ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<formatNumber>:
  endbr64
  push   %r14
  mov    %edi,%edx
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  mov    %rsi,%r12
  lea    D110(%rip),%rsi
  push   %rbp
  push   %rbx
  lea    D43(%rip),%rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  lea    D55(%rip),%rax
  lea    0x50(%rsp),%r13
  mov    %r13,%rdi
  mov    %rax,(%rsp)
  lea    D72(%rip),%rax
  mov    %rax,0x8(%rsp)
  lea    D68(%rip),%rax
  mov    %rax,0x10(%rsp)
  lea    D2(%rip),%rax
  mov    %rax,0x18(%rsp)
  lea    D57(%rip),%rax
  mov    %rax,0x20(%rsp)
  lea    D80(%rip),%rax
  mov    %rax,0x28(%rsp)
  lea    D116(%rip),%rax
  mov    %rax,0x30(%rsp)
  lea    D84(%rip),%rax
  mov    %rax,0x38(%rsp)
  lea    D31(%rip),%rax
  mov    %rax,0x40(%rsp)
  lea    D105(%rip),%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  call   <sprintf@plt>
  sub    $0x1,%eax
  lea    0x51(%rsp,%rax,1),%rbp
  nopl   0x0(%rax,%rax,1)
L18:
  movsbl 0x0(%r13),%eax
  movslq %r14d,%rdi
  mov    %rbx,%rsi
  add    $0x1,%r13
  add    %r12,%rdi
  sub    $0x30,%eax
  cltq
  mov    (%rsp,%rax,8),%rdx
  xor    %eax,%eax
  call   <sprintf@plt>
  add    %eax,%r14d
  cmp    %rbp,%r13
  jne    L18
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0xa0,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D110
type_infered: string
name: retrieve_data
arguments: label: D43
type_infered: string
name: retrieve_data
arguments: label: D55
type_infered: string
name: retrieve_data
arguments: label: D72
type_infered: string
name: retrieve_data
arguments: label: D68
type_infered: string
name: retrieve_data
arguments: label: D2
type_infered: string
name: retrieve_data
arguments: label: D57
type_infered: string
name: retrieve_data
arguments: label: D80
type_infered: string
name: retrieve_data
arguments: label: D116
type_infered: string
name: retrieve_data
arguments: label: D84
type_infered: string
name: retrieve_data
arguments: label: D31
type_infered: string
name: retrieve_data
arguments: label: D105
type_infered: string

