ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<input_timeout>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %esi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x24(%rbp),%rax
  mov    %rax,%rdi
  call   <FD_ZERO@plt>
  lea    -0x24(%rbp),%rdx
  mov    -0x34(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <FD_SET@plt>
  mov    -0x38(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  lea    -0x20(%rbp),%rdx
  lea    -0x24(%rbp),%rsi
  mov    %rdx,%r8
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    %eax,%edi
  call   <select@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
int input_timeout(int filedes, unsigned int seconds) {
  fd_set set;
  struct timeval timeout;
  FD_ZERO(&set);
  FD_SET(filedes, &set);
  timeout.tv_sec = seconds;
  timeout.tv_usec = 0;
  return (select(FD_SETSIZE, &set, NULL, NULL, &timeout));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<input_timeout>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %edi,%r12d
  mov    %esi,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rbx
  mov    %rbx,%rdi
  call   <FD_ZERO@plt>
  mov    %rbx,%rsi
  mov    %r12d,%edi
  call   <FD_SET@plt>
  mov    %ebp,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  lea    0x10(%rsp),%r8
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    %rbx,%rsi
  mov    D3(%rip),%rax
  mov    (%rax),%edi
  call   <select@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int input_timeout(int filedes, unsigned int seconds) {
  fd_set set;
  struct timeval timeout;
  FD_ZERO(&set);
  FD_SET(filedes, &set);
  timeout.tv_sec = seconds;
  timeout.tv_usec = 0;
  return (select(FD_SETSIZE, &set, NULL, NULL, &timeout));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<input_timeout>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rbp
  mov    %rbp,%rdi
  call   <FD_ZERO@plt>
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <FD_SET@plt>
  mov    D9(%rip),%rax
  xor    %ecx,%ecx
  xor    %edx,%edx
  lea    0x10(%rsp),%r8
  mov    %rbp,%rsi
  mov    %ebx,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  mov    (%rax),%edi
  call   <select@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int input_timeout(int filedes, unsigned int seconds) {
  fd_set set;
  struct timeval timeout;
  FD_ZERO(&set);
  FD_SET(filedes, &set);
  timeout.tv_sec = seconds;
  timeout.tv_usec = 0;
  return (select(FD_SETSIZE, &set, NULL, NULL, &timeout));
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<input_timeout>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rbp
  mov    %rbp,%rdi
  call   <FD_ZERO@plt>
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <FD_SET@plt>
  mov    D6(%rip),%rax
  xor    %ecx,%ecx
  xor    %edx,%edx
  lea    0x10(%rsp),%r8
  mov    %rbp,%rsi
  mov    %ebx,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  mov    (%rax),%edi
  call   <select@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int input_timeout(int filedes, unsigned int seconds) {
  fd_set set;
  struct timeval timeout;
  FD_ZERO(&set);
  FD_SET(filedes, &set);
  timeout.tv_sec = seconds;
  timeout.tv_usec = 0;
  return (select(FD_SETSIZE, &set, NULL, NULL, &timeout));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cmp_ull>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jae    L0
  mov    $0xffffffff,%eax
  jmp    L22
L0:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jae    L17
  mov    $0x1,%eax
  jmp    L22
L17:
  mov    $0x0,%eax
L22:
  pop    %rbp
  ret
```

assistant: ```c
int cmp_ull(const void *ap, const void *bp) {
  unsigned long long a = *(unsigned long long *)ap;
  unsigned long long b = *(unsigned long long *)bp;
  if (a < b)
    return -1;
  else if (b < a)
    return 1;
  else
    return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cmp_ull>:
