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

```asm
<die>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void die(char *msg) {
  perror(msg);
  exit(0);
}
```
user: What is the c source code of the assembly code below:

```asm
<die>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void die(char *msg) {
  perror(msg);
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<die>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void die(char *msg) {
  perror(msg);
  exit(0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fiat_p224_nonzero>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%ecx
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%esi
  mov    -0x20(%rbp),%rax
  add    $0xc,%rax
  mov    (%rax),%edi
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%r8d
  mov    -0x20(%rbp),%rax
  add    $0x14,%rax
  mov    (%rax),%r9d
  mov    -0x20(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%eax
  or     %r9d,%eax
  or     %r8d,%eax
  or     %edi,%eax
  or     %esi,%eax
  or     %ecx,%eax
  or     %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void fiat_p224_nonzero(uint32_t *out1, const uint32_t arg1[7]) {
  uint32_t x1;
  x1 = ((arg1[0]) | ((arg1[1]) | ((arg1[2]) | ((arg1[3]) | ((arg1[4]) | ((arg1[5]) | (arg1[6])))))));
  *out1 = x1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fiat_p224_nonzero>:
  endbr64
  mov    (%rsi),%eax
  or     0x4(%rsi),%eax
  or     0x8(%rsi),%eax
  or     0xc(%rsi),%eax
  or     0x10(%rsi),%eax
  or     0x14(%rsi),%eax
  or     0x18(%rsi),%eax
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void fiat_p224_nonzero(uint32_t *out1, const uint32_t arg1[7]) {
  uint32_t x1;
  x1 = ((arg1[0]) | ((arg1[1]) | ((arg1[2]) | ((arg1[3]) | ((arg1[4]) | ((arg1[5]) | (arg1[6])))))));
  *out1 = x1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<fiat_p224_nonzero>:
  endbr64
  mov    (%rsi),%eax
  or     0x4(%rsi),%eax
  or     0x8(%rsi),%eax
  or     0xc(%rsi),%eax
  or     0x10(%rsi),%eax
  or     0x14(%rsi),%eax
  or     0x18(%rsi),%eax
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void fiat_p224_nonzero(uint32_t *out1, const uint32_t arg1[7]) {
  uint32_t x1;
  x1 = ((arg1[0]) | ((arg1[1]) | ((arg1[2]) | ((arg1[3]) | ((arg1[4]) | ((arg1[5]) | (arg1[6])))))));
  *out1 = x1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fiat_p224_nonzero>:
  endbr64
  mov    (%rsi),%eax
  or     0x4(%rsi),%eax
  or     0x8(%rsi),%eax
  or     0xc(%rsi),%eax
  or     0x10(%rsi),%eax
  or     0x14(%rsi),%eax
  or     0x18(%rsi),%eax
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void fiat_p224_nonzero(uint32_t *out1, const uint32_t arg1[7]) {
  uint32_t x1;
  x1 = ((arg1[0]) | ((arg1[1]) | ((arg1[2]) | ((arg1[3]) | ((arg1[4]) | ((arg1[5]) | (arg1[6])))))));
  *out1 = x1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<validateChoiceEx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    -0x28(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x30(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L31
  mov    $0x1,%eax
  jmp    L17
L31:
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x18(%rbp)
L4:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L34
  mov    -0x30(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <isAbbreviatedPhrase@plt>
  test   %rax,%rax
  je     L11
  mov    -0x18(%rbp),%rax
  sub    -0x38(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x40(%rbp)
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x1,%eax
  jmp    L17
L11:
  mov    -0x40(%rbp),%rax
  add    %rax,-0x18(%rbp)
  jmp    L4
L34:
  nop
  mov    $0x0,%eax
L17:
  leave
  ret
```

assistant: ```c