tool: D110: %d

tool: D43:  %s

tool: D55: zero

tool: D72: one

tool: D68: two

tool: D2: three

tool: D57: four

tool: D80: five

tool: D116: six

tool: D84: seven

tool: D31: eight

tool: D105: nine

assistant: ```c
char *formatNumber(int n, char *outbuf) {
  int ochar = 0, ichar = 0;
  int i;
  char tmp[64];
  const char *NUMBERS[] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
  ichar = sprintf(tmp, "%d", n);
  for (i = 0; i < ichar; i++)
    ochar += sprintf(outbuf + ochar, " %s", NUMBERS[tmp[i] - '0']);
  return outbuf;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<formatNumber>:
  endbr64
  push   %r14
  lea    D96(%rip),%rcx
  mov    %edi,%edx
  xor    %r14d,%r14d
  push   %r13
  movq   %rcx,%xmm0
  lea    D111(%rip),%rcx
  push   %r12
  mov    %rsi,%r12
  lea    D91(%rip),%rsi
  push   %rbp
  push   %rbx
  lea    D68(%rip),%rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  lea    D62(%rip),%rax
  lea    0x50(%rsp),%r13
  mov    %r13,%rdi
  movq   %rax,%xmm1
  lea    D119(%rip),%rax
  punpcklqdq %xmm1,%xmm0
  movq   %rax,%xmm2
  lea    D97(%rip),%rax
  movaps %xmm0,(%rsp)
  movq   %rcx,%xmm0
  lea    D89(%rip),%rcx
  movq   %rax,%xmm3
  punpcklqdq %xmm2,%xmm0
  lea    D101(%rip),%rax
  movaps %xmm0,0x10(%rsp)
  movq   %rcx,%xmm0
  movq   %rax,%xmm4
  lea    D71(%rip),%rcx
  punpcklqdq %xmm3,%xmm0
  lea    D76(%rip),%rax
  movaps %xmm0,0x20(%rsp)
  movq   %rcx,%xmm0
  movq   %rax,%xmm5
  lea    D3(%rip),%rcx
  punpcklqdq %xmm4,%xmm0
  xor    %eax,%eax
  movaps %xmm0,0x30(%rsp)
  movq   %rcx,%xmm0
  punpcklqdq %xmm5,%xmm0
  movaps %xmm0,0x40(%rsp)
  call   <sprintf@plt>
  sub    $0x1,%eax
  lea    0x51(%rsp,%rax,1),%rbp
  nopl   (%rax)
L14:
  movsbl 0x0(%r13),%eax
  movslq %r14d,%rdi
  mov    %rbx,%rsi
  add    $0x1,%r13
  add    %r12,%rdi
  sub    $0x30,%eax
  cltq
  mov    (%rsp,%rax,8),%rdx
  xor    %eax,%eax
  call   <sprintf@plt>
  add    %eax,%r14d
  cmp    %r13,%rbp
  jne    L14
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0xa0,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D96
data_type: string
name: disassemble_data
arguments: data_label: D111
data_type: string
name: disassemble_data
arguments: data_label: D91
data_type: string
name: disassemble_data
arguments: data_label: D68
data_type: string
name: disassemble_data
arguments: data_label: D62
data_type: string
name: disassemble_data
arguments: data_label: D119
data_type: string
name: disassemble_data
arguments: data_label: D97
data_type: string
name: disassemble_data
arguments: data_label: D89
data_type: string
name: disassemble_data
arguments: data_label: D101
data_type: string
name: disassemble_data
arguments: data_label: D71
data_type: string
name: disassemble_data
arguments: data_label: D76
data_type: string
name: disassemble_data
arguments: data_label: D3
data_type: string

tool: D96: zero

tool: D111: two

tool: D91: %d

tool: D68:  %s

tool: D62: one

tool: D119: three

tool: D97: five

tool: D89: four

tool: D101: seven

tool: D71: six

tool: D76: nine

tool: D3: eight

assistant: ```c
char *formatNumber(int n, char *outbuf) {
  int ochar = 0, ichar = 0;
  int i;
  char tmp[64];
  const char *NUMBERS[] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
  ichar = sprintf(tmp, "%d", n);
  for (i = 0; i < ichar; i++)
    ochar += sprintf(outbuf + ochar, " %s", NUMBERS[tmp[i] - '0']);
  return outbuf;
}
```
user: What is the c source code of the assembly code below:

