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
<test001>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x4,%edi
  call   <bmalloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    $0xd,%edi
  call   <bmalloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <bfree@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <bfree@plt>
  mov    $0x0,%eax
  call   <print_buddy@plt>
  nop
  leave
  ret
```

assistant: ```c
void test001() {
  void *p1, *p2;
  p1 = bmalloc(4);
  p2 = bmalloc(13);
  bfree(p2);
  bfree(p1);
  print_buddy();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test001>:
  endbr64
  push   %rbx
  mov    $0x4,%edi
  call   <bmalloc@plt>
  mov    %rax,%rbx
  mov    $0xd,%edi
  call   <bmalloc@plt>
  mov    %rax,%rdi
  call   <bfree@plt>
  mov    %rbx,%rdi
  call   <bfree@plt>
  mov    $0x0,%eax
  call   <print_buddy@plt>
  pop    %rbx
  ret
```

assistant: ```c
void test001() {
  void *p1, *p2;
  p1 = bmalloc(4);
  p2 = bmalloc(13);
  bfree(p2);
  bfree(p1);
  print_buddy();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test001>:
  endbr64
  push   %rbp
  mov    $0x4,%edi
  call   <bmalloc@plt>
  mov    $0xd,%edi
  mov    %rax,%rbp
  call   <bmalloc@plt>
  mov    %rax,%rdi
  call   <bfree@plt>
  mov    %rbp,%rdi
  call   <bfree@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    1070 <print_buddy@plt>
```

assistant: ```c
void test001() {
  void *p1, *p2;
  p1 = bmalloc(4);
  p2 = bmalloc(13);
  bfree(p2);
  bfree(p1);
  print_buddy();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test001>:
  endbr64
  push   %rbp
  mov    $0x4,%edi
  call   <bmalloc@plt>
  mov    $0xd,%edi
  mov    %rax,%rbp
  call   <bmalloc@plt>
  mov    %rax,%rdi
  call   <bfree@plt>
  mov    %rbp,%rdi
  call   <bfree@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    1070 <print_buddy@plt>
```

assistant: ```c
void test001() {
  void *p1, *p2;
  p1 = bmalloc(4);
  p2 = bmalloc(13);
  bfree(p2);
  bfree(p1);
  print_buddy();
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<numDigits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0xa,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L1:
  addl   $0x1,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x8(%rbp)
L2:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  mov    %edx,%eax
  cmp    %eax,-0x14(%rbp)
  jne    L1
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  pop    %rbp
  ret
```

assistant: ```c
int numDigits(int num) {
  int factor = 10;
  int count = 0;
  while (num % factor != num) {
    count++;
    factor = factor * 10;
  }
  return count + 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<numDigits>:
  endbr64
  mov    %edi,%esi
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  mov    %edi,%edx
  sub    %eax,%edx
  cmp    %edx,%edi
  je     L14
  mov    $0x0,%edi
  mov    $0xa,%ecx
L18:
  add    $0x1,%edi
  lea    (%rcx,%rcx,4),%ecx
  add    %ecx,%ecx
  mov    %esi,%eax
  cltd
  idiv   %ecx
  cmp    %esi,%edx
  jne    L18
L28:
  lea    0x1(%rdi),%eax
  ret
L14:
  mov    $0x0,%edi
  jmp    L28
```

