms;
  if (year > 1900)
    year -= 1900;
  tms.tm_year = year;
  tms.tm_mon = month - 1;
  tms.tm_mday = day;
  tms.tm_hour = 0;
  tms.tm_min = tms.tm_sec = 0;
  tms.tm_isdst = 0;
  *tp = mktime(&tms);
  if (*tp == (time_t)(-1))
    return 0;
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mk_result_date>:
  endbr64
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  cmp    $0x76c,%edi
  lea    -0x76c(%rdi),%eax
  mov    %edx,0x8(%rsp)
  movq   $0x0,0x28(%rsp)
  cmova  %eax,%edi
  sub    $0x1,%esi
  mov    %esi,0x4(%rsp)
  mov    %edi,(%rsp)
  mov    %rsp,%rdi
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x10(%rsp)
  call   <mktime@plt>
  mov    %eax,(%rbx)
  cmp    $0xffffffff,%eax
  setne  %al
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x40,%rsp
  movzbl %al,%eax
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int mk_result_date(unsigned year, unsigned month, unsigned day, time_t *tp) {
  struct tm tms;
  if (year > 1900)
    year -= 1900;
  tms.tm_year = year;
  tms.tm_mon = month - 1;
  tms.tm_mday = day;
  tms.tm_hour = 0;
  tms.tm_min = tms.tm_sec = 0;
  tms.tm_isdst = 0;
  *tp = mktime(&tms);
  if (*tp == (time_t)(-1))
    return 0;
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<mk_result_date>:
  endbr64
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  cmp    $0x76c,%edi
  lea    -0x76c(%rdi),%eax
  mov    %edx,0x8(%rsp)
  cmova  %eax,%edi
  sub    $0x1,%esi
  movd   %esi,%xmm1
  movd   %edi,%xmm0
  mov    %rsp,%rdi
  punpckldq %xmm1,%xmm0
  movq   %xmm0,(%rsp)
  pxor   %xmm0,%xmm0
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  call   <mktime@plt>
  mov    %eax,(%rbx)
  cmp    $0xffffffff,%eax
  setne  %al
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x40,%rsp
  movzbl %al,%eax
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int mk_result_date(unsigned year, unsigned month, unsigned day, time_t *tp) {
  struct tm tms;
  if (year > 1900)
    year -= 1900;
  tms.tm_year = year;
  tms.tm_mon = month - 1;
  tms.tm_mday = day;
  tms.tm_hour = 0;
  tms.tm_min = tms.tm_sec = 0;
  tms.tm_isdst = 0;
  *tp = mktime(&tms);
  if (*tp == (time_t)(-1))
    return 0;
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<equal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L5
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L5:
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D17
datatype: string
name: disassemble_data
arguments: data_label: D9
datatype: string

tool: D17:   expected: '%s'


tool: D9:     actual: '%s'


assistant: ```c
void equal(char *a, char *b) {
  if (strcmp(a, b)) {
    printf("\n");
    printf("  expected: '%s'\n", a);
    printf("    actual: '%s'\n", b);
    printf("\n");
    exit(1);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<equal>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L7
L6:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbp,%rsi
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L6
```

tool_calls: name: unpack_data
arguments: data_tag: D12
type_guessed: string
name: unpack_data
arguments: data_tag: D2
type_guessed: string

tool: D12:   expected: '%s'


tool: D2:     actual: '%s'


assistant: ```c
void equal(char *a, char *b) {
  if (strcmp(a, b)) {
    printf("\n");
    printf("  expected: '%s'\n", a);
    printf("    actual: '%s'\n", b);
    printf("\n");
    exit(1);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<equal>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x8,%rsp
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L3
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r12,%rsi
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  mov    $0x1,%edi
  pop    %rbp
  pop    %r12
  jmp    10b0 <exit@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D1
data_type: string
name: parse_data
arguments: data_tag: D11
data_type: string

tool: D1:   expected: '%s'


tool: D11:     actual: '%s'


assistant: ```c
void equal(char *a, char *b) {
  if (strcmp(a, b)) {
    printf("\n");
    printf("  expected: '%s'\n", a);
    printf("    actual: '%s'\n", b);
    printf("\n");
    exit(1);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<equal>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x8,%rsp
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L7
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r12,%rsi
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  mov    $0x1,%edi
  pop    %rbp
  pop    %r12
  jmp    10b0 <exit@plt>
```

