ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xf,%esi
  lea    D48(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xf0,%esi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D51(%rip),%rdi
  call   <puts@plt>
  lea    D53(%rip),%rdi
  call   <puts@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xff,%esi
  lea    D59(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xf,%esi
  lea    D56(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xf0,%esi
  lea    D66(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D48
datatype: string
name: read_data
arguments: label: D4
datatype: string
name: read_data
arguments: label: D51
datatype: string
name: read_data
arguments: label: D53
datatype: string
name: read_data
arguments: label: D59
datatype: string
name: read_data
arguments: label: D56
datatype: string
name: read_data
arguments: label: D66
datatype: string

tool: D48: flag1=%02X


tool: D4: flag2=%02X


tool: D51: WRITE True in the flag1

tool: D53: WRITE False in the flag2

tool: D59: flag3 = %02X


tool: D56: flag3 & flag1 = %02X


tool: D66: flag3 & flag2 = %02X


assistant: ```c
int main() {
  int flag1, flag2, flag3;
  flag1 = 0X01 | 0x02 | 0x04 | 0x08;
  printf("flag1=%02X\n", flag1);
  flag2 = 0x10 | 0x20 | 0x40 | 0x80;
  printf("flag2=%02X\n", flag2);
  printf("\n");
  if (flag1 & 0x02)
    printf("WRITE True in the flag1\n");
  else
    printf("WRITE False in the flag1\n");
  if (flag2 & 0x02)
    printf("WRITE True in the flag2\n");
  else
    printf("WRITE False in the flag2\n");
  printf("\n");
  flag3 = flag1 | flag2;
  printf("flag3 = %02X\n", flag3);
  printf("flag3 & flag1 = %02X\n", flag3 & flag1);
  printf("flag3 & flag2 = %02X\n", flag3 & flag2);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xf,%esi
  lea    D12(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xf0,%esi
  lea    D45(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D58(%rip),%rdi
  call   <puts@plt>
  lea    D53(%rip),%rdi
  call   <puts@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xff,%esi
  lea    D36(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xf,%esi
  lea    D32(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xf0,%esi
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D12
data_type: string
name: disassemble_data
arguments: label: D45
data_type: string
name: disassemble_data
arguments: label: D58
data_type: string
name: disassemble_data
arguments: label: D53
data_type: string
name: disassemble_data
arguments: label: D36
data_type: string
name: disassemble_data
arguments: label: D32
data_type: string
name: disassemble_data
arguments: label: D24
data_type: string

tool: D12: flag1=%02X


tool: D45: flag2=%02X


tool: D58: WRITE True in the flag1

tool: D53: WRITE False in the flag2

tool: D36: flag3 = %02X


tool: D32: flag3 & flag1 = %02X


tool: D24: flag3 & flag2 = %02X


assistant: ```c
int main() {
  int flag1, flag2, flag3;
  flag1 = 0X01 | 0x02 | 0x04 | 0x08;
  printf("flag1=%02X\n", flag1);
  flag2 = 0x10 | 0x20 | 0x40 | 0x80;
  printf("flag2=%02X\n", flag2);
  printf("\n");
  if (flag1 & 0x02)
    printf("WRITE True in the flag1\n");
  else
    printf("WRITE False in the flag1\n");
  if (flag2 & 0x02)
    printf("WRITE True in the flag2\n");
  else
    printf("WRITE False in the flag2\n");
  printf("\n");
  flag3 = flag1 | flag2;
  printf("flag3 = %02X\n", flag3);
  printf("flag3 & flag1 = %02X\n", flag3 & flag1);
  printf("flag3 & flag2 = %02X\n", flag3 & flag2);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xf,%esi
  lea    D39(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xf0,%esi
  lea    D69(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D37(%rip),%rdi
  call   <puts@plt>
  lea    D13(%rip),%rdi
  call   <puts@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xff,%esi
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xf,%esi
  lea    D66(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xf0,%esi
  lea    D40(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D39
data_type: string
name: extract_symbolic_data
arguments: data_label: D69
data_type: string
name: extract_symbolic_data
arguments: data_label: D37
data_type: string
name: extract_symbolic_data
arguments: data_label: D13
data_type: string
name: extract_symbolic_data
arguments: data_label: D6
data_type: string
name: extract_symbolic_data
arguments: data_label: D66
data_type: string
name: extract_symbolic_data
arguments: data_label: D40
data_type: string

