ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<print_string>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  movslq %edx,%rbx
  sub    $0x8,%rsp
  test   %edx,%edx
  jns    L11
  mov    %rsi,%rdi
  call   <strlen@plt>
  movslq %eax,%rbx
L11:
  add    %r13,%rbx
  lea    D43(%rip),%r15
  lea    D25(%rip),%r14
  lea    D58(%rip),%rbp
  cmp    %rbx,%r13
  jae    L43
  nopw   0x0(%rax,%rax,1)
L32:
  movzbl 0x0(%r13),%edx
  lea    -0x7(%rdx),%eax
  cmp    $0x1b,%al
  ja     L13
  movzbl %al,%eax
  movslq 0x0(%rbp,%rax,4),%rax
  add    %rbp,%rax
  notrack jmp *%rax
  nopw   0x0(%rax,%rax,1)
L13:
  movsbl %dl,%edi
  call   <isprint@plt>
  movzbl 0x0(%r13),%edx
  mov    %r14,%rsi
  test   %eax,%eax
  jne    L22
  mov    %r15,%rsi
  nopw   0x0(%rax,%rax,1)
L22:
  movzbl %dl,%edx
  mov    %r12,%rdi
  add    $0x1,%r13
  call   <fprintf@plt>
  cmp    %r13,%rbx
  jne    L32
L43:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
  lea    D83(%rip),%rsi
  jmp    L22
  nopl   0x0(%rax)
  lea    D2(%rip),%rsi
  jmp    L22
  nopl   0x0(%rax)
  lea    D105(%rip),%rsi
  jmp    L22
  nopl   0x0(%rax)
  lea    D14(%rip),%rsi
  jmp    L22
  nopl   0x0(%rax)
  lea    D102(%rip),%rsi
  jmp    L22
  nopl   0x0(%rax)
  lea    D54(%rip),%rsi
  jmp    L22
  nopl   0x0(%rax)
  lea    D0(%rip),%rsi
  jmp    L22
  nopl   0x0(%rax)
  lea    D17(%rip),%rsi
  jmp    L22
```

tool_calls: name: parse_data
arguments: data_tag: D43
type_infered: string
name: parse_data
arguments: data_tag: D25
type_infered: string
name: parse_data
arguments: data_tag: D83
type_infered: string
name: parse_data
arguments: data_tag: D2
type_infered: string
name: parse_data
arguments: data_tag: D105
type_infered: string
name: parse_data
arguments: data_tag: D14
type_infered: string
name: parse_data
arguments: data_tag: D102
type_infered: string
name: parse_data
arguments: data_tag: D54
type_infered: string
name: parse_data
arguments: data_tag: D0
type_infered: string
name: parse_data
arguments: data_tag: D17
type_infered: string

tool: D43: \x%02x

tool: D25: %c

tool: D83: \"

tool: D2: \r

tool: D105: \f

tool: D14: \v

tool: D102: \n

tool: D54: \t

tool: D0: \a

tool: D17: \b

assistant: ```c
void print_string(FILE *file, const char *s, int len) {
  const char *end = s + (len < 0 ? (int)strlen(s) : len);
  const char *p;
  for (p = s; p < end; ++p) {
    const char *format;
    switch (*p) {
    case '\a':
      format = "\\a";
      break;
    case '\b':
      format = "\\b";
      break;
    case '\f':
      format = "\\f";
      break;
    case '\n':
      format = "\\n";
      break;
    case '\r':
      format = "\\r";
      break;
    case '\t':
      format = "\\t";
      break;
    case '\v':
      format = "\\v";
      break;
    case '\"':
      format = "\\\"";
      break;
    default:
      format = isprint(*p) ? "%c" : "\\x%02x";
      break;
    }
    fprintf(file, format, ((unsigned)*p) & 0xff);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<print_menu_hdr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D27
data_type: string
name: retrieve_data
arguments: data_label: D15
data_type: string
name: retrieve_data
arguments: data_label: D12
data_type: string

tool: D27: 
			*****************************************************

tool: D15: 				 EMPLOYEE %s MENU 


tool: D12: 			*****************************************************

assistant: ```c
int print_menu_hdr(char *header) {
  printf("\n\t\t\t*****************************************************\n");
  printf("\t\t\t\t EMPLOYEE %s MENU \n", header);
  printf("\t\t\t*****************************************************\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<print_menu_hdr>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D5(%rip),%rdi
  call   <puts@plt>
  mov    %rbx,%rsi
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D23(%rip),%rdi
  call   <puts@plt>
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D5
type: string
name: retrieve_data
arguments: data_label: D17
type: string
name: retrieve_data
arguments: data_label: D23
type: string

tool: D5: 
			*****************************************************

tool: D17: 				 EMPLOYEE %s MENU 


tool: D23: 			*****************************************************

assistant: ```c
int print_menu_hdr(char *header) {
  printf("\n\t\t\t*****************************************************\n");
  printf("\t\t\t\t EMPLOYEE %s MENU \n", header);
  printf("\t\t\t*****************************************************\n");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<print_menu_hdr>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  lea    D27(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D4(%rip),%rdi
  pop    %rbp
  jmp    1060 <puts@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D27
type_guessed: string
name: retrieve_data
arguments: tag: D15
type_guessed: string
name: retrieve_data
arguments: tag: D4
type_guessed: string

tool: D27: 
			*****************************************************

tool: D15: 				 EMPLOYEE %s MENU 


tool: D4: 			*****************************************************

assistant: ```c
int print_menu_hdr(char *header) {
  printf("\n\t\t\t*****************************************************\n");
  printf("\t\t\t\t EMPLOYEE %s MENU \n", header);
  printf("\t\t\t*****************************************************\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_menu_hdr>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  lea    D0(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D5(%rip),%rdi
  pop    %rbp
  jmp    1060 <puts@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D0
type_guessed: string
name: unpack_data
arguments: data_label: D7
type_guessed: string
name: unpack_data
arguments: data_label: D5
type_guessed: string

tool: D0: 
			*****************************************************

tool: D7: 				 EMPLOYEE %s MENU 


tool: D5: 			*****************************************************

assistant: ```c
int print_menu_hdr(char *header) {
  printf("\n\t\t\t*****************************************************\n");
  printf("\t\t\t\t EMPLOYEE %s MENU \n", header);
  printf("\t\t\t*****************************************************\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ldiv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jle    L3
  cmpq   $0x0,-0x20(%rbp)
  jns    L3
  negq   -0x18(%rbp)
  negq   -0x20(%rbp)
L3:
  mov    -0x18(%rbp),%rax
  cqto
  idivq  -0x20(%rbp)
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  cqto
  idivq  -0x20(%rbp)
  mov    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jns    L2
  cmpq   $0x0,-0x20(%rbp)
  jle    L2
  mov    -0x8(%rbp),%rax
  test   %rax,%rax
  jle    L2
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  sub    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
L2:
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  pop    %rbp
  ret
```

