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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  lea    -0xf4000(%rsp),%r11
L57:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L57
  sub    $0x278,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    -0xf4260(%rbp),%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0xf4260(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0xf4268(%rbp)
  cmpq   $0x3,-0xf4268(%rbp)
  ja     L59
  lea    -0xf4260(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  call   <pre_defined@plt>
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L60
L59:
  movl   $0x6,-0xf4274(%rbp)
  mov    $0x2,%edi
  call   <log10@plt>
  lea    (%rax,%rax,1),%ebx
  mov    $0x3,%edi
  call   <log10@plt>
  add    %ebx,%eax
  lea    (%rax,%rax,1),%ebx
  mov    $0x5,%edi
  call   <log10@plt>
  add    %ebx,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  %xmm0,-0xf4270(%rbp)
L82:
  addl   $0x1,-0xf4274(%rbp)
  mov    -0xf4274(%rbp),%eax
  mov    %eax,%edi
  call   <log10@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  -0xf4270(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0xf4270(%rbp)
  mov    -0xf4268(%rbp),%rax
  test   %rax,%rax
  js     L38
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  jmp    L79
L38:
  mov    %rax,%rdx
  shr    %rdx
  and    $0x1,%eax
  or     %rax,%rdx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdx,%xmm0
  addsd  %xmm0,%xmm0
L79:
  movsd  -0xf4270(%rbp),%xmm1
  comisd %xmm0,%xmm1
  jbe    L32
  mov    -0xf4274(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L60
L32:
  jmp    L82
L60:
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D22
type_infered: string
name: disassemble_data
arguments: label: D1
type_infered: string

tool: D22: %s

tool: D1: %d


assistant: ```c
int main() {
  char buffer[1000001];
  scanf("%s", buffer);
  size_t digits = strlen(buffer);
  if (digits < 4) {
    printf("%d\n", pre_defined(atoi(buffer)));
  } else {
    int number = 6;
    double log_sum = 4 * log10(2) + 2 * log10(3) + log10(5);
    while (1) {
      log_sum += log10(++number);
      if (log_sum > digits) {
        printf("%d\n", number - 1);
        break;
      }
    }
  }
  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
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xf4000(%rsp),%r11
L57:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L57
  sub    $0x268,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xf4258(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  cmp    $0x3,%rax
  ja     L39
  mov    %rbx,%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  call   <pre_defined@plt>
  mov    %eax,%esi
  lea    D23(%rip),%rdi
  call   <printf@plt>
L19:
  mov    0xf4258(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L49
  mov    $0x0,%eax
  add    $0xf4268,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L39:
  mov    %rax,%rbp
  mov    $0x2,%edi
  call   <log10@plt>
  mov    %eax,%r12d
  mov    $0x3,%edi
  call   <log10@plt>
  mov    %eax,%ebx
  mov    $0x5,%edi
  call   <log10@plt>
  lea    (%rbx,%r12,2),%edx
  lea    (%rax,%rdx,2),%eax
  pxor   %xmm3,%xmm3
  cvtsi2sd %eax,%xmm3
  movsd  %xmm3,0x8(%rsp)
  mov    $0x6,%ebx
  mov    %rbp,%r13
  and    $0x1,%r13d
  jmp    L37
L30:
  mov    %rbp,%rax
  shr    %rax
  or     %r13,%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  addsd  %xmm0,%xmm0
L73:
  movsd  0x8(%rsp),%xmm2
  comisd %xmm0,%xmm2
  ja     L55
L37:
  mov    %ebx,%r12d
  add    $0x1,%ebx
  mov    %ebx,%edi
  call   <log10@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  addsd  0x8(%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
  test   %rbp,%rbp
  js     L30
  pxor   %xmm0,%xmm0
  cvtsi2sd %rbp,%xmm0
  jmp    L73
L55:
  mov    %r12d,%esi
  lea    D23(%rip),%rdi
  call   <printf@plt>
  jmp    L19
L49:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D3
data_type: string
name: unpack_data
arguments: label: D23
data_type: string