tool: D39: flag1=%02X


tool: D69: flag2=%02X


tool: D37: WRITE True in the flag1

tool: D13: WRITE False in the flag2

tool: D6: flag3 = %02X


tool: D66: flag3 & flag1 = %02X


tool: D40: flag3 & flag2 = %02X


assistant: ```c
int main() {
  int flag1, flag2, flag3;
  flag1 = 0X01 | 0x02 | 0x04 | 0x08;
  printf("flag1=%02X\n", flag1);
  flag2 = 0x10 | 0x20 | 0x40 | 0x80;
  printf("flag2=%02X\n", flag2);
  printf("\n");
  if (flag1 & 0x02)
    printf("WRITE True in the flag1\n");
  else
    printf("WRITE False in the flag1\n");
  if (flag2 & 0x02)
    printf("WRITE True in the flag2\n");
  else
    printf("WRITE False in the flag2\n");
  printf("\n");
  flag3 = flag1 | flag2;
  printf("flag3 = %02X\n", flag3);
  printf("flag3 & flag1 = %02X\n", flag3 & flag1);
  printf("flag3 & flag2 = %02X\n", flag3 & flag2);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<tpMQTTCreate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    %r9d,-0x34(%rbp)
  mov    -0x34(%rbp),%r9d
  mov    -0x30(%rbp),%r8
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  push   0x28(%rbp)
  push   $0x1
  push   $0x0
  push   0x20(%rbp)
  mov    0x18(%rbp),%edi
  push   %rdi
  push   0x10(%rbp)
  mov    %rax,%rdi
  call   <handleMQTTCreate@plt>
  add    $0x30,%rsp
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int tpMQTTCreate(char *host, int port, int keepalive, char *userName, char *password, int enableServerCertAuth, char *sudirribeTopic[], int sudirribeTopicSize, char *publishTopic, char *clientID) {
  int rc = handleMQTTCreate(host, port, keepalive, userName, password, enableServerCertAuth, sudirribeTopic, sudirribeTopicSize, publishTopic, NULL, 1, clientID);
  return rc;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<tpMQTTCreate>:
  endbr64
  sub    $0x8,%rsp
  push   0x28(%rsp)
  push   $0x1
  push   $0x0
  push   0x38(%rsp)
  mov    0x38(%rsp),%eax
  push   %rax
  push   0x38(%rsp)
  call   <handleMQTTCreate@plt>
  add    $0x38,%rsp
  ret
```

assistant: ```c
int tpMQTTCreate(char *host, int port, int keepalive, char *userName, char *password, int enableServerCertAuth, char *sudirribeTopic[], int sudirribeTopicSize, char *publishTopic, char *clientID) {
  int rc = handleMQTTCreate(host, port, keepalive, userName, password, enableServerCertAuth, sudirribeTopic, sudirribeTopicSize, publishTopic, NULL, 1, clientID);
  return rc;
}
```
user: What is the c source code of the assembly code below:

```asm
<tpMQTTCreate>:
  endbr64
  sub    $0x8,%rsp
  push   0x28(%rsp)
  push   $0x1
  push   $0x0
  push   0x38(%rsp)
  mov    0x38(%rsp),%eax
  push   %rax
  push   0x38(%rsp)
  call   <handleMQTTCreate@plt>
  add    $0x38,%rsp
  ret
```