assistant: ```c
ldiv_t ldiv(long num, long denom) {
  ldiv_t r;
  if (num > 0 && denom < 0) {
    num = -num;
    denom = -denom;
  }
  r.quot = num / denom;
  r.rem = num % denom;
  if (num < 0 && denom > 0) {
    if (r.rem > 0) {
      r.quot++;
      r.rem -= denom;
    }
  }
  return r;
}
```
user: What is the c source code of the assembly code below:

```asm
<ldiv>:
  endbr64
  test   %rdi,%rdi
  jle    L25
  test   %rsi,%rsi
  jns    L25
  neg    %rdi
  mov    %rsi,%rcx
  neg    %rcx
  mov    %rdi,%rax
  cqto
  idiv   %rcx
  mov    %rax,%rcx
  mov    %rdi,%rax
  cqto
  idiv   %rsi
L2:
  mov    %rcx,%rax
  ret
L25:
  mov    %rdi,%rax
  cqto
  idiv   %rsi
  mov    %rax,%rcx
  shr    $0x3f,%rdi
  test   %rsi,%rsi
  setg   %r9b
  test   %dil,%r9b
  je     L2
  test   %rdx,%rdx
  jle    L2
  lea    0x1(%rax),%rcx
  sub    %rsi,%rdx
  jmp    L2
```

assistant: ```c
ldiv_t ldiv(long num, long denom) {
  ldiv_t r;
  if (num > 0 && denom < 0) {
    num = -num;
    denom = -denom;
  }
  r.quot = num / denom;
  r.rem = num % denom;
  if (num < 0 && denom > 0) {
    if (r.rem > 0) {
      r.quot++;
      r.rem -= denom;
    }
  }
  return r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ldiv>:
  endbr64
  test   %rdi,%rdi
  jle    L17
  test   %rsi,%rsi
  jns    L17
  neg    %rdi
  mov    %rsi,%rcx
  mov    %rdi,%rax
  neg    %rcx
  cqto
  idiv   %rcx
  mov    %rax,%r8
  mov    %rdi,%rax
  cqto
  idiv   %rsi
L47:
  mov    %r8,%rax
  ret
  xchg   %ax,%ax
L17:
  mov    %rdi,%rax
  cqto
  idiv   %rsi
  mov    %rax,%r8
  test   %rdi,%rdi
  jns    L47
  test   %rsi,%rsi
  jle    L47
  test   %rdx,%rdx
  jle    L47
  add    $0x1,%r8
  sub    %rsi,%rdx
  mov    %r8,%rax
  ret
```

