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

```asm
<getNuvotonTime>:
  endbr64
  push   %r12
  movzbl (%rdi),%esi
  movzbl 0x1(%rdi),%eax
  movzbl 0x2(%rdi),%ecx
  movzbl 0x3(%rdi),%r8d
  shl    $0x8,%esi
  movzbl 0x4(%rdi),%r9d
  lea    D5(%rip),%rdi
  imul   $0xe10,%ecx,%r12d
  add    %eax,%esi
  imul   $0x3c,%r8d,%edx
  imul   $0x15180,%esi,%eax
  add    %edx,%r12d
  add    $0x2efbbf80,%eax
  add    %r9d,%r12d
  add    %rax,%r12
  mov    %r12,%rdx
  call   <printf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
type_infered: string

tool: D5: MJD = %d epoch = %ld, time = %02d:%02d:%02d


assistant: ```c
unsigned long getNuvotonTime(char *nuvotonTimeString) {
  unsigned int mjd = ((nuvotonTimeString[0] & 0xFF) * 256) + (nuvotonTimeString[1] & 0xFF);
  unsigned long epoch = ((mjd - 40587) * 86400);
  unsigned int hour = nuvotonTimeString[2] & 0xFF;
  unsigned int min = nuvotonTimeString[3] & 0xFF;
  unsigned int sec = nuvotonTimeString[4] & 0xFF;
  epoch += (hour * 3600 + min * 60 + sec);
  printf("MJD = %d epoch = %ld, time = %02d:%02d:%02d\n", mjd, epoch, hour, min, sec);
  return epoch;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<getNuvotonTime>:
  endbr64
  push   %r12
  movzbl (%rdi),%esi
  movzbl 0x1(%rdi),%eax
  movzbl 0x2(%rdi),%ecx
  movzbl 0x3(%rdi),%r8d
  shl    $0x8,%esi
  movzbl 0x4(%rdi),%r9d
  lea    D3(%rip),%rdi
  imul   $0xe10,%ecx,%r12d
  add    %eax,%esi
  imul   $0x3c,%r8d,%edx
  imul   $0x15180,%esi,%eax
  add    %edx,%r12d
  add    $0x2efbbf80,%eax
  add    %r9d,%r12d
  add    %rax,%r12
  mov    %r12,%rdx
  call   <printf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D3
type_infered: string

tool: D3: MJD = %d epoch = %ld, time = %02d:%02d:%02d


assistant: ```c
unsigned long getNuvotonTime(char *nuvotonTimeString) {
  unsigned int mjd = ((nuvotonTimeString[0] & 0xFF) * 256) + (nuvotonTimeString[1] & 0xFF);
  unsigned long epoch = ((mjd - 40587) * 86400);
  unsigned int hour = nuvotonTimeString[2] & 0xFF;
  unsigned int min = nuvotonTimeString[3] & 0xFF;
  unsigned int sec = nuvotonTimeString[4] & 0xFF;
  epoch += (hour * 3600 + min * 60 + sec);
  printf("MJD = %d epoch = %ld, time = %02d:%02d:%02d\n", mjd, epoch, hour, min, sec);
  return epoch;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<sum_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L8
L18:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L8:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L18
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int sum_int(int data[], int n) {
  int s = 0;
  for (int i = 0; i < n; i++) {
    s += data[i];
  }
  return (s);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sum_int>:
  endbr64
  test   %esi,%esi
  jle    L5
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rcx
  mov    $0x0,%edx
L27:
  add    (%rax),%edx
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L27
L10:
  mov    %edx,%eax
  ret
L5:
  mov    $0x0,%edx
  jmp    L10
```

