ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<forward_drf>:
  endbr64
  test   %ecx,%ecx
  jle    L15
  mov    (%rsi),%eax
  add    $0x1,%eax
  mov    %eax,(%rdi)
  cmp    $0x1,%ecx
  je     L99
  mov    %rdx,%rax
  lea    0x8(%rsi),%r9
  sub    %rdi,%rax
  cmp    $0xc,%rax
  seta   %al
  cmp    $0x2,%ecx
  setne  %r8b
  and    %r8d,%eax
  lea    0x4(%rdi),%r8
  sub    %r9,%r8
  cmp    $0x8,%r8
  seta   %r8b
  test   %r8b,%al
  je     L17
  lea    0x4(%rdx),%rax
  sub    %r9,%rax
  cmp    $0x8,%rax
  jbe    L17
  lea    -0x2(%rcx),%eax
  lea    -0x1(%rcx),%r9d
  cmp    $0x2,%eax
  jbe    L16
  mov    %r9d,%eax
  movdqa D2(%rip),%xmm2
  movdqa D25(%rip),%xmm1
  shr    $0x2,%eax
  lea    -0x1(%rax),%r8d
  mov    $0x4,%eax
  shl    $0x4,%r8
  add    $0x14,%r8
  nopw   0x0(%rax,%rax,1)
L19:
  movdqu (%rsi,%rax,1),%xmm0
  paddd  %xmm2,%xmm0
  movups %xmm0,(%rdi,%rax,1)
  movdqu -0x4(%rdi,%rax,1),%xmm0
  paddd  %xmm1,%xmm0
  movups %xmm0,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %r8,%rax
  jne    L19
  mov    %r9d,%r10d
  and    $0xfffffffc,%r10d
  lea    0x1(%r10),%r8d
  mov    %r8d,%eax
  cmp    %r10d,%r9d
  je     L86
  sub    %r10d,%ecx
  lea    -0x1(%rcx),%r9d
  cmp    $0x2,%ecx
  je     L51
L59:
  movq   D2(%rip),%xmm1
  mov    %r8d,%ecx
  movq   (%rsi,%rcx,4),%xmm0
  paddd  %xmm1,%xmm0
  movq   D25(%rip),%xmm1
  movq   %xmm0,(%rdi,%rcx,4)
  movq   -0x4(%rdi,%rcx,4),%xmm0
  paddd  %xmm1,%xmm0
  movq   %xmm0,(%rdx,%rcx,4)
  mov    %r9d,%ecx
  and    $0xfffffffe,%ecx
  add    %ecx,%eax
  cmp    %ecx,%r9d
  je     L86
L51:
  cltq
  mov    (%rsi,%rax,4),%ecx
  add    $0x1,%ecx
  mov    %ecx,(%rdi,%rax,4)
  mov    -0x4(%rdi,%rax,4),%ecx
  add    $0x2,%ecx
  mov    %ecx,(%rdx,%rax,4)
  ret
  cs nopw 0x0(%rax,%rax,1)
L86:
  ret
  nopl   0x0(%rax)
L17:
  movslq %ecx,%rcx
  mov    $0x1,%eax
L44:
  mov    (%rsi,%rax,4),%r11d
  mov    -0x4(%rdi,%rax,4),%r10d
  lea    0x1(%r11),%r8d
  mov    %r8d,(%rdi,%rax,4)
  lea    0x2(%r10),%r8d
  mov    %r8d,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L44
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  ret
  nopl   0x0(%rax)
L99:
  ret
L16:
  mov    $0x1,%eax
  mov    $0x1,%r8d
  jmp    L59
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
type: byte[16]
name: extract_symbolic_data
arguments: data_label: D25
type: byte[16]

tool: {"D2": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: {"D25": [0x02,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x02,0x00,0x00,0x00]}

assistant: ```c
void forward_drf(int a[], int b[], int c[], int N) {
  for (int i = 0; i < N; i++) {
    a[i] = b[i] + 1;
  S2:
    if (i > 0) {
      c[i] = a[i - 1] + 2;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<POWI>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jns    L10
  mov    $0x0,%eax
  call   <POWNEG@plt>
  jmp    L22
L10:
  mov    -0x20(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L3
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L53
L3:
  movl   $0x1,-0x4(%rbp)
L53:
  sarl   -0x8(%rbp)
  jmp    L28
L31:
  mov    -0xc(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L23
  mov    -0x4(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L23:
  sarl   -0x8(%rbp)
L28:
  cmpl   $0x0,-0x8(%rbp)
  jne    L31
  mov    -0x4(%rbp),%eax
L22:
  leave
  ret
```

assistant: ```c
POWI(statlink, pow, num) int *statlink;
int pow, num;
{
  if (pow < 0)
    POWNEG();
  else {
    int n, p, r;