assistant: ```c
ldiv_t ldiv(long num, long denom) {
  ldiv_t r;
  if (num > 0 && denom < 0) {
    num = -num;
    denom = -denom;
  }
  r.quot = num / denom;
  r.rem = num % denom;
  if (num < 0 && denom > 0) {
    if (r.rem > 0) {
      r.quot++;
      r.rem -= denom;
    }
  }
  return r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ldiv>:
  endbr64
  test   %rdi,%rdi
  jle    L42
  test   %rsi,%rsi
  jns    L42
  neg    %rdi
  mov    %rsi,%rcx
  mov    %rdi,%rax
  neg    %rcx
  cqto
  idiv   %rcx
  mov    %rax,%r8
  mov    %rdi,%rax
  cqto
  idiv   %rsi
L18:
  mov    %r8,%rax
  ret
  xchg   %ax,%ax
L42:
  mov    %rdi,%rax
  cqto
  idiv   %rsi
  mov    %rax,%r8
  test   %rdi,%rdi
  jns    L18
  test   %rsi,%rsi
  jle    L18
  test   %rdx,%rdx
  jle    L18
  add    $0x1,%r8
  sub    %rsi,%rdx
  mov    %r8,%rax
  ret
```

assistant: ```c
ldiv_t ldiv(long num, long denom) {
  ldiv_t r;
  if (num > 0 && denom < 0) {
    num = -num;
    denom = -denom;
  }
  r.quot = num / denom;
  r.rem = num % denom;
  if (num < 0 && denom > 0) {
    if (r.rem > 0) {
      r.quot++;
      r.rem -= denom;
    }
  }
  return r;
}
```
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    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jle    L17
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x8(%rbp)
  jmp    L14
L17:
  movl   $0xa,-0x8(%rbp)
L14:
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <omp_set_num_threads@plt>
  mov    $0x0,%eax
  call   <omp_get_thread_num@plt>
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  call   <omp_get_num_threads@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D2
type_infered: string

tool: D2: 
Hola Mundo soy el hilo =%d de %d que somos en total.