tool_calls: name: unpack_data
arguments: label: D6
type_guessed: string
name: unpack_data
arguments: label: D11
type_guessed: string

tool: D6:   expected: '%s'


tool: D11:     actual: '%s'


assistant: ```c
void equal(char *a, char *b) {
  if (strcmp(a, b)) {
    printf("\n");
    printf("  expected: '%s'\n", a);
    printf("    actual: '%s'\n", b);
    printf("\n");
    exit(1);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<prefix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L21
  cmpq   $0x0,-0x20(%rbp)
  jne    L45
L21:
  mov    $0x0,%eax
  jmp    L59
L45:
  movl   $0x0,-0x4(%rbp)
  jmp    L23
L3:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L40
  mov    $0x0,%eax
  jmp    L59
L40:
  addl   $0x1,-0x4(%rbp)
L23:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L3
  mov    $0x1,%eax
L59:
  pop    %rbp
  ret
```

assistant: ```c
int prefix(char *pre, char *str) {
  if (!pre || !str)
    return 0;
  int i;
  for (i = 0; pre[i]; i++) {
    if (pre[i] != str[i]) {
      return 0;
    }
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<prefix>:
  endbr64
  test   %rdi,%rdi
  je     L13
  test   %rsi,%rsi
  je     L13
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L45
  mov    $0x0,%eax
L49:
  cmp    %dl,(%rsi,%rax,1)
  jne    L24
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  jne    L49
  mov    $0x1,%eax
  ret
L13:
  mov    $0x0,%eax
  ret
L45:
  mov    $0x1,%eax
  ret
L24:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int prefix(char *pre, char *str) {
  if (!pre || !str)
    return 0;
  int i;
  for (i = 0; pre[i]; i++) {
    if (pre[i] != str[i]) {
      return 0;
    }
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<prefix>:
  endbr64
  test   %rdi,%rdi
  je     L40
  test   %rsi,%rsi
  je     L40
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L2
  xor    %eax,%eax
  jmp    L4
  nopl   0x0(%rax)
L1:
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  je     L2
L4:
  cmp    %dl,(%rsi,%rax,1)
  je     L1
L40:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L2:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int prefix(char *pre, char *str) {
  if (!pre || !str)
    return 0;
  int i;
  for (i = 0; pre[i]; i++) {
    if (pre[i] != str[i]) {
      return 0;
    }
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<prefix>:
  endbr64
  test   %rdi,%rdi
  je     L24
  test   %rsi,%rsi
  je     L24
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L55
  xor    %eax,%eax
  jmp    L6
  nopl   0x0(%rax)
L39:
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  je     L55
L6:
  cmp    %dl,(%rsi,%rax,1)
  je     L39
L24:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L55:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int prefix(char *pre, char *str) {
  if (!pre || !str)
    return 0;
  int i;
  for (i = 0; pre[i]; i++) {
    if (pre[i] != str[i]) {
      return 0;
    }
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<DESIntStatus>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x3c,%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  je     L0
  mov    -0x18(%rbp),%rax
  add    $0x40,%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  and    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    $0xffff8038,%ecx
  add    %rcx,%rax
  mov    (%rax),%rax
  shl    $0x10,%rax
  or     %rdx,%rax
  jmp    L15
L0:
  mov    -0x18(%rbp),%rax
  mov    $0xffff8034,%edx
  add    %rdx,%rax
  mov    (%rax),%rax
  shl    $0x10,%rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  or     %rdx,%rax
L15:
  pop    %rbp
  ret
```

assistant: ```c
uint32_t DESIntStatus(uint32_t ui32Base, bool bMasked) {
  uint32_t ui32Status, ui32Enable;
  ;
  ui32Status = (*((volatile unsigned long *)(ui32Base + 0x0000003C)));
  if (bMasked) {
    ui32Enable = (*((volatile unsigned long *)(ui32Base + 0x00000040)));
    return ((ui32Status & ui32Enable) | ((*((volatile unsigned long *)(ui32Base + 0xFFFF8038))) << 16));
  } else {
    return (ui32Status | ((*((volatile unsigned long *)(ui32Base + 0xFFFF8034))) << 16));
  }
}
```