ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<P_evaluate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  movsd  %xmm0,-0x28(%rbp)
  mov    %rdi,-0x30(%rbp)
  mov    %esi,-0x34(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    $0x0,%ebx
  jmp    L6
L4:
  movslq %ebx,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  pxor   %xmm2,%xmm2
  cvtsi2sd %rax,%xmm2
  movsd  %xmm2,-0x40(%rbp)
  mov    -0x34(%rbp),%eax
  sub    %ebx,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%edi
  movq   %rax,%xmm0
  call   <intpow@plt>
  mulsd  -0x40(%rbp),%xmm0
  movsd  -0x18(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  add    $0x1,%ebx
L6:
  cmp    %ebx,-0x34(%rbp)
  jge    L4
  movsd  -0x18(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
double P_evaluate(double x, long *coefficients, int degree) {
  double result;
  register int i;
  result = 0.;
  for (i = 0; i < degree + 1; i++)
    result += (double)coefficients[i] * intpow(x, degree - i);
  return (result);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<P_evaluate>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %esi,%esi
  js     L20
  movq   %xmm0,%r12
  mov    %rdi,%r13
  mov    %esi,%ebp
  mov    $0x0,%ebx
  pxor   %xmm3,%xmm3
  movsd  %xmm3,(%rsp)
L21:
  pxor   %xmm1,%xmm1
  cvtsi2sdq 0x0(%r13,%rbx,8),%xmm1
  movsd  %xmm1,0x8(%rsp)
  mov    %ebp,%edi
  sub    %ebx,%edi
  movq   %r12,%xmm0
  call   <intpow@plt>
  mulsd  0x8(%rsp),%xmm0
  addsd  (%rsp),%xmm0
  movsd  %xmm0,(%rsp)
  add    $0x1,%rbx
  cmp    %ebx,%ebp
  jge    L21
L25:
  movsd  (%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L20:
  pxor   %xmm4,%xmm4
  movsd  %xmm4,(%rsp)
  jmp    L25
```

assistant: ```c
double P_evaluate(double x, long *coefficients, int degree) {
  double result;
  register int i;
  result = 0.;
  for (i = 0; i < degree + 1; i++)
    result += (double)coefficients[i] * intpow(x, degree - i);
  return (result);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<P_evaluate>:
  endbr64
  test   %esi,%esi
  js     L19
  push   %r14
  pxor   %xmm1,%xmm1
  movq   %xmm0,%r14
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  nopw   0x0(%rax,%rax,1)
L14:
  pxor   %xmm2,%xmm2
  mov    %ebp,%edi
  movq   %r14,%xmm0
  movsd  %xmm1,0x8(%rsp)
  cvtsi2sdq (%r12,%rbx,8),%xmm2
  sub    %ebx,%edi
  add    $0x1,%rbx
  movsd  %xmm2,(%rsp)
  call   <intpow@plt>
  mulsd  (%rsp),%xmm0
  movsd  0x8(%rsp),%xmm1
  addsd  %xmm0,%xmm1
  cmp    %ebx,%ebp
  jge    L14
  add    $0x18,%rsp
  movapd %xmm1,%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r14
  ret
  nopl   0x0(%rax)
L19:
  pxor   %xmm1,%xmm1
  movapd %xmm1,%xmm0
  ret
```

assistant: ```c
double P_evaluate(double x, long *coefficients, int degree) {
  double result;
  register int i;
  result = 0.;
  for (i = 0; i < degree + 1; i++)
    result += (double)coefficients[i] * intpow(x, degree - i);
  return (result);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<P_evaluate>:
  endbr64
  test   %esi,%esi
  js     L3
  push   %r14
  pxor   %xmm1,%xmm1
  movq   %xmm0,%r14
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  nopw   0x0(%rax,%rax,1)
L1:
  pxor   %xmm2,%xmm2
  mov    %ebp,%edi
  movq   %r14,%xmm0
  movsd  %xmm1,0x8(%rsp)
  cvtsi2sdq (%r12,%rbx,8),%xmm2
  sub    %ebx,%edi
  add    $0x1,%rbx
  movsd  %xmm2,(%rsp)
  call   <intpow@plt>
  mulsd  (%rsp),%xmm0
  movsd  0x8(%rsp),%xmm1
  addsd  %xmm0,%xmm1
  cmp    %ebx,%ebp
  jge    L1
  add    $0x18,%rsp
  movapd %xmm1,%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r14
  ret
  nopl   0x0(%rax)
L3:
  pxor   %xmm1,%xmm1
  movapd %xmm1,%xmm0
  ret
```

assistant: ```c
double P_evaluate(double x, long *coefficients, int degree) {
  double result;
  register int i;
  result = 0.;
  for (i = 0; i < degree + 1; i++)
    result += (double)coefficients[i] * intpow(x, degree - i);
  return (result);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<bsd_strlcat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L68
L45:
  addq   $0x1,-0x20(%rbp)
L68:
  mov    -0x10(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  test   %rax,%rax
  je     L12
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L45
L12:
  mov    -0x20(%rbp),%rax
  sub    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x38(%rbp),%rax
  sub    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L23
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    -0x8(%rbp),%rdx
  add    %rdx,%rax
  jmp    L51
L17:
  cmpq   $0x1,-0x10(%rbp)
  je     L66
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    -0x18(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  subq   $0x1,-0x10(%rbp)
L66:
  addq   $0x1,-0x18(%rbp)
L23:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L17
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  sub    -0x30(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
L51:
  leave
  ret
```

assistant: ```c
size_t bsd_strlcat(char *dst, const char *src, size_t siz) {
  char *d = dst;
  const char *s = src;
  size_t n = siz;
  size_t dlen;
  while (n-- != 0 && *d != '\0')
    d++;
  dlen = d - dst;
  n = siz - dlen;
  if (n == 0)
    return (dlen + strlen(s));
  while (*s != '\0') {
    if (n != 1) {
      *d++ = *s;
      n--;
    }
    s++;
  }
  *d = '\0';
  return (dlen + (s - src));
}
```
user: What is the c source code of the assembly code below?

