ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<brd_fpath>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x62,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x72,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x64,(%rax)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <mak_fpath@plt>
  nop
  leave
  ret
```

assistant: ```c
void brd_fpath(fpath, board, fname) char *fpath;
char *board;
char *fname;
{
  *fpath++ = 'b';
  *fpath++ = 'r';
  *fpath++ = 'd';
  mak_fpath(fpath, board, fname);
}
```
user: What is the c source code of the assembly code below:

```asm
<brd_fpath>:
  endbr64
  sub    $0x8,%rsp
  movb   $0x62,(%rdi)
  movb   $0x72,0x1(%rdi)
  movb   $0x64,0x2(%rdi)
  add    $0x3,%rdi
  mov    $0x0,%eax
  call   <mak_fpath@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void brd_fpath(fpath, board, fname) char *fpath;
char *board;
char *fname;
{
  *fpath++ = 'b';
  *fpath++ = 'r';
  *fpath++ = 'd';
  mak_fpath(fpath, board, fname);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<brd_fpath>:
  endbr64
  mov    $0x7262,%eax
  movb   $0x64,0x2(%rdi)
  add    $0x3,%rdi
  mov    %ax,-0x3(%rdi)
  xor    %eax,%eax
  jmp    1050 <mak_fpath@plt>
```

assistant: ```c
void brd_fpath(fpath, board, fname) char *fpath;
char *board;
char *fname;
{
  *fpath++ = 'b';
  *fpath++ = 'r';
  *fpath++ = 'd';
  mak_fpath(fpath, board, fname);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<brd_fpath>:
  endbr64
  mov    $0x7262,%eax
  movb   $0x64,0x2(%rdi)
  add    $0x3,%rdi
  mov    %ax,-0x3(%rdi)
  xor    %eax,%eax
  jmp    1050 <mak_fpath@plt>
```

assistant: ```c
void brd_fpath(fpath, board, fname) char *fpath;
char *board;
char *fname;
{
  *fpath++ = 'b';
  *fpath++ = 'r';
  *fpath++ = 'd';
  mak_fpath(fpath, board, fname);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<arithmatic_low_level>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <yyselect@plt>
  cmp    $0x25,%eax
  je     L17
  cmp    $0x26,%eax
  je     L7
  jmp    L21
L17:
  mov    $0x0,%eax
  call   <get_lexval@plt>
  jmp    L21
L7:
  mov    $0x0,%eax
  call   <get_lexval@plt>
  nop
L21:
  nop
  pop    %rbp
  ret
```

assistant: ```c
arithmatic_low_level() {
  switch (yyselect()) {
  case 37: {
    get_lexval();
  } break;
  case 38: {
    get_lexval();
  } break;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<arithmatic_low_level>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <yyselect@plt>
  cmp    $0x25,%eax
  je     L12
  cmp    $0x26,%eax
  je     L21
L4:
  add    $0x8,%rsp
  ret
L12:
  mov    $0x0,%eax
  call   <get_lexval@plt>
  jmp    L4
L21:
  mov    $0x0,%eax
  call   <get_lexval@plt>
  jmp    L4
```

assistant: ```c
arithmatic_low_level() {
  switch (yyselect()) {
  case 37: {
    get_lexval();
  } break;
  case 38: {
    get_lexval();
  } break;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<arithmatic_low_level>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <yyselect@plt>
  sub    $0x25,%eax
  cmp    $0x1,%eax
  ja     L8
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <get_lexval@plt>
  nopw   0x0(%rax,%rax,1)
L8:
  add    $0x8,%rsp
  ret
```

