ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<FastInverseSqrt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movss  %xmm0,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movss  -0x14(%rbp),%xmm1
  movss  D1(%rip),%xmm0
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  lea    -0x14(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  sar    %eax
  mov    %eax,%edx
  mov    $0x5f3759df,%eax
  sub    %edx,%eax
  mov    %eax,-0x10(%rbp)
  lea    -0x10(%rbp),%rax
  movss  (%rax),%xmm0
  movss  %xmm0,-0x14(%rbp)
  movss  -0x14(%rbp),%xmm0
  movaps %xmm0,%xmm1
  mulss  -0xc(%rbp),%xmm1
  movss  -0x14(%rbp),%xmm0
  movaps %xmm1,%xmm2
  mulss  %xmm0,%xmm2
  movss  D7(%rip),%xmm0
  movaps %xmm0,%xmm1
  subss  %xmm2,%xmm1
  movss  -0x14(%rbp),%xmm0
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0x14(%rbp)
  movss  -0x14(%rbp),%xmm0
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D1
type_infered: byte[4]
name: disassemble_data
arguments: data_label: D7
type_infered: byte[4]

tool: {"D1": [0x00,0x00,0x00,0x3f]}

tool: {"D7": [0x00,0x00,0xc0,0x3f]}

assistant: ```c
float FastInverseSqrt(float x) {
  float xhalf = 0.5f * x;
  int i = *(int *)&x;
  i = 0x5f3759df - (i >> 1);
  x = *(float *)&i;
  x = x * (1.5f - xhalf * x * x);
  return x;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<FastInverseSqrt>:
  endbr64
  movd   %xmm0,%edx
  sar    %edx
  mov    $0x5f3759df,%eax
  sub    %edx,%eax
  movd   %eax,%xmm2
  mulss  D18(%rip),%xmm0
  mulss  %xmm2,%xmm0
  mulss  %xmm2,%xmm0
  movss  D0(%rip),%xmm1
  subss  %xmm0,%xmm1
  mulss  %xmm2,%xmm1
  movaps %xmm1,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D18
type: byte[4]
name: unpack_data
arguments: data_tag: D0
type: byte[4]

tool: {"D18": [0x00,0x00,0x00,0x3f]}

tool: {"D0": [0x00,0x00,0xc0,0x3f]}

assistant: ```c
float FastInverseSqrt(float x) {
  float xhalf = 0.5f * x;
  int i = *(int *)&x;
  i = 0x5f3759df - (i >> 1);
  x = *(float *)&i;
  x = x * (1.5f - xhalf * x * x);
  return x;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<FastInverseSqrt>:
  endbr64
  movd   %xmm0,%edx
  mov    $0x5f3759df,%eax
  movss  D15(%rip),%xmm1
  mulss  D14(%rip),%xmm0
  sar    %edx
  sub    %edx,%eax
  movd   %eax,%xmm2
  mulss  %xmm2,%xmm0
  mulss  %xmm2,%xmm0
  subss  %xmm0,%xmm1
  mulss  %xmm2,%xmm1
  movaps %xmm1,%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D15
data_type: byte[4]
name: extract_symbolic_data
arguments: tag: D14
data_type: byte[4]

tool: {"D15": [0x00,0x00,0xc0,0x3f]}

tool: {"D14": [0x00,0x00,0x00,0x3f]}

assistant: ```c
float FastInverseSqrt(float x) {
  float xhalf = 0.5f * x;
  int i = *(int *)&x;
  i = 0x5f3759df - (i >> 1);
  x = *(float *)&i;
  x = x * (1.5f - xhalf * x * x);
  return x;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<FastInverseSqrt>:
  endbr64
  movd   %xmm0,%edx
  mov    $0x5f3759df,%eax
  movss  D8(%rip),%xmm1
  mulss  D19(%rip),%xmm0
  sar    %edx
  sub    %edx,%eax
  movd   %eax,%xmm2
  mulss  %xmm2,%xmm0
  mulss  %xmm2,%xmm0
  subss  %xmm0,%xmm1
  mulss  %xmm2,%xmm1
  movaps %xmm1,%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
type: byte[4]
name: extract_symbolic_data
arguments: tag: D19
type: byte[4]

