ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0xa,%edi
  call   <malloc@plt>
  lea    D16(%rip),%rdi
  mov    %rax,%rbp
  call   <printf@plt>
  lea    D12(%rip),%rdi
  call   <printf@plt>
  lea    D27(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D16
data_type: string
name: read_data
arguments: label: D12
data_type: string
name: read_data
arguments: label: D27
data_type: string

tool: D16: Probleme d'allocation memoire !


tool: D12: Probleme de multiplication !


tool: D27: Tout va bien !


assistant: ```c
int main(void) {
  char *s = NULL;
  char a, b;
  s = malloc(10 * sizeof(char));
  a = b = 2;
  printf("Probleme d'allocation memoire !\n");
  if (a + 1 * b + 2 != 12)
    printf("Probleme de multiplication !\n");
  else
    ;
  printf("Tout va bien !\n");
  free(s);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movw   $0x102,-0xa(%rbp)
  movzbl -0xa(%rbp),%eax
  cmp    $0x1,%al
  jne    L0
  movzbl -0x9(%rbp),%eax
  cmp    $0x2,%al
  jne    L0
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L29
L0:
  movzbl -0xa(%rbp),%eax
  cmp    $0x2,%al
  jne    L42
  movzbl -0x9(%rbp),%eax
  cmp    $0x1,%al
  jne    L42
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L29
L42:
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L29:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D12
data_type: string
name: disassemble_data
arguments: label: D13
data_type: string
name: disassemble_data
arguments: label: D0
data_type: string

tool: D12: big-endian

tool: D13: little-endian

tool: D0: unnkown

assistant: ```c
int main(int argc, char **argv) {
  union {
    short s;
    char c[sizeof(short)];
  } un;
  un.s = 0x0102;
  if (sizeof(short) == 2) {
    if (un.c[0] == 1 && un.c[1] == 2)
      printf("big-endian\n");
    else if (un.c[0] == 2 && un.c[1] == 1)
      printf("little-endian\n");
    else
      printf("unnkown\n");
  } else {
    printf("sizeof(short) = %lu\n", sizeof(short));
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D8: little-endian

assistant: ```c
int main(int argc, char **argv) {
  union {
    short s;
    char c[sizeof(short)];
  } un;
  un.s = 0x0102;
  if (sizeof(short) == 2) {
    if (un.c[0] == 1 && un.c[1] == 2)
      printf("big-endian\n");
    else if (un.c[0] == 2 && un.c[1] == 1)
      printf("little-endian\n");
    else
      printf("unnkown\n");
  } else {
    printf("sizeof(short) = %lu\n", sizeof(short));
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D6
datatype: string

tool: D6: little-endian

assistant: ```c
int main(int argc, char **argv) {
  union {
    short s;
    char c[sizeof(short)];
  } un;
  un.s = 0x0102;
  if (sizeof(short) == 2) {
    if (un.c[0] == 1 && un.c[1] == 2)
      printf("big-endian\n");
    else if (un.c[0] == 2 && un.c[1] == 1)
      printf("little-endian\n");
    else
      printf("unnkown\n");
  } else {
    printf("sizeof(short) = %lu\n", sizeof(short));
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D1
type_guessed: string

tool: D1: little-endian

assistant: ```c
int main(int argc, char **argv) {
  union {
    short s;
    char c[sizeof(short)];
  } un;
  un.s = 0x0102;
  if (sizeof(short) == 2) {
    if (un.c[0] == 1 && un.c[1] == 2)
      printf("big-endian\n");
    else if (un.c[0] == 2 && un.c[1] == 1)
      printf("little-endian\n");
    else
      printf("unnkown\n");
  } else {
    printf("sizeof(short) = %lu\n", sizeof(short));
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r14
  lea    D42(%rip),%rdi
  push   %r13
  push   %r12
  lea    D26(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D43(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  xchg   %ax,%ax
L72:
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  cmp    $0x1,%esi
  jbe    L83
  test   $0x1,%sil
  jne    L49
  shr    %esi
  mov    %esi,0x4(%rsp)
  jmp    L72
  nopl   0x0(%rax)
L49:
  lea    0x1(%rsi,%rsi,2),%esi
  mov    %esi,0x4(%rsp)
  jmp    L72
  nopw   0x0(%rax,%rax,1)
L83:
  lea    D31(%rip),%rdi
  xor    %eax,%eax
  xor    %r13d,%r13d
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  lea    D8(%rip),%r14
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  je     L3
  nopw   0x0(%rax,%rax,1)
L19:
  xor    %ebx,%ebx
  cmp    %r13d,%eax
  je     L57
  nopw   0x0(%rax,%rax,1)
L11:
  mov    $0x20,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  sub    %r13d,%eax
  cmp    %ebx,%eax
  ja     L11
L57:
  xor    %r12d,%r12d
  nopl   0x0(%rax,%rax,1)
L67:
  mov    %r13d,%edi
  call   <factorial@plt>
  mov    %r13d,%edi
  sub    %r12d,%edi
  mov    %eax,%ebx
  call   <factorial@plt>
  mov    %r12d,%edi
  add    $0x1,%r12d
  mov    %eax,%ebp
  call   <factorial@plt>
  mov    %ebp,%ecx
  mov    %r14,%rdi
  imul   %eax,%ecx
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %r13d,%r12d
  jle    L67
  mov    $0xa,%edi
  add    $0x1,%r13d
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  cmp    %r13d,%eax
  ja     L19
L3:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L74
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L74:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D42
data_type: string
name: retrieve_data
arguments: label: D26
data_type: string
name: retrieve_data
arguments: label: D43
data_type: string
name: retrieve_data
arguments: label: D31
data_type: string
name: retrieve_data
arguments: label: D8
data_type: string