  endbr64
  mov    (%rdi),%rcx
  mov    (%rsi),%rdx
  mov    $0xffffffff,%eax
  cmp    %rdx,%rcx
  jb     L5
  seta   %al
  movzbl %al,%eax
L5:
  ret
```

assistant: ```c
int cmp_ull(const void *ap, const void *bp) {
  unsigned long long a = *(unsigned long long *)ap;
  unsigned long long b = *(unsigned long long *)bp;
  if (a < b)
    return -1;
  else if (b < a)
    return 1;
  else
    return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cmp_ull>:
  endbr64
  mov    (%rsi),%rax
  cmp    %rax,(%rdi)
  mov    $0xffffffff,%edx
  seta   %al
  movzbl %al,%eax
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int cmp_ull(const void *ap, const void *bp) {
  unsigned long long a = *(unsigned long long *)ap;
  unsigned long long b = *(unsigned long long *)bp;
  if (a < b)
    return -1;
  else if (b < a)
    return 1;
  else
    return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cmp_ull>:
  endbr64
  mov    (%rsi),%rax
  cmp    %rax,(%rdi)
  mov    $0xffffffff,%edx
  seta   %al
  movzbl %al,%eax
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int cmp_ull(const void *ap, const void *bp) {
  unsigned long long a = *(unsigned long long *)ap;
  unsigned long long b = *(unsigned long long *)bp;
  if (a < b)
    return -1;
  else if (b < a)
    return 1;
  else
    return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<match_ipvfuture>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rcx
  lea    -0x10(%rbp),%rax
  mov    $0x76,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <match_char@plt>
  test   %rax,%rax
  je     L11
  mov    -0x20(%rbp),%rcx
  lea    -0x10(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <match_set@plt>
  test   %rax,%rax
  je     L11
  mov    -0x20(%rbp),%rcx
  lea    -0x10(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <match_xset@plt>
  mov    -0x20(%rbp),%rcx
  lea    -0x10(%rbp),%rax
  mov    $0x2e,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <match_char@plt>
  test   %rax,%rax
  je     L11
  mov    -0x20(%rbp),%rcx
  lea    -0x10(%rbp),%rax
  mov    $0x53,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <match_set@plt>
  test   %rax,%rax
  je     L11
  mov    -0x20(%rbp),%rcx
  lea    -0x10(%rbp),%rax
  mov    $0x53,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <match_xset@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  mov    $0x1,%eax
  jmp    L20
L11:
  mov    $0x0,%eax
L20:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L38
  call   <__stack_chk_fail@plt>
L38:
  leave
  ret
```

assistant: ```c
int match_ipvfuture(char const **str, char const *end) {
  char const *i = *str;
  if (match_char(&i, end, 'v')) {
    if (match_set(&i, end, (4))) {
      match_xset(&i, end, (4));
      if (match_char(&i, end, '.')) {
        if (match_set(&i, end, (1 | 2 | 16 | 64))) {
          match_xset(&i, end, (1 | 2 | 16 | 64));
          *str = i;
          return 1;
        }
      }
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<match_ipvfuture>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    (%rdi),%rax
  mov    %rax,(%rsp)
  mov    %rsp,%rdi
  mov    $0x76,%edx
  call   <match_char@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  jne    L6
L29:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L23
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  mov    %rsp,%rdi
  mov    $0x4,%edx
  mov    %rbx,%rsi
  call   <match_set@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  je     L29
  mov    %rsp,%r12
  mov    $0x4,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <match_xset@plt>
  mov    $0x2e,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <match_char@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  je     L29
  mov    %rsp,%rdi
  mov    $0x53,%edx
  mov    %rbx,%rsi