assistant: ```c
int sum_int(int data[], int n) {
  int s = 0;
  for (int i = 0; i < n; i++) {
    s += data[i];
  }
  return (s);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<sum_int>:
  endbr64
  test   %esi,%esi
  jle    L13
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L18:
  add    (%rdi),%eax
  add    $0x4,%rdi
  cmp    %rdx,%rdi
  jne    L18
  ret
  nopl   0x0(%rax)
L13:
  xor    %eax,%eax
  ret
```

assistant: ```c
int sum_int(int data[], int n) {
  int s = 0;
  for (int i = 0; i < n; i++) {
    s += data[i];
  }
  return (s);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sum_int>:
  endbr64
  mov    %rdi,%rdx
  test   %esi,%esi
  jle    L67
  lea    -0x1(%rsi),%eax
  cmp    $0x2,%eax
  jbe    L28
  mov    %esi,%ecx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    $0x2,%ecx
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdi,%rcx,1),%rcx
  nopl   0x0(%rax,%rax,1)
L25:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  cmp    %rcx,%rax
  jne    L25
  movdqa %xmm0,%xmm1
  mov    %esi,%ecx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%ecx
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  test   $0x3,%sil
  je     L26
L33:
  movslq %ecx,%rdi
  lea    0x0(,%rdi,4),%r8
  add    (%rdx,%rdi,4),%eax
  lea    0x1(%rcx),%edi
  cmp    %edi,%esi
  jle    L63
  add    $0x2,%ecx
  add    0x4(%rdx,%r8,1),%eax
  cmp    %ecx,%esi
  jle    L63
  add    0x8(%rdx,%r8,1),%eax
  ret
  nopl   0x0(%rax)
L67:
  xor    %eax,%eax
L63:
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  ret
L28:
  xor    %ecx,%ecx
  xor    %eax,%eax
  jmp    L33
```

assistant: ```c
int sum_int(int data[], int n) {
  int s = 0;
  for (int i = 0; i < n; i++) {
    s += data[i];
  }
  return (s);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<count_initial_substrings>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  jmp    L1
L2:
  addq   $0x1,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  add    %rax,-0x10(%rbp)
L1:
  mov    -0x30(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <initial_strstr@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L2
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
size_t count_initial_substrings(const char *str, const char *sub_str) {
  size_t n_strs = 0;
  const char *cur_str = str;
  size_t sub_str_len = strlen(sub_str);
  while ((cur_str = initial_strstr(cur_str, sub_str)) != NULL) {
    n_strs++;
    cur_str += sub_str_len;
  }
  return (n_strs);
}
```
user: Translate the below assembly code into C programming language:

```asm
<count_initial_substrings>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %rax,%r13
  mov    $0x0,%ebp
  jmp    L12
L14:
  add    $0x1,%rbp
  lea    (%rax,%r13,1),%rbx
L12:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <initial_strstr@plt>
  test   %rax,%rax
  jne    L14
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
size_t count_initial_substrings(const char *str, const char *sub_str) {
  size_t n_strs = 0;
  const char *cur_str = str;
  size_t sub_str_len = strlen(sub_str);
  while ((cur_str = initial_strstr(cur_str, sub_str)) != NULL) {
    n_strs++;
    cur_str += sub_str_len;
  }
  return (n_strs);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<count_initial_substrings>:
  endbr64
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rax,%r13
  jmp    L14
  nopl   0x0(%rax)
L4:
  add    $0x1,%r12
  lea    (%rax,%r13,1),%rbp
L14:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <initial_strstr@plt>
  test   %rax,%rax
  jne    L4
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
size_t count_initial_substrings(const char *str, const char *sub_str) {
  size_t n_strs = 0;
  const char *cur_str = str;
  size_t sub_str_len = strlen(sub_str);
  while ((cur_str = initial_strstr(cur_str, sub_str)) != NULL) {
    n_strs++;
    cur_str += sub_str_len;
  }
  return (n_strs);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<count_initial_substrings>:
  endbr64
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rax,%r13
  jmp    L19
  nopl   0x0(%rax)
L12:
  add    $0x1,%r12
  lea    (%rax,%r13,1),%rbp
L19:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <initial_strstr@plt>
  test   %rax,%rax
  jne    L12
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
size_t count_initial_substrings(const char *str, const char *sub_str) {
  size_t n_strs = 0;
  const char *cur_str = str;
  size_t sub_str_len = strlen(sub_str);
  while ((cur_str = initial_strstr(cur_str, sub_str)) != NULL) {
    n_strs++;
    cur_str += sub_str_len;
  }
  return (n_strs);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strspn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %rdi,-0x68(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x50(%rbp),%rax
  mov    $0x40,%esi
  mov    %rax,%rdi
  call   <memclr@plt>
  movq   $0x0,-0x58(%rbp)
  jmp    L42
L55:
  mov    -0x70(%rbp),%rdx
  mov    -0x58(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  shr    $0x4,%rax
  mov    -0x50(%rbp,%rax,4),%edx
  mov    -0x70(%rbp),%rcx
  mov    -0x58(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  and    $0xf,%eax
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    -0x70(%rbp),%rcx
  mov    -0x58(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  shr    $0x4,%rax
  or     %esi,%edx
  mov    %edx,-0x50(%rbp,%rax,4)
  addq   $0x1,-0x58(%rbp)
L42:
  mov    -0x70(%rbp),%rdx
  mov    -0x58(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L55
  movq   $0x0,-0x58(%rbp)
  jmp    L46
L36:
  mov    -0x68(%rbp),%rdx
  mov    -0x58(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  shr    $0x4,%rax
  mov    -0x50(%rbp,%rax,4),%edx
  mov    -0x68(%rbp),%rcx
  mov    -0x58(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  and    $0xf,%eax
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  not    %eax
  and    %edx,%eax
  test   %eax,%eax
  jne    L61
  addq   $0x1,-0x58(%rbp)
L46:
  mov    -0x68(%rbp),%rdx
  mov    -0x58(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L36
  jmp    L0
L61:
  nop
L0:
  mov    -0x58(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

