ylfeng/ReF-Decompile-dataset huggingface dataset:
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
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <insert@plt>
  mov    $0x3,%edi
  call   <insert@plt>
  xor    %eax,%eax
  call   <print@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  insert(2);
  insert(3);
  print();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getop>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  nop
L99:
  mov    $0x0,%eax
  call   <getch@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L99
  cmpl   $0x9,-0x4(%rbp)
  je     L99
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L6
  cmpl   $0x2e,-0x4(%rbp)
  je     L6
  mov    -0x4(%rbp),%eax
  jmp    L93
L6:
  movl   $0x0,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L58
  nop
L24:
  mov    $0x0,%eax
  call   <getch@plt>
  mov    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L24
L58:
  cmpl   $0x2e,-0x4(%rbp)
  jne    L43
  nop
L17:
  mov    $0x0,%eax
  call   <getch@plt>
  mov    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L17
L43:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x30,(%rax)
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  je     L57
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <ungetch@plt>
L57:
  mov    $0x30,%eax
L93:
  leave
  ret
```

assistant: ```c
int getop(char s[]) {
  int i, c;
  while ((s[0] = c = getch()) == ' ' || c == '\t')
    ;
  s[1] = '\0';
  if (!isdigit(c) && c != '.')
    return c;
  i = 0;
  if (isdigit(c))
    while (isdigit(s[++i] = c = getch()))
      ;
  if (c == '.')
    while (isdigit(s[++i] = c = getch()))
      ;
  s[i] = '/0';
  if (c != EOF)
    ungetch(c);
  return '0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<getop>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
L34:
  mov    $0x0,%eax
  call   <getch@plt>
  mov    %eax,%ebx
  mov    %al,0x0(%rbp)
  cmp    $0x9,%eax
  je     L34
  cmp    $0x20,%al
  je     L34
  movb   $0x0,0x1(%rbp)
  movsbl %al,%r12d
  mov    %r12d,%edi
  call   <isdigit@plt>
  cmp    $0x2e,%ebx
  je     L39
  test   %rax,%rax
  je     L66
L39:
  mov    %r12d,%edi
  call   <isdigit@plt>
  mov    $0x0,%r13d
  test   %rax,%rax
  je     L25
  mov    $0x1,%r12d
L22:
  mov    $0x0,%eax
  call   <getch@plt>
  mov    %eax,%ebx
  mov    %r12d,%r13d
  mov    %al,0x0(%rbp,%r12,1)
  movsbl %al,%edi
  call   <isdigit@plt>
  add    $0x1,%r12
  test   %rax,%rax
  jne    L22
L25:
  cmp    $0x2e,%ebx
  je     L42
L15:
  movslq %r13d,%r13
  movb   $0x30,0x0(%rbp,%r13,1)
  mov    D8(%rip),%rax
  cmp    %ebx,(%rax)
  jne    L4
  mov    $0x30,%ebx
L66:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L42:
  lea    0x1(%r13),%r12d
  movslq %r12d,%r12
L1:
  mov    $0x0,%eax
  call   <getch@plt>
  mov    %eax,%ebx
  mov    %r12d,%r13d
  mov    %al,0x0(%rbp,%r12,1)
  movsbl %al,%edi
  call   <isdigit@plt>
  add    $0x1,%r12
  test   %rax,%rax
  jne    L1
  jmp    L15
L4:
  mov    %ebx,%edi
  call   <ungetch@plt>
  mov    $0x30,%ebx
  jmp    L66
```

assistant: ```c
int getop(char s[]) {
  int i, c;
  while ((s[0] = c = getch()) == ' ' || c == '\t')
    ;
  s[1] = '\0';
  if (!isdigit(c) && c != '.')
    return c;
  i = 0;
  if (isdigit(c))
    while (isdigit(s[++i] = c = getch()))
      ;
  if (c == '.')
