ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<aes_encrypt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %rcx,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x2c(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L58
L73:
  mov    -0x2c(%rbp),%eax
  mov    -0x50(%rbp),%rdx
  shr    $0x4,%rdx
  cmp    %rdx,%rax
  jne    L30
  mov    -0x50(%rbp),%rax
  and    $0xf,%eax
  mov    %eax,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  je     L14
  lea    -0x20(%rbp),%rax
  mov    $0x10,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L17
L30:
  movl   $0x10,-0x24(%rbp)
L17:
  mov    -0x24(%rbp),%edx
  mov    -0x2c(%rbp),%eax
  shl    $0x4,%rax
  mov    %rax,%rcx
  mov    -0x40(%rbp),%rax
  add    %rax,%rcx
  lea    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movl   $0x0,-0x28(%rbp)
  jmp    L74
L1:
  mov    -0x2c(%rbp),%eax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%eax
  add    %rax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x28(%rbp),%ecx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  xor    %eax,%edx
  mov    -0x28(%rbp),%eax
  mov    %dl,-0x20(%rbp,%rax,1)
  addl   $0x1,-0x28(%rbp)
L74:
  mov    -0x28(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jb     L1
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <encrypt@plt>
  lea    -0x20(%rbp),%rcx
  mov    -0x38(%rbp),%rax
  mov    $0x10,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x2c(%rbp),%eax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x48(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  lea    -0x20(%rbp),%rax
  mov    $0x10,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  addl   $0x1,-0x2c(%rbp)
L58:
  mov    -0x2c(%rbp),%eax
  mov    -0x50(%rbp),%rdx
  shr    $0x4,%rdx
  cmp    %rdx,%rax
  jbe    L73
  jmp    L7
L14:
  nop
L7:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L63
  call   <__stack_chk_fail@plt>
L63:
  leave
  ret
```

assistant: ```c
void aes_encrypt(unsigned char *iv, unsigned char *inbuf, unsigned char *outbuf, unsigned long long len) {
  unsigned char block[16];
  unsigned int blockno = 0, i;
  for (blockno = 0; blockno <= (len / sizeof(block)); blockno++) {
    unsigned int fraction;
    if (blockno == (len / sizeof(block))) {
      fraction = len % sizeof(block);
      if (fraction == 0)
        break;
      memset(block, 0, sizeof(block));
    } else
      fraction = 16;
    memcpy(block, inbuf + blockno * sizeof(block), fraction);
    for (i = 0; i < fraction; i++)
      block[i] = inbuf[blockno * sizeof(block) + i] ^ iv[i];
    encrypt(block);
    memcpy(iv, block, sizeof(block));
    memcpy(outbuf + blockno * sizeof(block), block, sizeof(block));
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<aes_encrypt>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,%r12
  mov    %rsi,0x10(%rsp)
  mov    %rdx,0x8(%rsp)
  mov    %rcx,0x18(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rcx,%r15
  shr    $0x4,%r15
  mov    $0x0,%r13d
  mov    $0x0,%ebp
  mov    $0x10,%r14d
  jmp    L19
L0:
  mov    %r14d,%ebx
L69:
  shl    $0x4,%rbp
  mov    %rbp,%rax
  mov    %rbp,(%rsp)
  mov    0x10(%rsp),%rbp
  add    %rax,%rbp
  lea    0x20(%rsp),%rdi
  mov    %ebx,%edx
  mov    %rbp,%rsi
  call   <memcpy@plt>
  mov    $0x0,%eax
L40:
  movzbl 0x0(%rbp,%rax,1),%edx
  xor    (%r12,%rax,1),%dl
  mov    %dl,0x20(%rsp,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%ebx
  ja     L40
  lea    0x20(%rsp),%rbx