assistant: ```c
arithmatic_low_level() {
  switch (yyselect()) {
  case 37: {
    get_lexval();
  } break;
  case 38: {
    get_lexval();
  } break;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<arithmatic_low_level>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <yyselect@plt>
  sub    $0x25,%eax
  cmp    $0x1,%eax
  ja     L9
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <get_lexval@plt>
  nopw   0x0(%rax,%rax,1)
L9:
  add    $0x8,%rsp
  ret
```

assistant: ```c
arithmatic_low_level() {
  switch (yyselect()) {
  case 37: {
    get_lexval();
  } break;
  case 38: {
    get_lexval();
  } break;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L46
L2:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovs  %eax,%edx
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  movslq %eax,%rcx
  imul   $0x66666667,%rcx,%rcx
  shr    $0x20,%rcx
  sar    $0x2,%ecx
  sar    $0x1f,%eax
  mov    %eax,%esi
  mov    %ecx,%eax
  sub    %esi,%eax
  cmp    %eax,%edx
  jg     L44
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovs  %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edi
  mov    D8(%rip),%rax
  mov    (%rax),%esi
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovs  %eax,%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%r8d
  sar    $0x1f,%r8d
  sub    %r8d,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %esi,%eax
  sub    %ecx,%eax
  cmp    %eax,%edi
  jle    L36
L44:
  mov    $0x0,%eax
  jmp    L45
L36:
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edi
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  lea    (%rdi,%rcx,1),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
L46:
  cmpl   $0x0,-0x14(%rbp)
  jne    L2
  mov    -0x4(%rbp),%eax
L45:
  pop    %rbp
  ret
```

assistant: ```c
int reverse(int x) {
  int res = 0;
  while (x != 0) {
    if (abs(res) > INT_MAX / 10 || abs(res) * 10 > INT_MAX - abs(x) % 10)
      return 0;
    res = res * 10 + x % 10;
    x /= 10;
  }
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse>:
  endbr64
  mov    %edi,%edx
  test   %edi,%edi
  je     L20
  mov    D5(%rip),%rax
  mov    (%rax),%r8d
  movslq %r8d,%rdi
  imul   $0x66666667,%rdi,%rdi
  sar    $0x22,%rdi
  mov    %r8d,%eax
  sar    $0x1f,%eax
  sub    %eax,%edi
  mov    $0x0,%ecx
L47:
  mov    %ecx,%eax
  neg    %eax
  cmovs  %ecx,%eax
  cmp    %edi,%eax
  jg     L40
  lea    (%rax,%rax,4),%r9d
  add    %r9d,%r9d
  mov    %edx,%esi
  neg    %esi
  cmovs  %edx,%esi
  movslq %esi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %esi,%r10d
  sar    $0x1f,%r10d
  sub    %r10d,%eax
  lea    (%rax,%rax,4),%eax
  lea    (%r8,%rax,2),%eax
  sub    %esi,%eax
  cmp    %eax,%r9d
  jg     L63
  lea    (%rcx,%rcx,4),%esi
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rsi,2),%ecx
  mov    %eax,%edx
  test   %eax,%eax
  jne    L47
  jmp    L12
L20:
  mov    %edi,%ecx
  jmp    L12
L40:
  mov    $0x0,%ecx
L12:
  mov    %ecx,%eax