    n = num;
    p = pow;
    if ((p & 1) != 0) {
      r = n;
    } else {
      r = 1;
    }
    p >>= 1;
    while (p != 0) {
      n *= n;
      if ((p & 1) != 0)
        r *= n;
      p >>= 1;
    }
    return (r);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<POWI>:
  endbr64
  test   %esi,%esi
  js     L27
  test   $0x1,%sil
  mov    $0x1,%eax
  cmovne %edx,%eax
  sar    %esi
  jne    L42
  ret
L27:
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <POWNEG@plt>
  add    $0x8,%rsp
  ret
L30:
  sar    %esi
  je     L43
L42:
  imul   %edx,%edx
  test   $0x1,%sil
  je     L30
  imul   %edx,%eax
  jmp    L30
L43:
  ret
```

assistant: ```c
POWI(statlink, pow, num) int *statlink;
int pow, num;
{
  if (pow < 0)
    POWNEG();
  else {
    int n, p, r;
    n = num;
    p = pow;
    if ((p & 1) != 0) {
      r = n;
    } else {
      r = 1;
    }
    p >>= 1;
    while (p != 0) {
      n *= n;
      if ((p & 1) != 0)
        r *= n;
      p >>= 1;
    }
    return (r);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<POWI>:
  endbr64
  test   %esi,%esi
  js     L13
  test   $0x1,%sil
  mov    $0x1,%eax
  cmovne %edx,%eax
  nopl   0x0(%rax)
L1:
  sar    %esi
  je     L7
L2:
  imul   %edx,%edx
  test   $0x1,%sil
  je     L1
  imul   %edx,%eax
  sar    %esi
  jne    L2
L7:
  ret
  nopl   (%rax)
L13:
  xor    %eax,%eax
  jmp    1050 <POWNEG@plt>
```

assistant: ```c
POWI(statlink, pow, num) int *statlink;
int pow, num;
{
  if (pow < 0)
    POWNEG();
  else {
    int n, p, r;
    n = num;
    p = pow;
    if ((p & 1) != 0) {
      r = n;
    } else {
      r = 1;
    }
    p >>= 1;
    while (p != 0) {
      n *= n;
      if ((p & 1) != 0)
        r *= n;
      p >>= 1;
    }
    return (r);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<POWI>:
  endbr64
  test   %esi,%esi
  js     L8
  test   $0x1,%sil
  mov    $0x1,%eax
  cmovne %edx,%eax
  nopl   0x0(%rax)
L38:
  sar    %esi
  je     L26
L17:
  imul   %edx,%edx
  test   $0x1,%sil
  je     L38
  imul   %edx,%eax
  sar    %esi
  jne    L17
L26:
  ret
  nopl   (%rax)
L8:
  xor    %eax,%eax
  jmp    1050 <POWNEG@plt>
```

assistant: ```c
POWI(statlink, pow, num) int *statlink;
int pow, num;
{
  if (pow < 0)
    POWNEG();
  else {
    int n, p, r;
    n = num;
    p = pow;
    if ((p & 1) != 0) {
      r = n;
    } else {
      r = 1;
    }
    p >>= 1;
    while (p != 0) {
      n *= n;
      if ((p & 1) != 0)
        r *= n;
      p >>= 1;
    }
    return (r);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <foo@plt>
  mov    $0x0,%eax
  call   <bar@plt>
  mov    $0x0,%eax
  call   <baz@plt>
  movl   $0x0,-0x34(%rbp)
  jmp    L26
L9:
  mov    -0x34(%rbp),%eax
  cltq
  movl   $0x0,-0x30(%rbp,%rax,4)
  addl   $0x1,-0x34(%rbp)
L26:
  cmpl   $0x9,-0x34(%rbp)
  jle    L9
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <qux@plt>
  mov    $0x0,%eax
  call   <quux@plt>
  mov    $0x0,%eax
  call   <corge@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
int main() {
  int a[10] = {};
  foo(a);
  bar();
  baz();
  for (int i = 0; i < 10; i++)
    a[i] = 0;
  qux(a);
  quux();
  corge();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <foo@plt>
  mov    $0x0,%eax
  call   <bar@plt>
  mov    $0x0,%eax
  call   <baz@plt>
  mov    %rbx,%rax
  lea    0x28(%rsp),%rdx
L4:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L4
  mov    %rsp,%rdi
  call   <qux@plt>
  mov    $0x0,%eax
  call   <quux@plt>