int validateChoiceEx(unsigned int *value, const char *string, const void *choices, size_t size) {
  *value = 0;
  if (!*string)
    return 1;
  const void *choice = choices;
  while (1) {
    typedef struct {
      const char *name;
    } Entry;
    const Entry *entry = choice;
    const char *name = entry->name;
    if (!name)
      break;
    if (isAbbreviatedPhrase(name, string)) {
      *value = (choice - choices) / size;
      return 1;
    }
    choice += size;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<validateChoiceEx>:
  endbr64
  movl   $0x0,(%rdi)
  mov    $0x1,%eax
  cmpb   $0x0,(%rsi)
  je     L36
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%r14
  mov    %rdx,%r13
  mov    %rcx,%rbp
  mov    (%rdx),%rdi
  test   %rdi,%rdi
  je     L44
  mov    %rdx,%rbx
L54:
  mov    %r14,%rsi
  call   <isAbbreviatedPhrase@plt>
  test   %rax,%rax
  jne    L53
  add    %rbp,%rbx
  mov    (%rbx),%rdi
  test   %rdi,%rdi
  jne    L54
  mov    $0x0,%eax
  jmp    L49
L53:
  mov    %rbx,%rax
  sub    %r13,%rax
  mov    $0x0,%edx
  div    %rbp
  mov    %eax,(%r12)
  mov    $0x1,%eax
L49:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L44:
  mov    $0x0,%eax
  jmp    L49
L36:
  ret
```

assistant: ```c
int validateChoiceEx(unsigned int *value, const char *string, const void *choices, size_t size) {
  *value = 0;
  if (!*string)
    return 1;
  const void *choice = choices;
  while (1) {
    typedef struct {
      const char *name;
    } Entry;
    const Entry *entry = choice;
    const char *name = entry->name;
    if (!name)
      break;
    if (isAbbreviatedPhrase(name, string)) {
      *value = (choice - choices) / size;
      return 1;
    }
    choice += size;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<validateChoiceEx>:
  endbr64
  movl   $0x0,(%rdi)
  cmpb   $0x0,(%rsi)
  mov    $0x1,%eax
  je     L39
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    (%rdx),%rdi
  mov    %rdx,%rbx
  test   %rdi,%rdi
  jne    L26
  jmp    L18
  nopl   (%rax)
L6:
  add    %rbp,%rbx
  mov    (%rbx),%rdi
  test   %rdi,%rdi
  je     L18
L26:
  mov    %r14,%rsi
  call   <isAbbreviatedPhrase@plt>
  test   %rax,%rax
  je     L6
  mov    %rbx,%rax
  xor    %edx,%edx
  sub    %r13,%rax
  div    %rbp
  mov    %eax,(%r12)
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L39:
  ret
  nopl   0x0(%rax)
L18:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int validateChoiceEx(unsigned int *value, const char *string, const void *choices, size_t size) {
  *value = 0;
  if (!*string)
    return 1;
  const void *choice = choices;
  while (1) {
    typedef struct {
      const char *name;
    } Entry;
    const Entry *entry = choice;
    const char *name = entry->name;
    if (!name)
      break;
    if (isAbbreviatedPhrase(name, string)) {
      *value = (choice - choices) / size;
      return 1;
    }
    choice += size;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<validateChoiceEx>:
  endbr64
  movl   $0x0,(%rdi)
  cmpb   $0x0,(%rsi)
  mov    $0x1,%eax
  je     L31
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    (%rdx),%rdi
  mov    %rdx,%rbx
  test   %rdi,%rdi
  jne    L32
  jmp    L43
  nopl   (%rax)
L17:
  add    %rbp,%rbx
  mov    (%rbx),%rdi
  test   %rdi,%rdi
  je     L43
L32:
  mov    %r14,%rsi
  call   <isAbbreviatedPhrase@plt>
  test   %rax,%rax
  je     L17
  mov    %rbx,%rax
  xor    %edx,%edx
  sub    %r13,%rax
  div    %rbp
  mov    %eax,(%r12)
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L31:
  ret
  nopl   0x0(%rax)
L43:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int validateChoiceEx(unsigned int *value, const char *string, const void *choices, size_t size) {