  ret
L63:
  mov    $0x0,%ecx
  jmp    L12
```

assistant: ```c
int reverse(int x) {
  int res = 0;
  while (x != 0) {
    if (abs(res) > INT_MAX / 10 || abs(res) * 10 > INT_MAX - abs(x) % 10)
      return 0;
    res = res * 10 + x % 10;
    x /= 10;
  }
  return res;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<reverse>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  je     L34
  mov    D4(%rip),%rdx
  xor    %r8d,%r8d
  mov    $0xcccccccd,%r9d
  movslq (%rdx),%rsi
  mov    %rsi,%rdi
  imul   $0x66666667,%rsi,%rsi
  mov    %edi,%edx
  sar    $0x1f,%edx
  sar    $0x22,%rsi
  sub    %edx,%esi
  jmp    L32
  nopl   0x0(%rax,%rax,1)
L8:
  lea    (%rdx,%rdx,4),%r10d
  mov    %eax,%ecx
  add    %r10d,%r10d
  neg    %ecx
  cmovs  %eax,%ecx
  mov    %ecx,%edx
  imul   %r9,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%edx
  lea    (%rdi,%rdx,2),%edx
  sub    %ecx,%edx
  cmp    %edx,%r10d
  jg     L34
  movslq %eax,%rdx
  mov    %eax,%ecx
  lea    (%r8,%r8,4),%r8d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  lea    (%rax,%r8,2),%r8d
  mov    %edx,%eax
  test   %edx,%edx
  je     L10
L32:
  mov    %r8d,%edx
  neg    %edx
  cmovs  %r8d,%edx
  cmp    %esi,%edx
  jle    L8
L34:
  xor    %r8d,%r8d
L10:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int reverse(int x) {
  int res = 0;
  while (x != 0) {
    if (abs(res) > INT_MAX / 10 || abs(res) * 10 > INT_MAX - abs(x) % 10)
      return 0;
    res = res * 10 + x % 10;
    x /= 10;
  }
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  je     L35
  mov    D6(%rip),%rdx
  xor    %r8d,%r8d
  mov    $0xcccccccd,%r9d
  movslq (%rdx),%rsi
  mov    %rsi,%rdi
  imul   $0x66666667,%rsi,%rsi
  mov    %edi,%edx
  sar    $0x1f,%edx
  sar    $0x22,%rsi
  sub    %edx,%esi
  jmp    L43
  nopl   0x0(%rax,%rax,1)
L38:
  lea    (%rdx,%rdx,4),%r10d
  mov    %eax,%ecx
  add    %r10d,%r10d
  neg    %ecx
  cmovs  %eax,%ecx
  mov    %ecx,%edx
  imul   %r9,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%edx
  lea    (%rdi,%rdx,2),%edx
  sub    %ecx,%edx
  cmp    %edx,%r10d
  jg     L35
  movslq %eax,%rdx
  mov    %eax,%ecx
  lea    (%r8,%r8,4),%r8d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  lea    (%rax,%r8,2),%r8d
  mov    %edx,%eax
  test   %edx,%edx
  je     L49
L43:
  mov    %r8d,%edx
  neg    %edx
  cmovs  %r8d,%edx
  cmp    %esi,%edx
  jle    L38
L35:
  xor    %r8d,%r8d
L49:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int reverse(int x) {
  int res = 0;
  while (x != 0) {
    if (abs(res) > INT_MAX / 10 || abs(res) * 10 > INT_MAX - abs(x) % 10)
      return 0;
    res = res * 10 + x % 10;
    x /= 10;
  }
  return res;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<xcalloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L8
  cmpq   $0x0,-0x20(%rbp)
  jne    L1
L8:
  mov    $0x0,%eax
  jmp    L25
L1:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
L25:
  leave
  ret
```