  mov    $0x0,%eax
  call   <corge@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int a[10] = {};
  foo(a);
  bar();
  baz();
  for (int i = 0; i < 10; i++)
    a[i] = 0;
  qux(a);
  quux();
  corge();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  pxor   %xmm0,%xmm0
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movaps %xmm0,(%rsp)
  mov    %rbp,%rdi
  movaps %xmm0,0x10(%rsp)
  movq   $0x0,0x20(%rsp)
  call   <foo@plt>
  xor    %eax,%eax
  call   <bar@plt>
  xor    %eax,%eax
  call   <baz@plt>
  pxor   %xmm0,%xmm0
  mov    %rbp,%rdi
  movq   $0x0,0x20(%rsp)
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <qux@plt>
  xor    %eax,%eax
  call   <quux@plt>
  xor    %eax,%eax
  call   <corge@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int a[10] = {};
  foo(a);
  bar();
  baz();
  for (int i = 0; i < 10; i++)
    a[i] = 0;
  qux(a);
  quux();
  corge();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  pxor   %xmm0,%xmm0
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movaps %xmm0,(%rsp)
  mov    %rbp,%rdi
  movaps %xmm0,0x10(%rsp)
  movq   $0x0,0x20(%rsp)
  call   <foo@plt>
  xor    %eax,%eax
  call   <bar@plt>
  xor    %eax,%eax
  call   <baz@plt>
  pxor   %xmm0,%xmm0
  mov    %rbp,%rdi
  movq   $0x0,0x20(%rsp)
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <qux@plt>
  xor    %eax,%eax
  call   <quux@plt>
  xor    %eax,%eax
  call   <corge@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int a[10] = {};
  foo(a);
  bar();
  baz();
  for (int i = 0; i < 10; i++)
    a[i] = 0;
  qux(a);
  quux();
  corge();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<qsort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    %rcx,-0x38(%rbp)
  mov    -0x2c(%rbp),%eax
  imul   -0x30(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x2c(%rbp),%eax
  shr    %eax
  mov    %eax,-0x14(%rbp)
  jmp    L32
L14:
  mov    -0x14(%rbp),%eax
  imul   -0x30(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x30(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L70
L5:
  mov    -0x1c(%rbp),%eax
  sub    -0x8(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  jmp    L60
L2:
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x38(%rbp),%rdx
  mov    %ecx,%esi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   *%rdx
  test   %eax,%eax
  jle    L4
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %eax,%edx
  mov    -0x18(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x28(%rbp),%rax
  add    %rax,%rcx
  mov    -0x30(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <_swp@plt>
  mov    -0x18(%rbp),%eax
  sub    -0xc(%rbp),%eax
  mov    %eax,-0x18(%rbp)
L60:
  cmpl   $0x0,-0x18(%rbp)
  jns    L2
  jmp    L78
L4:
  nop
L78:
  mov    -0x1c(%rbp),%edx
  mov    -0x30(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x1c(%rbp)
L70:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jae    L5
  mov    -0x14(%rbp),%eax
  shr    %eax
  mov    %eax,-0x14(%rbp)
L32:
  cmpl   $0x0,-0x14(%rbp)
  jne    L14
  nop
  leave
  ret
```

assistant: ```c
qsort(base, nel, width, compar) char *base;
int (*compar)();
unsigned width, nel;
{
  int i, j;
  unsigned gap, ngap, t1;
  int jd, t2;
  t1 = nel * width;
  for (ngap = nel / 2; ngap > 0; ngap /= 2) {
    gap = ngap * width;
    t2 = gap + width;
    jd = base + gap;
    for (i = t2; i <= t1; i += width)
      for (j = i - t2; j >= 0; j -= gap) {
        if ((*compar)(base + j, jd + j) <= 0)