tool: D3: %s

tool: D23: %d


assistant: ```c
int main() {
  char buffer[1000001];
  scanf("%s", buffer);
  size_t digits = strlen(buffer);
  if (digits < 4) {
    printf("%d\n", pre_defined(atoi(buffer)));
  } else {
    int number = 6;
    double log_sum = 4 * log10(2) + 2 * log10(3) + log10(5);
    while (1) {
      log_sum += log10(++number);
      if (log_sum > digits) {
        printf("%d\n", number - 1);
        break;
      }
    }
  }
  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
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xf4000(%rsp),%r11
L46:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L46
  sub    $0x260,%rsp
  lea    D25(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0xf4258(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  mov    %r12,%rsi
  call   <scanf@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    $0x3,%rax
  ja     L24
  mov    %r12,%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  call   <pre_defined@plt>
  lea    D20(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
L15:
  mov    0xf4258(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L41
  add    $0xf4260,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L24:
  mov    $0x2,%edi
  mov    %rax,%rbp
  call   <log10@plt>
  mov    $0x3,%edi
  mov    %eax,%r12d
  call   <log10@plt>
  mov    $0x5,%edi
  mov    %eax,%ebx
  call   <log10@plt>
  lea    (%rbx,%r12,2),%edx
  pxor   %xmm3,%xmm3
  pxor   %xmm0,%xmm0
  cvtsi2sd %rbp,%xmm3
  lea    (%rax,%rdx,2),%eax
  mov    $0x6,%ebx
  cvtsi2sd %eax,%xmm0
  movsd  %xmm3,0x8(%rsp)
  nopl   0x0(%rax,%rax,1)
L10:
  mov    %ebx,%ebp
  add    $0x1,%ebx
  movsd  %xmm0,(%rsp)
  mov    %ebx,%edi
  call   <log10@plt>
  pxor   %xmm1,%xmm1
  movsd  (%rsp),%xmm0
  cvtsi2sd %eax,%xmm1
  addsd  %xmm1,%xmm0
  comisd 0x8(%rsp),%xmm0
  jbe    L10
  mov    %ebp,%esi
  lea    D20(%rip),%rdi
  call   <printf@plt>
  jmp    L15
L41:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D25
datatype: string
name: parse_data
arguments: data_label: D20
datatype: string

tool: D25: %s

tool: D20: %d


assistant: ```c
int main() {
  char buffer[1000001];
  scanf("%s", buffer);
  size_t digits = strlen(buffer);
  if (digits < 4) {
    printf("%d\n", pre_defined(atoi(buffer)));
  } else {
    int number = 6;
    double log_sum = 4 * log10(2) + 2 * log10(3) + log10(5);
    while (1) {
      log_sum += log10(++number);
      if (log_sum > digits) {
        printf("%d\n", number - 1);
        break;
      }
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xf4000(%rsp),%r11
L30:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L30
  sub    $0x260,%rsp
  lea    D24(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0xf4258(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  mov    %r12,%rsi
  call   <scanf@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    $0x3,%rax
  ja     L12
  mov    %r12,%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  call   <pre_defined@plt>
  lea    D25(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
L22:
  mov    0xf4258(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0xf4260,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  mov    $0x2,%edi
  mov    %rax,%rbp
  call   <log10@plt>
  mov    $0x3,%edi
  mov    %eax,%r12d
  call   <log10@plt>
  mov    $0x5,%edi
  mov    %eax,%ebx
  call   <log10@plt>
  lea    (%rbx,%r12,2),%edx
  pxor   %xmm3,%xmm3
  pxor   %xmm0,%xmm0
  cvtsi2sd %rbp,%xmm3
  lea    (%rax,%rdx,2),%eax
  mov    $0x6,%ebx
  cvtsi2sd %eax,%xmm0
  movsd  %xmm3,0x8(%rsp)
  nopl   0x0(%rax,%rax,1)
L13:
  mov    %ebx,%ebp
  add    $0x1,%ebx
  movsd  %xmm0,(%rsp)
  mov    %ebx,%edi
  call   <log10@plt>
  pxor   %xmm1,%xmm1
  movsd  (%rsp),%xmm0
  cvtsi2sd %eax,%xmm1
  addsd  %xmm1,%xmm0
  comisd 0x8(%rsp),%xmm0
  jbe    L13
  mov    %ebp,%esi
  lea    D25(%rip),%rdi
  call   <printf@plt>
  jmp    L22
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D24
datatype: string
name: unpack_data
arguments: tag: D25
datatype: string

