ylfeng/ReF-Decompile-dataset huggingface dataset:
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
  jmp    L74
L103:
  mov    -0x14(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x10(%rbp)
  jmp    L124
L118:
  movl   $0x0,-0xc(%rbp)
  jmp    L48
L26:
  cmpl   $0x19,-0x10(%rbp)
  jle    L90
  mov    -0x10(%rbp),%eax
  add    $0x47,%eax
  jmp    L86
L90:
  mov    -0x10(%rbp),%eax
  add    $0x41,%eax
L86:
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0xc(%rbp)
L48:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jl     L26
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x10(%rbp)
L124:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jl     L118
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x0,-0x10(%rbp)
  jmp    L19
L72:
  movl   $0x0,-0xc(%rbp)
  jmp    L73
L43:
  cmpl   $0x19,-0xc(%rbp)
  jle    L2
  mov    -0xc(%rbp),%eax
  add    $0x47,%eax
  jmp    L111
L2:
  mov    -0xc(%rbp),%eax
  add    $0x41,%eax
L111:
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0xc(%rbp)
L73:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jl     L43
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x10(%rbp)
L19:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jl     L72
L74:
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmp    $0xffffffff,%eax
  jne    L103
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L110
  call   <__stack_chk_fail@plt>
L110:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
type_infered: string
name: extract_symbolic_data
arguments: label: D6
type_infered: string

tool: D7: 2 %d %d


tool: D6: %d

assistant: ```c
int main() {
  int n, i, j;
  while (~scanf("%d", &n)) {
    printf("2 %d %d\n", n, n);
    for (i = 0; i < n; ++i) {
      for (j = 0; j < n; ++j)
        printf("%c", i > 25 ? 'a' + i - 26 : 'A' + i);
      printf("\n");
    }
    printf("\n");
    for (i = 0; i < n; ++i) {
      for (j = 0; j < n; ++j)
        printf("%c", j > 25 ? 'a' + j - 26 : 'A' + j);
      printf("\n");
    }
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  lea    D2(%rip),%r12
  lea    D9(%rip),%r14
L17:
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmp    $0xffffffff,%eax
  je     L58
  mov    0x4(%rsp),%esi
  mov    %esi,%edx
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%r15d
  cmpl   $0x0,0x4(%rsp)
  jg     L46
L99:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%ebp
  cmpl   $0x0,0x4(%rsp)
  jg     L89
  jmp    L17
L102:
  lea    0x41(%r15),%ebp
  lea    0x47(%r15),%eax
  cmp    $0x19,%r15d
  cmovg  %eax,%ebp
  mov    $0x0,%ebx
L64:
  mov    %ebp,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    %ebx,0x4(%rsp)
  jg     L64
L45:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r15d
  cmp    %r15d,0x4(%rsp)
  jle    L99
L46:
  cmpl   $0x0,0x4(%rsp)
  jg     L102
  jmp    L45
L94:
  lea    0x47(%rbx),%eax
  lea    0x41(%rbx),%edi
  cmp    $0x1a,%ebx
  cmovge %eax,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    %ebx,0x4(%rsp)
  jg     L94
L52:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%ebp
  cmp    %ebp,0x4(%rsp)
  jle    L17
L89:
  mov    $0x0,%ebx
  cmpl   $0x0,0x4(%rsp)
  jg     L94
  jmp    L52
L58:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L47
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L47:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
type: string
name: extract_symbolic_data
arguments: data_label: D9
type: string