assistant: ```c
int numDigits(int num) {
  int factor = 10;
  int count = 0;
  while (num % factor != num) {
    count++;
    factor = factor * 10;
  }
  return count + 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<numDigits>:
  endbr64
  movslq %edi,%rax
  mov    %rax,%rsi
  imul   $0x66666667,%rax,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  mov    %esi,%edx
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%edx
  cmp    %edx,%esi
  je     L3
  xor    %edi,%edi
  mov    $0xa,%ecx
L15:
  mov    %esi,%eax
  lea    (%rcx,%rcx,4),%ecx
  mov    %edi,%r8d
  add    $0x1,%edi
  add    %ecx,%ecx
  cltd
  idiv   %ecx
  cmp    %esi,%edx
  jne    L15
  lea    0x2(%r8),%eax
  ret
  nopl   0x0(%rax)
L3:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int numDigits(int num) {
  int factor = 10;
  int count = 0;
  while (num % factor != num) {
    count++;
    factor = factor * 10;
  }
  return count + 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<numDigits>:
  endbr64
  movslq %edi,%rax
  mov    %rax,%rsi
  imul   $0x66666667,%rax,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  mov    %esi,%edx
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%edx
  cmp    %edx,%esi
  je     L13
  xor    %edi,%edi
  mov    $0xa,%ecx
L11:
  mov    %esi,%eax
  lea    (%rcx,%rcx,4),%ecx
  mov    %edi,%r8d
  add    $0x1,%edi
  add    %ecx,%ecx
  cltd
  idiv   %ecx
  cmp    %esi,%edx
  jne    L11
  lea    0x2(%r8),%eax
  ret
  nopl   0x0(%rax)
L13:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int numDigits(int num) {
  int factor = 10;
  int count = 0;
  while (num % factor != num) {
    count++;
    factor = factor * 10;
  }
  return count + 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<YM2612_writeReg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %esi,%ecx
  mov    %ax,-0x14(%rbp)
  mov    %ecx,%eax
  mov    %al,-0x18(%rbp)
  mov    %edx,%eax
  mov    %al,-0x1c(%rbp)
  movq   $0xa04000,-0x8(%rbp)
  movzwl -0x14(%rbp),%eax
  add    %eax,%eax
  and    $0x2,%eax
  mov    %ax,-0xa(%rbp)
  movzwl -0xa(%rbp),%edx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x18(%rbp),%eax
  mov    %al,(%rdx)
  movzwl -0xa(%rbp),%eax
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x1c(%rbp),%eax
  mov    %al,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void YM2612_writeReg(const unsigned short part, const unsigned char reg, const unsigned char data) {
  volatile unsigned char *pb;
  unsigned short port;
  pb = (unsigned char *)0xA04000;
  port = (part << 1) & 2;
  pb[port + 0] = reg;
  pb[port + 1] = data;
}
```
user: What is the c source code of the assembly code below:

```asm
<YM2612_writeReg>:
  endbr64
  add    %edi,%edi
  and    $0x2,%edi
  mov    %sil,0xa04000(%rdi)
  mov    %dl,0xa04001(%rdi)
  ret
```

assistant: ```c
void YM2612_writeReg(const unsigned short part, const unsigned char reg, const unsigned char data) {
  volatile unsigned char *pb;
  unsigned short port;
  pb = (unsigned char *)0xA04000;
  port = (part << 1) & 2;
  pb[port + 0] = reg;
  pb[port + 1] = data;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<YM2612_writeReg>:
  endbr64
  add    %edi,%edi
  and    $0x2,%edi
  mov    %sil,0xa04000(%rdi)
  mov    %dl,0xa04001(%rdi)
  ret
```