  *value = 0;
  if (!*string)
    return 1;
  const void *choice = choices;
  while (1) {
    typedef struct {
      const char *name;
    } Entry;
    const Entry *entry = choice;
    const char *name = entry->name;
    if (!name)
      break;
    if (isAbbreviatedPhrase(name, string)) {
      *value = (choice - choices) / size;
      return 1;
    }
    choice += size;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x41,-0x4(%rbp)
  jmp    L11
L0:
  mov    -0x4(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <lower@plt>
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0x4(%rbp)
L11:
  cmpl   $0x5a,-0x4(%rbp)
  jle    L0
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
main() {
  int l;
  for (l = 'A'; l <= 'Z'; l++) {
    putchar(l);
    putchar(lower(l));
  }
  putchar('\n');
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x41,%ebx
L5:
  mov    %ebx,%edi
  call   <putchar@plt>
  mov    %ebx,%edi
  call   <lower@plt>
  movsbl %al,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    $0x5b,%ebx
  jne    L5
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
main() {
  int l;
  for (l = 'A'; l <= 'Z'; l++) {
    putchar(l);
    putchar(lower(l));
  }
  putchar('\n');
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x41,%ebx
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %ebx,%edi
  call   <putchar@plt>
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <lower@plt>
  movsbl %al,%edi
  call   <putchar@plt>
  cmp    $0x5b,%ebx
  jne    L6
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
main() {
  int l;
  for (l = 'A'; l <= 'Z'; l++) {
    putchar(l);
    putchar(lower(l));
  }
  putchar('\n');
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x41,%ebx
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %ebx,%edi
  call   <putchar@plt>
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <lower@plt>
  movsbl %al,%edi
  call   <putchar@plt>
  cmp    $0x5b,%ebx
  jne    L7
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
main() {
  int l;
  for (l = 'A'; l <= 'Z'; l++) {
    putchar(l);
    putchar(lower(l));
  }
  putchar('\n');
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<USBSTK5505_I2C_reset>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <USBSTK5505_I2C_close@plt>
  mov    $0x0,%eax
  call   <USBSTK5505_I2C_init@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
short USBSTK5505_I2C_reset() {
  USBSTK5505_I2C_close();
  USBSTK5505_I2C_init();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<USBSTK5505_I2C_reset>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <USBSTK5505_I2C_close@plt>
  mov    $0x0,%eax
  call   <USBSTK5505_I2C_init@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
short USBSTK5505_I2C_reset() {
  USBSTK5505_I2C_close();
  USBSTK5505_I2C_init();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<USBSTK5505_I2C_reset>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <USBSTK5505_I2C_close@plt>
  xor    %eax,%eax
  call   <USBSTK5505_I2C_init@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
short USBSTK5505_I2C_reset() {
  USBSTK5505_I2C_close();
  USBSTK5505_I2C_init();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<USBSTK5505_I2C_reset>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <USBSTK5505_I2C_close@plt>
  xor    %eax,%eax
  call   <USBSTK5505_I2C_init@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
short USBSTK5505_I2C_reset() {
  USBSTK5505_I2C_close();
  USBSTK5505_I2C_init();
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<opcode_to_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  movzbl -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  lea    D8(%rip),%rax