  mov    %rbx,%rdi
  call   <encrypt@plt>
  mov    %r14d,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  mov    (%rsp),%rdi
  add    0x8(%rsp),%rdi
  mov    %r14d,%edx
  mov    %rbx,%rsi
  call   <memcpy@plt>
  add    $0x1,%r13d
  mov    %r13d,%ebp
  cmp    %r15,%rbp
  ja     L61
L19:
  cmp    %rbp,%r15
  jne    L0
  mov    0x18(%rsp),%ebx
  and    $0xf,%ebx
  je     L61
  lea    0x20(%rsp),%rdi
  mov    %r14d,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  jmp    L69
L61:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L33:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void aes_encrypt(unsigned char *iv, unsigned char *inbuf, unsigned char *outbuf, unsigned long long len) {
  unsigned char block[16];
  unsigned int blockno = 0, i;
  for (blockno = 0; blockno <= (len / sizeof(block)); blockno++) {
    unsigned int fraction;
    if (blockno == (len / sizeof(block))) {
      fraction = len % sizeof(block);
      if (fraction == 0)
        break;
      memset(block, 0, sizeof(block));
    } else
      fraction = 16;
    memcpy(block, inbuf + blockno * sizeof(block), fraction);
    for (i = 0; i < fraction; i++)
      block[i] = inbuf[blockno * sizeof(block) + i] ^ iv[i];
    encrypt(block);
    memcpy(iv, block, sizeof(block));
    memcpy(outbuf + blockno * sizeof(block), block, sizeof(block));
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<aes_encrypt>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rsi,0x8(%rsp)
  lea    0x20(%rsp),%r15
  mov    %rdx,0x10(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  mov    %rcx,%rax
  shr    $0x4,%rax
  mov    %rax,(%rsp)
  mov    %ecx,%eax
  and    $0xf,%eax
  mov    %eax,0x1c(%rsp)
  nopl   0x0(%rax,%rax,1)
L23:
  cmp    (%rsp),%rbp
  jne    L36
  mov    0x1c(%rsp),%r13d
  mov    %r13d,%ebx
  test   %r13d,%r13d
  je     L53
  mov    $0x10,%edx
  xor    %esi,%esi
  mov    %r15,%rdi
  call   <memset@plt>
  mov    %r13d,%edx
L58:
  mov    0x8(%rsp),%rax
  shl    $0x4,%rbp
  mov    %r15,%rdi
  lea    (%rax,%rbp,1),%r13
  mov    %r13,%rsi
  call   <memcpy@plt>
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L20:
  movzbl 0x0(%r13,%rax,1),%edx
  xor    (%r14,%rax,1),%dl
  mov    %dl,(%r15,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rbx
  jne    L20
  mov    %r15,%rdi
  call   <encrypt@plt>
  mov    $0x10,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    0x10(%rsp),%rax
  mov    $0x10,%edx
  mov    %r15,%rsi
  lea    (%rax,%rbp,1),%rdi
  lea    0x1(%r12),%ebp
  call   <memcpy@plt>
  mov    %rbp,%r12
  cmp    (%rsp),%rbp
  jbe    L23
L53:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L36:
  mov    $0x10,%edx
  mov    $0x10,%ebx
  jmp    L58
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void aes_encrypt(unsigned char *iv, unsigned char *inbuf, unsigned char *outbuf, unsigned long long len) {
  unsigned char block[16];
  unsigned int blockno = 0, i;
  for (blockno = 0; blockno <= (len / sizeof(block)); blockno++) {
    unsigned int fraction;
    if (blockno == (len / sizeof(block))) {
      fraction = len % sizeof(block);
      if (fraction == 0)
        break;
      memset(block, 0, sizeof(block));
    } else
      fraction = 16;
    memcpy(block, inbuf + blockno * sizeof(block), fraction);
    for (i = 0; i < fraction; i++)
      block[i] = inbuf[blockno * sizeof(block) + i] ^ iv[i];
    encrypt(block);
    memcpy(iv, block, sizeof(block));
    memcpy(outbuf + blockno * sizeof(block), block, sizeof(block));