assistant: ```c
void YM2612_writeReg(const unsigned short part, const unsigned char reg, const unsigned char data) {
  volatile unsigned char *pb;
  unsigned short port;
  pb = (unsigned char *)0xA04000;
  port = (part << 1) & 2;
  pb[port + 0] = reg;
  pb[port + 1] = data;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<YM2612_writeReg>:
  endbr64
  add    %edi,%edi
  and    $0x2,%edi
  mov    %sil,0xa04000(%rdi)
  mov    %dl,0xa04001(%rdi)
  ret
```

assistant: ```c
void YM2612_writeReg(const unsigned short part, const unsigned char reg, const unsigned char data) {
  volatile unsigned char *pb;
  unsigned short port;
  pb = (unsigned char *)0xA04000;
  port = (part << 1) & 2;
  pb[port + 0] = reg;
  pb[port + 1] = data;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```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
  mov    (%rax),%eax
  shl    $0x7,%eax
  and    $0x4334b500,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  nop
  mov    -0xc(%rbp),%eax
  and    $0x30f04006,%eax
  or     $0x70533f9,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  char copy11;
  unsigned short copy12;
  char copy13;
  char copy14;
  unsigned short copy16;
  {
    state[0UL] = (input[0UL] & 914778474UL) << 7U;
    local1 = 0UL;
    while (local1 < 0U) {
      if (state[0UL] > local1) {
        if (state[0UL] > local1) {
          copy11 = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = copy11;
          copy12 = *((unsigned short *)(&state[local1]) + 0);
          *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 1);
          *((unsigned short *)(&state[local1]) + 1) = copy12;
        } else {
          copy13 = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy13;
          state[local1] >>= ((state[local1] >> 3U) & 7U) | 1UL;
        }
      } else if (state[0UL] == local1) {
        copy14 = *((char *)(&state[local1]) + 0);
        *((char *)(&state[local1]) + 0) = *((char *)(&state[local1]) + 2);
        *((char *)(&state[local1]) + 2) = copy14;
      } else {
        state[0UL] = state[local1] | state[0UL];
        copy16 = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = copy16;
      }
      local1++;
    }
    output[0UL] = (state[0UL] & 837828902UL) | 117781497U;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  shl    $0x7,%eax
  and    $0x300000,%eax
  or     $0x70533f9,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  char copy11;
  unsigned short copy12;
  char copy13;
  char copy14;
  unsigned short copy16;
  {
    state[0UL] = (input[0UL] & 914778474UL) << 7U;
    local1 = 0UL;
    while (local1 < 0U) {
      if (state[0UL] > local1) {
        if (state[0UL] > local1) {
          copy11 = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = copy11;
          copy12 = *((unsigned short *)(&state[local1]) + 0);
          *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 1);
          *((unsigned short *)(&state[local1]) + 1) = copy12;
        } else {
          copy13 = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy13;
          state[local1] >>= ((state[local1] >> 3U) & 7U) | 1UL;
        }
      } else if (state[0UL] == local1) {
        copy14 = *((char *)(&state[local1]) + 0);
        *((char *)(&state[local1]) + 0) = *((char *)(&state[local1]) + 2);
        *((char *)(&state[local1]) + 2) = copy14;
      } else {
        state[0UL] = state[local1] | state[0UL];
        copy16 = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = copy16;
      }
      local1++;
    }
    output[0UL] = (state[0UL] & 837828902UL) | 117781497U;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  shl    $0x7,%eax
  and    $0x300000,%eax
  or     $0x70533f9,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  char copy11;
  unsigned short copy12;
  char copy13;
  char copy14;
  unsigned short copy16;
  {
    state[0UL] = (input[0UL] & 914778474UL) << 7U;
    local1 = 0UL;
    while (local1 < 0U) {
      if (state[0UL] > local1) {
        if (state[0UL] > local1) {
          copy11 = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = copy11;
          copy12 = *((unsigned short *)(&state[local1]) + 0);
          *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 1);
          *((unsigned short *)(&state[local1]) + 1) = copy12;
        } else {
          copy13 = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy13;
          state[local1] >>= ((state[local1] >> 3U) & 7U) | 1UL;
        }
      } else if (state[0UL] == local1) {
        copy14 = *((char *)(&state[local1]) + 0);
        *((char *)(&state[local1]) + 0) = *((char *)(&state[local1]) + 2);
        *((char *)(&state[local1]) + 2) = copy14;
      } else {
        state[0UL] = state[local1] | state[0UL];
        copy16 = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = copy16;
      }
      local1++;
    }
    output[0UL] = (state[0UL] & 837828902UL) | 117781497U;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  shl    $0x7,%eax
  and    $0x300000,%eax
  or     $0x70533f9,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  char copy11;
  unsigned short copy12;
  char copy13;
  char copy14;
  unsigned short copy16;
  {
    state[0UL] = (input[0UL] & 914778474UL) << 7U;
    local1 = 0UL;
    while (local1 < 0U) {
      if (state[0UL] > local1) {
        if (state[0UL] > local1) {
          copy11 = *((char *)(&state[0UL]) + 3);
          *((char *)(&state[0UL]) + 3) = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = copy11;
          copy12 = *((unsigned short *)(&state[local1]) + 0);
          *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 1);
          *((unsigned short *)(&state[local1]) + 1) = copy12;
        } else {
          copy13 = *((char *)(&state[local1]) + 3);
          *((char *)(&state[local1]) + 3) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy13;
          state[local1] >>= ((state[local1] >> 3U) & 7U) | 1UL;
        }
      } else if (state[0UL] == local1) {
        copy14 = *((char *)(&state[local1]) + 0);
        *((char *)(&state[local1]) + 0) = *((char *)(&state[local1]) + 2);
        *((char *)(&state[local1]) + 2) = copy14;
      } else {
        state[0UL] = state[local1] | state[0UL];
        copy16 = *((unsigned short *)(&state[local1]) + 1);
        *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = copy16;
      }
      local1++;
    }
    output[0UL] = (state[0UL] & 837828902UL) | 117781497U;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<oui_append_3o>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  shr    $0x10,%rax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x10(%rbp),%rax
  shr    $0x8,%rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  mov    -0x10(%rbp),%rdx
  mov    %dl,(%rax)
  mov    $0x3,%eax
  pop    %rbp
  ret
```

