ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<strncmp>:
  endbr64
  test   %edx,%edx
  jle    L32
  lea    -0x1(%rdx),%r8d
  xor    %eax,%eax
  jmp    L31
L28:
  cmp    %cl,%dl
  jne    L27
  lea    0x1(%rax),%rdx
  cmp    %rax,%r8
  je     L32
  mov    %rdx,%rax
L31:
  movzbl (%rdi,%rax,1),%edx
  movsbl (%rsi,%rax,1),%ecx
  test   %dl,%dl
  jne    L28
  neg    %cl
  sbb    %eax,%eax
L35:
  ret
  nopl   0x0(%rax)
L27:
  mov    $0x1,%eax
  test   %cl,%cl
  je     L35
  movsbl %dl,%eax
  sub    %ecx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L32:
  xor    %eax,%eax
  ret
```

assistant: ```c
int strncmp(const char *s1, const char *s2, size_t n) {
  const char *scan1;
  const char *scan2;
  int count;
  scan1 = s1;
  scan2 = s2;
  count = n;
  while (--count >= 0 && *scan1 != '\0' && *scan1 == *scan2) {
    scan1++;
    scan2++;
  }
  if (count < 0)
    return (0);
  if (*scan1 == '\0' && *scan2 == '\0')
    return (0);
  else if (*scan1 == '\0')
    return (-1);
  else if (*scan2 == '\0')
    return (1);
  else
    return (*scan1 - *scan2);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<init_serial>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%esi
  mov    $0x3f9,%edi
  call   <outb@plt>
  mov    $0x80,%esi
  mov    $0x3fb,%edi
  call   <outb@plt>
  mov    $0x3,%esi
  mov    $0x3f8,%edi
  call   <outb@plt>
  mov    $0x0,%esi
  mov    $0x3f9,%edi
  call   <outb@plt>
  mov    $0x3,%esi
  mov    $0x3fb,%edi
  call   <outb@plt>
  mov    $0xc7,%esi
  mov    $0x3fa,%edi
  call   <outb@plt>
  mov    $0xb,%esi
  mov    $0x3fc,%edi
  call   <outb@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void init_serial() {
  outb(0x3f8 + 1, 0x00);
  outb(0x3f8 + 3, 0x80);
  outb(0x3f8 + 0, 0x03);
  outb(0x3f8 + 1, 0x00);
  outb(0x3f8 + 3, 0x03);
  outb(0x3f8 + 2, 0xC7);
  outb(0x3f8 + 4, 0x0B);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<init_serial>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  mov    $0x3f9,%edi
  call   <outb@plt>
  mov    $0x80,%esi
  mov    $0x3fb,%edi
  call   <outb@plt>
  mov    $0x3,%esi
  mov    $0x3f8,%edi
  call   <outb@plt>
  mov    $0x0,%esi
  mov    $0x3f9,%edi
  call   <outb@plt>
  mov    $0x3,%esi
  mov    $0x3fb,%edi
  call   <outb@plt>
  mov    $0xc7,%esi
  mov    $0x3fa,%edi
  call   <outb@plt>
  mov    $0xb,%esi
  mov    $0x3fc,%edi
  call   <outb@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void init_serial() {
  outb(0x3f8 + 1, 0x00);
  outb(0x3f8 + 3, 0x80);
  outb(0x3f8 + 0, 0x03);
  outb(0x3f8 + 1, 0x00);
  outb(0x3f8 + 3, 0x03);
  outb(0x3f8 + 2, 0xC7);
  outb(0x3f8 + 4, 0x0B);
}
```
user: Translate the below assembly code into C programming language:

```asm
<init_serial>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  mov    $0x3f9,%edi
  call   <outb@plt>
  mov    $0x80,%esi
  mov    $0x3fb,%edi
  call   <outb@plt>
  mov    $0x3,%esi
  mov    $0x3f8,%edi
  call   <outb@plt>
  xor    %esi,%esi
  mov    $0x3f9,%edi
  call   <outb@plt>
  mov    $0x3,%esi
  mov    $0x3fb,%edi
  call   <outb@plt>
  mov    $0xc7,%esi
  mov    $0x3fa,%edi
  call   <outb@plt>
  mov    $0xb,%esi
  mov    $0x3fc,%edi
  add    $0x8,%rsp
  jmp    1050 <outb@plt>
```