tool: D42: Conjetura Collatz: 

tool: D26: %u

tool: D43: %u


tool: D31: Pascal Triangle: 

tool: D8: %d 

assistant: ```c
int main(int argc, char const *argv[]) {
  unsigned int start;
  printf("Conjetura Collatz: ");
  scanf("%u", &start);
  printf("%u\n", start);
  while (start != 1 && start) {
    if (!(start % 2)) {
      start /= 2;
      printf("%u\n", start);
    } else {
      start = start * 3 + 1;
      printf("%u\n", start);
    }
  }
  printf("Pascal Triangle: ");
  scanf("%u", &start);
  for (int i = 0; i < start; i++) {
    for (int j = 0; j < start - i; j++) {
      printf(" ");
    }
    for (int j = 1; j <= i + 1; j++) {
      printf("%d ", factorial(i) / (factorial(i - j + 1) * factorial(j - 1)));
    }
    printf("\n");
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<cli_val_address_parse_binary>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    $0x0,%eax
  call   <cli_val_address_sizeof@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jbe    L7
  mov    $0x0,%eax
  jmp    L5
L7:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <cli_val_address_data@plt>
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %ecx,%edi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rax
L5:
  leave
  ret
```

assistant: ```c
int cli_val_address_parse_binary(struct cli_val_address *v, const void *b, size_t length) {
  size_t size = cli_val_address_sizeof();
  if (size > length) {
    return 0;
  }
  memcpy(cli_val_address_data(v), b, size);
  return size;
}
```
user: Translate the below assembly code into C programming language:

```asm
<cli_val_address_parse_binary>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    $0x0,%eax
  call   <cli_val_address_sizeof@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  cmp    %rbp,%rbx
  jbe    L6
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  mov    %r13,%rdi
  call   <cli_val_address_data@plt>
  mov    %eax,%edi
  mov    %rbx,%rdx
  mov    %r12,%rsi
  call   <memcpy@plt>
  mov    %ebx,%eax
  jmp    L2
```

assistant: ```c
int cli_val_address_parse_binary(struct cli_val_address *v, const void *b, size_t length) {
  size_t size = cli_val_address_sizeof();
  if (size > length) {
    return 0;
  }
  memcpy(cli_val_address_data(v), b, size);
  return size;
}
```
user: What is the c source code of the assembly code below:

```asm
<cli_val_address_parse_binary>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %rdi,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <cli_val_address_sizeof@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  cmp    %rbp,%rbx
  ja     L4
  mov    %r13,%rdi
  call   <cli_val_address_data@plt>
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %eax,%edi
  call   <memcpy@plt>
  mov    %ebx,%eax
L4:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int cli_val_address_parse_binary(struct cli_val_address *v, const void *b, size_t length) {
  size_t size = cli_val_address_sizeof();
  if (size > length) {
    return 0;
  }
  memcpy(cli_val_address_data(v), b, size);
  return size;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<cli_val_address_parse_binary>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %rdi,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <cli_val_address_sizeof@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  cmp    %rbp,%rbx
  ja     L2
  mov    %r13,%rdi
  call   <cli_val_address_data@plt>
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %eax,%edi
  call   <memcpy@plt>
  mov    %ebx,%eax
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int cli_val_address_parse_binary(struct cli_val_address *v, const void *b, size_t length) {
  size_t size = cli_val_address_sizeof();
  if (size > length) {
    return 0;
  }
  memcpy(cli_val_address_data(v), b, size);
  return size;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x1,-0xc(%rbp)
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x10(%rbp)
  jmp    L35
L33:
  cmpl   $0x0,-0x10(%rbp)
  jne    L43
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x14(%rbp)
  jmp    L26
L43:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x14(%rbp)
L26:
  mov    -0x14(%rbp),%eax
  imul   -0xc(%rbp),%eax
  add    %eax,-0x18(%rbp)
  shll   -0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L35:
  cmpl   $0x4,-0x10(%rbp)
  jle    L33
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
type: string
name: extract_symbolic_data
arguments: data_tag: D23
type: string
name: extract_symbolic_data
arguments: data_tag: D20
type: string