assistant: ```c
int tpMQTTCreate(char *host, int port, int keepalive, char *userName, char *password, int enableServerCertAuth, char *sudirribeTopic[], int sudirribeTopicSize, char *publishTopic, char *clientID) {
  int rc = handleMQTTCreate(host, port, keepalive, userName, password, enableServerCertAuth, sudirribeTopic, sudirribeTopicSize, publishTopic, NULL, 1, clientID);
  return rc;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<tpMQTTCreate>:
  endbr64
  sub    $0x8,%rsp
  push   0x28(%rsp)
  push   $0x1
  push   $0x0
  push   0x38(%rsp)
  mov    0x38(%rsp),%eax
  push   %rax
  push   0x38(%rsp)
  call   <handleMQTTCreate@plt>
  add    $0x38,%rsp
  ret
```

assistant: ```c
int tpMQTTCreate(char *host, int port, int keepalive, char *userName, char *password, int enableServerCertAuth, char *sudirribeTopic[], int sudirribeTopicSize, char *publishTopic, char *clientID) {
  int rc = handleMQTTCreate(host, port, keepalive, userName, password, enableServerCertAuth, sudirribeTopic, sudirribeTopicSize, publishTopic, NULL, 1, clientID);
  return rc;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<multiply>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x1c(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void multiply(int *number, int multiplier) {
  int mul = (*number) * multiplier;
  *number = mul;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<multiply>:
  endbr64
  imul   (%rdi),%esi
  mov    %esi,(%rdi)
  ret
```

assistant: ```c
void multiply(int *number, int multiplier) {
  int mul = (*number) * multiplier;
  *number = mul;
}
```
user: What is the c source code of the assembly code below:

```asm
<multiply>:
  endbr64
  imul   (%rdi),%esi
  mov    %esi,(%rdi)
  ret
```

assistant: ```c
void multiply(int *number, int multiplier) {
  int mul = (*number) * multiplier;
  *number = mul;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<multiply>:
  endbr64
  imul   (%rdi),%esi
  mov    %esi,(%rdi)
  ret
```