assistant: ```c
void init_serial() {
  outb(0x3f8 + 1, 0x00);
  outb(0x3f8 + 3, 0x80);
  outb(0x3f8 + 0, 0x03);
  outb(0x3f8 + 1, 0x00);
  outb(0x3f8 + 3, 0x03);
  outb(0x3f8 + 2, 0xC7);
  outb(0x3f8 + 4, 0x0B);
}
```
user: Translate the below assembly code into C programming language:

```asm
<maximum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L23
L28:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jge    L17
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
L17:
  addl   $0x1,-0x4(%rbp)
L23:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L28
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int maximum(int num[], int n) {
  int max = INT_MIN;
  for (int i = 0; i < n; ++i) {
    if (num[i] > max) {
      max = num[i];
    }
  }
  return max;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<maximum>:
  endbr64
  mov    D3(%rip),%rax
  mov    (%rax),%edx
  test   %esi,%esi
  jle    L14
  mov    %rdi,%rax
  lea    -0x1(%rsi),%ecx
  lea    0x4(%rdi,%rcx,4),%rsi
L18:
  mov    (%rax),%ecx
  cmp    %ecx,%edx
  cmovl  %ecx,%edx
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L18
L14:
  mov    %edx,%eax
  ret
```

assistant: ```c
int maximum(int num[], int n) {
  int max = INT_MIN;
  for (int i = 0; i < n; ++i) {
    if (num[i] > max) {
      max = num[i];
    }
  }
  return max;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<maximum>:
  endbr64
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  test   %esi,%esi
  jle    L2
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rcx
  nopl   0x0(%rax)
L4:
  mov    (%rdi),%edx
  cmp    %edx,%eax
  cmovl  %edx,%eax
  add    $0x4,%rdi
  cmp    %rcx,%rdi
  jne    L4
L2:
  ret
```