tool: {"D8": [0x00,0x00,0xc0,0x3f]}

tool: {"D19": [0x00,0x00,0x00,0x3f]}

assistant: ```c
float FastInverseSqrt(float x) {
  float xhalf = 0.5f * x;
  int i = *(int *)&x;
  i = 0x5f3759df - (i >> 1);
  x = *(float *)&i;
  x = x * (1.5f - xhalf * x * x);
  return x;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x989680,-0x4(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jle    L2
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp)
L2:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <time_map_grow@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <time_map_grow_predicted@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <time_map_replace@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <time_map_fetch_random@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <time_map_fetch_sequential@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <time_map_fetch_empty@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <time_map_remove@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <time_map_toggle@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int iters = 10000000;
  if (argc > 1) {
    iters = atoi(argv[1]);
  }
  time_map_grow(iters);
  time_map_grow_predicted(iters);
  time_map_replace(iters);
  time_map_fetch_random(iters);
  time_map_fetch_sequential(iters);
  time_map_fetch_empty(iters);
  time_map_remove(iters);
  time_map_toggle(iters);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x989680,%ebx
  cmp    $0x1,%edi
  jg     L18
L4:
  mov    %ebx,%edi
  call   <time_map_grow@plt>
  mov    %ebx,%edi
  call   <time_map_grow_predicted@plt>
  mov    %ebx,%edi
  call   <time_map_replace@plt>
  mov    %ebx,%edi
  call   <time_map_fetch_random@plt>
  mov    %ebx,%edi
  call   <time_map_fetch_sequential@plt>
  mov    %ebx,%edi
  call   <time_map_fetch_empty@plt>
  mov    %ebx,%edi
  call   <time_map_remove@plt>
  mov    %ebx,%edi
  call   <time_map_toggle@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
L18:
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%ebx
  jmp    L4
```

assistant: ```c
int main(int argc, char **argv) {
  int iters = 10000000;
  if (argc > 1) {
    iters = atoi(argv[1]);
  }
  time_map_grow(iters);
  time_map_grow_predicted(iters);
  time_map_replace(iters);
  time_map_fetch_random(iters);
  time_map_fetch_sequential(iters);
  time_map_fetch_empty(iters);
  time_map_remove(iters);
  time_map_toggle(iters);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x989680,%ebp
  cmp    $0x1,%edi
  jle    L9
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%ebp
L9:
  mov    %ebp,%edi
  call   <time_map_grow@plt>
  mov    %ebp,%edi
  call   <time_map_grow_predicted@plt>
  mov    %ebp,%edi
  call   <time_map_replace@plt>
  mov    %ebp,%edi
  call   <time_map_fetch_random@plt>
  mov    %ebp,%edi
  call   <time_map_fetch_sequential@plt>
  mov    %ebp,%edi
  call   <time_map_fetch_empty@plt>
  mov    %ebp,%edi
  call   <time_map_remove@plt>
  mov    %ebp,%edi
  call   <time_map_toggle@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int iters = 10000000;
  if (argc > 1) {
    iters = atoi(argv[1]);
  }
  time_map_grow(iters);
  time_map_grow_predicted(iters);
  time_map_replace(iters);
  time_map_fetch_random(iters);