          break;
        _swp(width, base + j, jd + j);
      }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<qsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,(%rsp)
  mov    %edx,%r13d
  mov    %rcx,%r15
  mov    %esi,%ecx
  imul   %edx,%ecx
  mov    %ecx,0x14(%rsp)
  mov    %esi,%ecx
  shr    %ecx
  mov    %ecx,0x1c(%rsp)
  cmp    $0x1,%esi
  ja     L43
L3:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L25:
  add    %r13d,0x10(%rsp)
  mov    0x10(%rsp),%eax
  mov    %eax,%ebx
  cmp    0x14(%rsp),%eax
  ja     L66
L63:
  sub    0x18(%rsp),%ebx
  js     L25
L68:
  lea    (%r14,%rbx,1),%r12d
  movslq %ebx,%rbp
  add    (%rsp),%rbp
  mov    %r12d,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   *%r15
  test   %eax,%eax
  jle    L25
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  mov    $0x0,%eax
  call   <_swp@plt>
  sub    0xc(%rsp),%ebx
  jns    L68
  jmp    L25
L66:
  mov    0x1c(%rsp),%ecx
  mov    %ecx,%eax
  shr    %eax
  cmp    $0x1,%ecx
  jbe    L3
  mov    %eax,0x1c(%rsp)
L43:
  mov    0x1c(%rsp),%eax
  imul   %r13d,%eax
  mov    %eax,0xc(%rsp)
  lea    0x0(%r13,%rax,1),%ecx
  mov    %ecx,0x10(%rsp)
  mov    %ecx,0x18(%rsp)
  mov    (%rsp),%rdx
  lea    (%rax,%rdx,1),%r14d
  cmp    0x14(%rsp),%ecx
  ja     L66
  mov    %ecx,%ebx
  jmp    L63
```

assistant: ```c
qsort(base, nel, width, compar) char *base;
int (*compar)();
unsigned width, nel;
{
  int i, j;
  unsigned gap, ngap, t1;
  int jd, t2;
  t1 = nel * width;
  for (ngap = nel / 2; ngap > 0; ngap /= 2) {
    gap = ngap * width;
    t2 = gap + width;
    jd = base + gap;
    for (i = t2; i <= t1; i += width)
      for (j = i - t2; j >= 0; j -= gap) {
        if ((*compar)(base + j, jd + j) <= 0)
          break;
        _swp(width, base + j, jd + j);
      }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<qsort>:
  endbr64
  push   %r15
  mov    %esi,%eax
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  mov    %esi,%edx
  push   %rbp
  imul   %r12d,%edx
  push   %rbx
  sub    $0x28,%rsp
  shr    %eax
  mov    %rcx,(%rsp)
  mov    %edx,0x14(%rsp)
  mov    %eax,0x1c(%rsp)
  je     L34
  mov    %rdi,%r13
  nopl   0x0(%rax)
L58:
  mov    0x1c(%rsp),%eax
  imul   %r12d,%eax
  lea    (%r12,%rax,1),%edx
  mov    %eax,0xc(%rsp)
  lea    (%rax,%r13,1),%r14d
  mov    %edx,0x10(%rsp)
  mov    %edx,0x18(%rsp)
  cmp    0x14(%rsp),%edx
  ja     L46
  nopw   0x0(%rax,%rax,1)
L22:
  mov    0x10(%rsp),%ebx
  sub    0x18(%rsp),%ebx
  jns    L65
  jmp    L28
  nopl   0x0(%rax)
L17:
  xor    %eax,%eax
  mov    %r15d,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <_swp@plt>
  sub    0xc(%rsp),%ebx
  js     L28
L65:
  movslq %ebx,%r8
  lea    (%r14,%rbx,1),%r15d
  xor    %eax,%eax
  mov    (%rsp),%rcx
  lea    0x0(%r13,%r8,1),%rbp
  mov    %r15d,%esi
  mov    %rbp,%rdi
  call   *%rcx
  test   %eax,%eax
  jg     L17
L28:
  add    %r12d,0x10(%rsp)
  mov    0x10(%rsp),%eax
  cmp    0x14(%rsp),%eax
  jbe    L22
L46:
  shrl   0x1c(%rsp)
  jne    L58
L34:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
qsort(base, nel, width, compar) char *base;
int (*compar)();
unsigned width, nel;
{
  int i, j;
  unsigned gap, ngap, t1;
  int jd, t2;
  t1 = nel * width;
  for (ngap = nel / 2; ngap > 0; ngap /= 2) {
    gap = ngap * width;
    t2 = gap + width;
    jd = base + gap;
    for (i = t2; i <= t1; i += width)
      for (j = i - t2; j >= 0; j -= gap) {