assistant: ```c
size_t oui_append_3o(unsigned char *cp, const unsigned long data) {
  *cp = (data >> 16) & 0xff;
  *(cp + 1) = (data >> 8) & 0xff;
  *(cp + 2) = data & 0xff;
  return 3;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<oui_append_3o>:
  endbr64
  mov    %rsi,%rax
  mov    %rsi,%rdx
  shr    $0x10,%rdx
  mov    %dl,(%rdi)
  movzbl %ah,%ecx
  mov    %cl,0x1(%rdi)
  mov    %sil,0x2(%rdi)
  mov    $0x3,%eax
  ret
```

assistant: ```c
size_t oui_append_3o(unsigned char *cp, const unsigned long data) {
  *cp = (data >> 16) & 0xff;
  *(cp + 1) = (data >> 8) & 0xff;
  *(cp + 2) = data & 0xff;
  return 3;
}
```
user: What is the c source code of the assembly code below?

```asm
<oui_append_3o>:
  endbr64
  mov    %rsi,%rax
  rol    $0x8,%si
  shr    $0x10,%rax
  mov    %si,0x1(%rdi)
  mov    %al,(%rdi)
  mov    $0x3,%eax
  ret
```

assistant: ```c
size_t oui_append_3o(unsigned char *cp, const unsigned long data) {
  *cp = (data >> 16) & 0xff;
  *(cp + 1) = (data >> 8) & 0xff;
  *(cp + 2) = data & 0xff;
  return 3;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<oui_append_3o>:
  endbr64
  mov    %rsi,%rax
  rol    $0x8,%si
  shr    $0x10,%rax
  mov    %si,0x1(%rdi)
  mov    %al,(%rdi)
  mov    $0x3,%eax
  ret
```

assistant: ```c
size_t oui_append_3o(unsigned char *cp, const unsigned long data) {
  *cp = (data >> 16) & 0xff;
  *(cp + 1) = (data >> 8) & 0xff;
  *(cp + 2) = data & 0xff;
  return 3;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<cvt_flip>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jne    L136
  mov    -0x4(%rbp),%eax
  jmp    L19
L136:
  mov    -0x4(%rbp),%eax
  sub    $0x7,%eax
  cmp    $0x25,%eax
  ja     L127
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D18(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D18(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0xa,%eax
  jmp    L19
  mov    $0xb,%eax
  jmp    L19
  mov    $0xc,%eax
  jmp    L19
  mov    $0x10,%eax
  jmp    L19
  mov    $0x19,%eax
  jmp    L19
  mov    $0x1c,%eax
  jmp    L19
  mov    $0x1f,%eax
  jmp    L19
  mov    $0x2b,%eax
  jmp    L19
  mov    $0x7,%eax
  jmp    L19
  mov    $0x8,%eax
  jmp    L19
  mov    $0x9,%eax
  jmp    L19
  mov    $0xf,%eax
  jmp    L19
  mov    $0x1a,%eax
  jmp    L19
  mov    $0x1d,%eax
  jmp    L19
  mov    $0x20,%eax
  jmp    L19
  mov    $0x2c,%eax
  jmp    L19
  mov    $0x23,%eax
  jmp    L19
  mov    $0x22,%eax
  jmp    L19
  mov    $0x26,%eax
  jmp    L19
  mov    $0x25,%eax
  jmp    L19
L127:
  mov    -0x4(%rbp),%eax
L19:
  pop    %rbp
  ret
```