  call   <match_set@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  je     L29
  mov    %rsp,%rdi
  mov    $0x53,%edx
  mov    %rbx,%rsi
  call   <match_xset@plt>
  mov    (%rsp),%rax
  mov    %rax,0x0(%rbp)
  mov    $0x1,%eax
  jmp    L29
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int match_ipvfuture(char const **str, char const *end) {
  char const *i = *str;
  if (match_char(&i, end, 'v')) {
    if (match_set(&i, end, (4))) {
      match_xset(&i, end, (4));
      if (match_char(&i, end, '.')) {
        if (match_set(&i, end, (1 | 2 | 16 | 64))) {
          match_xset(&i, end, (1 | 2 | 16 | 64));
          *str = i;
          return 1;
        }
      }
    }
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<match_ipvfuture>:
  endbr64
  push   %r12
  mov    $0x76,%edx
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    (%rdi),%rax
  mov    %rsp,%r12
  mov    %r12,%rdi
  mov    %rax,(%rsp)
  call   <match_char@plt>
  test   %rax,%rax
  jne    L24
L30:
  xor    %eax,%eax
L54:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L40
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L24:
  mov    $0x4,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <match_set@plt>
  test   %rax,%rax
  je     L30
  mov    $0x4,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <match_xset@plt>
  mov    $0x2e,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <match_char@plt>
  test   %rax,%rax
  je     L30
  mov    $0x53,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <match_set@plt>
  test   %rax,%rax
  je     L30
  mov    $0x53,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <match_xset@plt>
  mov    (%rsp),%rax
  mov    %rax,(%rbx)
  mov    $0x1,%eax
  jmp    L54
L40:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int match_ipvfuture(char const **str, char const *end) {
  char const *i = *str;
  if (match_char(&i, end, 'v')) {
    if (match_set(&i, end, (4))) {
      match_xset(&i, end, (4));
      if (match_char(&i, end, '.')) {
        if (match_set(&i, end, (1 | 2 | 16 | 64))) {
          match_xset(&i, end, (1 | 2 | 16 | 64));
          *str = i;
          return 1;
        }
      }
    }
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<match_ipvfuture>:
  endbr64
  push   %r12
  mov    $0x76,%edx
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    (%rdi),%rax
  mov    %rsp,%r12
  mov    %r12,%rdi
  mov    %rax,(%rsp)
  call   <match_char@plt>
  test   %rax,%rax
  jne    L15
L58:
  xor    %eax,%eax
L23:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L28
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  mov    $0x4,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <match_set@plt>
  test   %rax,%rax
  je     L58
  mov    $0x4,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <match_xset@plt>
  mov    $0x2e,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <match_char@plt>
  test   %rax,%rax
  je     L58
  mov    $0x53,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <match_set@plt>
  test   %rax,%rax
  je     L58
  mov    $0x53,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <match_xset@plt>
  mov    (%rsp),%rax
  mov    %rax,(%rbx)
  mov    $0x1,%eax
  jmp    L23
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int match_ipvfuture(char const **str, char const *end) {
  char const *i = *str;
  if (match_char(&i, end, 'v')) {
    if (match_set(&i, end, (4))) {
      match_xset(&i, end, (4));
      if (match_char(&i, end, '.')) {
        if (match_set(&i, end, (1 | 2 | 16 | 64))) {
          match_xset(&i, end, (1 | 2 | 16 | 64));
          *str = i;
          return 1;
        }
      }
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <foo@plt>
  mov    $0x0,%eax
  call   <bar@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  foo();