  time_map_fetch_sequential(iters);
  time_map_fetch_empty(iters);
  time_map_remove(iters);
  time_map_toggle(iters);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x989680,%ebp
  cmp    $0x1,%edi
  jle    L1
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%ebp
L1:
  mov    %ebp,%edi
  call   <time_map_grow@plt>
  mov    %ebp,%edi
  call   <time_map_grow_predicted@plt>
  mov    %ebp,%edi
  call   <time_map_replace@plt>
  mov    %ebp,%edi
  call   <time_map_fetch_random@plt>
  mov    %ebp,%edi
  call   <time_map_fetch_sequential@plt>
  mov    %ebp,%edi
  call   <time_map_fetch_empty@plt>
  mov    %ebp,%edi
  call   <time_map_remove@plt>
  mov    %ebp,%edi
  call   <time_map_toggle@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int iters = 10000000;
  if (argc > 1) {
    iters = atoi(argv[1]);
  }
  time_map_grow(iters);
  time_map_grow_predicted(iters);
  time_map_replace(iters);
  time_map_fetch_random(iters);
  time_map_fetch_sequential(iters);
  time_map_fetch_empty(iters);
  time_map_remove(iters);
  time_map_toggle(iters);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<xunsetenv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <unsetenv@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int xunsetenv(const char *name) {
  unsetenv(name);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xunsetenv>:
  endbr64
  sub    $0x8,%rsp
  call   <unsetenv@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int xunsetenv(const char *name) {
  unsetenv(name);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<xunsetenv>:
  endbr64
  sub    $0x8,%rsp
  call   <unsetenv@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int xunsetenv(const char *name) {
  unsetenv(name);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<xunsetenv>:
  endbr64
  sub    $0x8,%rsp
  call   <unsetenv@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int xunsetenv(const char *name) {
  unsetenv(name);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<zstrncate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  cmpq   $0x0,-0x38(%rbp)
  jne    L71
  mov    -0x28(%rbp),%rax
  jmp    L59
L71:
  subq   $0x1,-0x38(%rbp)
  cmpq   $0x0,-0x40(%rbp)
  je     L22
  mov    -0x40(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  je     L22
  mov    -0x40(%rbp),%rax
  sub    -0x28(%rbp),%rax
  add    $0x1,%rax
  cmp    %rax,-0x38(%rbp)
  jae    L46
L22:
  movl   $0x0,-0x14(%rbp)
  jmp    L60
L121:
  addq   $0x1,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
L60:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L80
  mov    -0x14(%rbp),%eax
  cltq
  cmp    %rax,-0x38(%rbp)
  ja     L121
  jmp    L80
L46:
  mov    -0x40(%rbp),%rax
  sub    -0x28(%rbp),%rax
  add    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L25
L80:
  nop
L25:
  mov    -0x14(%rbp),%eax
  cltq
  cmp    %rax,-0x38(%rbp)
  ja     L61
  mov    -0x28(%rbp),%rax
  jmp    L59
L65:
  addl   $0x1,-0x14(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L61:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L93
  mov    -0x14(%rbp),%eax
  cltq
  cmp    %rax,-0x38(%rbp)
  ja     L65
L93:
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x28(%rbp),%rax
L59:
  pop    %rbp
  ret
```