assistant: ```c
size_t strspn(const char *s, const char *accept) {
  uint16_t mask[16];
  size_t i;
  memclr(mask, sizeof(uint16_t) * 16);
  for (i = 0; accept[i]; i++) {
    mask[(size_t)accept[i] >> 4] |= (1 << ((size_t)accept[i] & 0xF));
  }
  for (i = 0; s[i]; i++) {
    if (mask[(size_t)s[i] >> 4] & ~(1 << ((size_t)s[i] & 0xF))) {
      break;
    }
  }
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strspn>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x40,%esi
  call   <memclr@plt>
  movzbl 0x0(%rbp),%ecx
  test   %cl,%cl
  je     L30
  lea    0x1(%rbp),%rsi
  mov    $0x1,%edi
L0:
  movsbq %cl,%rax
  shr    $0x4,%rax
  and    $0xf,%ecx
  mov    %edi,%edx
  shl    %cl,%edx
  or     %edx,(%rsp,%rax,4)
  add    $0x1,%rsi
  movzbl -0x1(%rsi),%ecx
  test   %cl,%cl
  jne    L0
L30:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L41
  mov    $0x0,%esi
  mov    $0x1,%edi
L5:
  mov    %eax,%ecx
  and    $0xf,%ecx
  mov    %edi,%edx
  shl    %cl,%edx
  not    %edx
  movsbq %al,%rax
  shr    $0x4,%rax
  and    (%rsp,%rax,4),%edx
  jne    L31
  add    $0x1,%rsi
  movzbl (%rbx,%rsi,1),%eax
  test   %al,%al
  jne    L5
L31:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  mov    %rsi,%rax
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  ret
L41:
  mov    $0x0,%esi
  jmp    L31
L45:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t strspn(const char *s, const char *accept) {
  uint16_t mask[16];
  size_t i;
  memclr(mask, sizeof(uint16_t) * 16);
  for (i = 0; accept[i]; i++) {
    mask[(size_t)accept[i] >> 4] |= (1 << ((size_t)accept[i] & 0xF));
  }
  for (i = 0; s[i]; i++) {
    if (mask[(size_t)s[i] >> 4] & ~(1 << ((size_t)s[i] & 0xF))) {
      break;
    }
  }
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strspn>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x40,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <memclr@plt>
  movzbl 0x0(%rbp),%ecx
  test   %cl,%cl
  je     L33
  lea    0x1(%rbp),%rsi
  mov    $0x1,%edi
  xchg   %ax,%ax
L7:
  movsbq %cl,%rax
  mov    %edi,%edx
  and    $0xf,%ecx
  add    $0x1,%rsi
  shl    %cl,%edx
  movzbl -0x1(%rsi),%ecx
  shr    $0x4,%rax
  or     %edx,(%rsp,%rax,4)
  test   %cl,%cl
  jne    L7
L33:
  movsbq (%rbx),%rax
  xor    %r8d,%r8d
  test   %al,%al
  je     L67
  mov    $0x1,%esi
  jmp    L6
L59:
  add    $0x1,%r8
  movsbq (%rbx,%r8,1),%rax
  test   %al,%al
  je     L67
L6:
  mov    %eax,%ecx
  mov    %esi,%edx
  shr    $0x4,%rax
  and    $0xf,%ecx
  shl    %cl,%edx
  not    %edx
  and    (%rsp,%rax,4),%edx
  je     L59
L67:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x58,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t strspn(const char *s, const char *accept) {
  uint16_t mask[16];
  size_t i;
  memclr(mask, sizeof(uint16_t) * 16);
  for (i = 0; accept[i]; i++) {
    mask[(size_t)accept[i] >> 4] |= (1 << ((size_t)accept[i] & 0xF));
  }
  for (i = 0; s[i]; i++) {
    if (mask[(size_t)s[i] >> 4] & ~(1 << ((size_t)s[i] & 0xF))) {
      break;
    }
  }
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strspn>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x40,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <memclr@plt>
  movzbl 0x0(%rbp),%ecx
  test   %cl,%cl
  je     L40
  lea    0x1(%rbp),%rsi
  mov    $0x1,%edi
  xchg   %ax,%ax
L51:
  movsbq %cl,%rax
  mov    %edi,%edx
  and    $0xf,%ecx
  add    $0x1,%rsi
  shl    %cl,%edx
  movzbl -0x1(%rsi),%ecx
  shr    $0x4,%rax
  or     %edx,(%rsp,%rax,4)
  test   %cl,%cl
  jne    L51
L40:
  movsbq (%rbx),%rax
  xor    %r8d,%r8d
  test   %al,%al
  je     L7
  mov    $0x1,%esi
  jmp    L2
L62:
  add    $0x1,%r8
  movsbq (%rbx,%r8,1),%rax
  test   %al,%al
  je     L7
L2:
  mov    %eax,%ecx
  mov    %esi,%edx
  shr    $0x4,%rax
  and    $0xf,%ecx
  shl    %cl,%edx
  not    %edx
  and    (%rsp,%rax,4),%edx
  je     L62
L7:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L52
  add    $0x58,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
L52:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t strspn(const char *s, const char *accept) {
  uint16_t mask[16];
  size_t i;
  memclr(mask, sizeof(uint16_t) * 16);
  for (i = 0; accept[i]; i++) {
    mask[(size_t)accept[i] >> 4] |= (1 << ((size_t)accept[i] & 0xF));
  }
  for (i = 0; s[i]; i++) {
    if (mask[(size_t)s[i] >> 4] & ~(1 << ((size_t)s[i] & 0xF))) {
      break;
    }
  }
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<apply_e_mod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rdx
  mov    -0x38(%rbp),%rcx
  mov    -0x30(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <copy_str@plt>
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rcx
  mov    -0x30(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <split_ext@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x30(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memmove@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
int apply_e_mod(const char *path, char *buf, size_t buf_len) {
  int root_len;
  const char *ext_pos;
  copy_str(buf, buf_len, path);
  split_ext(buf, &root_len, &ext_pos);
  memmove(buf, ext_pos, strlen(ext_pos) + 1);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<apply_e_mod>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rsi,%rbx
  mov    %rdx,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rdi,%rdx
  mov    %rbx,%rdi
  call   <copy_str@plt>
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  mov    %rbx,%rdi
  call   <split_ext@plt>
  mov    0x10(%rsp),%rbp
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <memmove@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int apply_e_mod(const char *path, char *buf, size_t buf_len) {
  int root_len;
  const char *ext_pos;
  copy_str(buf, buf_len, path);
  split_ext(buf, &root_len, &ext_pos);
  memmove(buf, ext_pos, strlen(ext_pos) + 1);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<apply_e_mod>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  mov    %rdx,%rsi
  mov    %rdi,%rdx
  mov    %rbp,%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <copy_str@plt>
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <split_ext@plt>
  mov    0x10(%rsp),%r12
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  lea    0x1(%rax),%rdx
  call   <memmove@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int apply_e_mod(const char *path, char *buf, size_t buf_len) {
  int root_len;
  const char *ext_pos;
  copy_str(buf, buf_len, path);
  split_ext(buf, &root_len, &ext_pos);
  memmove(buf, ext_pos, strlen(ext_pos) + 1);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<apply_e_mod>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  mov    %rdx,%rsi
  mov    %rdi,%rdx
  mov    %rbp,%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <copy_str@plt>
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <split_ext@plt>
  mov    0x10(%rsp),%r12
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  lea    0x1(%rax),%rdx
  call   <memmove@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int apply_e_mod(const char *path, char *buf, size_t buf_len) {
  int root_len;
  const char *ext_pos;
  copy_str(buf, buf_len, path);
  split_ext(buf, &root_len, &ext_pos);
  memmove(buf, ext_pos, strlen(ext_pos) + 1);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x1,-0x18(%rbp)
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
L7:
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  je     L18
  mov    -0x18(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  cmp    %eax,%edx
  jne    L20
  addl   $0x1,-0xc(%rbp)
  addl   $0x1,-0x14(%rbp)
  jmp    L2
L20:
  movl   $0x0,-0x14(%rbp)
L2:
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L7
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L7
L18:
  nop
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D22
type_infered: string
name: retrieve_data
arguments: label: D7
type_infered: string