tool: D2: %d

tool: D9: 2 %d %d


assistant: ```c
int main() {
  int n, i, j;
  while (~scanf("%d", &n)) {
    printf("2 %d %d\n", n, n);
    for (i = 0; i < n; ++i) {
      for (j = 0; j < n; ++j)
        printf("%c", i > 25 ? 'a' + i - 26 : 'A' + i);
      printf("\n");
    }
    printf("\n");
    for (i = 0; i < n; ++i) {
      for (j = 0; j < n; ++j)
        printf("%c", j > 25 ? 'a' + j - 26 : 'A' + j);
      printf("\n");
    }
  }
  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   %r15
  push   %r14
  lea    D4(%rip),%r14
  push   %r13
  push   %r12
  lea    D3(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  nopl   (%rax)
L83:
  xor    %eax,%eax
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmp    $0xffffffff,%eax
  je     L24
  mov    0x4(%rsp),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  xor    %r15d,%r15d
  mov    %esi,%edx
  call   <printf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L70
  nopl   0x0(%rax)
L22:
  cmp    $0x19,%r15d
  lea    0x47(%r15),%ebp
  lea    0x41(%r15),%eax
  cmovle %eax,%ebp
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L42:
  mov    %ebp,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L42
  mov    $0xa,%edi
  add    $0x1,%r15d
  call   <putchar@plt>
  cmp    %r15d,0x4(%rsp)
  jg     L22
L70:
  mov    $0xa,%edi
  xor    %ebp,%ebp
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L83
  nopl   (%rax)
L38:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L53:
  lea    0x47(%rbx),%eax
  cmp    $0x1a,%ebx
  lea    0x41(%rbx),%edi
  cmovge %eax,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L53
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  cmp    %ebp,0x4(%rsp)
  jg     L38
  jmp    L83
L24:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L79
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L79:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
type: string
name: retrieve_data
arguments: data_tag: D3
type: string

tool: D4: 2 %d %d


tool: D3: %d

assistant: ```c
int main() {
  int n, i, j;
  while (~scanf("%d", &n)) {
    printf("2 %d %d\n", n, n);
    for (i = 0; i < n; ++i) {
      for (j = 0; j < n; ++j)
        printf("%c", i > 25 ? 'a' + i - 26 : 'A' + i);
      printf("\n");
    }
    printf("\n");
    for (i = 0; i < n; ++i) {
      for (j = 0; j < n; ++j)
        printf("%c", j > 25 ? 'a' + j - 26 : 'A' + j);
      printf("\n");
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  lea    D14(%rip),%r14
  push   %r13
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  nopl   (%rax)
L49:
  xor    %eax,%eax
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmp    $0xffffffff,%eax
  je     L8
  mov    0x4(%rsp),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  xor    %r15d,%r15d
  mov    %esi,%edx
  call   <printf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L119
  nopl   0x0(%rax)
L110:
  cmp    $0x19,%r15d
  lea    0x41(%r15),%ebp
  lea    0x47(%r15),%eax
  cmovg  %eax,%ebp
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L83:
  mov    %ebp,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L83
  mov    $0xa,%edi
  add    $0x1,%r15d
  call   <putchar@plt>
  cmp    %r15d,0x4(%rsp)
  jg     L110
L119:
  mov    $0xa,%edi
  xor    %ebp,%ebp
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L49
  nopl   (%rax)
L53:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L92:
  cmp    $0x19,%ebx
  jle    L86
  lea    0x47(%rbx),%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    0x4(%rsp),%ebx
  jl     L92
L17:
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  cmp    %ebp,0x4(%rsp)
  jg     L53
  jmp    L49
  xchg   %ax,%ax
L86:
  lea    0x41(%rbx),%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L92
  jmp    L17
L8:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D14
datatype: string
name: disassemble_data
arguments: label: D0
datatype: string