assistant: ```c
char *zstrncate(char *dst, const char *src, size_t len, char *endpoint) {
  char *dp = dst;
  const char *sp = src;
  int cc = 0;
  if (len == 0)
    return dst;
  len--;
  if ((!endpoint) || (endpoint == dst) || ((endpoint - dst + 1) > len)) {
    cc = 0;
    while ((*dp != '\0') && (cc < len)) {
      dp++;
      cc++;
    }
  } else {
    cc = endpoint - dst + 1;
    dp = endpoint;
  }
  if (cc >= len)
    return dst;
  while ((*sp) && (cc < len)) {
    cc++;
    *dp = *sp;
    dp++;
    sp++;
  }
  *dp = '\0';
  return dst;
}
```
user: Translate the below assembly code into C programming language:

```asm
<zstrncate>:
  endbr64
  mov    %rdi,%r8
  mov    %rsi,%rax
  test   %rdx,%rdx
  je     L69
  sub    $0x1,%rdx
  test   %rcx,%rcx
  je     L45
  cmp    %rcx,%rdi
  je     L45
  mov    %rcx,%rsi
  sub    %rdi,%rsi
  lea    0x1(%rsi),%rdi
  cmp    %rdx,%rdi
  ja     L45
  add    $0x1,%esi
L44:
  movslq %esi,%rdi
  cmp    %rdx,%rdi
  jae    L69
  movzbl (%rax),%esi
  test   %sil,%sil
  je     L15
  sub    %rdi,%rdx
  add    %rcx,%rdx
L111:
  mov    %sil,(%rcx)
  add    $0x1,%rcx
  add    $0x1,%rax
  movzbl (%rax),%esi
  test   %sil,%sil
  je     L15
  cmp    %rdx,%rcx
  jne    L111
  mov    %rdx,%rcx
L15:
  movb   $0x0,(%rcx)
L69:
  mov    %r8,%rax
  ret
L45:
  cmpb   $0x0,(%r8)
  je     L5
  lea    (%r8,%rdx,1),%rdi
  mov    %r8,%rcx
  mov    $0x0,%esi
  test   %rdx,%rdx
  je     L69
L21:
  add    $0x1,%rcx
  add    $0x1,%esi
  cmpb   $0x0,(%rcx)
  je     L44
  cmp    %rdi,%rcx
  jne    L21
  jmp    L69
L5:
  mov    %r8,%rcx
  mov    $0x0,%esi
  jmp    L44
```

assistant: ```c
char *zstrncate(char *dst, const char *src, size_t len, char *endpoint) {
  char *dp = dst;
  const char *sp = src;
  int cc = 0;
  if (len == 0)
    return dst;
  len--;
  if ((!endpoint) || (endpoint == dst) || ((endpoint - dst + 1) > len)) {
    cc = 0;
    while ((*dp != '\0') && (cc < len)) {
      dp++;
      cc++;
    }
  } else {
    cc = endpoint - dst + 1;
    dp = endpoint;
  }
  if (cc >= len)
    return dst;
  while ((*sp) && (cc < len)) {
    cc++;
    *dp = *sp;
    dp++;
    sp++;
  }
  *dp = '\0';
  return dst;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<zstrncate>:
  endbr64
  mov    %rdi,%r8
  mov    %rsi,%rdi
  test   %rdx,%rdx
  je     L4
  sub    $0x1,%rdx
  test   %rcx,%rcx
  je     L12
  cmp    %rcx,%r8
  je     L12
  mov    %rcx,%rax
  sub    %r8,%rax
  lea    0x1(%rax),%rsi
  cmp    %rdx,%rsi
  jbe    L51
L12:
  cmpb   $0x0,(%r8)
  je     L78
  lea    (%r8,%rdx,1),%r9
  mov    %r8,%rcx
  mov    $0x1,%esi
  test   %rdx,%rdx
  jne    L67
  jmp    L4
  nopl   (%rax)
L96:
  add    $0x1,%rsi
  cmp    %r9,%rcx
  je     L4
L67:
  add    $0x1,%rcx
  mov    %esi,%eax
  cmpb   $0x0,(%rcx)
  jne    L96
L35:
  cmp    %rdx,%rsi
  jae    L4
  movzbl (%rdi),%esi
  test   %sil,%sil
  je     L91
  cltq
  sub    %rax,%rdi
  jmp    L133
L20:
  add    $0x1,%rax
  cmp    %rax,%rdx
  jbe    L91
L133:
  mov    %sil,(%rcx)
  movzbl 0x1(%rdi,%rax,1),%esi
  add    $0x1,%rcx
  test   %sil,%sil
  jne    L20
L91:
  movb   $0x0,(%rcx)
L4:
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L51:
  add    $0x1,%eax
  movslq %eax,%rsi
  jmp    L35
L78:
  mov    %r8,%rcx
  xor    %esi,%esi
  xor    %eax,%eax
  jmp    L35
```