  bar();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int main() {
  foo();
  bar();
  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
  xor    %eax,%eax
  call   <foo@plt>
  xor    %eax,%eax
  call   <bar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  foo();
  bar();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

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

assistant: ```c
int main() {
  foo();
  bar();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_interleave>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %rdi,-0x88(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x60(%rbp)
  movl   $0x6,-0x5c(%rbp)
  movl   $0xc,-0x58(%rbp)
  movl   $0x14,-0x54(%rbp)
  movl   $0x1a,-0x50(%rbp)
  movl   $0x20,-0x4c(%rbp)
  movl   $0x0,-0x78(%rbp)
  jmp    L55
L50:
  mov    -0x78(%rbp),%eax
  cltq
  mov    -0x60(%rbp,%rax,4),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x88(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x68(%rbp)
  mov    -0x68(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <_reverse@plt>
  mov    -0x68(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x68(%rbp),%rax
  add    $0x1c,%rax
  mov    (%rax),%eax
  mov    -0x68(%rbp),%rdx
  lea    0x1c(%rdx),%rbx
  mov    %eax,%edi
  call   <_reverse@plt>
  mov    %eax,(%rbx)
  movl   $0x0,-0x74(%rbp)
  movl   $0x80,-0x6c(%rbp)
  jmp    L48
L9:
  mov    -0x74(%rbp),%eax
  cltq
  movl   $0x0,-0x40(%rbp,%rax,4)
  movl   $0x0,-0x70(%rbp)
  jmp    L51
L68:
  mov    -0x74(%rbp),%eax
  cltq
  mov    -0x40(%rbp,%rax,4),%esi
  mov    -0x70(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x68(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  and    -0x6c(%rbp),%eax
  test   %eax,%eax
  setne  %al
  movzbl %al,%edx
  mov    -0x70(%rbp),%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     %eax,%esi
  mov    %esi,%edx
  mov    -0x74(%rbp),%eax
  cltq
  mov    %edx,-0x40(%rbp,%rax,4)
  addl   $0x1,-0x70(%rbp)
L51:
  cmpl   $0x7,-0x70(%rbp)
  jle    L68
  addl   $0x1,-0x74(%rbp)
  sarl   -0x6c(%rbp)
L48:
  cmpl   $0x7,-0x74(%rbp)
  jle    L9
  lea    -0x40(%rbp),%rcx
  mov    -0x68(%rbp),%rax
  mov    $0x8,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  addl   $0x1,-0x78(%rbp)
L55:
  cmpl   $0x5,-0x78(%rbp)
  jle    L50
  nop
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L69
  call   <__stack_chk_fail@plt>
L69:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void _interleave(uint8_t *frame) {
  int b, i, j;
  int offset[6] = {0, 6, 12, 20, 26, 32};
  uint8_t r[8];
  uint8_t m;
  for (b = 0; b < 6; b++) {
    uint8_t *s = frame + offset[b];
    s[0] = _reverse(s[0]);
    s[7] = _reverse(s[7]);
    for (i = 0, m = 0x80; i < 8; i++, m >>= 1) {
      r[i] = 0x00;
      for (j = 0; j < 8; j++) {
        r[i] |= ((m & s[j]) ? 1 : 0) << j;
      }
    }
    memcpy(s, r, 8);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_interleave>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x50,%rsp
  mov    %rdi,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  movl   $0x6,0x4(%rsp)
  movl   $0xc,0x8(%rsp)
  movl   $0x14,0xc(%rsp)
  movl   $0x1a,0x10(%rsp)
  movl   $0x20,0x14(%rsp)
  mov    %rsp,%r12
  lea    0x18(%rsp),%r14
  lea    0x40(%rsp),%rbp
L22:
  movslq (%r12),%rax
  lea    0x0(%r13,%rax,4),%rbx
  mov    (%rbx),%edi
  call   <_reverse@plt>
  mov    %eax,(%rbx)
  mov    0x1c(%rbx),%edi
  call   <_reverse@plt>
  mov    %eax,0x1c(%rbx)
  lea    0x20(%rsp),%rdi
  mov    $0x80,%esi
L25:
  mov    %rdi,%r8
  mov    $0x0,%ecx
  mov    $0x0,%edx
L23:
  mov    %esi,%eax
  and    (%rbx,%rcx,4),%eax
  setne  %al
  movzbl %al,%eax
  shl    %cl,%eax
  or     %eax,%edx
  add    $0x1,%rcx
  cmp    $0x8,%rcx
  jne    L23
  mov    %edx,(%r8)
  sar    %esi
  add    $0x4,%rdi
  cmp    %rbp,%rdi
  jne    L25