  mov    (%rdx,%rax,1),%rax
  pop    %rbp
  ret
```

assistant: ```c
const char *opcode_to_string(unsigned char op) {
  static char *mnemonic_lookup[256] = {"BRK", "ORA", "???", "???", "???", "ORA", "ASL", "???", "PHP", "ORA", "ASL", "???", "???", "ORA", "ASL", "???", "BPL", "ORA", "???", "???", "???", "ORA", "ASL", "???", "CLC", "ORA", "???", "???", "???", "ORA", "ASL", "???", "JSR", "AND", "???", "???", "BIT", "AND", "ROL", "???", "PLP", "AND", "ROL", "???", "BIT", "AND", "ROL", "???", "BMI", "AND", "???", "???", "???", "AND", "ROL", "???", "SEC", "AND", "???", "???", "???", "AND", "ROL", "???", "RTI", "EOR", "???", "???", "???", "EOR", "LSR", "???", "PHA", "EOR", "LSR", "???", "JMP", "EOR", "LSR", "???", "BVC", "EOR", "???", "???", "???", "EOR", "LSR", "???", "CLI", "EOR", "???", "???", "???", "EOR", "LSR", "???", "RTS", "ADC", "???", "???", "???", "ADC", "ROR", "???", "PLA", "ADC", "ROR", "???", "JMP", "ADC", "ROR", "???", "BVS", "ADC", "???", "???", "???", "ADC", "ROR", "???", "SEI", "ADC", "???", "???", "???", "ADC", "ROR", "???",
                                       "???", "STA", "???", "???", "STY", "STA", "STX", "???", "DEY", "???", "TXA", "???", "STY", "STA", "STX", "???", "BCC", "STA", "???", "???", "STY", "STA", "STX", "???", "TYA", "STA", "TXS", "???", "???", "STA", "???", "???", "LDY", "LDA", "LDX", "???", "LDY", "LDA", "LDX", "???", "TAY", "LDA", "TAX", "???", "LDY", "LDA", "LDX", "???", "BCS", "LDA", "???", "???", "LDY", "LDA", "LDX", "???", "CLV", "LDA", "TSX", "???", "LDY", "LDA", "LDX", "???", "CPY", "CMP", "???", "???", "CPY", "CMP", "DEC", "???", "INY", "CMP", "DEX", "???", "CPY", "CMP", "DEC", "???", "BNE", "CMP", "???", "???", "???", "CMP", "DEC", "???", "CLD", "CMP", "???", "???", "???", "CMP", "DEC", "???", "CPX", "SBC", "???", "???", "CPX", "SBC", "INC", "???", "INX", "SBC", "NOP", "???", "CPX", "SBC", "INC", "???", "BEQ", "SBC", "???", "???", "???", "SBC", "INC", "???", "SED", "SBC", "???", "???", "???", "SBC", "INC", "???"};
  return mnemonic_lookup[op];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<opcode_to_string>:
  endbr64
  movzbl %dil,%edi
  lea    D9(%rip),%rax
  mov    (%rax,%rdi,8),%rax
  ret
```

assistant: ```c
const char *opcode_to_string(unsigned char op) {
  static char *mnemonic_lookup[256] = {"BRK", "ORA", "???", "???", "???", "ORA", "ASL", "???", "PHP", "ORA", "ASL", "???", "???", "ORA", "ASL", "???", "BPL", "ORA", "???", "???", "???", "ORA", "ASL", "???", "CLC", "ORA", "???", "???", "???", "ORA", "ASL", "???", "JSR", "AND", "???", "???", "BIT", "AND", "ROL", "???", "PLP", "AND", "ROL", "???", "BIT", "AND", "ROL", "???", "BMI", "AND", "???", "???", "???", "AND", "ROL", "???", "SEC", "AND", "???", "???", "???", "AND", "ROL", "???", "RTI", "EOR", "???", "???", "???", "EOR", "LSR", "???", "PHA", "EOR", "LSR", "???", "JMP", "EOR", "LSR", "???", "BVC", "EOR", "???", "???", "???", "EOR", "LSR", "???", "CLI", "EOR", "???", "???", "???", "EOR", "LSR", "???", "RTS", "ADC", "???", "???", "???", "ADC", "ROR", "???", "PLA", "ADC", "ROR", "???", "JMP", "ADC", "ROR", "???", "BVS", "ADC", "???", "???", "???", "ADC", "ROR", "???", "SEI", "ADC", "???", "???", "???", "ADC", "ROR", "???",