assistant: ```c
int main(int argc, char *argv[]) {
  int nthreads;
  int thread;
  if (argc > 1) {
    nthreads = atoi(argv[1]);
  } else
    nthreads = 10;
  omp_set_num_threads(nthreads);
#pragma omp parallel
  {
    thread = omp_get_thread_num();
    nthreads = omp_get_num_threads();
    printf("\nHola Mundo soy el hilo =%d de %d que somos en total.\n", thread, nthreads);
  }
  printf("\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    %edi,%eax
  mov    $0xa,%edi
  cmp    $0x1,%eax
  jg     L17
L14:
  call   <omp_set_num_threads@plt>
  mov    $0x0,%eax
  call   <omp_get_thread_num@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <omp_get_num_threads@plt>
  mov    %eax,%edx
  mov    %ebx,%esi
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
L17:
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  jmp    L14
```

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

tool: D5: 
Hola Mundo soy el hilo =%d de %d que somos en total.


assistant: ```c
int main(int argc, char *argv[]) {
  int nthreads;
  int thread;
  if (argc > 1) {
    nthreads = atoi(argv[1]);
  } else
    nthreads = 10;
  omp_set_num_threads(nthreads);
#pragma omp parallel
  {
    thread = omp_get_thread_num();
    nthreads = omp_get_num_threads();
    printf("\nHola Mundo soy el hilo =%d de %d que somos en total.\n", thread, nthreads);
  }
  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
<main>:
  endbr64
  mov    %edi,%r8d
  push   %rbp
  mov    $0xa,%edi
  cmp    $0x1,%r8d
  jle    L1
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
L1:
  call   <omp_set_num_threads@plt>
  xor    %eax,%eax
  call   <omp_get_thread_num@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <omp_get_num_threads@plt>
  mov    %ebp,%esi
  lea    D3(%rip),%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D3
type: string

tool: D3: 
Hola Mundo soy el hilo =%d de %d que somos en total.


assistant: ```c
int main(int argc, char *argv[]) {
  int nthreads;
  int thread;
  if (argc > 1) {
    nthreads = atoi(argv[1]);
  } else
    nthreads = 10;
  omp_set_num_threads(nthreads);
#pragma omp parallel
  {
    thread = omp_get_thread_num();
    nthreads = omp_get_num_threads();
    printf("\nHola Mundo soy el hilo =%d de %d que somos en total.\n", thread, nthreads);
  }
  printf("\n");
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  mov    %edi,%r8d
  push   %rbp
  mov    $0xa,%edi
  cmp    $0x1,%r8d
  jle    L1
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
L1:
  call   <omp_set_num_threads@plt>
  xor    %eax,%eax
  call   <omp_get_thread_num@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <omp_get_num_threads@plt>
  mov    %ebp,%esi
  lea    D2(%rip),%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

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

tool: D2: 
Hola Mundo soy el hilo =%d de %d que somos en total.


assistant: ```c
int main(int argc, char *argv[]) {
  int nthreads;
  int thread;
  if (argc > 1) {
    nthreads = atoi(argv[1]);
  } else
    nthreads = 10;
  omp_set_num_threads(nthreads);
#pragma omp parallel
  {
    thread = omp_get_thread_num();
    nthreads = omp_get_num_threads();
    printf("\nHola Mundo soy el hilo =%d de %d que somos en total.\n", thread, nthreads);
  }
  printf("\n");
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<konk_delay>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L16:
  mov    $0x0,%eax
  call   <barrier@plt>
  addl   $0x1,-0x4(%rbp)
L7:
  mov    -0x14(%rbp),%eax
  imul   $0x3e8,%eax,%eax
  cmp    %eax,-0x4(%rbp)
  jl     L16
  nop
  nop
  leave
  ret
```

assistant: ```c
void konk_delay(int delay) {
  int i;
  for (i = 0; i < (1000 * delay); i++) {
    barrier();
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<konk_delay>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  imul   $0x3e8,%edi,%ebp
  test   %ebp,%ebp
  jle    L2
  mov    $0x0,%ebx
L14:
  mov    $0x0,%eax
  call   <barrier@plt>
  add    $0x1,%ebx
  cmp    %ebp,%ebx
  jne    L14
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void konk_delay(int delay) {
  int i;
  for (i = 0; i < (1000 * delay); i++) {
    barrier();
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<konk_delay>:
  endbr64
  push   %rbp
  imul   $0x3e8,%edi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  test   %ebp,%ebp
  jle    L0
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L5:
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <barrier@plt>
  cmp    %ebp,%ebx
  jne    L5
L0:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void konk_delay(int delay) {
  int i;
  for (i = 0; i < (1000 * delay); i++) {
    barrier();
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<konk_delay>:
  endbr64
  push   %rbp
  imul   $0x3e8,%edi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  test   %ebp,%ebp
  jle    L16
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <barrier@plt>
  cmp    %ebp,%ebx
  jne    L4
L16:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void konk_delay(int delay) {
  int i;
  for (i = 0; i < (1000 * delay); i++) {
    barrier();
  }
}
```
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
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x10(%rbp)
  movl   $0x2,-0xc(%rbp)
  lea    -0x10(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
main() {
  int *p = (int[2]){
      1,
      2,
  };
  return p[1];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  mov    $0x2,%eax
  ret
```

assistant: ```c
main() {
  int *p = (int[2]){
      1,
      2,
  };
  return p[1];
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  mov    $0x2,%eax
  ret
```

assistant: ```c
main() {
  int *p = (int[2]){
      1,
      2,
  };
  return p[1];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  mov    $0x2,%eax
  ret
```