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x1(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  movzbl -0x1(%rbp),%edx
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void swap(char *a, char *b) {
  char tmp = *b;
  *b = *a;
  *a = tmp;
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<swap>:
  endbr64
  movzbl (%rsi),%eax
  movzbl (%rdi),%edx
  mov    %dl,(%rsi)
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void swap(char *a, char *b) {
  char tmp = *b;
  *b = *a;
  *a = tmp;
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<swap>:
  endbr64
  movzbl (%rsi),%eax
  movzbl (%rdi),%edx
  mov    %dl,(%rsi)
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void swap(char *a, char *b) {
  char tmp = *b;
  *b = *a;
  *a = tmp;
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<swap>:
  endbr64
  movzbl (%rsi),%eax
  movzbl (%rdi),%edx
  mov    %dl,(%rsi)
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void swap(char *a, char *b) {
  char tmp = *b;
  *b = *a;
  *a = tmp;
  return;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<print_hex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L0:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  and    $0xf,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  shr    $0x4,%al
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L12:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L0
  nop
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D9
datatype: string

tool: D9: %x%x

assistant: ```c
void print_hex(unsigned char *buf, int len) {
  for (int i = 0; i < len; i++) {
    printf("%x%x", buf[i] / 16, buf[i] & 15);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<print_hex>:
  endbr64
  test   %esi,%esi
  jle    L15
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%r12
  lea    D7(%rip),%rbp
L19:
  movzbl (%rbx),%esi
  mov    %esi,%edx
  and    $0xf,%edx
  shr    $0x4,%sil
  movzbl %sil,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L19
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  ret
```

tool_calls: name: parse_data
arguments: label: D7
type_guessed: string

tool: D7: %x%x

assistant: ```c
void print_hex(unsigned char *buf, int len) {
  for (int i = 0; i < len; i++) {
    printf("%x%x", buf[i] / 16, buf[i] & 15);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<print_hex>:
  endbr64
  test   %esi,%esi
  jle    L10
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x1(%rdi,%rax,1),%r12
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  xchg   %ax,%ax
L16:
  movzbl (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  mov    %esi,%edx
  shr    $0x4,%sil
  and    $0xf,%edx
  movzbl %sil,%esi
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L16
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L10:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
type_guessed: string

tool: D6: %x%x

assistant: ```c
void print_hex(unsigned char *buf, int len) {
  for (int i = 0; i < len; i++) {
    printf("%x%x", buf[i] / 16, buf[i] & 15);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<print_hex>:
  endbr64
  test   %esi,%esi
  jle    L11
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x1(%rdi,%rax,1),%r12
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  xchg   %ax,%ax
L9:
  movzbl (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  mov    %esi,%edx
  shr    $0x4,%sil
  and    $0xf,%edx
  movzbl %sil,%esi