    while (isdigit(s[++i] = c = getch()))
      ;
  s[i] = '/0';
  if (c != EOF)
    ungetch(c);
  return '0';
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<getop>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L82:
  xor    %eax,%eax
  call   <getch@plt>
  mov    %al,(%rbx)
  mov    %eax,%r12d
  cmp    $0x9,%eax
  je     L82
  cmp    $0x20,%al
  je     L82
  movb   $0x0,0x1(%rbx)
  movsbl %al,%ebp
  mov    %ebp,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L86
  cmp    $0x2e,%r12d
  je     L87
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L87:
  mov    %ebp,%edi
  xor    %r13d,%r13d
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L17
L2:
  lea    0x1(%rbx,%r13,1),%rbx
  nopl   0x0(%rax)
L48:
  xor    %eax,%eax
  mov    %rbx,%rbp
  add    $0x1,%rbx
  call   <getch@plt>
  mov    %al,-0x1(%rbx)
  movsbl %al,%edi
  mov    %eax,%r12d
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L48
  mov    D2(%rip),%rax
  movb   $0x30,0x0(%rbp)
  cmp    %r12d,(%rax)
  jne    L45
L52:
  add    $0x8,%rsp
  mov    $0x30,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L86:
  mov    %ebp,%edi
  xor    %r13d,%r13d
  call   <isdigit@plt>
  test   %rax,%rax
  je     L66
L17:
  mov    $0x1,%ebp
  nopl   0x0(%rax)
L75:
  xor    %eax,%eax
  movslq %ebp,%r13
  call   <getch@plt>
  mov    %al,(%rbx,%rbp,1)
  movsbl %al,%edi
  mov    %eax,%r12d
  add    $0x1,%rbp
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L75
L66:
  movslq %r13d,%rbp
  add    %rbx,%rbp
  cmp    $0x2e,%r12d
  je     L2
  mov    D2(%rip),%rax
  movb   $0x30,0x0(%rbp)
  cmp    %r12d,(%rax)
  je     L52
L45:
  mov    %r12d,%edi
  mov    $0x30,%r12d
  call   <ungetch@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int getop(char s[]) {
  int i, c;
  while ((s[0] = c = getch()) == ' ' || c == '\t')
    ;
  s[1] = '\0';
  if (!isdigit(c) && c != '.')
    return c;
  i = 0;
  if (isdigit(c))
    while (isdigit(s[++i] = c = getch()))
      ;
  if (c == '.')
    while (isdigit(s[++i] = c = getch()))
      ;
  s[i] = '/0';
  if (c != EOF)
    ungetch(c);
  return '0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<getop>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   (%rax)
L28:
  xor    %eax,%eax
  call   <getch@plt>
  mov    %al,(%rbx)
  mov    %eax,%r12d
  cmp    $0x9,%eax
  je     L28
  cmp    $0x20,%al
  je     L28
  movb   $0x0,0x1(%rbx)
  movsbl %al,%ebp
  mov    %ebp,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L2
  cmp    $0x2e,%r12d
  je     L128
L76:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L128:
  mov    %ebp,%edi
  xor    %r14d,%r14d
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L6
L46:
  movslq %r14d,%r14
  lea    0x1(%rbx,%r14,1),%rbx
  nopw   0x0(%rax,%rax,1)
L98:
  xor    %eax,%eax
  mov    %rbx,%r13
  add    $0x1,%rbx
  call   <getch@plt>
  mov    %al,-0x1(%rbx)
  movsbl %al,%edi
  mov    %eax,%r12d
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L98
  jmp    L88
  nopl   0x0(%rax,%rax,1)
L2:
  mov    %ebp,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L73
L6:
  mov    %ebx,%r15d
  lea    0x1(%rbx),%rbp
  neg    %r15d
  xchg   %ax,%ax
L125:
  xor    %eax,%eax
  lea    (%r15,%rbp,1),%r14d
  mov    %rbp,%r13
  add    $0x1,%rbp
  call   <getch@plt>
  mov    %al,-0x1(%rbp)
  movsbl %al,%edi
  mov    %eax,%r12d
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L125
L0:
  cmp    $0x2e,%r12d
  je     L46
L88:
  mov    D1(%rip),%rax
  movb   $0x30,0x0(%r13)
  cmp    %r12d,(%rax)
  je     L110
  mov    %r12d,%edi
  mov    $0x30,%r12d
  call   <ungetch@plt>
  jmp    L76
  nop
L110:
  mov    $0x30,%r12d
  jmp    L76
  nopl   0x0(%rax,%rax,1)
L73:
  mov    %rbx,%r13
  xor    %r14d,%r14d
  jmp    L0
```

assistant: ```c
int getop(char s[]) {
  int i, c;
  while ((s[0] = c = getch()) == ' ' || c == '\t')
    ;
  s[1] = '\0';
  if (!isdigit(c) && c != '.')
    return c;
  i = 0;
  if (isdigit(c))
    while (isdigit(s[++i] = c = getch()))
      ;
  if (c == '.')