assistant: ```c
void multiply(int *number, int multiplier) {
  int mul = (*number) * multiplier;
  *number = mul;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<current_time_ns>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  lea    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <clock_gettime@plt>
  mov    -0x18(%rbp),%rax
  imul   $0x3b9aca00,%rax,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
unsigned long long current_time_ns() {
  struct timespec t = {0, 0};
  clock_gettime(CLOCK_MONOTONIC, &t);
  unsigned long long s = 1000000000ULL * (unsigned long long)t.tv_sec;
  return (((unsigned long long)t.tv_nsec)) + s;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<current_time_ns>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  mov    %rsp,%rsi
  mov    D9(%rip),%rax
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  imul   $0x3b9aca00,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long current_time_ns() {
  struct timespec t = {0, 0};
  clock_gettime(CLOCK_MONOTONIC, &t);
  unsigned long long s = 1000000000ULL * (unsigned long long)t.tv_sec;
  return (((unsigned long long)t.tv_nsec)) + s;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<current_time_ns>:
  endbr64
  sub    $0x28,%rsp
  pxor   %xmm0,%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D6(%rip),%rax
  mov    %rsp,%rsi
  movaps %xmm0,(%rsp)
  movq   $0x0,0x10(%rsp)
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  imul   $0x3b9aca00,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long current_time_ns() {
  struct timespec t = {0, 0};
  clock_gettime(CLOCK_MONOTONIC, &t);
  unsigned long long s = 1000000000ULL * (unsigned long long)t.tv_sec;
  return (((unsigned long long)t.tv_nsec)) + s;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<current_time_ns>:
  endbr64
  sub    $0x28,%rsp
  pxor   %xmm0,%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D3(%rip),%rax
  mov    %rsp,%rsi
  movaps %xmm0,(%rsp)
  movq   $0x0,0x10(%rsp)
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  imul   $0x3b9aca00,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long current_time_ns() {
  struct timespec t = {0, 0};
  clock_gettime(CLOCK_MONOTONIC, &t);
  unsigned long long s = 1000000000ULL * (unsigned long long)t.tv_sec;
  return (((unsigned long long)t.tv_nsec)) + s;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<random_byte_lfsr32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x8(%rbp)
  jmp    L24
L29:
  movl   $0x0,-0x4(%rbp)
  mov    D81(%rip),%eax
  and    $0x2,%eax
  test   %eax,%eax
  je     L18
  xorl   $0x1,-0x4(%rbp)
L18:
  mov    D81(%rip),%eax
  and    $0x20,%eax
  test   %eax,%eax
  je     L27
  xorl   $0x1,-0x4(%rbp)
L27:
  mov    D81(%rip),%eax
  and    $0x40,%eax
  test   %eax,%eax
  je     L19
  xorl   $0x1,-0x4(%rbp)
L19:
  mov    D81(%rip),%eax
  test   %eax,%eax
  jns    L34
  xorl   $0x1,-0x4(%rbp)
L34:
  mov    D81(%rip),%eax
  sar    %eax
  mov    %eax,D18(%rip)
  mov    -0x4(%rbp),%eax
  shl    $0x1f,%eax
  mov    %eax,%edx
  mov    D18(%rip),%eax
  or     %edx,%eax
  mov    %eax,D18(%rip)
  addl   $0x1,-0x8(%rbp)
L24:
  cmpl   $0x8,-0x8(%rbp)
  jle    L29
  mov    D18(%rip),%eax
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
uint8_t random_byte_lfsr32(void) {
  static uint32_t lfsr = 0xDEADBEEF;
  uint8_t state;
  for (state = 0; state <= 8; state++) {
    uint8_t new_bit = 0;
    if (lfsr & ((uint32_t)(1) << (1)))
      new_bit ^= 1;
    if (lfsr & ((uint32_t)(1) << (5)))
      new_bit ^= 1;
    if (lfsr & ((uint32_t)(1) << (6)))
      new_bit ^= 1;
    if (lfsr & ((uint32_t)(1) << (31)))
      new_bit ^= 1;
    lfsr >>= 1;
    lfsr |= (uint32_t)(new_bit) << 31;
  };
  return (uint8_t)(lfsr & 0xFF);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<random_byte_lfsr32>:
  endbr64
  mov    D8(%rip),%eax
  mov    $0x9,%ecx
  jmp    L16
L14:
  test   %edx,%edx
  sete   %dl
  movzbl %dl,%edx
L38:
  mov    %edx,%esi
  xor    $0x1,%esi
  test   $0x40,%al
  cmovne %esi,%edx
  mov    %edx,%esi
  xor    $0x1,%esi
  test   %eax,%eax
  cmovs  %esi,%edx
  sar    %eax
  shl    $0x1f,%edx
  or     %edx,%eax
  sub    $0x1,%ecx
  je     L20
L16:
  mov    %eax,%edx
  and    $0x2,%edx
  test   $0x20,%al
  jne    L14
  test   %edx,%edx
  setne  %dl
  movzbl %dl,%edx
  jmp    L38
L20:
  mov    %eax,D9(%rip)
  movzbl %al,%eax
  ret
```

assistant: ```c
uint8_t random_byte_lfsr32(void) {
  static uint32_t lfsr = 0xDEADBEEF;
  uint8_t state;
  for (state = 0; state <= 8; state++) {
    uint8_t new_bit = 0;
    if (lfsr & ((uint32_t)(1) << (1)))
      new_bit ^= 1;
    if (lfsr & ((uint32_t)(1) << (5)))
      new_bit ^= 1;
    if (lfsr & ((uint32_t)(1) << (6)))
      new_bit ^= 1;
    if (lfsr & ((uint32_t)(1) << (31)))
      new_bit ^= 1;
    lfsr >>= 1;
    lfsr |= (uint32_t)(new_bit) << 31;
  };
  return (uint8_t)(lfsr & 0xFF);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<random_byte_lfsr32>:
  endbr64
  mov    D8(%rip),%eax
  mov    $0x9,%ecx
  jmp    L33
  nopl   0x0(%rax)
L3:
  test   %edx,%edx
  setne  %dl
  movzbl %dl,%edx
L34:
  mov    %edx,%esi
  xor    $0x1,%esi
  test   $0x40,%al
  cmovne %esi,%edx
  mov    %edx,%esi
  xor    $0x1,%esi
  test   %eax,%eax
  cmovs  %esi,%edx
  sar    %eax
  shl    $0x1f,%edx
  or     %edx,%eax
  sub    $0x1,%ecx
  je     L25
L33:
  mov    %eax,%edx
  and    $0x2,%edx
  test   $0x20,%al
  je     L3
  test   %edx,%edx
  sete   %dl
  movzbl %dl,%edx
  jmp    L34
  nopl   0x0(%rax,%rax,1)
L25:
  mov    %eax,D7(%rip)
  movzbl %al,%eax
  ret
```