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L9
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L11:
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D8
datatype: string

tool: D8: %x%x

assistant: ```c
void print_hex(unsigned char *buf, int len) {
  for (int i = 0; i < len; i++) {
    printf("%x%x", buf[i] / 16, buf[i] & 15);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<factorCnt2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L46
L52:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  imul   %eax,%eax
  cmp    %eax,-0x14(%rbp)
  jge    L23
  shll   -0xc(%rbp)
  jmp    L11
L23:
  movl   $0x0,-0x8(%rbp)
  jmp    L13
L44:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0x8(%rbp)
L13:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  je     L44
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0xc(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L46:
  cmpl   $0x1,-0x14(%rbp)
  je     L11
  mov    -0x10(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L52
L11:
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int factorCnt2(int n, int *primes, int primes_n) {
  int i, cnt;
  for (i = 0, cnt = 1; n != 1 && i < primes_n; i++) {
    int p = primes[i], f;
    if (p * p > n) {
      cnt *= 2;
      break;
    }
    for (f = 0; n % p == 0; f++) {
      n /= p;
    }
    cnt *= f + 1;
  }
  return cnt;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<factorCnt2>:
  endbr64
  mov    %edi,%ecx
  mov    %rsi,%r11
  mov    %edx,%r10d
  cmp    $0x1,%edi
  je     L72
  test   %edx,%edx
  jle    L72
  mov    $0x1,%r8d
  mov    $0x1,%r9d
  jmp    L38
L69:
  mov    $0x0,%edi
L63:
  add    $0x1,%edi
  imul   %edi,%r9d
  cmp    $0x1,%ecx
  setne  %dl
  cmp    %r8d,%r10d
  setg   %al
  add    $0x1,%r8
  test   %al,%dl
  je     L88
L38:
  mov    -0x4(%r11,%r8,4),%esi
  mov    %esi,%eax
  imul   %esi,%eax
  cmp    %ecx,%eax
  jg     L21
  mov    %ecx,%eax
  cltd
  idiv   %esi
  mov    %edx,%edi
  test   %edx,%edx
  jne    L69
L19:
  mov    %ecx,%eax
  cltd
  idiv   %esi
  mov    %eax,%ecx
  add    $0x1,%edi
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L19
  jmp    L63
L21:
  add    %r9d,%r9d
L88:
  mov    %r9d,%eax
  ret
L72:
  mov    $0x1,%r9d
  jmp    L88
```

assistant: ```c
int factorCnt2(int n, int *primes, int primes_n) {
  int i, cnt;
  for (i = 0, cnt = 1; n != 1 && i < primes_n; i++) {
    int p = primes[i], f;
    if (p * p > n) {
      cnt *= 2;
      break;
    }
    for (f = 0; n % p == 0; f++) {
      n /= p;
    }
    cnt *= f + 1;
  }
  return cnt;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<factorCnt2>:
  endbr64
  cmp    $0x1,%edi
  je     L28
  mov    %edx,%r11d
  test   %edx,%edx
  jle    L28
  push   %rbx
  mov    %edi,%ecx
  mov    %rsi,%rbx
  mov    $0x1,%r9d
  mov    $0x1,%r10d
  nopw   0x0(%rax,%rax,1)
L1:
  mov    -0x4(%rbx,%r9,4),%esi
  mov    %esi,%eax
  imul   %esi,%eax
  cmp    %ecx,%eax
  jg     L33
  mov    %ecx,%eax
  cltd
  idiv   %esi
  mov    %edx,%edi
  test   %edx,%edx
  jne    L14
  nopl   0x0(%rax)
L44:
  mov    %ecx,%eax
  mov    %edi,%r8d
  add    $0x1,%edi
  cltd
  idiv   %esi
  cltd
  mov    %eax,%ecx
  idiv   %esi
  test   %edx,%edx
  je     L44
  add    $0x2,%r8d
  imul   %r8d,%r10d
L14:
  cmp    $0x1,%ecx
  setne  %dl
  cmp    %r9d,%r11d
  setg   %al
  add    $0x1,%r9
  test   %al,%dl
  jne    L1
  mov    %r10d,%eax
  pop    %rbx
  ret
L33:
  add    %r10d,%r10d
  pop    %rbx