assistant: ```c
void *xcalloc(size_t count, size_t sz) {
  void *ptr = NULL;
  if (count == 0 || sz == 0)
    return (NULL);
  ptr = calloc(count, sz);
  return (ptr);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xcalloc>:
  endbr64
  test   %rdi,%rdi
  je     L8
  test   %rsi,%rsi
  je     L8
  sub    $0x8,%rsp
  call   <calloc@plt>
  add    $0x8,%rsp
  ret
L8:
  mov    $0x0,%eax
  ret
```

assistant: ```c
void *xcalloc(size_t count, size_t sz) {
  void *ptr = NULL;
  if (count == 0 || sz == 0)
    return (NULL);
  ptr = calloc(count, sz);
  return (ptr);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<xcalloc>:
  endbr64
  test   %rdi,%rdi
  je     L5
  test   %rsi,%rsi
  je     L5
  jmp    1050 <calloc@plt>
  nopl   0x0(%rax,%rax,1)
L5:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *xcalloc(size_t count, size_t sz) {
  void *ptr = NULL;
  if (count == 0 || sz == 0)
    return (NULL);
  ptr = calloc(count, sz);
  return (ptr);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<xcalloc>:
  endbr64
  test   %rdi,%rdi
  je     L15
  test   %rsi,%rsi
  je     L15
  jmp    1050 <calloc@plt>
  nopl   0x0(%rax,%rax,1)
L15:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *xcalloc(size_t count, size_t sz) {
  void *ptr = NULL;
  if (count == 0 || sz == 0)
    return (NULL);
  ptr = calloc(count, sz);
  return (ptr);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<mutate_part>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %ax,-0x20(%rbp)
  movzbl -0x1c(%rbp),%eax
  and    $0x3,%eax
  cmp    $0x2,%eax
  je     L85
  cmp    $0x2,%eax
  jg     L56
  test   %eax,%eax
  je     L14
  cmp    $0x1,%eax
  je     L42
  jmp    L56
L14:
  movw   $0x80,-0xc(%rbp)
  movw   $0x0,-0xa(%rbp)
  jmp    L113
L42:
  movw   $0x80,-0xc(%rbp)
  movzwl -0x20(%rbp),%eax
  mov    %ax,-0xa(%rbp)
  jmp    L113
L85:
  movzwl -0x20(%rbp),%eax
  mov    %ax,-0xc(%rbp)
  movw   $0x0,-0xa(%rbp)
  nop
L113:
  movw   $0x0,-0x10(%rbp)
  jmp    L87
L52:
  movzwl -0xc(%rbp),%eax
  mov    %ax,-0xe(%rbp)
  jmp    L74
L32:
  mov    $0x0,%eax
  call   <Random@plt>
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movsd  D15(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  D13(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  D6(%rip),%xmm0
  comisd -0x8(%rbp),%xmm0
  jbe    L5
  movswq -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %eax,%edx
  movzwl -0xe(%rbp),%eax
  mov    %edx,%ecx
  xor    %eax,%ecx
  movswq -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L5:
  sarw   -0xe(%rbp)
L74:
  movzwl -0xe(%rbp),%eax
  cmp    -0xa(%rbp),%ax
  jne    L32
  movzwl -0x10(%rbp),%eax
  add    $0x1,%eax
  mov    %ax,-0x10(%rbp)
L87:
  cmpw   $0xf,-0x10(%rbp)
  jle    L52
  jmp    L23
L56:
  nop
L23:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D15
type_infered: f64
name: read_data
arguments: data_tag: D13
type_infered: f64
name: read_data
arguments: data_tag: D6
type_infered: f64

tool: D15: 32767

tool: D13: 65534

tool: D6: 0.05

assistant: ```c
void mutate_part(unsigned char *p, unsigned char protect, short boundary) {
  short i, j, jf, jt;
  double r;
  switch (protect & 3) {
  case 0:
    jf = 0x80;
    jt = 0;
    break;
  case 1:
    jf = 0x80;
    jt = boundary;
    break;
  case 2:
    jf = boundary;
    jt = 0;
    break;
  default:
    return;
  }
  for (i = 0; i < 16; i++) {
    for (j = jf; j != jt; j >>= 1) {
      r = (Random() + 32767.) / 65534.;
      if (r < 0.05)
        p[i] ^= j;
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<mutate_part>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  and    $0x3,%esi
  cmp    $0x1,%sil
  je     L8
  mov    %edx,%r13d
  cmp    $0x2,%sil
  je     L34
  mov    $0x0,%r12d
  mov    $0x80,%r13d
  test   %sil,%sil
  je     L71
L76:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L8:
  mov    %edx,%r12d
  mov    $0x80,%r13d
  jmp    L71
L34:
  mov    $0x0,%r12d
L71:
  mov    %rdi,%rbp
  lea    0x10(%rdi),%r14
  jmp    L80
L53:
  sar    %bx
  cmp    %bx,%r12w
  je     L103
L97:
  mov    $0x0,%eax
  call   <Random@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  addsd  D15(%rip),%xmm0
  divsd  D26(%rip),%xmm0
  movsd  D11(%rip),%xmm1
  comisd %xmm0,%xmm1
  jbe    L53
  xor    %bl,0x0(%rbp)
  jmp    L53
L103:
  add    $0x1,%rbp
  cmp    %r14,%rbp
  je     L76
L80:
  mov    %r13d,%ebx
  cmp    %r12w,%r13w
  jne    L97
  jmp    L103
```