```asm
<RandomFunc>:
  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
  movzbl (%rax),%eax
  sub    $0x6b,%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  shr    $0x3,%al
  movzbl %al,%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L51
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L28
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L33
  movzbl -0x9(%rbp),%eax
  mov    %eax,%edx
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  shl    $0x3,%eax
  and    $0x78,%eax
  or     %edx,%eax
  mov    %al,-0x9(%rbp)
  jmp    L16
L33:
  movzbl -0x9(%rbp),%eax
  mov    %eax,%edx
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  shl    $0x2,%eax
  and    $0x1c,%eax
  or     %edx,%eax
  mov    %al,-0x9(%rbp)
  jmp    L16
L28:
  movzbl -0x9(%rbp),%eax
  mov    %eax,%edx
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  shl    $0x3,%eax
  and    $0x78,%eax
  or     %edx,%eax
  mov    %al,-0x9(%rbp)
  jmp    L16
L51:
  movzbl -0x9(%rbp),%eax
  mov    %eax,%edi
  movzbl -0x9(%rbp),%eax
  movzbl %al,%edx
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  and    $0xe,%eax
  or     $0x1,%eax
  mov    %edx,%esi
  mov    %eax,%ecx
  sar    %cl,%esi
  movzbl -0x9(%rbp),%eax
  movzbl %al,%edx
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  and    $0xe,%eax
  or     $0x1,%eax
  mov    %eax,%ecx
  mov    $0x8,%eax
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     %esi,%eax
  shl    $0x4,%eax
  and    $0xfffffff0,%eax
  or     %edi,%eax
  mov    %al,-0x9(%rbp)
L16:
  movzbl -0x9(%rbp),%eax
  xor    $0xffffffde,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  {
    state[0UL] = (input[0UL] - 51238316UL) - (unsigned char)191;
    if ((state[0UL] >> (unsigned char)3) & (unsigned char)1) {
      if (state[0UL] & (unsigned char)1) {
        if (state[0UL] & (unsigned char)1) {
          state[0UL] |= (state[0UL] & (unsigned char)15) << 3UL;
        } else {
          state[0UL] |= (state[0UL] & (unsigned char)7) << 2UL;
        }
      } else {
        state[0UL] |= (state[0UL] & (unsigned char)15) << 3UL;
      }
    } else {
      state[0UL] |= (((state[0UL] >> ((state[0UL] & (unsigned char)15) | 1UL)) | (state[0UL] << (8 - ((state[0UL] & (unsigned char)15) | 1UL)))) & (unsigned char)31) << 4UL;
    }
    output[0UL] = state[0UL] ^ 306719454UL;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<RandomFunc>:
  endbr64
  mov    %rsi,%rdx
  movzbl (%rdi),%eax
  sub    $0x6b,%eax
  test   $0x8,%al
  je     L16
  movzbl %al,%ecx
  test   $0x1,%al
  je     L0
  shl    $0x3,%ecx
  and    $0x78,%ecx
  or     %ecx,%eax
  jmp    L5
L0:
  shl    $0x3,%ecx
  and    $0x78,%ecx
  or     %ecx,%eax
  jmp    L5
L16:
  movzbl %al,%edi
  mov    %eax,%r8d
  and    $0xe,%r8d
  or     $0x1,%r8d
  mov    $0x8,%ecx
  sub    %r8d,%ecx
  mov    %edi,%esi
  shl    %cl,%esi
  mov    %r8d,%ecx
  sar    %cl,%edi
  or     %edi,%esi
  shl    $0x4,%esi
  or     %esi,%eax
L5:
  xor    $0xffffffde,%eax
  mov    %al,(%rdx)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  {
    state[0UL] = (input[0UL] - 51238316UL) - (unsigned char)191;
    if ((state[0UL] >> (unsigned char)3) & (unsigned char)1) {
      if (state[0UL] & (unsigned char)1) {
        if (state[0UL] & (unsigned char)1) {
          state[0UL] |= (state[0UL] & (unsigned char)15) << 3UL;
        } else {
          state[0UL] |= (state[0UL] & (unsigned char)7) << 2UL;
        }
      } else {
        state[0UL] |= (state[0UL] & (unsigned char)15) << 3UL;
      }
    } else {
      state[0UL] |= (((state[0UL] >> ((state[0UL] & (unsigned char)15) | 1UL)) | (state[0UL] << (8 - ((state[0UL] & (unsigned char)15) | 1UL)))) & (unsigned char)31) << 4UL;
    }
    output[0UL] = state[0UL] ^ 306719454UL;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  sub    $0x6b,%eax
  movzbl %al,%edx
  test   $0x8,%al
  je     L5
  shl    $0x3,%edx
  and    $0x78,%edx
  or     %edx,%eax
  xor    $0xffffffde,%eax
  mov    %al,(%rsi)
  ret
  nop
L5:
  mov    %eax,%r8d
  mov    $0x8,%ecx
  and    $0xe,%r8d
  or     $0x1,%r8d
  movzbl %r8b,%edi
  sub    %edi,%ecx
  mov    %edx,%edi
  shl    %cl,%edi
  mov    %r8d,%ecx
  sar    %cl,%edx
  or     %edx,%edi
  shl    $0x4,%edi
  or     %edi,%eax
  xor    $0xffffffde,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  {
    state[0UL] = (input[0UL] - 51238316UL) - (unsigned char)191;