assistant: ```c
int cvt_flip(int type, int flip) {
  if (flip == 0)
    return type;
  switch (type) {
  case 7:
    return 10;
  case 8:
    return 11;
  case 9:
    return 12;
  case 15:
    return 16;
  case 26:
    return 25;
  case 29:
    return 28;
  case 32:
    return 31;
  case 44:
    return 43;
  case 10:
    return 7;
  case 11:
    return 8;
  case 12:
    return 9;
  case 16:
    return 15;
  case 25:
    return 26;
  case 28:
    return 29;
  case 31:
    return 32;
  case 43:
    return 44;
  case 34:
    return 35;
  case 35:
    return 34;
  case 37:
    return 38;
  case 38:
    return 37;
  default:
    return type;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<cvt_flip>:
  endbr64
  mov    %edi,%eax
  test   %esi,%esi
  je     L14
  lea    -0x7(%rdi),%eax
  cmp    $0x25,%eax
  ja     L0
  mov    %eax,%eax
  lea    D6(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0xb,%eax
  ret
  mov    $0xc,%eax
  ret
  mov    $0x10,%eax
  ret
  mov    $0x19,%eax
  ret
  mov    $0x1c,%eax
  ret
  mov    $0x1f,%eax
  ret
  mov    $0x2b,%eax
  ret
  mov    $0x7,%eax
  ret
  mov    $0x8,%eax
  ret
  mov    $0x9,%eax
  ret
  mov    $0xf,%eax
  ret
  mov    $0x1a,%eax
  ret
  mov    $0x1d,%eax
  ret
  mov    $0x20,%eax
  ret
  mov    $0x2c,%eax
  ret
  mov    $0x23,%eax
  ret
  mov    $0x22,%eax
  ret
  mov    $0x26,%eax
  ret
  mov    $0x25,%eax
  ret
L0:
  mov    %edi,%eax
  ret
  mov    $0xa,%eax
L14:
  ret
```

assistant: ```c
int cvt_flip(int type, int flip) {
  if (flip == 0)
    return type;
  switch (type) {
  case 7:
    return 10;
  case 8:
    return 11;
  case 9:
    return 12;
  case 15:
    return 16;
  case 26:
    return 25;
  case 29:
    return 28;
  case 32:
    return 31;
  case 44:
    return 43;
  case 10:
    return 7;
  case 11:
    return 8;
  case 12:
    return 9;
  case 16:
    return 15;
  case 25:
    return 26;
  case 28:
    return 29;
  case 31:
    return 32;
  case 43:
    return 44;
  case 34:
    return 35;
  case 35:
    return 34;
  case 37:
    return 38;
  case 38:
    return 37;
  default:
    return type;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<_wtoi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  lea    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0xa,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <wcstol@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int _wtoi(const wchar_t *w) {
  wchar_t *endptr = 0;
  return wcstol(w, &endptr, 10);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_wtoi>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%rsi
  mov    $0xa,%edx
  call   <wcstol@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int _wtoi(const wchar_t *w) {
  wchar_t *endptr = 0;
  return wcstol(w, &endptr, 10);
}
```
user: Translate the below assembly code into C programming language:

```asm
<_wtoi>:
  endbr64
  sub    $0x18,%rsp
  mov    $0xa,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%rsp)
  call   <wcstol@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int _wtoi(const wchar_t *w) {
  wchar_t *endptr = 0;
  return wcstol(w, &endptr, 10);
}
```
user: Translate the below assembly code into C programming language:

```asm
<_wtoi>:
  endbr64
  sub    $0x18,%rsp
  mov    $0xa,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%rsp)
  call   <wcstol@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int _wtoi(const wchar_t *w) {
  wchar_t *endptr = 0;
  return wcstol(w, &endptr, 10);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<is_num_pld>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1c0,%rsp
  mov    %edi,-0x1b4(%rbp)
  mov    %esi,-0x1b8(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1a4(%rbp)
  jmp    L0
L28:
  mov    -0x1a4(%rbp),%eax
  cltq
  movl   $0xffffffff,-0x1a0(%rbp,%rax,4)
  addl   $0x1,-0x1a4(%rbp)
L0:
  cmpl   $0x63,-0x1a4(%rbp)
  jle    L28
  subl   $0x1,-0x1a4(%rbp)
  jmp    L5
L7:
  mov    -0x1b4(%rbp),%eax
  cltd
  idivl  -0x1b8(%rbp)
  mov    -0x1a4(%rbp),%eax
  cltq
  mov    %edx,-0x1a0(%rbp,%rax,4)
  mov    -0x1a4(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1b4(%rbp),%eax
  sub    %edx,%eax
  cltd
  idivl  -0x1b8(%rbp)
  mov    %eax,-0x1b4(%rbp)
  subl   $0x1,-0x1a4(%rbp)
L5:
  cmpl   $0x0,-0x1b4(%rbp)
  jne    L7
  lea    -0x1a0(%rbp),%rax
  mov    %rax,%rdi
  call   <is_pld@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  leave
  ret
```

assistant: ```c
int is_num_pld(int n, int hex) {
  int i, a[100];
  for (i = 0; i < 100; i++)
    a[i] = -1;
  for (i--; n != 0; i--) {
    a[i] = n % hex;
    n = (n - a[i]) / hex;
  }
  return is_pld(a);
}
```