tool_calls: name: read_data
arguments: label: D15
type_guessed: f64
name: read_data
arguments: label: D26
type_guessed: f64
name: read_data
arguments: label: D11
type_guessed: f64

tool: D15: 32767

tool: D26: 65534

tool: D11: 0.05

assistant: ```c
void mutate_part(unsigned char *p, unsigned char protect, short boundary) {
  short i, j, jf, jt;
  double r;
  switch (protect & 3) {
  case 0:
    jf = 0x80;
    jt = 0;
    break;
  case 1:
    jf = 0x80;
    jt = boundary;
    break;
  case 2:
    jf = boundary;
    jt = 0;
    break;
  default:
    return;
  }
  for (i = 0; i < 16; i++) {
    for (j = jf; j != jt; j >>= 1) {
      r = (Random() + 32767.) / 65534.;
      if (r < 0.05)
        p[i] ^= j;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<mutate_part>:
  endbr64
  and    $0x3,%esi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x1,%sil
  je     L72
  cmp    $0x2,%sil
  je     L51
  xor    %ebp,%ebp
  mov    $0x80,%r12d
  test   %sil,%sil
  je     L7
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L51:
  mov    %edx,%r12d
  xor    %ebp,%ebp
L7:
  mov    %rdi,%rbx
  lea    0x10(%rdi),%r13
  nopl   0x0(%rax)
L58:
  mov    %r12d,%r14d
  cmp    %bp,%r12w
  je     L60
  nopl   0x0(%rax)
L39:
  xor    %eax,%eax
  call   <Random@plt>
  movsd  D21(%rip),%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  addsd  D13(%rip),%xmm0
  divsd  D10(%rip),%xmm0
  comisd %xmm0,%xmm1
  jbe    L30
  xor    %r14b,(%rbx)
L30:
  sar    %r14w
  cmp    %r14w,%bp
  jne    L39
L60:
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L58
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L72:
  mov    %edx,%ebp
  mov    $0x80,%r12d
  jmp    L7
```

tool_calls: name: read_data
arguments: tag: D21
data_type: f64
name: read_data
arguments: tag: D13
data_type: f64
name: read_data
arguments: tag: D10
data_type: f64

tool: D21: 0.05

tool: D13: 32767

tool: D10: 65534

assistant: ```c
void mutate_part(unsigned char *p, unsigned char protect, short boundary) {
  short i, j, jf, jt;
  double r;
  switch (protect & 3) {
  case 0:
    jf = 0x80;
    jt = 0;
    break;
  case 1:
    jf = 0x80;
    jt = boundary;
    break;
  case 2:
    jf = boundary;
    jt = 0;
    break;
  default:
    return;
  }
  for (i = 0; i < 16; i++) {
    for (j = jf; j != jt; j >>= 1) {
      r = (Random() + 32767.) / 65534.;
      if (r < 0.05)
        p[i] ^= j;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mutate_part>:
  endbr64
  and    $0x3,%esi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x1,%sil
  je     L86
  cmp    $0x2,%sil
  je     L24
  test   %sil,%sil
  je     L20
L31:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L24:
  mov    %edx,%r12d
  xor    %ebp,%ebp
L80:
  cmp    %bp,%r12w
  je     L31
L7:
  mov    %rdi,%rbx
  lea    0x10(%rdi),%r13
  nop
L59:
  mov    %r12d,%r14d
  nopl   0x0(%rax,%rax,1)
L57:
  xor    %eax,%eax
  call   <Random@plt>
  movsd  D29(%rip),%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  addsd  D1(%rip),%xmm0
  divsd  D7(%rip),%xmm0
  comisd %xmm0,%xmm1
  jbe    L78
  xor    %r14b,(%rbx)
L78:
  sar    %r14w
  cmp    %r14w,%bp
  jne    L57
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L59
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L20:
  xor    %ebp,%ebp
  mov    $0x80,%r12d
  jmp    L7
L86:
  mov    %edx,%ebp
  mov    $0x80,%r12d
  jmp    L80
```