tool: D14: 2 %d %d


tool: D0: %d

assistant: ```c
int main() {
  int n, i, j;
  while (~scanf("%d", &n)) {
    printf("2 %d %d\n", n, n);
    for (i = 0; i < n; ++i) {
      for (j = 0; j < n; ++j)
        printf("%c", i > 25 ? 'a' + i - 26 : 'A' + i);
      printf("\n");
    }
    printf("\n");
    for (i = 0; i < n; ++i) {
      for (j = 0; j < n; ++j)
        printf("%c", j > 25 ? 'a' + j - 26 : 'A' + j);
      printf("\n");
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<check_sum_of_primes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x2,-0x8(%rbp)
  jmp    L13
L34:
  mov    -0x14(%rbp),%eax
  sub    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <check_prime@plt>
  test   %rax,%rax
  je     L43
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <check_prime@plt>
  test   %rax,%rax
  je     L43
  mov    $0x1,%eax
  jmp    L20
L43:
  nop
  addl   $0x1,-0x8(%rbp)
L13:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L34
  mov    $0x0,%eax
L20:
  leave
  ret
```

assistant: ```c
int check_sum_of_primes(int a) {
  int m;
  int n;
  for (m = 2; m < a; m++) {
    n = a - m;
    if (check_prime(m) && check_prime(n))
      return 1;
    else
      continue;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<check_sum_of_primes>:
  endbr64
  cmp    $0x2,%edi
  jle    L52
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    $0x2,%ebx
  jmp    L43
L16:
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  je     L26
L43:
  mov    %ebx,%edi
  call   <check_prime@plt>
  test   %rax,%rax
  je     L16
  mov    %ebp,%edi
  sub    %ebx,%edi
  call   <check_prime@plt>
  test   %rax,%rax
  je     L16
  mov    $0x1,%eax
L47:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L26:
  mov    $0x0,%eax
  jmp    L47
L52:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int check_sum_of_primes(int a) {
  int m;
  int n;
  for (m = 2; m < a; m++) {
    n = a - m;
    if (check_prime(m) && check_prime(n))
      return 1;
    else
      continue;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<check_sum_of_primes>:
  endbr64
  cmp    $0x2,%edi
  jle    L2
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    $0x2,%ebx
  sub    $0x8,%rsp
L32:
  mov    %ebx,%edi
  call   <check_prime@plt>
  test   %rax,%rax
  je     L13
  mov    %ebp,%edi
  sub    %ebx,%edi
  call   <check_prime@plt>
  test   %rax,%rax
  je     L13
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L13:
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L32
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L2:
  xor    %eax,%eax
  ret
```