    while (isdigit(s[++i] = c = getch()))
      ;
  s[i] = '/0';
  if (c != EOF)
    ungetch(c);
  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
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xa0,%rsp
  mov    %edi,-0x1094(%rbp)
  mov    %rsi,-0x10a0(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x1010(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x1088(%rbp)
  jmp    L56
L1:
  mov    -0x1088(%rbp),%eax
  cltq
  movl   $0x0,-0x1080(%rbp,%rax,4)
  addl   $0x1,-0x1088(%rbp)
L56:
  cmpl   $0x19,-0x1088(%rbp)
  jle    L1
  lea    -0x1010(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x1084(%rbp)
  movl   $0x0,-0x1088(%rbp)
  jmp    L63
L65:
  mov    -0x1088(%rbp),%eax
  cltq
  movzbl -0x1010(%rbp,%rax,1),%eax
  mov    %al,-0x1089(%rbp)
  movsbl -0x1089(%rbp),%eax
  mov    %eax,%edi
  call   <isalpha@plt>
  test   %rax,%rax
  je     L60
  movsbl -0x1089(%rbp),%eax
  mov    %eax,%edi
  call   <toupper@plt>
  movsbl %al,%eax
  sub    $0x41,%eax
  movslq %eax,%rdx
  mov    -0x1080(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x1080(%rbp,%rax,4)
L60:
  addl   $0x1,-0x1088(%rbp)
L63:
  mov    -0x1088(%rbp),%eax
  cmp    -0x1084(%rbp),%eax
  jl     L65
  movl   $0x0,-0x1088(%rbp)
  jmp    L81
L76:
  mov    -0x1088(%rbp),%eax
  cltq
  mov    -0x1080(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L24
  mov    -0x1088(%rbp),%eax
  cltq
  mov    -0x1080(%rbp,%rax,4),%eax
  mov    -0x1088(%rbp),%edx
  lea    0x41(%rdx),%ecx
  mov    %eax,%edx
  mov    %ecx,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L24:
  addl   $0x1,-0x1088(%rbp)
L81:
  cmpl   $0x19,-0x1088(%rbp)
  jle    L76
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L61
  call   <__stack_chk_fail@plt>
L61:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D15
data_type: string
name: extract_symbolic_data
arguments: data_tag: D19
data_type: string

tool: D15: %[^
]

tool: D19: %c: %d


assistant: ```c
int main(int argc, const char *argv[]) {
  char input_chars[4096 + 1];
  scanf("%[^\n]", input_chars);
  int idx;
  int counter[26];
  for (idx = 0; idx < 26; idx++) {
    counter[idx] = 0;
  }
  int input_len = strlen(input_chars);
  for (idx = 0; idx < input_len; idx++) {
    char chr = input_chars[idx];
    if (isalpha(chr)) {
      counter[toupper(chr) - 'A']++;
    }
  }
  for (idx = 0; idx < 26; idx++) {
    if (counter[idx] != 0) {
      printf("%c: %d\n", idx + 'A', counter[idx]);
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1078(%rsp)
  xor    %eax,%eax
  lea    0x70(%rsp),%rsi
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  mov    %rsp,%rbx
  lea    0x68(%rsp),%r12
  mov    %rbx,%rax
L46:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %r12,%rax
  jne    L46
  lea    0x70(%rsp),%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L66
  lea    0x70(%rsp),%rbp
  lea    -0x1(%rax),%eax
  lea    0x71(%rsp,%rax,1),%r13
  jmp    L98
L75:
  add    $0x1,%rbp
  cmp    %r13,%rbp
  je     L66
L98:
  movsbl 0x0(%rbp),%r14d
  mov    %r14d,%edi
  call   <isalpha@plt>
  test   %rax,%rax
  je     L75
  mov    %r14d,%edi
  call   <toupper@plt>
  movsbl %al,%eax
  sub    $0x41,%eax
  cltq
  addl   $0x1,(%rsp,%rax,4)
  jmp    L75
L66:
  mov    $0x41,%ebp
  lea    D2(%rip),%r13
  jmp    L62
L3:
  add    $0x4,%rbx
  add    $0x1,%ebp
  cmp    %r12,%rbx
  je     L48
L62:
  mov    (%rbx),%edx
  test   %edx,%edx
  je     L3
  mov    %ebp,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  jmp    L3
L48:
  mov    0x1078(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L100
  mov    $0x0,%eax
  add    $0x1080,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L100:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
type_guessed: string
name: retrieve_data
arguments: data_tag: D2
type_guessed: string

tool: D7: %[^
]

tool: D2: %c: %d


assistant: ```c
int main(int argc, const char *argv[]) {
  char input_chars[4096 + 1];
  scanf("%[^\n]", input_chars);
  int idx;
  int counter[26];
  for (idx = 0; idx < 26; idx++) {
    counter[idx] = 0;
  }
  int input_len = strlen(input_chars);
  for (idx = 0; idx < input_len; idx++) {
    char chr = input_chars[idx];
    if (isalpha(chr)) {