assistant: ```c
uint8_t random_byte_lfsr32(void) {
  static uint32_t lfsr = 0xDEADBEEF;
  uint8_t state;
  for (state = 0; state <= 8; state++) {
    uint8_t new_bit = 0;
    if (lfsr & ((uint32_t)(1) << (1)))
      new_bit ^= 1;
    if (lfsr & ((uint32_t)(1) << (5)))
      new_bit ^= 1;
    if (lfsr & ((uint32_t)(1) << (6)))
      new_bit ^= 1;
    if (lfsr & ((uint32_t)(1) << (31)))
      new_bit ^= 1;
    lfsr >>= 1;
    lfsr |= (uint32_t)(new_bit) << 31;
  };
  return (uint8_t)(lfsr & 0xFF);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<strrev>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L42
  mov    $0x1,%eax
  jmp    L40
L42:
  movq   $0x0,-0x10(%rbp)
  jmp    L54
L22:
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addq   $0x1,-0x10(%rbp)
L54:
  mov    -0x10(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L22
  movq   $0x1,-0x10(%rbp)
  jmp    L10
L16:
  mov    -0x20(%rbp),%rax
  sub    -0x10(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%rdx
  lea    -0x1(%rdx),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addq   $0x1,-0x10(%rbp)
L10:
  mov    -0x10(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jbe    L16
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  movq   $0x0,-0x8(%rbp)
  mov    $0x0,%eax
L40:
  leave
  ret
```

assistant: ```c
int strrev(unsigned char *str, size_t s) {
  size_t i;
  unsigned char *tmp;
  if ((tmp = malloc(sizeof(unsigned char *) * s)) == NULL)
    return 1;
  for (i = 0; i < s; i++) {
    tmp[i] = str[i];
  }
  for (i = 1; i <= s; i++) {
    str[i - 1] = tmp[s - i];
  }
  free(tmp);
  tmp = NULL;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<strrev>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  lea    0x0(,%rsi,8),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L8
  mov    %rax,%rdi
  test   %rbp,%rbp
  je     L31
  mov    $0x0,%edx
L44:
  movzbl (%rbx,%rdx,1),%ecx
  mov    %cl,(%rdi,%rdx,1)
  mov    %rdx,%rcx
  add    $0x1,%rdx
  cmp    %rdx,%rbp
  jne    L44
  add    %rdi,%rcx
  mov    $0x1,%esi
L48:
  movzbl (%rcx),%r8d
  mov    %r8b,-0x1(%rbx,%rsi,1)
  add    $0x1,%rsi
  sub    $0x1,%rcx
  cmp    %rdx,%rsi
  jbe    L48
L31:
  call   <free@plt>
  mov    $0x0,%eax
L36:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  mov    $0x1,%eax
  jmp    L36
```