assistant: ```c
int maximum(int num[], int n) {
  int max = INT_MIN;
  for (int i = 0; i < n; ++i) {
    if (num[i] > max) {
      max = num[i];
    }
  }
  return max;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<maximum>:
  endbr64
  mov    D5(%rip),%rax
  mov    %esi,%edx
  mov    (%rax),%eax
  test   %esi,%esi
  jle    L29
  lea    -0x1(%rsi),%ecx
  cmp    $0x2,%ecx
  jbe    L66
  mov    %esi,%ecx
  movd   %eax,%xmm3
  mov    %rdi,%rax
  shr    $0x2,%ecx
  pshufd $0x0,%xmm3,%xmm0
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdi,%rcx,1),%rcx
L26:
  movdqu (%rax),%xmm1
  add    $0x10,%rax
  movdqa %xmm1,%xmm2
  pcmpgtd %xmm0,%xmm2
  pand   %xmm2,%xmm1
  pandn  %xmm0,%xmm2
  movdqa %xmm2,%xmm0
  por    %xmm1,%xmm0
  cmp    %rcx,%rax
  jne    L26
  movdqa %xmm0,%xmm2
  mov    %edx,%ecx
  psrldq $0x8,%xmm2
  and    $0xfffffffc,%ecx
  movdqa %xmm2,%xmm1
  pcmpgtd %xmm0,%xmm1
  pand   %xmm1,%xmm2
  pandn  %xmm0,%xmm1
  por    %xmm2,%xmm1
  movdqa %xmm1,%xmm2
  psrldq $0x4,%xmm2
  movdqa %xmm2,%xmm0
  pcmpgtd %xmm1,%xmm0
  pand   %xmm0,%xmm2
  pandn  %xmm1,%xmm0
  por    %xmm2,%xmm0
  movd   %xmm0,%eax
  test   $0x3,%dl
  je     L30
L10:
  movslq %ecx,%rsi
  lea    0x0(,%rsi,4),%r8
  mov    (%rdi,%rsi,4),%esi
  cmp    %esi,%eax
  cmovl  %esi,%eax
  lea    0x1(%rcx),%esi
  cmp    %esi,%edx
  jle    L29
  mov    0x4(%rdi,%r8,1),%esi
  cmp    %esi,%eax
  cmovl  %esi,%eax
  add    $0x2,%ecx
  cmp    %ecx,%edx
  jle    L29
  mov    0x8(%rdi,%r8,1),%edx
  cmp    %edx,%eax
  cmovl  %edx,%eax
L29:
  ret
  nopl   0x0(%rax,%rax,1)
L30:
  ret
L66:
  xor    %ecx,%ecx
  jmp    L10
```

assistant: ```c
int maximum(int num[], int n) {
  int max = INT_MIN;
  for (int i = 0; i < n; ++i) {
    if (num[i] > max) {
      max = num[i];
    }
  }
  return max;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fft_size>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
L53:
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L27
L47:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x14(%rbp)
L27:
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L47
  jmp    L17
L48:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
L17:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  add    %ecx,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L48
  jmp    L91
L2:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
L91:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    %eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x2,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L2
  jmp    L1
L72:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
L1:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x3,%ecx
  sub    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L72
  cmpl   $0x1,-0x14(%rbp)
  je     L69
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L53
L69:
  nop
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int fft_size(int n) {
  int m;
  while (1) {
    m = n;
    while (!(n % 2))
      n /= 2;
    while (!(n % 3))
      n /= 3;
    while (!(n % 5))
      n /= 5;
    while (!(n % 7))
      n /= 7;
    if (1 == n)
      break;
    else
      n = m + 1;
  }
  return m;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fft_size>:
  endbr64
  mov    %edi,%esi
L54:
  test   $0x1,%sil
  jne    L83
  mov    %esi,%eax
L105:
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  sar    %edx
  mov    %edx,%eax
  test   $0x1,%dl
  je     L105
L104:
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,2),%edx
  cmp    %edx,%eax
  jne    L82
L18:
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x1f,%eax
  sub    %eax,%edx
  mov    %edx,%eax
  movslq %edx,%rcx
  imul   $0x55555556,%rcx,%rcx
  shr    $0x20,%rcx
  mov    %edx,%edi
  sar    $0x1f,%edi
  sub    %edi,%ecx
  lea    (%rcx,%rcx,2),%ecx
  cmp    %ecx,%edx
  je     L18
L82:
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x21,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  cmp    %edx,%eax
  jne    L57
L29:
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x21,%rdx
  sar    $0x1f,%eax
  sub    %eax,%edx
  mov    %edx,%eax
  movslq %edx,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x21,%rcx
  mov    %edx,%edi
  sar    $0x1f,%edi
  sub    %edi,%ecx
  lea    (%rcx,%rcx,4),%ecx
  cmp    %ecx,%edx
  je     L29
L57:
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x2,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    0x0(,%rdx,8),%ecx
  sub    %edx,%ecx
  cmp    %ecx,%eax
  jne    L88
L4:
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  sub    %eax,%edx
  mov    %edx,%eax
  movslq %edx,%rcx
  imul   $0xffffffff92492493,%rcx,%rcx
  shr    $0x20,%rcx
  add    %edx,%ecx
  sar    $0x2,%ecx
  mov    %edx,%edi
  sar    $0x1f,%edi
  sub    %edi,%ecx
  lea    0x0(,%rcx,8),%edi
  sub    %ecx,%edi
  cmp    %edi,%edx
  je     L4
L88:
  cmp    $0x1,%eax
  je     L89
  add    $0x1,%esi
  jmp    L54
L83:
  mov    %esi,%eax
  jmp    L104
L89:
  mov    %esi,%eax
  ret
```