        if ((*compar)(base + j, jd + j) <= 0)
          break;
        _swp(width, base + j, jd + j);
      }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<qsort>:
  endbr64
  push   %r15
  mov    %esi,%eax
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  mov    %esi,%edx
  push   %rbp
  imul   %r12d,%edx
  push   %rbx
  sub    $0x28,%rsp
  shr    %eax
  mov    %rcx,(%rsp)
  mov    %edx,0x14(%rsp)
  mov    %eax,0x1c(%rsp)
  je     L27
  mov    %rdi,%r13
  nopl   0x0(%rax)
L3:
  mov    0x1c(%rsp),%eax
  imul   %r12d,%eax
  lea    (%r12,%rax,1),%edx
  mov    %eax,0xc(%rsp)
  lea    (%rax,%r13,1),%r14d
  mov    %edx,0x10(%rsp)
  mov    %edx,0x18(%rsp)
  cmp    0x14(%rsp),%edx
  ja     L47
  nopw   0x0(%rax,%rax,1)
L70:
  mov    0x10(%rsp),%ebx
  sub    0x18(%rsp),%ebx
  jns    L43
  jmp    L19
  nopl   0x0(%rax)
L46:
  xor    %eax,%eax
  mov    %r15d,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <_swp@plt>
  sub    0xc(%rsp),%ebx
  js     L19
L43:
  movslq %ebx,%r8
  lea    (%r14,%rbx,1),%r15d
  xor    %eax,%eax
  mov    (%rsp),%rcx
  lea    0x0(%r13,%r8,1),%rbp
  mov    %r15d,%esi
  mov    %rbp,%rdi
  call   *%rcx
  test   %eax,%eax
  jg     L46
L19:
  add    %r12d,0x10(%rsp)
  mov    0x10(%rsp),%eax
  cmp    0x14(%rsp),%eax
  jbe    L70
L47:
  shrl   0x1c(%rsp)
  jne    L3
L27:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
qsort(base, nel, width, compar) char *base;
int (*compar)();
unsigned width, nel;
{
  int i, j;
  unsigned gap, ngap, t1;
  int jd, t2;
  t1 = nel * width;
  for (ngap = nel / 2; ngap > 0; ngap /= 2) {
    gap = ngap * width;
    t2 = gap + width;
    jd = base + gap;
    for (i = t2; i <= t1; i += width)
      for (j = i - t2; j >= 0; j -= gap) {
        if ((*compar)(base + j, jd + j) <= 0)
          break;
        _swp(width, base + j, jd + j);
      }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<power>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  jne    L45
  mov    $0x1,%eax
  jmp    L44
L45:
  cmpl   $0x1,-0x18(%rbp)
  jne    L34
  mov    -0x14(%rbp),%eax
  jmp    L44
L34:
  mov    -0x18(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <power>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  jne    L33
  mov    -0x4(%rbp),%eax
  imul   -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L33:
  mov    -0x4(%rbp),%eax
L44:
  leave
  ret
```

assistant: ```c
int power(int num, int exp) {
  if (exp == 0)
    return (1);
  if (exp == 1)
    return (num);
  int temp = power(num, exp / 2);
  temp *= temp;
  if (exp % 2 == 1)
    temp *= num;
  return (temp);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<power>:
  endbr64
  mov    $0x1,%eax
  test   %esi,%esi
  je     L26
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    %edi,%eax
  cmp    $0x1,%esi
  je     L16
  shr    $0x1f,%esi
  add    %ebx,%esi
  sar    %esi
  call   <power>
  imul   %eax,%eax
  mov    %ebx,%edx
  shr    $0x1f,%edx
  add    %edx,%ebx
  and    $0x1,%ebx
  sub    %edx,%ebx
  cmp    $0x1,%ebx
  je     L34
L16:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L34:
  imul   %ebp,%eax
  jmp    L16
L26:
  ret
```

assistant: ```c
int power(int num, int exp) {
  if (exp == 0)
    return (1);
  if (exp == 1)
    return (num);
  int temp = power(num, exp / 2);
  temp *= temp;
  if (exp % 2 == 1)
    temp *= num;
  return (temp);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<power>:
  endbr64
  mov    $0x1,%eax
  test   %esi,%esi