tool: D24: %s

tool: D25: %d


assistant: ```c
int main() {
  char buffer[1000001];
  scanf("%s", buffer);
  size_t digits = strlen(buffer);
  if (digits < 4) {
    printf("%d\n", pre_defined(atoi(buffer)));
  } else {
    int number = 6;
    double log_sum = 4 * log10(2) + 2 * log10(3) + log10(5);
    while (1) {
      log_sum += log10(++number);
      if (log_sum > digits) {
        printf("%d\n", number - 1);
        break;
      }
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<freqawav>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  movsd  %xmm0,-0x18(%rbp)
  mov    %edi,-0x1c(%rbp)
  mov    %esi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    %r9,-0x40(%rbp)
  mov    -0x40(%rbp),%r9
  mov    -0x38(%rbp),%r8
  mov    -0x30(%rbp),%rcx
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%esi
  mov    -0x1c(%rbp),%edi
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  mov    $0x1,%eax
  call   <freqwave@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L16
  mov    -0x4(%rbp),%eax
  jmp    L1
L16:
  mov    -0x40(%rbp),%r9
  mov    -0x38(%rbp),%r8
  mov    -0x38(%rbp),%rcx
  mov    -0x24(%rbp),%edx
  mov    -0x24(%rbp),%esi
  mov    -0x1c(%rbp),%edi
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  mov    $0x1,%eax
  call   <waveawav@plt>
L1:
  leave
  ret
```

assistant: ```c
int freqawav(dummy, nfreq, sfreq, sawav, freq, awav, stat)
double dummy;
int nfreq, sfreq, sawav;
const double freq[];
double awav[];
int stat[];
{
  int status;
  if ((status = freqwave(dummy, nfreq, sfreq, sawav, freq, awav, stat))) {
    return status;
  }
  return waveawav(dummy, nfreq, sawav, sawav, awav, awav, stat);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<freqawav>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movq   %xmm0,%r12
  mov    %edi,%r13d
  mov    %edx,%ebx
  mov    %r8,%rbp
  mov    %r9,%r14
  mov    $0x1,%eax
  call   <freqwave@plt>
  test   %eax,%eax
  je     L0
L3:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  mov    %r14,%r9
  mov    %rbp,%r8
  mov    %rbp,%rcx
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %r13d,%edi
  movq   %r12,%xmm0
  mov    $0x1,%eax
  call   <waveawav@plt>
  jmp    L3
```

assistant: ```c
int freqawav(dummy, nfreq, sfreq, sawav, freq, awav, stat)
double dummy;
int nfreq, sfreq, sawav;
const double freq[];
double awav[];
int stat[];
{
  int status;
  if ((status = freqwave(dummy, nfreq, sfreq, sawav, freq, awav, stat))) {
    return status;
  }
  return waveawav(dummy, nfreq, sawav, sawav, awav, awav, stat);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<freqawav>:
  endbr64
  push   %r13
  mov    $0x1,%eax
  mov    %edi,%r13d
  push   %r12
  mov    %r8,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %r9,%rbx
  sub    $0x18,%rsp
  movsd  %xmm0,0x8(%rsp)
  call   <freqwave@plt>
  test   %eax,%eax
  je     L3
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  mov    %rbx,%r9
  mov    %r12,%r8
  pop    %rbx
  mov    %r12,%rcx
  mov    %ebp,%edx
  mov    %ebp,%esi
  mov    %r13d,%edi
  pop    %rbp
  mov    $0x1,%eax
  pop    %r12
  pop    %r13
  jmp    1060 <waveawav@plt>
```

assistant: ```c
int freqawav(dummy, nfreq, sfreq, sawav, freq, awav, stat)
double dummy;
int nfreq, sfreq, sawav;
const double freq[];
double awav[];
int stat[];
{
  int status;
  if ((status = freqwave(dummy, nfreq, sfreq, sawav, freq, awav, stat))) {
    return status;
  }
  return waveawav(dummy, nfreq, sawav, sawav, awav, awav, stat);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<freqawav>:
  endbr64
  push   %r13
  mov    $0x1,%eax
  mov    %edi,%r13d
  push   %r12
  mov    %r8,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %r9,%rbx
  sub    $0x18,%rsp
  movsd  %xmm0,0x8(%rsp)
  call   <freqwave@plt>
  test   %eax,%eax
  je     L7
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  mov    %rbx,%r9
  mov    %r12,%r8
  pop    %rbx
  mov    %r12,%rcx
  mov    %ebp,%edx
  mov    %ebp,%esi
  mov    %r13d,%edi
  pop    %rbp
  mov    $0x1,%eax
  pop    %r12
  pop    %r13
  jmp    1060 <waveawav@plt>
```