assistant: ```c
char *zstrncate(char *dst, const char *src, size_t len, char *endpoint) {
  char *dp = dst;
  const char *sp = src;
  int cc = 0;
  if (len == 0)
    return dst;
  len--;
  if ((!endpoint) || (endpoint == dst) || ((endpoint - dst + 1) > len)) {
    cc = 0;
    while ((*dp != '\0') && (cc < len)) {
      dp++;
      cc++;
    }
  } else {
    cc = endpoint - dst + 1;
    dp = endpoint;
  }
  if (cc >= len)
    return dst;
  while ((*sp) && (cc < len)) {
    cc++;
    *dp = *sp;
    dp++;
    sp++;
  }
  *dp = '\0';
  return dst;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<zstrncate>:
  endbr64
  mov    %rdi,%r8
  mov    %rsi,%rdi
  test   %rdx,%rdx
  je     L28
  sub    $0x1,%rdx
  test   %rcx,%rcx
  je     L52
  cmp    %rcx,%r8
  je     L52
  mov    %rcx,%rax
  sub    %r8,%rax
  lea    0x1(%rax),%rsi
  cmp    %rdx,%rsi
  jbe    L151
L52:
  cmpb   $0x0,(%r8)
  je     L143
  lea    (%r8,%rdx,1),%r9
  mov    %r8,%rcx
  mov    $0x1,%esi
  test   %rdx,%rdx
  jne    L122
  jmp    L28
  nopl   (%rax)
L27:
  add    $0x1,%rsi
  cmp    %r9,%rcx
  je     L28
L122:
  add    $0x1,%rcx
  mov    %esi,%eax
  cmpb   $0x0,(%rcx)
  jne    L27
L30:
  cmp    %rdx,%rsi
  jae    L28
  movzbl (%rdi),%esi
  test   %sil,%sil
  je     L73
  cltq
  sub    %rax,%rdi
  jmp    L148
L38:
  add    $0x1,%rax
  cmp    %rax,%rdx
  jbe    L73
L148:
  mov    %sil,(%rcx)
  movzbl 0x1(%rdi,%rax,1),%esi
  add    $0x1,%rcx
  test   %sil,%sil
  jne    L38
L73:
  movb   $0x0,(%rcx)
L28:
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L151:
  add    $0x1,%eax
  movslq %eax,%rsi
  jmp    L30
L143:
  mov    %r8,%rcx
  xor    %esi,%esi
  xor    %eax,%eax
  jmp    L30
```

assistant: ```c
char *zstrncate(char *dst, const char *src, size_t len, char *endpoint) {
  char *dp = dst;
  const char *sp = src;
  int cc = 0;
  if (len == 0)
    return dst;
  len--;
  if ((!endpoint) || (endpoint == dst) || ((endpoint - dst + 1) > len)) {
    cc = 0;
    while ((*dp != '\0') && (cc < len)) {
      dp++;
      cc++;
    }
  } else {
    cc = endpoint - dst + 1;
    dp = endpoint;
  }
  if (cc >= len)
    return dst;
  while ((*sp) && (cc < len)) {
    cc++;
    *dp = *sp;
    dp++;
    sp++;
  }
  *dp = '\0';
  return dst;
}
```
user: What is the c source code of the assembly code below?

```asm
<minalpher_P>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %esi,%eax
  mov    %al,-0x4c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x3c(%rbp)
  jmp    L82
L74:
  mov    -0x3c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x3c(%rbp),%eax
  cltq
  mov    %dl,-0x30(%rbp,%rax,1)
  addl   $0x1,-0x3c(%rbp)
L82:
  cmpl   $0x1f,-0x3c(%rbp)
  jle    L74
  movl   $0x11,-0x40(%rbp)
  jmp    L57
L47:
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <S@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <T@plt>
  cmpb   $0x0,-0x4c(%rbp)
  jne    L15
  mov    $0x11,%eax
  sub    -0x40(%rbp),%eax
  mov    %eax,-0x34(%rbp)
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <M@plt>
  mov    -0x34(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <E@plt>
  jmp    L4
L15:
  cmpb   $0x1,-0x4c(%rbp)
  jne    L4
  mov    -0x40(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x34(%rbp)
  mov    -0x34(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <E@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <M@plt>
L4:
  subl   $0x1,-0x40(%rbp)
L57:
  cmpl   $0x0,-0x40(%rbp)
  jg     L47
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <S@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <T@plt>
  movl   $0x0,-0x38(%rbp)
  jmp    L97
L18:
  mov    -0x38(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rax,%rdx
  mov    -0x38(%rbp),%eax
  cltq
  movzbl -0x30(%rbp,%rax,1),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x38(%rbp)
L97:
  cmpl   $0x1f,-0x38(%rbp)
  jle    L18
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L71
  call   <__stack_chk_fail@plt>
L71:
  leave
  ret
```