  je     L31
  push   %r14
  mov    %edi,%eax
  push   %r13
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x18,%rsp
  cmp    $0x1,%esi
  je     L37
  mov    %esi,%ebp
  mov    $0x1,%eax
  shr    $0x1f,%ebp
  add    %esi,%ebp
  sar    %ebp
  je     L15
  cmp    $0x1,%ebp
  je     L91
  mov    %ebp,%r13d
  mov    $0x1,%eax
  shr    $0x1f,%r13d
  add    %ebp,%r13d
  mov    %r13d,%r14d
  sar    %r14d
  je     L35
  cmp    $0x1,%r14d
  je     L62
  mov    %r14d,%esi
  shr    $0x1f,%r13d
  mov    %edi,0xc(%rsp)
  shr    $0x1f,%esi
  add    %r14d,%esi
  add    %r13d,%r14d
  sar    %esi
  and    $0x1,%r14d
  call   <power>
  sub    %r13d,%r14d
  mov    0xc(%rsp),%edi
  imul   %eax,%eax
  cmp    $0x1,%r14d
  jne    L36
  imul   %edi,%eax
L36:
  imul   %eax,%eax
L35:
  mov    %ebp,%edx
  shr    $0x1f,%edx
  add    %edx,%ebp
  and    $0x1,%ebp
  sub    %edx,%ebp
  cmp    $0x1,%ebp
  jne    L33
  imul   %edi,%eax
L33:
  imul   %eax,%eax
L15:
  mov    %ebx,%edx
  shr    $0x1f,%edx
  add    %edx,%ebx
  and    $0x1,%ebx
  sub    %edx,%ebx
  cmp    $0x1,%ebx
  je     L48
L37:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L31:
  ret
  nopl   0x0(%rax)
L48:
  add    $0x18,%rsp
  imul   %edi,%eax
  pop    %rbx
  pop    %rbp
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L91:
  mov    %edi,%eax
  imul   %edi,%eax
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L62:
  mov    %edi,%eax
  imul   %edi,%eax
  jmp    L35
```

assistant: ```c
int power(int num, int exp) {
  if (exp == 0)
    return (1);
  if (exp == 1)
    return (num);
  int temp = power(num, exp / 2);
  temp *= temp;
  if (exp % 2 == 1)
    temp *= num;
  return (temp);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<power>:
  endbr64
  mov    $0x1,%eax
  test   %esi,%esi
  je     L97
  push   %r14
  mov    %edi,%eax
  push   %r13
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x18,%rsp
  cmp    $0x1,%esi
  je     L38
  mov    %esi,%ebp
  mov    $0x1,%eax
  shr    $0x1f,%ebp
  add    %esi,%ebp
  sar    %ebp
  je     L70
  cmp    $0x1,%ebp
  je     L9
  mov    %ebp,%r13d
  mov    $0x1,%eax
  shr    $0x1f,%r13d
  add    %ebp,%r13d
  mov    %r13d,%r14d
  sar    %r14d
  je     L24
  cmp    $0x1,%r14d
  je     L2
  mov    %r14d,%esi
  shr    $0x1f,%r13d
  mov    %edi,0xc(%rsp)
  shr    $0x1f,%esi
  add    %r14d,%esi
  add    %r13d,%r14d
  sar    %esi
  and    $0x1,%r14d
  call   <power>
  sub    %r13d,%r14d
  mov    0xc(%rsp),%edi
  imul   %eax,%eax
  cmp    $0x1,%r14d
  jne    L83
  imul   %edi,%eax
L83:
  imul   %eax,%eax
L24:
  mov    %ebp,%edx
  shr    $0x1f,%edx
  add    %edx,%ebp
  and    $0x1,%ebp
  sub    %edx,%ebp
  cmp    $0x1,%ebp
  jne    L101
  imul   %edi,%eax
L101:
  imul   %eax,%eax
L70:
  mov    %ebx,%edx
  shr    $0x1f,%edx
  add    %edx,%ebx
  and    $0x1,%ebx
  sub    %edx,%ebx
  cmp    $0x1,%ebx
  je     L6
L38:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L97:
  ret
  nopl   0x0(%rax)
L6:
  add    $0x18,%rsp
  imul   %edi,%eax
  pop    %rbx
  pop    %rbp
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L9:
  mov    %edi,%eax
  imul   %edi,%eax
  jmp    L70
  nopw   0x0(%rax,%rax,1)
L2:
  mov    %edi,%eax
  imul   %edi,%eax
  jmp    L24
```

assistant: ```c
int power(int num, int exp) {
  if (exp == 0)
    return (1);
  if (exp == 1)
    return (num);
  int temp = power(num, exp / 2);
  temp *= temp;
  if (exp % 2 == 1)
    temp *= num;
  return (temp);
}
```