    if ((state[0UL] >> (unsigned char)3) & (unsigned char)1) {
      if (state[0UL] & (unsigned char)1) {
        if (state[0UL] & (unsigned char)1) {
          state[0UL] |= (state[0UL] & (unsigned char)15) << 3UL;
        } else {
          state[0UL] |= (state[0UL] & (unsigned char)7) << 2UL;
        }
      } else {
        state[0UL] |= (state[0UL] & (unsigned char)15) << 3UL;
      }
    } else {
      state[0UL] |= (((state[0UL] >> ((state[0UL] & (unsigned char)15) | 1UL)) | (state[0UL] << (8 - ((state[0UL] & (unsigned char)15) | 1UL)))) & (unsigned char)31) << 4UL;
    }
    output[0UL] = state[0UL] ^ 306719454UL;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  sub    $0x6b,%eax
  movzbl %al,%edx
  test   $0x8,%al
  je     L0
  shl    $0x3,%edx
  and    $0x78,%edx
  or     %edx,%eax
  xor    $0xffffffde,%eax
  mov    %al,(%rsi)
  ret
  nop
L0:
  mov    %eax,%r8d
  mov    $0x8,%ecx
  and    $0xe,%r8d
  or     $0x1,%r8d
  movzbl %r8b,%edi
  sub    %edi,%ecx
  mov    %edx,%edi
  shl    %cl,%edi
  mov    %r8d,%ecx
  sar    %cl,%edx
  or     %edx,%edi
  shl    $0x4,%edi
  or     %edi,%eax
  xor    $0xffffffde,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  {
    state[0UL] = (input[0UL] - 51238316UL) - (unsigned char)191;
    if ((state[0UL] >> (unsigned char)3) & (unsigned char)1) {
      if (state[0UL] & (unsigned char)1) {
        if (state[0UL] & (unsigned char)1) {
          state[0UL] |= (state[0UL] & (unsigned char)15) << 3UL;
        } else {
          state[0UL] |= (state[0UL] & (unsigned char)7) << 2UL;
        }
      } else {
        state[0UL] |= (state[0UL] & (unsigned char)15) << 3UL;
      }
    } else {
      state[0UL] |= (((state[0UL] >> ((state[0UL] & (unsigned char)15) | 1UL)) | (state[0UL] << (8 - ((state[0UL] & (unsigned char)15) | 1UL)))) & (unsigned char)31) << 4UL;
    }
    output[0UL] = state[0UL] ^ 306719454UL;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<pfftw_permutation_512>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x24(%rbp),%eax
  lea    0x3(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x2,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x2,-0x18(%rbp)
  jg     L6
  mov    -0x18(%rbp),%eax
  shl    $0x7,%eax
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <pfftw_permutation_128@plt>
  add    %ebx,%eax
  jmp    L3
L6:
  mov    -0x18(%rbp),%eax
  shl    $0x7,%eax
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  cltd
  shr    $0x19,%edx
  add    %edx,%eax
  and    $0x7f,%eax
  sub    %edx,%eax
  mov    %eax,%edi
  call   <pfftw_permutation_128@plt>
  add    %ebx,%eax
L3:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int pfftw_permutation_512(int i) {
  int i1 = i % 4;
  int i2 = i / 4;
  if (i1 <= 2)
    return (i1 * 128 + pfftw_permutation_128(i2));
  else
    return (i1 * 128 + pfftw_permutation_128((i2 + 1) % 128));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<pfftw_permutation_512>:
  endbr64
  push   %rbx
  mov    %edi,%edx
  mov    %edi,%ecx
  sar    $0x1f,%ecx
  shr    $0x1e,%ecx
  lea    (%rdi,%rcx,1),%ebx
  and    $0x3,%ebx
  sub    %ecx,%ebx
  lea    0x3(%rdi),%edi
  test   %edx,%edx
  cmovns %edx,%edi
  sar    $0x2,%edi
  cmp    $0x2,%ebx
  jg     L6
  call   <pfftw_permutation_128@plt>
  shl    $0x7,%ebx
  add    %ebx,%eax
L8:
  pop    %rbx
  ret
L6:
  add    $0x1,%edi
  mov    %edi,%eax
  sar    $0x1f,%eax
  shr    $0x19,%eax
  add    %eax,%edi
  and    $0x7f,%edi
  sub    %eax,%edi
  call   <pfftw_permutation_128@plt>
  add    $0x180,%eax
  jmp    L8
```

assistant: ```c
int pfftw_permutation_512(int i) {
  int i1 = i % 4;
  int i2 = i / 4;
  if (i1 <= 2)
    return (i1 * 128 + pfftw_permutation_128(i2));
  else
    return (i1 * 128 + pfftw_permutation_128((i2 + 1) % 128));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<pfftw_permutation_512>:
  endbr64
  mov    %edi,%eax
  push   %rbx
  cltd
  shr    $0x1e,%edx
  lea    (%rdi,%rdx,1),%ebx
  add    $0x3,%edi
  and    $0x3,%ebx
  sub    %edx,%ebx
  test   %eax,%eax
  cmovns %eax,%edi
  sar    $0x2,%edi
  cmp    $0x3,%ebx
  je     L5
  call   <pfftw_permutation_128@plt>
  shl    $0x7,%ebx
  add    %ebx,%eax
  pop    %rbx
  ret
  nop
L5:
  add    $0x1,%edi
  mov    %edi,%eax
  sar    $0x1f,%eax
  shr    $0x19,%eax
  add    %eax,%edi
  and    $0x7f,%edi
  sub    %eax,%edi
  call   <pfftw_permutation_128@plt>
  pop    %rbx
  add    $0x180,%eax
  ret
```