assistant: ```c
int strrev(unsigned char *str, size_t s) {
  size_t i;
  unsigned char *tmp;
  if ((tmp = malloc(sizeof(unsigned char *) * s)) == NULL)
    return 1;
  for (i = 0; i < s; i++) {
    tmp[i] = str[i];
  }
  for (i = 1; i <= s; i++) {
    str[i - 1] = tmp[s - i];
  }
  free(tmp);
  tmp = NULL;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strrev>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rsi,8),%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  test   %rax,%rax
  je     L31
  mov    %rax,%rdi
  test   %rbp,%rbp
  je     L26
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L22:
  movzbl (%rbx,%rdx,1),%ecx
  mov    %cl,(%rdi,%rdx,1)
  mov    %rdx,%rcx
  add    $0x1,%rdx
  cmp    %rdx,%rbp
  jne    L22
  add    %rdi,%rcx
  mov    $0x1,%esi
  nopl   0x0(%rax,%rax,1)
L29:
  movzbl (%rcx),%r8d
  sub    $0x1,%rcx
  mov    %r8b,-0x1(%rbx,%rsi,1)
  add    $0x1,%rsi
  cmp    %rdx,%rsi
  jbe    L29
L26:
  call   <free@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L31:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int strrev(unsigned char *str, size_t s) {
  size_t i;
  unsigned char *tmp;
  if ((tmp = malloc(sizeof(unsigned char *) * s)) == NULL)
    return 1;
  for (i = 0; i < s; i++) {
    tmp[i] = str[i];
  }
  for (i = 1; i <= s; i++) {
    str[i - 1] = tmp[s - i];
  }
  free(tmp);
  tmp = NULL;
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strrev>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rsi,8),%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  test   %rax,%rax
  je     L180
  mov    %rax,%rdi
  test   %rbp,%rbp
  je     L98
  lea    0x1(%rbx),%rcx
  lea    -0x1(%rbp),%rdx
  sub    %rcx,%rax
  cmp    $0xe,%rax
  jbe    L129
  cmp    $0x6,%rdx
  jbe    L129
  mov    %rbp,%rcx
  cmp    $0xe,%rdx
  jbe    L105
  and    $0xfffffffffffffff0,%rcx
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L50:
  movdqu (%rbx,%rax,1),%xmm0
  movups %xmm0,(%rdi,%rax,1)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L50
  mov    %rbp,%rax
  and    $0xfffffffffffffff0,%rax
  test   $0xf,%bpl
  je     L118
  mov    %rbp,%rcx
  sub    %rax,%rcx
  lea    -0x1(%rcx),%rsi
  cmp    $0x6,%rsi
  jbe    L177
L83:
  mov    (%rbx,%rax,1),%rsi
  mov    %rsi,(%rdi,%rax,1)
  mov    %rcx,%rsi
  and    $0xfffffffffffffff8,%rsi
  add    %rsi,%rax
  cmp    %rsi,%rcx
  je     L118
L177:
  movzbl (%rbx,%rax,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  lea    0x1(%rax),%rcx
  cmp    %rcx,%rbp
  jbe    L118
  movzbl 0x1(%rbx,%rax,1),%ecx
  mov    %cl,0x1(%rdi,%rax,1)
  lea    0x2(%rax),%rcx
  cmp    %rcx,%rbp
  jbe    L118
  movzbl 0x2(%rbx,%rax,1),%ecx
  mov    %cl,0x2(%rdi,%rax,1)
  lea    0x3(%rax),%rcx
  cmp    %rcx,%rbp
  jbe    L118
  movzbl 0x3(%rbx,%rax,1),%ecx
  mov    %cl,0x3(%rdi,%rax,1)
  lea    0x4(%rax),%rcx
  cmp    %rcx,%rbp
  jbe    L118
  movzbl 0x4(%rbx,%rax,1),%ecx
  mov    %cl,0x4(%rdi,%rax,1)
  lea    0x5(%rax),%rcx
  cmp    %rcx,%rbp
  jbe    L118
  movzbl 0x5(%rbx,%rax,1),%ecx
  mov    %cl,0x5(%rdi,%rax,1)
  lea    0x6(%rax),%rcx
  cmp    %rcx,%rbp
  jbe    L118
  movzbl 0x6(%rbx,%rax,1),%ecx
  mov    %cl,0x6(%rdi,%rax,1)
L118:
  add    %rdi,%rdx
  mov    $0x1,%eax
  nopl   (%rax)
L155:
  movzbl (%rdx),%ecx
  sub    $0x1,%rdx
  mov    %cl,-0x1(%rbx,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rbp
  jae    L155
L98:
  call   <free@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L180:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L129:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L135:
  movzbl (%rbx,%rax,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rbp
  jne    L135
  jmp    L118
L105:
  xor    %eax,%eax
  jmp    L83
```

assistant: ```c
int strrev(unsigned char *str, size_t s) {
  size_t i;
  unsigned char *tmp;
  if ((tmp = malloc(sizeof(unsigned char *) * s)) == NULL)
    return 1;
  for (i = 0; i < s; i++) {
    tmp[i] = str[i];
  }
  for (i = 1; i <= s; i++) {
    str[i - 1] = tmp[s - i];
  }
  free(tmp);
  tmp = NULL;
  return 0;
}
```