assistant: ```c
int check_sum_of_primes(int a) {
  int m;
  int n;
  for (m = 2; m < a; m++) {
    n = a - m;
    if (check_prime(m) && check_prime(n))
      return 1;
    else
      continue;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<check_sum_of_primes>:
  endbr64
  cmp    $0x2,%edi
  jle    L30
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    $0x2,%ebx
  sub    $0x8,%rsp
L9:
  mov    %ebx,%edi
  call   <check_prime@plt>
  test   %rax,%rax
  je     L2
  mov    %ebp,%edi
  sub    %ebx,%edi
  call   <check_prime@plt>
  test   %rax,%rax
  je     L2
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L2:
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L9
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L30:
  xor    %eax,%eax
  ret
```

assistant: ```c
int check_sum_of_primes(int a) {
  int m;
  int n;
  for (m = 2; m < a; m++) {
    n = a - m;
    if (check_prime(m) && check_prime(n))
      return 1;
    else
      continue;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<___sane_connect>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rdx
  mov    D26(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L18
  movl   $0x1,-0xc(%rbp)
  mov    D24(%rip),%rax
  mov    (%rax),%edx
  mov    D14(%rip),%rax
  mov    (%rax),%esi
  lea    -0xc(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0x4,%r8d
  mov    %eax,%edi
  call   <setsockopt@plt>
L18:
  mov    -0x18(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <connect@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int ___sane_connect(int sock, struct sockaddr *sa, socklen_t len) {
  if (sa->sa_family == AF_INET) {
    int on = 1;
    (void)setsockopt(sock, SOL_SOCKET, SO_KEEPALIVE, (char *)&on, sizeof(on));
  }
  return (connect(sock, sa, len));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<___sane_connect>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%r12d
  mov    %rsi,%rbx
  mov    %edx,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D20(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rsi)
  je     L7
L4:
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    %r12d,%edi
  call   <connect@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L11
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L7:
  movl   $0x1,0x4(%rsp)
  lea    0x4(%rsp),%rcx
  mov    $0x4,%r8d
  mov    D1(%rip),%rax
  mov    (%rax),%edx
  mov    D8(%rip),%rax
  mov    (%rax),%esi
  call   <setsockopt@plt>
  jmp    L4
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ___sane_connect(int sock, struct sockaddr *sa, socklen_t len) {
  if (sa->sa_family == AF_INET) {
    int on = 1;
    (void)setsockopt(sock, SOL_SOCKET, SO_KEEPALIVE, (char *)&on, sizeof(on));
  }
  return (connect(sock, sa, len));
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<___sane_connect>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  mov    D5(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rsi)
  je     L18
L8:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x28,%rsp
  mov    %r12d,%edx
  mov    %rbp,%rsi
  pop    %rbp
  pop    %r12
  jmp    1090 <connect@plt>
L18:
  mov    D9(%rip),%rax
  lea    0x14(%rsp),%rcx
  mov    %edi,0xc(%rsp)
  mov    $0x4,%r8d
  movl   $0x1,0x14(%rsp)
  mov    (%rax),%edx
  mov    D21(%rip),%rax
  mov    (%rax),%esi
  call   <setsockopt@plt>
  mov    0xc(%rsp),%edi
  jmp    L8
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ___sane_connect(int sock, struct sockaddr *sa, socklen_t len) {
  if (sa->sa_family == AF_INET) {
    int on = 1;
    (void)setsockopt(sock, SOL_SOCKET, SO_KEEPALIVE, (char *)&on, sizeof(on));
  }
  return (connect(sock, sa, len));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<___sane_connect>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  mov    D21(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rsi)
  je     L20
L21:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x28,%rsp
  mov    %r12d,%edx
  mov    %rbp,%rsi
  pop    %rbp
  pop    %r12
  jmp    1090 <connect@plt>
L20:
  mov    D12(%rip),%rax
  lea    0x14(%rsp),%rcx
  mov    %edi,0xc(%rsp)
  mov    $0x4,%r8d
  movl   $0x1,0x14(%rsp)
  mov    (%rax),%edx
  mov    D14(%rip),%rax
  mov    (%rax),%esi
  call   <setsockopt@plt>
  mov    0xc(%rsp),%edi
  jmp    L21
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ___sane_connect(int sock, struct sockaddr *sa, socklen_t len) {
  if (sa->sa_family == AF_INET) {
    int on = 1;
    (void)setsockopt(sock, SOL_SOCKET, SO_KEEPALIVE, (char *)&on, sizeof(on));
  }
  return (connect(sock, sa, len));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xf4000(%rsp),%r11
L24:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L24
  sub    $0x278,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xf4268(%rsp)
  xor    %eax,%eax
  mov    $0x1,%r14d
  lea    D57(%rip),%r15
  lea    D40(%rip),%r13