tool: D0: please Enter binary number at most 5 digits:

tool: D23: %d

tool: D20: %d


assistant: ```c
int main(void) {
  int binary_number = 0;
  int decimal_number = 0;
  int digit = 0;
  int counter = 0;
  int weight = 1;
  printf("please Enter binary number at most 5 digits:");
  scanf("%d", &binary_number);
  for (counter = 0; counter < 5; counter++) {
    if (counter == 0) {
      digit = binary_number % 10;
    } else {
      binary_number = binary_number / 10;
      digit = binary_number % 10;
    }
    decimal_number = decimal_number + digit * weight;
    weight = weight * 2;
  }
  printf("%d\n", decimal_number);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    D7(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D29(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  mov    $0x1,%esi
  mov    $0x1,%edi
  mov    $0x0,%r8d
  jmp    L1
L10:
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ecx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  mov    %ecx,%eax
  sub    %edx,%eax
  imul   %edi,%eax
  add    %eax,%r8d
  add    %edi,%edi
L21:
  add    $0x1,%esi
L1:
  cmp    $0x1,%esi
  je     L10
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%r9d
  sar    $0x1f,%r9d
  sub    %r9d,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  imul   %edi,%eax
  add    %eax,%r8d
  cmp    $0x4,%esi
  jg     L0
  add    %edi,%edi
  jmp    L21
L0:
  mov    %ecx,0x4(%rsp)
  mov    %r8d,%esi
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D7
type_infered: string
name: parse_data
arguments: label: D29
type_infered: string
name: parse_data
arguments: label: D28
type_infered: string

tool: D7: please Enter binary number at most 5 digits:

tool: D29: %d

tool: D28: %d


assistant: ```c
int main(void) {
  int binary_number = 0;
  int decimal_number = 0;
  int digit = 0;
  int counter = 0;
  int weight = 1;
  printf("please Enter binary number at most 5 digits:");
  scanf("%d", &binary_number);
  for (counter = 0; counter < 5; counter++) {
    if (counter == 0) {
      digit = binary_number % 10;
    } else {
      binary_number = binary_number / 10;
      digit = binary_number % 10;
    }
    decimal_number = decimal_number + digit * weight;
    weight = weight * 2;
  }
  printf("%d\n", decimal_number);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D26(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D17(%rip),%rdi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rax
  xor    %esi,%esi
  xor    %r9d,%r9d
  mov    $0x1,%edi
  add    $0x1,%esi
  mov    %rax,%rdx
  lea    (%rdi,%rdi,1),%r10d
  cmp    $0x1,%esi
  je     L37
L7:
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  mov    %eax,%edx
  cltq
  imul   $0x66666667,%rax,%rcx
  mov    %edx,%r8d
  sar    $0x1f,%r8d
  sar    $0x22,%rcx
  sub    %r8d,%ecx
  lea    (%rcx,%rcx,4),%r8d
  mov    %edx,%ecx
  add    %r8d,%r8d
  sub    %r8d,%ecx
  imul   %edi,%ecx
  add    %ecx,%r9d
  cmp    $0x5,%esi
  je     L28
L17:
  mov    %r10d,%edi
  add    $0x1,%esi
  lea    (%rdi,%rdi,1),%r10d
  cmp    $0x1,%esi
  jne    L7
L37:
  imul   $0x66666667,%rax,%rcx
  mov    %edx,%r8d
  sar    $0x1f,%r8d
  sar    $0x22,%rcx
  sub    %r8d,%ecx
  lea    (%rcx,%rcx,4),%r8d
  mov    %edx,%ecx
  add    %r8d,%r8d
  sub    %r8d,%ecx
  imul   %edi,%ecx
  add    %ecx,%r9d
  jmp    L17
  nopl   0x0(%rax,%rax,1)
L28:
  xor    %eax,%eax
  mov    %r9d,%esi
  lea    D21(%rip),%rdi
  mov    %edx,0x4(%rsp)
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L36:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D26
type: string
name: extract_symbolic_data
arguments: data_label: D17
type: string
name: extract_symbolic_data
arguments: data_label: D21
type: string

tool: D26: please Enter binary number at most 5 digits:

tool: D17: %d

tool: D21: %d


assistant: ```c
int main(void) {
  int binary_number = 0;
  int decimal_number = 0;
  int digit = 0;
  int counter = 0;
  int weight = 1;
  printf("please Enter binary number at most 5 digits:");
  scanf("%d", &binary_number);
  for (counter = 0; counter < 5; counter++) {
    if (counter == 0) {
      digit = binary_number % 10;
    } else {
      binary_number = binary_number / 10;
      digit = binary_number % 10;
    }
    decimal_number = decimal_number + digit * weight;
    weight = weight * 2;
  }
  printf("%d\n", decimal_number);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D7(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D17(%rip),%rdi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rax
  mov    %rax,%rcx
  imul   $0x66666667,%rax,%rax
  mov    %ecx,%edx
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  movslq %eax,%rdx
  mov    %eax,%esi
  mov    %eax,%edi
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%esi
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%ecx
  sar    $0x22,%rdx
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%esi
  movslq %edx,%rax
  add    %esi,%esi
  imul   $0x66666667,%rax,%rax
  sub    %esi,%edi
  lea    (%rcx,%rdi,2),%esi
  mov    %edx,%ecx
  lea    D19(%rip),%rdi
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  lea    (%rsi,%rdx,4),%ecx
  movslq %eax,%rsi
  cltd
  imul   $0x66666667,%rsi,%rsi
  sar    $0x22,%rsi
  sub    %edx,%esi
  lea    (%rsi,%rsi,4),%edx
  mov    %esi,0x4(%rsp)
  add    %edx,%edx
  sub    %edx,%eax
  lea    (%rcx,%rax,8),%edx
  movslq %esi,%rax
  mov    %esi,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  xor    %eax,%eax
  shl    $0x4,%esi
  add    %edx,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D7
type_infered: string
name: unpack_data
arguments: data_label: D17
type_infered: string
name: unpack_data
arguments: data_label: D19
type_infered: string

tool: D7: please Enter binary number at most 5 digits:

tool: D17: %d

tool: D19: %d


assistant: ```c
int main(void) {
  int binary_number = 0;
  int decimal_number = 0;
  int digit = 0;
  int counter = 0;
  int weight = 1;
  printf("please Enter binary number at most 5 digits:");
  scanf("%d", &binary_number);
  for (counter = 0; counter < 5; counter++) {
    if (counter == 0) {
      digit = binary_number % 10;
    } else {
      binary_number = binary_number / 10;
      digit = binary_number % 10;
    }
    decimal_number = decimal_number + digit * weight;
    weight = weight * 2;
  }
  printf("%d\n", decimal_number);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<encode_base64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  movq   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  add    %rax,%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L30
  mov    $0x0,%eax
  jmp    L3
L30:
  mov    -0x20(%rbp),%rax
  lea    (%rax,%rax,1),%rcx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <b64_ntop@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L1
  mov    -0x20(%rbp),%rax
  lea    (%rax,%rax,1),%rdx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <_explicit_bzero@plt>
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x28(%rbp),%rax
  movq   $0x0,(%rax)
  mov    $0x0,%eax
  jmp    L3
L1:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  mov    $0x1,%eax
L3:
  leave
  ret
```

assistant: ```c
int encode_base64(const char *src, size_t src_len, char **dst, size_t *dst_len) {
  int sz;
  *dst = NULL;
  if ((*dst = (char *)malloc(src_len * 2 + 1)) == NULL)
    return 0;
  if ((sz = b64_ntop((unsigned char *)src, src_len, *dst, src_len * 2)) <= 0) {
    _explicit_bzero(*dst, src_len * 2);
    free(*dst);
    *dst = NULL;
    return 0;
  }
  (*dst)[sz] = 0;
  *dst_len = (size_t)sz;
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<encode_base64>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    %rcx,%r13
  movq   $0x0,(%rdx)
  lea    (%rsi,%rsi,1),%r14
  lea    0x1(%r14),%rdi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    %rax,(%rbx)
  mov    $0x0,%eax
  test   %rdx,%rdx
  je     L13
  mov    %r14,%rcx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <b64_ntop@plt>
  test   %eax,%eax
  jle    L1
  cltq
  mov    (%rbx),%rdx
  movb   $0x0,(%rdx,%rax,1)
  mov    %rax,0x0(%r13)
  mov    $0x1,%eax
L13:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L1:
  mov    %r14,%rsi
  mov    (%rbx),%rdi
  call   <_explicit_bzero@plt>
  mov    (%rbx),%rdi
  call   <free@plt>
  movq   $0x0,(%rbx)
  mov    $0x0,%eax
  jmp    L13
```

assistant: ```c
int encode_base64(const char *src, size_t src_len, char **dst, size_t *dst_len) {
  int sz;
  *dst = NULL;
  if ((*dst = (char *)malloc(src_len * 2 + 1)) == NULL)
    return 0;
  if ((sz = b64_ntop((unsigned char *)src, src_len, *dst, src_len * 2)) <= 0) {
    _explicit_bzero(*dst, src_len * 2);
    free(*dst);
    *dst = NULL;
    return 0;
  }
  (*dst)[sz] = 0;
  *dst_len = (size_t)sz;
  return 1;
}
```