assistant: ```c
int freqawav(dummy, nfreq, sfreq, sawav, freq, awav, stat)
double dummy;
int nfreq, sfreq, sawav;
const double freq[];
double awav[];
int stat[];
{
  int status;
  if ((status = freqwave(dummy, nfreq, sfreq, sawav, freq, awav, stat))) {
    return status;
  }
  return waveawav(dummy, nfreq, sawav, sawav, awav, awav, stat);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<unbase64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x2c(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <BIO_new_mem_buf@plt>
  mov    %rax,-0x20(%rbp)
  mov    D17(%rip),%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <BIO_set_close@plt>
  mov    $0x0,%eax
  call   <BIO_f_base64@plt>
  mov    %eax,%edi
  call   <BIO_new@plt>
  mov    %rax,-0x18(%rbp)
  mov    D12(%rip),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <BIO_set_flags@plt>
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <BIO_push@plt>
  mov    -0x2c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x2c(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <BIO_read@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <BIO_free_all@plt>
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *unbase64(const char *in, int in_len, int *out_len) {
  BIO *buf, *b64;
  char *out;
  long len;
  buf = BIO_new_mem_buf(in, in_len);
  BIO_set_close(buf, BIO_CLOSE);
  b64 = BIO_new(BIO_f_base64());
  BIO_set_flags(b64, BIO_FLAGS_BASE64_NO_NL);
  BIO_push(b64, buf);
  out = malloc((in_len + 1) * sizeof(char));
  len = BIO_read(b64, out, in_len);
  out[len] = '\0';
  *out_len = len;
  BIO_free_all(b64);
  return out;
}
```
user: What is the c source code of the assembly code below?