assistant: ```c
int fft_size(int n) {
  int m;
  while (1) {
    m = n;
    while (!(n % 2))
      n /= 2;
    while (!(n % 3))
      n /= 3;
    while (!(n % 5))
      n /= 5;
    while (!(n % 7))
      n /= 7;
    if (1 == n)
      break;
    else
      n = m + 1;
  }
  return m;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fft_size>:
  endbr64
  mov    %edi,%r8d
  nopw   0x0(%rax,%rax,1)
L23:
  mov    %r8d,%eax
  test   $0x1,%r8b
  jne    L90
  nopl   0x0(%rax)
L54:
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  mov    %edx,%eax
  sar    %eax
  and    $0x2,%edx
  je     L54
  jmp    L90
  nopw   0x0(%rax,%rax,1)
L85:
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x55555556,%rdx,%rdx
  mov    %eax,%ecx
  shr    $0x20,%rdx
  mov    %edx,%eax
  sub    %ecx,%eax
L90:
  imul   $0xaaaaaaab,%eax,%edx
  add    $0x2aaaaaaa,%edx
  cmp    $0x55555554,%edx
  jbe    L85
  jmp    L45
  nopl   (%rax)
L53:
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x66666667,%rdx,%rdx
  mov    %eax,%ecx
  sar    $0x21,%rdx
  mov    %edx,%eax
  sub    %ecx,%eax
L45:
  imul   $0xcccccccd,%eax,%edx
  add    $0x19999999,%edx
  cmp    $0x33333332,%edx
  jbe    L53
  jmp    L52
  nopl   (%rax)
L43:
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x1f,%eax
  sar    $0x2,%edx
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
L52:
  imul   $0xb6db6db7,%eax,%edx
  add    $0x12492492,%edx
  cmp    $0x24924924,%edx
  jbe    L43
  cmp    $0x1,%eax
  je     L47
  add    $0x1,%r8d
  jmp    L23
  cs nopw 0x0(%rax,%rax,1)
L47:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int fft_size(int n) {
  int m;
  while (1) {
    m = n;
    while (!(n % 2))
      n /= 2;
    while (!(n % 3))
      n /= 3;
    while (!(n % 5))
      n /= 5;
    while (!(n % 7))
      n /= 7;
    if (1 == n)
      break;
    else
      n = m + 1;
  }
  return m;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fft_size>:
  endbr64
  mov    %edi,%r8d
  nopw   0x0(%rax,%rax,1)
L64:
  mov    %r8d,%eax
  test   $0x1,%r8b
  jne    L37
  nopl   0x0(%rax)
L60:
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  mov    %edx,%eax
  sar    %eax
  and    $0x2,%edx
  je     L60
  jmp    L37
  nopw   0x0(%rax,%rax,1)
L57:
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x55555556,%rdx,%rdx
  mov    %eax,%ecx
  shr    $0x20,%rdx
  mov    %edx,%eax
  sub    %ecx,%eax
L37:
  imul   $0xaaaaaaab,%eax,%edx
  add    $0x2aaaaaaa,%edx
  cmp    $0x55555554,%edx
  jbe    L57
  jmp    L12
  nopl   (%rax)
L41:
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x66666667,%rdx,%rdx
  mov    %eax,%ecx
  sar    $0x21,%rdx
  mov    %edx,%eax
  sub    %ecx,%eax
L12:
  imul   $0xcccccccd,%eax,%edx
  add    $0x19999999,%edx
  cmp    $0x33333332,%edx
  jbe    L41
  jmp    L48
  nopl   (%rax)
L20:
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x1f,%eax
  sar    $0x2,%edx
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
L48:
  imul   $0xb6db6db7,%eax,%edx
  add    $0x12492492,%edx
  cmp    $0x24924924,%edx
  jbe    L20
  cmp    $0x1,%eax
  je     L42
  add    $0x1,%r8d
  jmp    L64
  cs nopw 0x0(%rax,%rax,1)
L42:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int fft_size(int n) {
  int m;
  while (1) {
    m = n;
    while (!(n % 2))
      n /= 2;
    while (!(n % 3))
      n /= 3;
    while (!(n % 5))
      n /= 5;
    while (!(n % 7))
      n /= 7;
    if (1 == n)
      break;
    else
      n = m + 1;
  }
  return m;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test_irc_message>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D0(%rip),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <irc_message@plt>
  lea    -0x210(%rbp),%rax
  mov    %rax,%rdi
  call   <read_mock@plt>
  lea    -0x210(%rbp),%rax
  lea    D38(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L16
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <mu_fail@plt>
L16:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
datatype: string
name: disassemble_data
arguments: data_tag: D29
datatype: string
name: disassemble_data
arguments: data_tag: D38
datatype: string
name: disassemble_data
arguments: data_tag: D15
datatype: string