tool_calls: name: unpack_data
arguments: data_tag: D29
type_guessed: f64
name: unpack_data
arguments: data_tag: D1
type_guessed: f64
name: unpack_data
arguments: data_tag: D7
type_guessed: f64

tool: D29: 0.05

tool: D1: 32767

tool: D7: 65534

assistant: ```c
void mutate_part(unsigned char *p, unsigned char protect, short boundary) {
  short i, j, jf, jt;
  double r;
  switch (protect & 3) {
  case 0:
    jf = 0x80;
    jt = 0;
    break;
  case 1:
    jf = 0x80;
    jt = boundary;
    break;
  case 2:
    jf = boundary;
    jt = 0;
    break;
  default:
    return;
  }
  for (i = 0; i < 16; i++) {
    for (j = jf; j != jt; j >>= 1) {
      r = (Random() + 32767.) / 65534.;
      if (r < 0.05)
        p[i] ^= j;
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<waitfor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    $0x0,%eax
  call   <get_ticks@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L7
L38:
  mov    $0x1f7,%edi
  call   <in_byte@plt>
  and    -0x14(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jne    L7
  mov    $0x1,%eax
  jmp    L22
L7:
  mov    $0x0,%eax
  call   <get_ticks@plt>
  sub    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  cmp    %eax,-0x1c(%rbp)
  jg     L38
  mov    $0x0,%eax
L22:
  leave
  ret
```

assistant: ```c
int waitfor(int mask, int val, int timeout) {
  int t = get_ticks();
  while (((get_ticks() - t) * 1000 / 100) < timeout) {
    if ((in_byte(0x1f7) & mask) == val) {
      return 1;
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<waitfor>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r13d
  mov    %esi,%r12d
  mov    %edx,%ebp
  mov    $0x0,%eax
  call   <get_ticks@plt>
  mov    %eax,%ebx
L29:
  mov    $0x0,%eax
  call   <get_ticks@plt>
  sub    %ebx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  cmp    %ebp,%eax
  jge    L3
  mov    $0x1f7,%edi
  call   <in_byte@plt>
  and    %r13d,%eax
  cmp    %r12d,%eax
  jne    L29
  mov    $0x1,%eax
  jmp    L0
L3:
  mov    $0x0,%eax
L0:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int waitfor(int mask, int val, int timeout) {
  int t = get_ticks();
  while (((get_ticks() - t) * 1000 / 100) < timeout) {
    if ((in_byte(0x1f7) & mask) == val) {
      return 1;
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<waitfor>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %edi,%r13d
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <get_ticks@plt>
  mov    %eax,%ebx
  jmp    L19
  nopl   0x0(%rax)
L29:
  mov    $0x1f7,%edi
  call   <in_byte@plt>
  and    %r13d,%eax
  cmp    %r12d,%eax
  je     L10
L19:
  xor    %eax,%eax
  call   <get_ticks@plt>
  sub    %ebx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  cmp    %ebp,%eax
  jl     L29
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L10:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int waitfor(int mask, int val, int timeout) {
  int t = get_ticks();
  while (((get_ticks() - t) * 1000 / 100) < timeout) {
    if ((in_byte(0x1f7) & mask) == val) {
      return 1;
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<waitfor>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %edi,%r13d
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <get_ticks@plt>
  mov    %eax,%ebx
  jmp    L23
  nopl   0x0(%rax)
L22:
  mov    $0x1f7,%edi
  call   <in_byte@plt>
  and    %r13d,%eax
  cmp    %r12d,%eax
  je     L17
L23:
  xor    %eax,%eax
  call   <get_ticks@plt>
  sub    %ebx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  cmp    %ebp,%eax
  jl     L22
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L17:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int waitfor(int mask, int val, int timeout) {
  int t = get_ticks();
  while (((get_ticks() - t) * 1000 / 100) < timeout) {
    if ((in_byte(0x1f7) & mask) == val) {
      return 1;
    }
  }
  return 0;
}
```