assistant: ```c
int pfftw_permutation_512(int i) {
  int i1 = i % 4;
  int i2 = i / 4;
  if (i1 <= 2)
    return (i1 * 128 + pfftw_permutation_128(i2));
  else
    return (i1 * 128 + pfftw_permutation_128((i2 + 1) % 128));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pfftw_permutation_512>:
  endbr64
  mov    %edi,%eax
  push   %rbx
  cltd
  shr    $0x1e,%edx
  lea    (%rdi,%rdx,1),%ebx
  add    $0x3,%edi
  and    $0x3,%ebx
  sub    %edx,%ebx
  test   %eax,%eax
  cmovns %eax,%edi
  sar    $0x2,%edi
  cmp    $0x3,%ebx
  je     L3
  call   <pfftw_permutation_128@plt>
  shl    $0x7,%ebx
  add    %ebx,%eax
  pop    %rbx
  ret
  nop
L3:
  add    $0x1,%edi
  mov    %edi,%eax
  sar    $0x1f,%eax
  shr    $0x19,%eax
  add    %eax,%edi
  and    $0x7f,%edi
  sub    %eax,%edi
  call   <pfftw_permutation_128@plt>
  pop    %rbx
  add    $0x180,%eax
  ret
```

assistant: ```c
int pfftw_permutation_512(int i) {
  int i1 = i % 4;
  int i2 = i / 4;
  if (i1 <= 2)
    return (i1 * 128 + pfftw_permutation_128(i2));
  else
    return (i1 * 128 + pfftw_permutation_128((i2 + 1) % 128));
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<CountFiles>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x430,%rsp
  mov    %rdi,-0x428(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x41c(%rbp)
  mov    D0(%rip),%rax
  mov    (%rax),%edx
  mov    -0x428(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <fseek@plt>
L11:
  mov    -0x428(%rbp),%rdx
  lea    -0x410(%rbp),%rax
  mov    $0x400,%esi
  mov    %rax,%rdi
  call   <fgets@plt>
  mov    %rax,-0x418(%rbp)
  cmpq   $0x0,-0x418(%rbp)
  je     L31
  lea    -0x410(%rbp),%rax
  mov    %rax,%rdi
  call   <TrimBuffer@plt>
  movzbl -0x410(%rbp),%eax
  test   %al,%al
  je     L11
  addl   $0x1,-0x41c(%rbp)
  jmp    L11
L31:
  nop
  mov    -0x41c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L32
  call   <__stack_chk_fail@plt>
L32:
  leave
  ret
```