      counter[toupper(chr) - 'A']++;
    }
  }
  for (idx = 0; idx < 26; idx++) {
    if (counter[idx] != 0) {
      printf("%c: %d\n", idx + 'A', counter[idx]);
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x88,%rsp
  lea    D16(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x1078(%rsp)
  xor    %eax,%eax
  lea    0x70(%rsp),%rbp
  mov    %rsp,%rbx
  mov    %rbp,%rsi
  call   <scanf@plt>
  xor    %eax,%eax
  mov    $0xd,%ecx
  mov    %rbx,%rdi
  rep stos %rax,%es:(%rdi)
  mov    %rbp,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L3
  sub    $0x1,%eax
  lea    0x71(%rsp,%rax,1),%r12
  jmp    L70
  nopw   0x0(%rax,%rax,1)
L60:
  add    $0x1,%rbp
  cmp    %r12,%rbp
  je     L3
L70:
  movsbl 0x0(%rbp),%r13d
  mov    %r13d,%edi
  call   <isalpha@plt>
  test   %rax,%rax
  je     L60
  mov    %r13d,%edi
  add    $0x1,%rbp
  call   <toupper@plt>
  movsbl %al,%eax
  sub    $0x41,%eax
  cltq
  addl   $0x1,(%rsp,%rax,4)
  cmp    %r12,%rbp
  jne    L70
L3:
  lea    0x68(%rsp),%r12
  mov    $0x41,%ebp
  lea    D7(%rip),%r13
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L23:
  add    $0x4,%rbx
  add    $0x1,%ebp
  cmp    %rbx,%r12
  je     L2
L33:
  mov    (%rbx),%edx
  test   %edx,%edx
  je     L23
  mov    %ebp,%esi
  mov    %r13,%rdi
  add    $0x4,%rbx
  add    $0x1,%ebp
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L33
L2:
  mov    0x1078(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  add    $0x1088,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L64:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D16
type_infered: string
name: unpack_data
arguments: data_tag: D7
type_infered: string

tool: D16: %[^
]

tool: D7: %c: %d


assistant: ```c
int main(int argc, const char *argv[]) {
  char input_chars[4096 + 1];
  scanf("%[^\n]", input_chars);
  int idx;
  int counter[26];
  for (idx = 0; idx < 26; idx++) {
    counter[idx] = 0;
  }
  int input_len = strlen(input_chars);
  for (idx = 0; idx < input_len; idx++) {
    char chr = input_chars[idx];
    if (isalpha(chr)) {
      counter[toupper(chr) - 'A']++;
    }
  }
  for (idx = 0; idx < 26; idx++) {
    if (counter[idx] != 0) {
      printf("%c: %d\n", idx + 'A', counter[idx]);
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x88,%rsp
  lea    D14(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x1078(%rsp)
  xor    %eax,%eax
  lea    0x70(%rsp),%rbp
  mov    %rsp,%rbx
  mov    %rbp,%rsi
  call   <scanf@plt>
  xor    %eax,%eax
  mov    $0xd,%ecx
  mov    %rbx,%rdi
  rep stos %rax,%es:(%rdi)
  mov    %rbp,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L50
  sub    $0x1,%eax
  lea    0x71(%rsp,%rax,1),%r12
  jmp    L31
  nopw   0x0(%rax,%rax,1)
L93:
  add    $0x1,%rbp
  cmp    %r12,%rbp
  je     L50
L31:
  movsbl 0x0(%rbp),%r13d
  mov    %r13d,%edi
  call   <isalpha@plt>
  test   %rax,%rax
  je     L93
  mov    %r13d,%edi
  add    $0x1,%rbp
  call   <toupper@plt>
  movsbl %al,%eax
  sub    $0x41,%eax
  cltq
  addl   $0x1,(%rsp,%rax,4)
  cmp    %r12,%rbp
  jne    L31
L50:
  lea    0x68(%rsp),%r12
  mov    $0x41,%ebp
  lea    D10(%rip),%r13
  jmp    L73
  nopl   0x0(%rax,%rax,1)
L63:
  add    $0x4,%rbx
  add    $0x1,%ebp
  cmp    %rbx,%r12
  je     L19
L73:
  mov    (%rbx),%edx
  test   %edx,%edx
  je     L63
  mov    %ebp,%esi
  mov    %r13,%rdi
  add    $0x4,%rbx
  add    $0x1,%ebp
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L73
L19:
  mov    0x1078(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x1088,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D14
datatype: string
name: retrieve_data
arguments: data_label: D10
datatype: string

tool: D14: %[^
]

tool: D10: %c: %d


assistant: ```c
int main(int argc, const char *argv[]) {
  char input_chars[4096 + 1];
  scanf("%[^\n]", input_chars);
  int idx;
  int counter[26];
  for (idx = 0; idx < 26; idx++) {
    counter[idx] = 0;
  }
  int input_len = strlen(input_chars);
  for (idx = 0; idx < input_len; idx++) {
    char chr = input_chars[idx];
    if (isalpha(chr)) {
      counter[toupper(chr) - 'A']++;
    }
  }
  for (idx = 0; idx < 26; idx++) {
    if (counter[idx] != 0) {
      printf("%c: %d\n", idx + 'A', counter[idx]);