  mov    %r10d,%eax
  ret
L28:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int factorCnt2(int n, int *primes, int primes_n) {
  int i, cnt;
  for (i = 0, cnt = 1; n != 1 && i < primes_n; i++) {
    int p = primes[i], f;
    if (p * p > n) {
      cnt *= 2;
      break;
    }
    for (f = 0; n % p == 0; f++) {
      n /= p;
    }
    cnt *= f + 1;
  }
  return cnt;
}
```
user: Translate the below assembly code into C programming language:

```asm
<factorCnt2>:
  endbr64
  cmp    $0x1,%edi
  je     L1
  mov    %edx,%r11d
  test   %edx,%edx
  jle    L1
  push   %rbx
  mov    %edi,%ecx
  mov    %rsi,%rbx
  mov    $0x1,%r9d
  mov    $0x1,%r10d
  nopw   0x0(%rax,%rax,1)
L59:
  mov    -0x4(%rbx,%r9,4),%esi
  mov    %esi,%eax
  imul   %esi,%eax
  cmp    %ecx,%eax
  jg     L9
  mov    %ecx,%eax
  cltd
  idiv   %esi
  mov    %edx,%edi
  test   %edx,%edx
  jne    L13
  nopl   0x0(%rax)
L40:
  mov    %ecx,%eax
  mov    %edi,%r8d
  add    $0x1,%edi
  cltd
  idiv   %esi
  cltd
  mov    %eax,%ecx
  idiv   %esi
  test   %edx,%edx
  je     L40
  add    $0x2,%r8d
  imul   %r8d,%r10d
L13:
  cmp    $0x1,%ecx
  setne  %dl
  cmp    %r9d,%r11d
  setg   %al
  add    $0x1,%r9
  test   %al,%dl
  jne    L59
  mov    %r10d,%eax
  pop    %rbx
  ret
L9:
  add    %r10d,%r10d
  pop    %rbx
  mov    %r10d,%eax
  ret
L1:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int factorCnt2(int n, int *primes, int primes_n) {
  int i, cnt;
  for (i = 0, cnt = 1; n != 1 && i < primes_n; i++) {
    int p = primes[i], f;
    if (p * p > n) {
      cnt *= 2;
      break;
    }
    for (f = 0; n % p == 0; f++) {
      n /= p;
    }
    cnt *= f + 1;
  }
  return cnt;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<pushresult>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L3
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <lua_pushusertag@plt>
  jmp    L14
L3:
  mov    $0x0,%eax
  call   <lua_pushnil@plt>
  mov    D16(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <strerror@plt>
  mov    %eax,%edi
  call   <lua_pushstring@plt>
  mov    D16(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <lua_pushnumber@plt>
L14:
  nop
  leave
  ret
```

assistant: ```c
void pushresult(int i) {
  if (i)
    lua_pushusertag(NULL, 0);
  else {
    lua_pushnil();
    lua_pushstring(strerror(errno));
    lua_pushnumber(errno);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<pushresult>:
  endbr64
  push   %rbx
  test   %edi,%edi
  je     L3
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <lua_pushusertag@plt>
L15:
  pop    %rbx
  ret
L3:
  mov    $0x0,%eax
  call   <lua_pushnil@plt>
  mov    D6(%rip),%rbx
  mov    (%rbx),%edi
  call   <strerror@plt>
  mov    %eax,%edi
  call   <lua_pushstring@plt>
  mov    (%rbx),%edi
  call   <lua_pushnumber@plt>
  jmp    L15
```

assistant: ```c
void pushresult(int i) {
  if (i)
    lua_pushusertag(NULL, 0);
  else {
    lua_pushnil();
    lua_pushstring(strerror(errno));
    lua_pushnumber(errno);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<pushresult>:
  endbr64
  test   %edi,%edi
  je     L4
  xor    %esi,%esi
  xor    %edi,%edi
  jmp    1090 <lua_pushusertag@plt>
  nopl   0x0(%rax)
L4:
  push   %rbx
  xor    %eax,%eax
  call   <lua_pushnil@plt>
  mov    D2(%rip),%rbx
  mov    (%rbx),%edi
  call   <strerror@plt>
  mov    %eax,%edi
  call   <lua_pushstring@plt>
  mov    (%rbx),%edi
  pop    %rbx
  jmp    10a0 <lua_pushnumber@plt>
```

assistant: ```c