assistant: ```c
void minalpher_P(unsigned char *_msg_, unsigned char _dir) {
  int r, rn;
  unsigned char dat[32];
  do {
    int u;
    for (u = 0; u < 32; u++)
      (dat)[u] = (_msg_)[u];
  } while (0);
  r = 17;
  while (r > 0) {
    S(dat);
    T(dat);
    if (_dir == 0) {
      rn = 17 - r;
      M(dat);
      E(dat, rn);
    } else if (_dir == 1) {
      rn = r - 1;
      E(dat, rn);
      M(dat);
    }
    r--;
  }
  S(dat);
  T(dat);
  do {
    int u;
    for (u = 0; u < 32; u++)
      (_msg_)[u] = (dat)[u];
  } while (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<minalpher_P>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%rbx
  mov    %esi,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
L41:
  movzbl (%rbx,%rax,1),%edx
  mov    %dl,(%rsp,%rax,1)
  add    $0x1,%rax
  cmp    $0x20,%rax
  jne    L41
  mov    $0x0,%ebp
  mov    %rsp,%r12
  jmp    L6
L48:
  mov    %r12,%rdi
  call   <M@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <E@plt>
L14:
  add    $0x1,%ebp
  cmp    $0x11,%ebp
  je     L62
L6:
  mov    %r12,%rdi
  call   <S@plt>
  mov    %r12,%rdi
  call   <T@plt>
  test   %r13b,%r13b
  je     L48
  cmp    $0x1,%r13b
  jne    L14
  mov    $0x10,%esi
  sub    %ebp,%esi
  mov    %r12,%rdi
  call   <E@plt>
  mov    %r12,%rdi
  call   <M@plt>
  jmp    L14
L62:
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <S@plt>
  mov    %rbp,%rdi
  call   <T@plt>
  mov    $0x0,%eax
L77:
  movzbl (%rsp,%rax,1),%edx
  mov    %dl,(%rbx,%rax,1)
  add    $0x1,%rax
  cmp    $0x20,%rax
  jne    L77
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L72
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L72:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void minalpher_P(unsigned char *_msg_, unsigned char _dir) {
  int r, rn;
  unsigned char dat[32];
  do {
    int u;
    for (u = 0; u < 32; u++)
      (dat)[u] = (_msg_)[u];
  } while (0);
  r = 17;
  while (r > 0) {
    S(dat);
    T(dat);
    if (_dir == 0) {
      rn = 17 - r;
      M(dat);
      E(dat, rn);
    } else if (_dir == 1) {
      rn = r - 1;
      E(dat, rn);
      M(dat);
    }
    r--;
  }
  S(dat);
  T(dat);
  do {
    int u;
    for (u = 0; u < 32; u++)
      (_msg_)[u] = (dat)[u];
  } while (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<minalpher_P>:
  endbr64
  push   %r14
  mov    $0x10,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  mov    $0x11,%ebx
  sub    $0x30,%rsp
  movdqu (%rdi),%xmm0
  movdqu 0x10(%rdi),%xmm1
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movaps %xmm0,(%rsp)
  movaps %xmm1,0x10(%rsp)
  jmp    L57
  nopl   0x0(%rax,%rax,1)
L18:
  cmp    $0x1,%r12b
  je     L64
L33:
  test   %ebx,%ebx
  je     L36
L57:
  mov    %rbp,%rdi
  sub    $0x1,%ebx
  call   <S@plt>
  mov    %rbp,%rdi
  call   <T@plt>
  test   %r12b,%r12b
  jne    L18
  mov    %rbp,%rdi
  call   <M@plt>
  mov    %r14d,%esi
  mov    %rbp,%rdi
  sub    %ebx,%esi
  call   <E@plt>
  test   %ebx,%ebx
  jne    L57
L36:
  mov    %rbp,%rdi
  call   <S@plt>
  mov    %rbp,%rdi
  call   <T@plt>
  movdqa (%rsp),%xmm2
  movdqa 0x10(%rsp),%xmm3
  movups %xmm2,0x0(%r13)
  movups %xmm3,0x10(%r13)
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L64:
  mov    %rbp,%rdi
  mov    %ebx,%esi
  call   <E@plt>
  mov    %rbp,%rdi
  call   <M@plt>
  jmp    L33
L53:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void minalpher_P(unsigned char *_msg_, unsigned char _dir) {
  int r, rn;
  unsigned char dat[32];
  do {
    int u;
    for (u = 0; u < 32; u++)
      (dat)[u] = (_msg_)[u];
  } while (0);
  r = 17;
  while (r > 0) {
    S(dat);
    T(dat);
    if (_dir == 0) {
      rn = 17 - r;
      M(dat);
      E(dat, rn);
    } else if (_dir == 1) {
      rn = r - 1;
      E(dat, rn);
      M(dat);
    }
    r--;
  }
  S(dat);
  T(dat);
  do {
    int u;
    for (u = 0; u < 32; u++)
      (_msg_)[u] = (dat)[u];
  } while (0);
}
```