tool: D0: Foo bar

tool: D29: #circus

tool: D38: PRIVMSG #circus :Foo bar


tool: D15: test_irc_message: msg should be 'PRIVMSG #circus :Foo bar\r\n'

assistant: ```c
void test_irc_message() {
  char msg[(512 + 1)];
  irc_message("#circus", "Foo bar");
  read_mock(msg);
  do {
    if (!((strcmp(msg, "PRIVMSG #circus :Foo bar\r\n") == 0))) {
      mu_fail("test_irc_message: msg should be 'PRIVMSG #circus :Foo bar\\r\\n'");
      return;
    }
  } while (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test_irc_message>:
  endbr64
  push   %rbx
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  lea    D27(%rip),%rsi
  lea    D30(%rip),%rdi
  call   <irc_message@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <read_mock@plt>
  lea    D23(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L21
L22:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x210,%rsp
  pop    %rbx
  ret
L21:
  lea    D11(%rip),%rdi
  call   <mu_fail@plt>
  jmp    L22
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D27
type_guessed: string
name: read_data
arguments: tag: D30
type_guessed: string
name: read_data
arguments: tag: D23
type_guessed: string
name: read_data
arguments: tag: D11
type_guessed: string

tool: D27: Foo bar

tool: D30: #circus

tool: D23: PRIVMSG #circus :Foo bar


tool: D11: test_irc_message: msg should be 'PRIVMSG #circus :Foo bar\r\n'

assistant: ```c
void test_irc_message() {
  char msg[(512 + 1)];
  irc_message("#circus", "Foo bar");
  read_mock(msg);
  do {
    if (!((strcmp(msg, "PRIVMSG #circus :Foo bar\r\n") == 0))) {
      mu_fail("test_irc_message: msg should be 'PRIVMSG #circus :Foo bar\\r\\n'");
      return;
    }
  } while (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<test_irc_message>:
  endbr64
  push   %rbp
  lea    D0(%rip),%rsi
  lea    D33(%rip),%rdi
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <irc_message@plt>
  mov    %rbp,%rdi
  call   <read_mock@plt>
  lea    D13(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L22
L25:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x210,%rsp
  pop    %rbp
  ret
  nopl   (%rax)
L22:
  lea    D38(%rip),%rdi
  call   <mu_fail@plt>
  jmp    L25
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D0
type_infered: string
name: read_data
arguments: data_label: D33
type_infered: string
name: read_data
arguments: data_label: D13
type_infered: string
name: read_data
arguments: data_label: D38
type_infered: string

tool: D0: Foo bar

tool: D33: #circus

tool: D13: PRIVMSG #circus :Foo bar


tool: D38: test_irc_message: msg should be 'PRIVMSG #circus :Foo bar\r\n'