void pushresult(int i) {
  if (i)
    lua_pushusertag(NULL, 0);
  else {
    lua_pushnil();
    lua_pushstring(strerror(errno));
    lua_pushnumber(errno);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<pushresult>:
  endbr64
  test   %edi,%edi
  je     L1
  xor    %esi,%esi
  xor    %edi,%edi
  jmp    1090 <lua_pushusertag@plt>
  nopl   0x0(%rax)
L1:
  push   %rbx
  xor    %eax,%eax
  call   <lua_pushnil@plt>
  mov    D3(%rip),%rbx
  mov    (%rbx),%edi
  call   <strerror@plt>
  mov    %eax,%edi
  call   <lua_pushstring@plt>
  mov    (%rbx),%edi
  pop    %rbx
  jmp    10a0 <lua_pushnumber@plt>
```

assistant: ```c
void pushresult(int i) {
  if (i)
    lua_pushusertag(NULL, 0);
  else {
    lua_pushnil();
    lua_pushstring(strerror(errno));
    lua_pushnumber(errno);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<_prtl8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x0,-0x14(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L7
L78:
  mov    -0x28(%rbp),%rax
  and    $0x7,%eax
  add    $0x30,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  cltq
  mov    %dl,-0x14(%rbp,%rax,1)
  sarq   $0x3,-0x28(%rbp)
  addl   $0x1,-0x18(%rbp)
L7:
  cmpl   $0xb,-0x18(%rbp)
  jle    L78
  movzbl -0x9(%rbp),%eax
  and    $0x33,%eax
  mov    %al,-0x9(%rbp)
  movl   $0xb,-0x18(%rbp)
  jmp    L2
L19:
  subl   $0x1,-0x18(%rbp)
L2:
  mov    -0x18(%rbp),%eax
  cltq
  movzbl -0x14(%rbp,%rax,1),%eax
  cmp    $0x30,%al
  je     L19
  cmpl   $0x0,-0x18(%rbp)
  jne    L12
  addl   $0x1,-0x18(%rbp)
  jmp    L12
L23:
  mov    -0x18(%rbp),%edx
  lea    -0x1(%rdx),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x30(%rbp)
  movslq %edx,%rdx
  movzbl -0x14(%rbp,%rdx,1),%edx
  mov    %dl,(%rax)
L12:
  cmpl   $0x0,-0x18(%rbp)
  jns    L23
  nop
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

assistant: ```c
_prtl8(num, str) long num;
char *str;
{
  int i;
  char c, temp[12];
  temp[0] = '\0';
  for (i = 1; i <= 11; i++) {
    temp[i] = (num & 07) + '0';
    num = num >> 3;
  }
  temp[11] &= '3';
  for (i = 11; temp[i] == '0'; i--)
    ;
  if (i == 0)
    i++;
  while (i >= 0)
    *str++ = temp[i--];
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<_prtl8>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movb   $0x0,0xc(%rsp)
  lea    0xd(%rsp),%rax
  lea    0x18(%rsp),%rcx
L7:
  mov    %edi,%edx
  and    $0x7,%edx
  add    $0x30,%edx
  mov    %dl,(%rax)
  sar    $0x3,%rdi
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L7
  movzbl 0x17(%rsp),%eax
  and    $0x33,%eax
  mov    %al,0x17(%rsp)
  cmp    $0x30,%al
  jne    L62
  mov    $0xa,%edx
  lea    0xd(%rsp),%rcx
L68:
  mov    %edx,%eax
  sub    $0x1,%rdx
  cmpb   $0x30,(%rcx,%rdx,1)
  je     L68
  test   %eax,%eax
  jne    L44
  mov    $0x1,%eax
L47:
  cltq
L83:
  add    $0x1,%rsi
  movzbl 0xc(%rsp,%rax,1),%edx
  mov    %dl,-0x1(%rsi)
  sub    $0x1,%rax
  test   %eax,%eax
  jns    L83
L8:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L77
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L44:
  jns    L47
  jmp    L8
L62:
  mov    $0xb,%eax
  jmp    L47
L77:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
_prtl8(num, str) long num;
char *str;
{
  int i;
  char c, temp[12];
  temp[0] = '\0';
  for (i = 1; i <= 11; i++) {
    temp[i] = (num & 07) + '0';
    num = num >> 3;
  }
  temp[11] &= '3';
  for (i = 11; temp[i] == '0'; i--)
    ;
  if (i == 0)
    i++;
  while (i >= 0)
    *str++ = temp[i--];
}
```