  jmp    L106
L3:
  mov    $0x0,%eax
  call   <getchar@plt>
  add    $0x1,%r14
L106:
  lea    0x20(%rsp),%rsi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L86
  lea    0x18(%rsp),%rsi
  lea    D49(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    %r14,%rsi
  lea    D41(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpq   $0x0,0x18(%rsp)
  jle    L3
  mov    $0x1,%ebp
  lea    D36(%rip),%r12
  lea    0x20(%rsp),%rbx
  jmp    L102
L86:
  mov    0xf4268(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  mov    $0x0,%eax
  add    $0xf4278,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L4:
  mov    %r13,%rdi
  call   <puts@plt>
L27:
  add    $0x1,%rbp
  cmp    %rbp,0x18(%rsp)
  jl     L3
L102:
  lea    0x8(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  mov    0x10(%rsp),%rdx
  cmp    %rdx,%rax
  jle    L105
  mov    %rdx,0x8(%rsp)
  mov    %rax,0x10(%rsp)
L105:
  mov    0x8(%rsp),%rax
  mov    0x10(%rsp),%rcx
L76:
  cmp    %rax,%rcx
  jle    L4
  movzbl (%rax,%rbx,1),%edx
  add    $0x1,%rax
  xor    (%rax,%rbx,1),%dl
  cmp    $0x1,%dl
  jne    L76
  lea    D18(%rip),%rdi
  call   <puts@plt>
  jmp    L27
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D57
data_type: string
name: retrieve_data
arguments: tag: D40
data_type: string
name: retrieve_data
arguments: tag: D49
data_type: string
name: retrieve_data
arguments: tag: D41
data_type: string
name: retrieve_data
arguments: tag: D36
data_type: string
name: retrieve_data
arguments: tag: D18
data_type: string

tool: D57: %s

tool: D40: Yes

tool: D49: %ld

tool: D41: Case %ld:


tool: D36: %ld %ld

tool: D18: No

assistant: ```c
int main() {
  char s[1000001];
  ;
  long c = 1, i, j, k, l, x, min, max, t;
  while (scanf("%s", s) == 1) {
    scanf("%ld", &t);
    printf("Case %ld:\n", c);
    for (i = 1; i <= t; i++) {
      j = 0;
      scanf("%ld %ld", &max, &min);
      if (min > max) {
        x = min;
        min = max;
        max = x;
      }
      k = 0;
      for (j = min; j < max; j++) {
        if ((s[j] ^ s[j + 1]) == 1) {
          k = 1;
          break;
        }
      }
      if (k == 0)
        printf("Yes\n");
      else
        printf("No\n");
    }
    getchar();
    c++;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xf4000(%rsp),%r11
L84:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L84
  sub    $0x288,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xf4278(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%r15
  movq   $0x1,0x8(%rsp)
  nop
L58:
  xor    %eax,%eax
  mov    %r15,%rsi
  lea    D47(%rip),%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L100
  lea    0x28(%rsp),%rsi
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%rsi
  xor    %eax,%eax
  lea    D39(%rip),%rdi
  call   <printf@plt>
  cmpq   $0x0,0x28(%rsp)
  jle    L71
  mov    $0x1,%ebx
  lea    0x18(%rsp),%r13
  lea    0x20(%rsp),%r12
  lea    D10(%rip),%rbp
  lea    D56(%rip),%r14
  nopl   0x0(%rax)
L54:
  mov    %r12,%rsi
  mov    %r13,%rdx
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x18(%rsp),%rax
  mov    0x20(%rsp),%rsi
  cmp    %rsi,%rax
  jle    L93
  mov    %rsi,%rdx
  mov    %rsi,0x18(%rsp)
  mov    %rax,%rsi
  mov    %rax,0x20(%rsp)
  mov    %rdx,%rax
  jmp    L93
  nopl   0x0(%rax)
L39:
  movzbl (%r15,%rax,1),%edx
  add    $0x1,%rax
  xor    (%r15,%rax,1),%dl
  cmp    $0x1,%dl
  je     L38
L93:
  cmp    %rsi,%rax
  jl     L39
  lea    D36(%rip),%rdi
  call   <puts@plt>
L13:
  add    $0x1,%rbx
  cmp    %rbx,0x28(%rsp)
  jge    L54
L71:
  xor    %eax,%eax
  call   <getchar@plt>
  addq   $0x1,0x8(%rsp)
  jmp    L58
  nopl   0x0(%rax,%rax,1)
L38:
  mov    %r14,%rdi
  call   <puts@plt>
  jmp    L13
L100:
  mov    0xf4278(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0xf4288,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D47
type_infered: string
name: parse_data
arguments: label: D38
type_infered: string
name: parse_data
arguments: label: D39
type_infered: string
name: parse_data
arguments: label: D10
type_infered: string
name: parse_data
arguments: label: D56
type_infered: string
name: parse_data
arguments: label: D36
type_infered: string

tool: D47: %s

tool: D38: %ld

tool: D39: Case %ld:


tool: D10: %ld %ld

tool: D56: No

tool: D36: Yes

assistant: ```c
int main() {
  char s[1000001];
  ;
  long c = 1, i, j, k, l, x, min, max, t;
  while (scanf("%s", s) == 1) {
    scanf("%ld", &t);
    printf("Case %ld:\n", c);
    for (i = 1; i <= t; i++) {
      j = 0;
      scanf("%ld %ld", &max, &min);
      if (min > max) {
        x = min;
        min = max;
        max = x;
      }
      k = 0;
      for (j = min; j < max; j++) {
        if ((s[j] ^ s[j + 1]) == 1) {
          k = 1;
          break;
        }
      }
      if (k == 0)
        printf("Yes\n");
      else
        printf("No\n");
    }
    getchar();
    c++;
  }
  return 0;
}
```