assistant: ```c
void test_irc_message() {
  char msg[(512 + 1)];
  irc_message("#circus", "Foo bar");
  read_mock(msg);
  do {
    if (!((strcmp(msg, "PRIVMSG #circus :Foo bar\r\n") == 0))) {
      mu_fail("test_irc_message: msg should be 'PRIVMSG #circus :Foo bar\\r\\n'");
      return;
    }
  } while (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
<test_irc_message>:
  endbr64
  push   %rbp
  lea    D9(%rip),%rsi
  lea    D15(%rip),%rdi
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <irc_message@plt>
  mov    %rbp,%rdi
  call   <read_mock@plt>
  lea    D30(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L21
L9:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x210,%rsp
  pop    %rbp
  ret
  nopl   (%rax)
L21:
  lea    D1(%rip),%rdi
  call   <mu_fail@plt>
  jmp    L9
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D9
type_guessed: string
name: unpack_data
arguments: label: D15
type_guessed: string
name: unpack_data
arguments: label: D30
type_guessed: string
name: unpack_data
arguments: label: D1
type_guessed: string

tool: D9: Foo bar

tool: D15: #circus

tool: D30: PRIVMSG #circus :Foo bar


tool: D1: test_irc_message: msg should be 'PRIVMSG #circus :Foo bar\r\n'

assistant: ```c
void test_irc_message() {
  char msg[(512 + 1)];
  irc_message("#circus", "Foo bar");
  read_mock(msg);
  do {
    if (!((strcmp(msg, "PRIVMSG #circus :Foo bar\r\n") == 0))) {
      mu_fail("test_irc_message: msg should be 'PRIVMSG #circus :Foo bar\\r\\n'");
      return;
    }
  } while (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<unboxString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <hExtractTag@plt>
  cmp    $0x40000000,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  and    $0x1fffffff,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x24(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  setg   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  movl   $0x0,-0x10(%rbp)
  jmp    L18
L2:
  mov    -0x10(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  addl   $0x1,-0x10(%rbp)
L18:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L2
  mov    -0x30(%rbp),%rax
  mov    -0xc(%rbp),%edx
  mov    %edx,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void unboxString(uint32_t *hptr, char *buf, uint32_t buf_len, uint32_t *str_len) {
  uint32_t i;
  uint32_t string_len;
  uint32_t *boxed_char;
  assert(hExtractTag(hptr) == 0x40000000);
  string_len = ((*hptr) & 0x1fffffff);
  assert(string_len <= (buf_len - 1));
  for (i = 0; i < string_len; i++) {
    boxed_char = (((*((uint32_t **)((hptr) + 1 + i)))));
    buf[i] = (char)((*((boxed_char) + 1)));
  }
  *str_len = string_len;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<unboxString>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %edx,%r14d
  mov    %rcx,%r13
  call   <hExtractTag@plt>
  cmp    $0x40000000,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    (%rbx),%r12d
  and    $0x1fffffff,%r12d
  cmp    %r14d,%r12d
  setl   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  test   %r12d,%r12d
  jle    L2
  lea    0x1(%r12),%ecx
  mov    %ecx,%ecx
  mov    $0x1,%eax
L17:
  mov    (%rbx,%rax,4),%rdx
  mov    0x4(%rdx),%edx
  mov    %dl,-0x1(%rbp,%rax,1)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L17
L2:
  mov    %r12d,0x0(%r13)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void unboxString(uint32_t *hptr, char *buf, uint32_t buf_len, uint32_t *str_len) {
  uint32_t i;
  uint32_t string_len;
  uint32_t *boxed_char;
  assert(hExtractTag(hptr) == 0x40000000);
  string_len = ((*hptr) & 0x1fffffff);
  assert(string_len <= (buf_len - 1));
  for (i = 0; i < string_len; i++) {
    boxed_char = (((*((uint32_t **)((hptr) + 1 + i)))));
    buf[i] = (char)((*((boxed_char) + 1)));
  }
  *str_len = string_len;
}
```