assistant: ```c
main() {
  int *p = (int[2]){
      1,
      2,
  };
  return p[1];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<askfp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x24(%rbp)
  jmp    L5
L18:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x2(%rax),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    -0x24(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x38(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x24(%rbp)
L5:
  cmpl   $0xc,-0x24(%rbp)
  jle    L18
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <getpass@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x38(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0xb(%rbp)
  mov    -0x38(%rbp),%rax
  movzbl 0x1(%rax),%eax
  mov    %al,-0xa(%rbp)
  lea    -0xb(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <crypt@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L2
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <smsg@plt>
  jmp    L44
L2:
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L44:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L47
  call   <__stack_chk_fail@plt>
L47:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D11
data_type: string
name: read_data
arguments: data_label: D7
data_type: string
name: read_data
arguments: data_label: D27
data_type: string

tool: D11: Introduceti Parola Supervizor (root) :

tool: D7: .p PASSWDOK


tool: D27: Password Not Accept!


assistant: ```c
void askfp(char *oldp) {
  char key[3], *newpass, *mlp;
  int i;
  for (i = 0; i < 13; i++)
    oldp[i] = oldp[i + 2];
  oldp[i] = '\0';
  newpass = getpass("Introduceti Parola Supervizor (root) :");
  key[0] = oldp[0];
  key[1] = oldp[1];
  mlp = crypt(newpass, key);
  if (strcmp(mlp, oldp) == 0)
    smsg(".p PASSWDOK\n");
  else
    printf("Password Not Accept!\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<askfp>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rdi,%rax
  lea    0xd(%rdi),%rcx
L39:
  movzbl 0x2(%rax),%edx
  mov    %dl,(%rax)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L39
  movb   $0x0,0xd(%rbx)
  lea    D29(%rip),%rdi
  call   <getpass@plt>
  mov    %rax,%rdi
  movzbl (%rbx),%eax
  mov    %al,0x5(%rsp)
  movzbl 0x1(%rbx),%eax
  mov    %al,0x6(%rsp)
  lea    0x5(%rsp),%rsi
  call   <crypt@plt>
  mov    %rax,%rdi
  mov    %rbx,%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L7
  lea    D0(%rip),%rdi
  call   <smsg@plt>
L15:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x10,%rsp
  pop    %rbx
  ret
L7:
  lea    D15(%rip),%rdi
  call   <printf@plt>
  jmp    L15
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D29
data_type: string
name: read_data
arguments: data_label: D0
data_type: string
name: read_data
arguments: data_label: D15
data_type: string

tool: D29: Introduceti Parola Supervizor (root) :

tool: D0: .p PASSWDOK


tool: D15: Password Not Accept!


assistant: ```c
void askfp(char *oldp) {
  char key[3], *newpass, *mlp;
  int i;
  for (i = 0; i < 13; i++)
    oldp[i] = oldp[i + 2];
  oldp[i] = '\0';
  newpass = getpass("Introduceti Parola Supervizor (root) :");
  key[0] = oldp[0];
  key[1] = oldp[1];
  mlp = crypt(newpass, key);
  if (strcmp(mlp, oldp) == 0)
    smsg(".p PASSWDOK\n");
  else
    printf("Password Not Accept!\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<askfp>:
  endbr64
  push   %rbp
  lea    0x2(%rdi),%rsi
  mov    %rdi,%rbp
  mov    $0xd,%edx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <memmove@plt>
  movb   $0x0,0xd(%rbp)
  lea    D7(%rip),%rdi
  call   <getpass@plt>
  lea    0x5(%rsp),%rsi
  mov    %rax,%rdi
  movzwl 0x0(%rbp),%eax
  mov    %ax,0x5(%rsp)
  call   <crypt@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L7
  lea    D14(%rip),%rdi
  call   <smsg@plt>
L5:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x10,%rsp
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  lea    D12(%rip),%rdi
  call   <printf@plt>
  jmp    L5
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D14
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D12
type_guessed: string

tool: D7: Introduceti Parola Supervizor (root) :

tool: D14: .p PASSWDOK


tool: D12: Password Not Accept!


assistant: ```c
void askfp(char *oldp) {
  char key[3], *newpass, *mlp;
  int i;
  for (i = 0; i < 13; i++)
    oldp[i] = oldp[i + 2];
  oldp[i] = '\0';
  newpass = getpass("Introduceti Parola Supervizor (root) :");
  key[0] = oldp[0];
  key[1] = oldp[1];
  mlp = crypt(newpass, key);
  if (strcmp(mlp, oldp) == 0)
    smsg(".p PASSWDOK\n");
  else
    printf("Password Not Accept!\n");
}
```