                                       "???", "STA", "???", "???", "STY", "STA", "STX", "???", "DEY", "???", "TXA", "???", "STY", "STA", "STX", "???", "BCC", "STA", "???", "???", "STY", "STA", "STX", "???", "TYA", "STA", "TXS", "???", "???", "STA", "???", "???", "LDY", "LDA", "LDX", "???", "LDY", "LDA", "LDX", "???", "TAY", "LDA", "TAX", "???", "LDY", "LDA", "LDX", "???", "BCS", "LDA", "???", "???", "LDY", "LDA", "LDX", "???", "CLV", "LDA", "TSX", "???", "LDY", "LDA", "LDX", "???", "CPY", "CMP", "???", "???", "CPY", "CMP", "DEC", "???", "INY", "CMP", "DEX", "???", "CPY", "CMP", "DEC", "???", "BNE", "CMP", "???", "???", "???", "CMP", "DEC", "???", "CLD", "CMP", "???", "???", "???", "CMP", "DEC", "???", "CPX", "SBC", "???", "???", "CPX", "SBC", "INC", "???", "INX", "SBC", "NOP", "???", "CPX", "SBC", "INC", "???", "BEQ", "SBC", "???", "???", "???", "SBC", "INC", "???", "SED", "SBC", "???", "???", "???", "SBC", "INC", "???"};
  return mnemonic_lookup[op];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<opcode_to_string>:
  endbr64
  movzbl %dil,%edi
  lea    D8(%rip),%rax
  mov    (%rax,%rdi,8),%rax
  ret
```

assistant: ```c
const char *opcode_to_string(unsigned char op) {
  static char *mnemonic_lookup[256] = {"BRK", "ORA", "???", "???", "???", "ORA", "ASL", "???", "PHP", "ORA", "ASL", "???", "???", "ORA", "ASL", "???", "BPL", "ORA", "???", "???", "???", "ORA", "ASL", "???", "CLC", "ORA", "???", "???", "???", "ORA", "ASL", "???", "JSR", "AND", "???", "???", "BIT", "AND", "ROL", "???", "PLP", "AND", "ROL", "???", "BIT", "AND", "ROL", "???", "BMI", "AND", "???", "???", "???", "AND", "ROL", "???", "SEC", "AND", "???", "???", "???", "AND", "ROL", "???", "RTI", "EOR", "???", "???", "???", "EOR", "LSR", "???", "PHA", "EOR", "LSR", "???", "JMP", "EOR", "LSR", "???", "BVC", "EOR", "???", "???", "???", "EOR", "LSR", "???", "CLI", "EOR", "???", "???", "???", "EOR", "LSR", "???", "RTS", "ADC", "???", "???", "???", "ADC", "ROR", "???", "PLA", "ADC", "ROR", "???", "JMP", "ADC", "ROR", "???", "BVS", "ADC", "???", "???", "???", "ADC", "ROR", "???", "SEI", "ADC", "???", "???", "???", "ADC", "ROR", "???",
                                       "???", "STA", "???", "???", "STY", "STA", "STX", "???", "DEY", "???", "TXA", "???", "STY", "STA", "STX", "???", "BCC", "STA", "???", "???", "STY", "STA", "STX", "???", "TYA", "STA", "TXS", "???", "???", "STA", "???", "???", "LDY", "LDA", "LDX", "???", "LDY", "LDA", "LDX", "???", "TAY", "LDA", "TAX", "???", "LDY", "LDA", "LDX", "???", "BCS", "LDA", "???", "???", "LDY", "LDA", "LDX", "???", "CLV", "LDA", "TSX", "???", "LDY", "LDA", "LDX", "???", "CPY", "CMP", "???", "???", "CPY", "CMP", "DEC", "???", "INY", "CMP", "DEX", "???", "CPY", "CMP", "DEC", "???", "BNE", "CMP", "???", "???", "???", "CMP", "DEC", "???", "CLD", "CMP", "???", "???", "???", "CMP", "DEC", "???", "CPX", "SBC", "???", "???", "CPX", "SBC", "INC", "???", "INX", "SBC", "NOP", "???", "CPX", "SBC", "INC", "???", "BEQ", "SBC", "???", "???", "???", "SBC", "INC", "???", "SED", "SBC", "???", "???", "???", "SBC", "INC", "???"};
  return mnemonic_lookup[op];
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<opcode_to_string>:
  endbr64
  movzbl %dil,%edi
  lea    D9(%rip),%rax
  mov    (%rax,%rdi,8),%rax
  ret
```

assistant: ```c
const char *opcode_to_string(unsigned char op) {
  static char *mnemonic_lookup[256] = {"BRK", "ORA", "???", "???", "???", "ORA", "ASL", "???", "PHP", "ORA", "ASL", "???", "???", "ORA", "ASL", "???", "BPL", "ORA", "???", "???", "???", "ORA", "ASL", "???", "CLC", "ORA", "???", "???", "???", "ORA", "ASL", "???", "JSR", "AND", "???", "???", "BIT", "AND", "ROL", "???", "PLP", "AND", "ROL", "???", "BIT", "AND", "ROL", "???", "BMI", "AND", "???", "???", "???", "AND", "ROL", "???", "SEC", "AND", "???", "???", "???", "AND", "ROL", "???", "RTI", "EOR", "???", "???", "???", "EOR", "LSR", "???", "PHA", "EOR", "LSR", "???", "JMP", "EOR", "LSR", "???", "BVC", "EOR", "???", "???", "???", "EOR", "LSR", "???", "CLI", "EOR", "???", "???", "???", "EOR", "LSR", "???", "RTS", "ADC", "???", "???", "???", "ADC", "ROR", "???", "PLA", "ADC", "ROR", "???", "JMP", "ADC", "ROR", "???", "BVS", "ADC", "???", "???", "???", "ADC", "ROR", "???", "SEI", "ADC", "???", "???", "???", "ADC", "ROR", "???",