```asm
<unbase64>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%r12d
  mov    %rdx,%r13
  call   <BIO_new_mem_buf@plt>
  mov    %rax,%rbp
  mov    D2(%rip),%rax
  mov    (%rax),%esi
  mov    %rbp,%rdi
  call   <BIO_set_close@plt>
  mov    $0x0,%eax
  call   <BIO_f_base64@plt>
  mov    %eax,%edi
  call   <BIO_new@plt>
  mov    %rax,%rbx
  mov    D13(%rip),%rax
  mov    (%rax),%esi
  mov    %rbx,%rdi
  call   <BIO_set_flags@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <BIO_push@plt>
  lea    0x1(%r12),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %r12d,%edx
  mov    %rax,%rsi
  mov    %rbx,%rdi
  call   <BIO_read@plt>
  movb   $0x0,0x0(%rbp,%rax,1)
  mov    %eax,0x0(%r13)
  mov    %rbx,%rdi
  call   <BIO_free_all@plt>
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *unbase64(const char *in, int in_len, int *out_len) {
  BIO *buf, *b64;
  char *out;
  long len;
  buf = BIO_new_mem_buf(in, in_len);
  BIO_set_close(buf, BIO_CLOSE);
  b64 = BIO_new(BIO_f_base64());
  BIO_set_flags(b64, BIO_FLAGS_BASE64_NO_NL);
  BIO_push(b64, buf);
  out = malloc((in_len + 1) * sizeof(char));
  len = BIO_read(b64, out, in_len);
  out[len] = '\0';
  *out_len = len;
  BIO_free_all(b64);
  return out;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<unbase64>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <BIO_new_mem_buf@plt>
  mov    %rax,%r12
  mov    D0(%rip),%rax
  mov    %r12,%rdi
  mov    (%rax),%esi
  call   <BIO_set_close@plt>
  xor    %eax,%eax
  call   <BIO_f_base64@plt>
  mov    %eax,%edi
  call   <BIO_new@plt>
  mov    %rax,%rbp
  mov    D17(%rip),%rax
  mov    %rbp,%rdi
  mov    (%rax),%esi
  call   <BIO_set_flags@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <BIO_push@plt>
  lea    0x1(%r13),%edi
  call   <malloc@plt>
  mov    %r13d,%edx
  mov    %rbp,%rdi
  mov    %rax,%rsi
  mov    %rax,%r12
  call   <BIO_read@plt>
  mov    %rbp,%rdi
  movb   $0x0,(%r12,%rax,1)
  mov    %eax,(%rbx)
  call   <BIO_free_all@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *unbase64(const char *in, int in_len, int *out_len) {
  BIO *buf, *b64;
  char *out;
  long len;
  buf = BIO_new_mem_buf(in, in_len);
  BIO_set_close(buf, BIO_CLOSE);
  b64 = BIO_new(BIO_f_base64());
  BIO_set_flags(b64, BIO_FLAGS_BASE64_NO_NL);
  BIO_push(b64, buf);
  out = malloc((in_len + 1) * sizeof(char));
  len = BIO_read(b64, out, in_len);
  out[len] = '\0';
  *out_len = len;
  BIO_free_all(b64);
  return out;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<unbase64>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <BIO_new_mem_buf@plt>
  mov    %rax,%r12
  mov    D3(%rip),%rax
  mov    %r12,%rdi
  mov    (%rax),%esi
  call   <BIO_set_close@plt>
  xor    %eax,%eax
  call   <BIO_f_base64@plt>
  mov    %eax,%edi
  call   <BIO_new@plt>
  mov    %rax,%rbp
  mov    D0(%rip),%rax
  mov    %rbp,%rdi
  mov    (%rax),%esi
  call   <BIO_set_flags@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <BIO_push@plt>
  lea    0x1(%r13),%edi
  call   <malloc@plt>
  mov    %r13d,%edx
  mov    %rbp,%rdi
  mov    %rax,%rsi
  mov    %rax,%r12
  call   <BIO_read@plt>
  mov    %rbp,%rdi
  movb   $0x0,(%r12,%rax,1)
  mov    %eax,(%rbx)
  call   <BIO_free_all@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *unbase64(const char *in, int in_len, int *out_len) {
  BIO *buf, *b64;
  char *out;
  long len;
  buf = BIO_new_mem_buf(in, in_len);
  BIO_set_close(buf, BIO_CLOSE);
  b64 = BIO_new(BIO_f_base64());
  BIO_set_flags(b64, BIO_FLAGS_BASE64_NO_NL);
  BIO_push(b64, buf);
  out = malloc((in_len + 1) * sizeof(char));
  len = BIO_read(b64, out, in_len);
  out[len] = '\0';
  *out_len = len;
  BIO_free_all(b64);
  return out;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fb_CVDFROMLONGINT>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rcx
  lea    -0x10(%rbp),%rax
  mov    $0x8,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movsd  -0x10(%rbp),%xmm0
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double fb_CVDFROMLONGINT(long long ll) {
  do {
    if ((8) == sizeof(ll) && (8) == sizeof(double)) {
      double to;
      memcpy(&to, &ll, 8);
      return to;
    } else {
      return (double)0;
    }
  } while (0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<fb_CVDFROMLONGINT>:
  endbr64
  sub    $0x28,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  mov    $0x8,%edx
  call   <memcpy@plt>
  movsd  0x10(%rsp),%xmm0
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double fb_CVDFROMLONGINT(long long ll) {
  do {
    if ((8) == sizeof(ll) && (8) == sizeof(double)) {
      double to;
      memcpy(&to, &ll, 8);
      return to;
    } else {
      return (double)0;
    }
  } while (0);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<fb_CVDFROMLONGINT>:
  endbr64
  sub    $0x28,%rsp
  mov    $0x8,%edx
  mov    %rdi,0x8(%rsp)
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <memcpy@plt>
  movsd  0x10(%rsp),%xmm0
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double fb_CVDFROMLONGINT(long long ll) {
  do {
    if ((8) == sizeof(ll) && (8) == sizeof(double)) {
      double to;
      memcpy(&to, &ll, 8);
      return to;
    } else {
      return (double)0;
    }
  } while (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fb_CVDFROMLONGINT>:
  endbr64
  sub    $0x28,%rsp
  mov    $0x8,%edx
  mov    %rdi,0x8(%rsp)
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <memcpy@plt>
  movsd  0x10(%rsp),%xmm0
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x28,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double fb_CVDFROMLONGINT(long long ll) {
  do {
    if ((8) == sizeof(ll) && (8) == sizeof(double)) {
      double to;
      memcpy(&to, &ll, 8);
      return to;
    } else {
      return (double)0;
    }
  } while (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<skipname>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,-0x18(%rbp)
  andl   $0x3,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L3
  mov    $0x4,%eax
  sub    -0x4(%rbp),%eax
  cltq
  add    %rax,-0x18(%rbp)
L3:
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *skipname(char *p) {
  int n = strlen(p) + 1;
  p += n;
  if (n &= 3)
    p += 4 - n;
  return (p);
}
```
user: Translate the below assembly code into C programming language:

```asm
<skipname>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%edx
  movslq %edx,%rax
  add    %rbx,%rax
  and    $0x3,%edx
  je     L1
  mov    $0x4,%ecx
  sub    %edx,%ecx
  movslq %ecx,%rdx
  add    %rdx,%rax
L1:
  pop    %rbx
  ret
```

assistant: ```c
char *skipname(char *p) {
  int n = strlen(p) + 1;
  p += n;
  if (n &= 3)
    p += 4 - n;
  return (p);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<skipname>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%edx
  movslq %edx,%rax
  add    %rbx,%rax
  and    $0x3,%edx
  je     L4
  mov    $0x4,%ecx
  sub    %edx,%ecx
  movslq %ecx,%rdx
  add    %rdx,%rax
L4:
  pop    %rbx
  ret
```

assistant: ```c
char *skipname(char *p) {
  int n = strlen(p) + 1;
  p += n;
  if (n &= 3)
    p += 4 - n;
  return (p);
}
```