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rm_whitespace>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L18
L2:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %eax,%eax
  jne    L18
  addq   $0x1,-0x10(%rbp)
L18:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L2
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void rm_whitespace(char *source) {
  char *i = source;
  char *j = source;
  while (*j != 0) {
    *i = *j++;
    if (!isspace(*i))
      i++;
  }
  *i = 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rm_whitespace>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L2
  mov    %rbx,%rbp
L9:
  add    $0x1,%rbp
  mov    %dil,(%rbx)
  movsbl %dil,%edi
  call   <isspace@plt>
  cmp    $0x1,%eax
  adc    $0x0,%rbx
  movzbl 0x0(%rbp),%edi
  test   %dil,%dil
  jne    L9
L2:
  movb   $0x0,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void rm_whitespace(char *source) {
  char *i = source;
  char *j = source;
  while (*j != 0) {
    *i = *j++;
    if (!isspace(*i))
      i++;
  }
  *i = 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rm_whitespace>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rbx,%rbp
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L9
  nopl   0x0(%rax,%rax,1)
L0:
  mov    %dil,0x0(%rbp)
  add    $0x1,%rbx
  call   <isspace@plt>
  movsbl (%rbx),%edi
  cmp    $0x1,%eax
  adc    $0x0,%rbp
  test   %dil,%dil
  jne    L0
L9:
  movb   $0x0,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void rm_whitespace(char *source) {
  char *i = source;
  char *j = source;
  while (*j != 0) {
    *i = *j++;
    if (!isspace(*i))
      i++;
  }
  *i = 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<rm_whitespace>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rbx,%rbp
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L13
  nopl   0x0(%rax,%rax,1)
L9:
  mov    %dil,0x0(%rbp)
  add    $0x1,%rbx
  call   <isspace@plt>
  movsbl (%rbx),%edi
  cmp    $0x1,%eax
  adc    $0x0,%rbp
  test   %dil,%dil
  jne    L9
L13:
  movb   $0x0,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void rm_whitespace(char *source) {
  char *i = source;
  char *j = source;
  while (*j != 0) {
    *i = *j++;
    if (!isspace(*i))
      i++;
  }
  *i = 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parrt_malloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D8
type_infered: string

tool: D8: malloc %zu -> %p


assistant: ```c
void *parrt_malloc(size_t n) {
  void *p = malloc(n);
  printf("malloc %zu -> %p\n", n, p);
  return p;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<parrt_malloc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rax,%rdx
  mov    %rbp,%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
type_infered: string

tool: D6: malloc %zu -> %p


assistant: ```c
void *parrt_malloc(size_t n) {
  void *p = malloc(n);
  printf("malloc %zu -> %p\n", n, p);
  return p;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<parrt_malloc>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rbp,%rsi
  lea    D3(%rip),%rdi
  mov    %rax,%rdx
  mov    %rax,%r12
  call   <printf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
data_type: string

tool: D3: malloc %zu -> %p


assistant: ```c
void *parrt_malloc(size_t n) {
  void *p = malloc(n);
  printf("malloc %zu -> %p\n", n, p);
  return p;
}
```
user: What is the c source code of the assembly code below?

```asm
<parrt_malloc>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rbp,%rsi
  lea    D0(%rip),%rdi
  mov    %rax,%rdx
  mov    %rax,%r12
  call   <printf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D0
datatype: string

tool: D0: malloc %zu -> %p


assistant: ```c
void *parrt_malloc(size_t n) {
  void *p = malloc(n);
  printf("malloc %zu -> %p\n", n, p);
  return p;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <gmx_lie@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  gmx_lie(argc, argv);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  call   <gmx_lie@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  gmx_lie(argc, argv);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  call   <gmx_lie@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  gmx_lie(argc, argv);
  return 0;
}
```