  lea    0x20(%rsp),%rsi
  mov    $0x8,%edx
  mov    %rbx,%rdi
  call   <memcpy@plt>
  add    $0x4,%r12
  cmp    %r14,%r12
  jne    L22
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x50,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void _interleave(uint8_t *frame) {
  int b, i, j;
  int offset[6] = {0, 6, 12, 20, 26, 32};
  uint8_t r[8];
  uint8_t m;
  for (b = 0; b < 6; b++) {
    uint8_t *s = frame + offset[b];
    s[0] = _reverse(s[0]);
    s[7] = _reverse(s[7]);
    for (i = 0, m = 0x80; i < 8; i++, m >>= 1) {
      r[i] = 0x00;
      for (j = 0; j < 8; j++) {
        r[i] |= ((m & s[j]) ? 1 : 0) << j;
      }
    }
    memcpy(s, r, 8);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<_interleave>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x18(%rsp),%r12
  lea    0x20(%rsp),%r15
  movabs $0x140000000c,%rax
  movl   $0x6,0x4(%rsp)
  lea    0x40(%rsp),%r14
  mov    %rax,0x8(%rsp)
  movabs $0x200000001a,%rax
  mov    %rax,0x10(%rsp)
  xor    %eax,%eax
  nop
L31:
  lea    0x0(%rbp,%rax,4),%r13
  mov    0x0(%r13),%edi
  call   <_reverse@plt>
  mov    0x1c(%r13),%edi
  mov    %eax,0x0(%r13)
  call   <_reverse@plt>
  pxor   %xmm0,%xmm0
  mov    %r15,%rdi
  mov    $0x80,%esi
  mov    %eax,0x1c(%r13)
  movaps %xmm0,(%r15)
  movaps %xmm0,0x10(%r15)
  nopl   0x0(%rax)
L44:
  xor    %ecx,%ecx
  xor    %edx,%edx
  nopl   0x0(%rax)
L19:
  mov    %esi,%eax
  and    0x0(%r13,%rcx,4),%eax
  setne  %al
  movzbl %al,%eax
  shl    %cl,%eax
  add    $0x1,%rcx
  or     %eax,%edx
  cmp    $0x8,%rcx
  jne    L19
  mov    %edx,(%rdi)
  add    $0x4,%rdi
  sar    %esi
  cmp    %rdi,%r14
  jne    L44
  mov    $0x8,%edx
  mov    %r15,%rsi
  mov    %r13,%rdi
  add    $0x4,%rbx
  call   <memcpy@plt>
  cmp    %rbx,%r12
  je     L36
  movslq (%rbx),%rax
  jmp    L31
L36:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L37:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void _interleave(uint8_t *frame) {
  int b, i, j;
  int offset[6] = {0, 6, 12, 20, 26, 32};
  uint8_t r[8];
  uint8_t m;
  for (b = 0; b < 6; b++) {
    uint8_t *s = frame + offset[b];
    s[0] = _reverse(s[0]);
    s[7] = _reverse(s[7]);
    for (i = 0, m = 0x80; i < 8; i++, m >>= 1) {
      r[i] = 0x00;
      for (j = 0; j < 8; j++) {
        r[i] |= ((m & s[j]) ? 1 : 0) << j;
      }
    }
    memcpy(s, r, 8);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffff91a2b3c5,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0xb,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%edx
  movslq %edx,%rax
  imul   $0xffffffff91a2b3c5,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0xb,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0xe10,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  movslq %eax,%rdx
  imul   $0xffffffff88888889,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffff88888889,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x5,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  imul   $0x3c,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  mov    -0xc(%rbp),%esi
  mov    -0x10(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    %esi,%r8d
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D21
datatype: string
name: extract_symbolic_data
arguments: data_tag: D19
datatype: string
name: extract_symbolic_data
arguments: data_tag: D23
datatype: string

tool: D21: Masukkan durasi dalam detik : 

tool: D19: %d

tool: D23: %d detik sama dengan %d jam, %d menit, %d detik.


assistant: ```c
int main() {
  int durasi;
  int jam;
  int menit;
  int detik;
  printf("Masukkan durasi dalam detik : ");
  scanf("%d", &durasi);
  jam = durasi / 3600;
  menit = (durasi % 3600) / 60;
  detik = durasi % 60;
  printf("%d detik sama dengan %d jam, %d menit, %d detik.\n", durasi, jam, menit, detik);
  return 0;
}
```