tool: D22: %d

tool: D7: %d
%d

assistant: ```c
int main() {
  int currentCount = 0, maxCount = 0, stacking = 0, interest, input = 1;
  scanf("%d", &interest);
  while (1) {
    scanf("%d", &input);
    if (!input) {
      break;
    }
    if (input == interest) {
      stacking++;
      currentCount++;
    } else {
      currentCount = 0;
    }
    if (maxCount < currentCount) {
      maxCount = currentCount;
    }
  }
  printf("%d\n%d", maxCount, stacking);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x1,0x4(%rsp)
  mov    %rsp,%rsi
  lea    D5(%rip),%rdi
  call   <scanf@plt>
  mov    $0x0,%r13d
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  lea    D5(%rip),%r12
  jmp    L34
L27:
  add    $0x1,%r13d
  add    $0x1,%ebx
L40:
  cmp    %ebx,%ebp
  cmovl  %ebx,%ebp
L34:
  lea    0x4(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  je     L22
  cmp    (%rsp),%eax
  je     L27
  mov    $0x0,%ebx
  jmp    L40
L22:
  mov    %r13d,%edx
  mov    %ebp,%esi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D5
type_guessed: string
name: unpack_data
arguments: tag: D10
type_guessed: string

tool: D5: %d

tool: D10: %d
%d

assistant: ```c
int main() {
  int currentCount = 0, maxCount = 0, stacking = 0, interest, input = 1;
  scanf("%d", &interest);
  while (1) {
    scanf("%d", &input);
    if (!input) {
      break;
    }
    if (input == interest) {
      stacking++;
      currentCount++;
    } else {
      currentCount = 0;
    }
    if (maxCount < currentCount) {
      maxCount = currentCount;
    }
  }
  printf("%d\n%d", maxCount, stacking);
  return 0;
}
```