assistant: ```c
int CountFiles(FILE *fptr) {
  char *result;
  int counter = 0;
  char buffer[1024];
  fseek(fptr, 0, SEEK_SET);
  for (;;) {
    result = fgets(buffer, 1024, fptr);
    if (!result)
      break;
    TrimBuffer(buffer);
    if (buffer[0] != '\0')
      counter++;
  }
  return counter;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<CountFiles>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %rdi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    D2(%rip),%rax
  mov    (%rax),%edx
  mov    $0x0,%esi
  call   <fseek@plt>
  mov    $0x0,%ebx
  mov    %rsp,%rbp
  jmp    L6
L16:
  mov    %rbp,%rdi
  call   <TrimBuffer@plt>
  cmpb   $0x1,(%rsp)
  sbb    $0xffffffff,%ebx
L6:
  mov    %r12,%rdx
  mov    $0x400,%esi
  mov    %rbp,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L16
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  mov    %ebx,%eax
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int CountFiles(FILE *fptr) {
  char *result;
  int counter = 0;
  char buffer[1024];
  fseek(fptr, 0, SEEK_SET);
  for (;;) {
    result = fgets(buffer, 1024, fptr);
    if (!result)
      break;
    TrimBuffer(buffer);
    if (buffer[0] != '\0')
      counter++;
  }
  return counter;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<CountFiles>:
  endbr64
  push   %r12
  xor    %esi,%esi
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  mov    D9(%rip),%rax
  mov    %rsp,%rbx
  mov    (%rax),%edx
  call   <fseek@plt>
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L26:
  mov    %rbx,%rdi
  call   <TrimBuffer@plt>
  cmpb   $0x1,(%rsp)
  sbb    $0xffffffff,%r12d
L18:
  mov    %rbp,%rdx
  mov    $0x400,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L26
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x410,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int CountFiles(FILE *fptr) {
  char *result;
  int counter = 0;
  char buffer[1024];
  fseek(fptr, 0, SEEK_SET);
  for (;;) {
    result = fgets(buffer, 1024, fptr);
    if (!result)
      break;
    TrimBuffer(buffer);
    if (buffer[0] != '\0')
      counter++;
  }
  return counter;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<CountFiles>:
  endbr64
  push   %r12
  xor    %esi,%esi
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  mov    D8(%rip),%rax
  mov    %rsp,%rbx
  mov    (%rax),%edx
  call   <fseek@plt>
  jmp    L27
  nopl   0x0(%rax,%rax,1)
L29:
  mov    %rbx,%rdi
  call   <TrimBuffer@plt>
  cmpb   $0x1,(%rsp)
  sbb    $0xffffffff,%r12d
L27:
  mov    %rbp,%rdx
  mov    $0x400,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L29
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x410,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int CountFiles(FILE *fptr) {
  char *result;
  int counter = 0;
  char buffer[1024];
  fseek(fptr, 0, SEEK_SET);
  for (;;) {
    result = fgets(buffer, 1024, fptr);
    if (!result)
      break;
    TrimBuffer(buffer);
    if (buffer[0] != '\0')
      counter++;
  }
  return counter;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ExtractFileBase>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L12
L33:
  subq   $0x1,-0x8(%rbp)
L12:
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  je     L52
  mov    -0x8(%rbp),%rax
  sub    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L33
  jmp    L52
L20:
  mov    -0x8(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x20(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L52:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L11
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L20
L11:
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void ExtractFileBase(const char *path, char *dest) {
  const char *src;
  src = path + strlen(path) - 1;
  while (src != path && *(src - 1) != '/')
    src--;
  while (*src && *src != '.') {
    *dest++ = *src++;
  }
  *dest = 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ExtractFileBase>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbp,%rax,1),%rax
  cmp    %rax,%rbp
  je     L40
L8:
  cmpb   $0x2f,-0x1(%rax)
  je     L40
  sub    $0x1,%rax
  cmp    %rax,%rbp
  jne    L8
L40:
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L63
  cmp    $0x2e,%dl
  je     L63
L9:
  add    $0x1,%rax
  add    $0x1,%rbx
  mov    %dl,-0x1(%rbx)
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L63
  cmp    $0x2e,%dl
  jne    L9
L63:
  movb   $0x0,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void ExtractFileBase(const char *path, char *dest) {
  const char *src;
  src = path + strlen(path) - 1;
  while (src != path && *(src - 1) != '/')
    src--;
  while (*src && *src != '.') {
    *dest++ = *src++;
  }
  *dest = 0;
}
```