                                       "???", "STA", "???", "???", "STY", "STA", "STX", "???", "DEY", "???", "TXA", "???", "STY", "STA", "STX", "???", "BCC", "STA", "???", "???", "STY", "STA", "STX", "???", "TYA", "STA", "TXS", "???", "???", "STA", "???", "???", "LDY", "LDA", "LDX", "???", "LDY", "LDA", "LDX", "???", "TAY", "LDA", "TAX", "???", "LDY", "LDA", "LDX", "???", "BCS", "LDA", "???", "???", "LDY", "LDA", "LDX", "???", "CLV", "LDA", "TSX", "???", "LDY", "LDA", "LDX", "???", "CPY", "CMP", "???", "???", "CPY", "CMP", "DEC", "???", "INY", "CMP", "DEX", "???", "CPY", "CMP", "DEC", "???", "BNE", "CMP", "???", "???", "???", "CMP", "DEC", "???", "CLD", "CMP", "???", "???", "???", "CMP", "DEC", "???", "CPX", "SBC", "???", "???", "CPX", "SBC", "INC", "???", "INX", "SBC", "NOP", "???", "CPX", "SBC", "INC", "???", "BEQ", "SBC", "???", "???", "???", "SBC", "INC", "???", "SED", "SBC", "???", "???", "???", "SBC", "INC", "???"};
  return mnemonic_lookup[op];
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ili9325_DisplayOff>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%esi
  mov    $0x10,%edi
  call   <ili9325_WriteReg@plt>
  mov    $0x0,%esi
  mov    $0x11,%edi
  call   <ili9325_WriteReg@plt>
  mov    $0x0,%esi
  mov    $0x12,%edi
  call   <ili9325_WriteReg@plt>
  mov    $0x0,%esi
  mov    $0x13,%edi
  call   <ili9325_WriteReg@plt>
  mov    $0x0,%esi
  mov    $0x29,%edi
  call   <ili9325_WriteReg@plt>
  mov    $0x0,%esi
  mov    $0x7,%edi
  call   <ili9325_WriteReg@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ili9325_DisplayOff(void) {
  ili9325_WriteReg(0x10, 0x0000);
  ili9325_WriteReg(0x11, 0x0000);
  ili9325_WriteReg(0x12, 0x0000);
  ili9325_WriteReg(0x13, 0x0000);
  ili9325_WriteReg(0x29, 0x0000);
  ili9325_WriteReg(0x07, 0x0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ili9325_DisplayOff>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  mov    $0x10,%edi
  call   <ili9325_WriteReg@plt>
  mov    $0x0,%esi
  mov    $0x11,%edi
  call   <ili9325_WriteReg@plt>
  mov    $0x0,%esi
  mov    $0x12,%edi
  call   <ili9325_WriteReg@plt>
  mov    $0x0,%esi
  mov    $0x13,%edi
  call   <ili9325_WriteReg@plt>
  mov    $0x0,%esi
  mov    $0x29,%edi
  call   <ili9325_WriteReg@plt>
  mov    $0x0,%esi
  mov    $0x7,%edi
  call   <ili9325_WriteReg@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void ili9325_DisplayOff(void) {
  ili9325_WriteReg(0x10, 0x0000);
  ili9325_WriteReg(0x11, 0x0000);
  ili9325_WriteReg(0x12, 0x0000);
  ili9325_WriteReg(0x13, 0x0000);
  ili9325_WriteReg(0x29, 0x0000);
  ili9325_WriteReg(0x07, 0x0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ili9325_DisplayOff>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  mov    $0x10,%edi
  call   <ili9325_WriteReg@plt>
  xor    %esi,%esi
  mov    $0x11,%edi
  call   <ili9325_WriteReg@plt>
  xor    %esi,%esi
  mov    $0x12,%edi
  call   <ili9325_WriteReg@plt>
  xor    %esi,%esi
  mov    $0x13,%edi
  call   <ili9325_WriteReg@plt>
  xor    %esi,%esi
  mov    $0x29,%edi
  call   <ili9325_WriteReg@plt>
  xor    %esi,%esi
  mov    $0x7,%edi
  add    $0x8,%rsp
  jmp    1050 <ili9325_WriteReg@plt>
```

assistant: ```c
void ili9325_DisplayOff(void) {
  ili9325_WriteReg(0x10, 0x0000);
  ili9325_WriteReg(0x11, 0x0000);
  ili9325_WriteReg(0x12, 0x0000);
  ili9325_WriteReg(0x13, 0x0000);
  ili9325_WriteReg(0x29, 0x0000);
  ili9325_WriteReg(0x07, 0x0);
}
```