```asm
<bsd_strlcat>:
  endbr64
  push   %rbx
  test   %rdx,%rdx
  je     L89
  mov    %rdi,%r8
  lea    (%rdi,%rdx,1),%rcx
  mov    %rdi,%rax
L21:
  cmpb   $0x0,(%rax)
  je     L7
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L21
  mov    %rcx,%rax
L7:
  mov    %rax,%rbx
  sub    %r8,%rbx
  sub    %rbx,%rdx
  je     L79
  movzbl (%rsi),%r8d
  mov    %rsi,%rcx
  test   %r8b,%r8b
  jne    L28
L68:
  movb   $0x0,(%rax)
  sub    %rsi,%rcx
  lea    (%rcx,%rbx,1),%rax
L9:
  pop    %rbx
  ret
L89:
  mov    %rdx,%rbx
L79:
  mov    %rsi,%rdi
  call   <strlen@plt>
  add    %rbx,%rax
  jmp    L9
L63:
  add    $0x1,%rcx
  movzbl (%rcx),%r8d
  test   %r8b,%r8b
  je     L68
L28:
  cmp    $0x1,%rdx
  je     L63
  mov    %r8b,(%rax)
  sub    $0x1,%rdx
  lea    0x1(%rax),%rax
  jmp    L63
```

assistant: ```c
size_t bsd_strlcat(char *dst, const char *src, size_t siz) {
  char *d = dst;
  const char *s = src;
  size_t n = siz;
  size_t dlen;
  while (n-- != 0 && *d != '\0')
    d++;
  dlen = d - dst;
  n = siz - dlen;
  if (n == 0)
    return (dlen + strlen(s));
  while (*s != '\0') {
    if (n != 1) {
      *d++ = *s;
      n--;
    }
    s++;
  }
  *d = '\0';
  return (dlen + (s - src));
}
```
user: Translate the below assembly code into C programming language:

```asm
<bsd_strlcat>:
  endbr64
  push   %r12
  test   %rdx,%rdx
  je     L70
  mov    %rdi,%r8
  lea    (%rdi,%rdx,1),%rcx
  mov    %rdi,%rax
  jmp    L71
  nopl   0x0(%rax,%rax,1)
L61:
  add    $0x1,%rax
  cmp    %rcx,%rax
  je     L28
L71:
  cmpb   $0x0,(%rax)
  jne    L61
L28:
  mov    %rax,%r12
  sub    %r8,%r12
  sub    %r12,%rdx
  je     L10
  movzbl (%rsi),%ecx
  mov    %rsi,%r8
  test   %cl,%cl
  jne    L45
  jmp    L73
  nopl   (%rax)
L101:
  mov    %cl,(%rax)
  sub    $0x1,%rdx
  mov    %r8,%rcx
  add    $0x1,%rax
L93:
  lea    0x1(%rcx),%r8
  movzbl 0x1(%rcx),%ecx
  test   %cl,%cl
  je     L11
L45:
  cmp    $0x1,%rdx
  jne    L101
  cmpb   $0x0,0x1(%r8)
  lea    0x1(%r8),%rcx
  jne    L93
  sub    %rsi,%rcx
  add    %rcx,%r12
L73:
  movb   $0x0,(%rax)
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L11:
  sub    %rsi,%r8
  add    %r8,%r12
  jmp    L73
L70:
  xor    %r12d,%r12d
L10:
  mov    %rsi,%rdi
  call   <strlen@plt>
  add    %rax,%r12
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
size_t bsd_strlcat(char *dst, const char *src, size_t siz) {
  char *d = dst;
  const char *s = src;
  size_t n = siz;
  size_t dlen;
  while (n-- != 0 && *d != '\0')
    d++;
  dlen = d - dst;
  n = siz - dlen;
  if (n == 0)
    return (dlen + strlen(s));
  while (*s != '\0') {
    if (n != 1) {
      *d++ = *s;
      n--;
    }
    s++;
  }
  *d = '\0';
  return (dlen + (s - src));
}
```
user: Translate the below assembly code into C programming language:

```asm
<bsd_strlcat>:
  endbr64
  push   %r12
  test   %rdx,%rdx
  je     L65
  mov    %rdi,%r8
  lea    (%rdi,%rdx,1),%rcx
  mov    %rdi,%rax
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L52:
  add    $0x1,%rax
  cmp    %rcx,%rax
  je     L55
L25:
  cmpb   $0x0,(%rax)
  jne    L52
L55:
  mov    %rax,%r12
  sub    %r8,%r12
  sub    %r12,%rdx
  je     L68
  movzbl (%rsi),%ecx
  mov    %rsi,%r8
  test   %cl,%cl
  jne    L63
  jmp    L108
  nopl   (%rax)
L98:
  mov    %cl,(%rax)
  sub    $0x1,%rdx
  mov    %r8,%rcx
  add    $0x1,%rax
L34:
  lea    0x1(%rcx),%r8
  movzbl 0x1(%rcx),%ecx
  test   %cl,%cl
  je     L32
L63:
  cmp    $0x1,%rdx
  jne    L98
  cmpb   $0x0,0x1(%r8)
  lea    0x1(%r8),%rcx
  jne    L34
  sub    %rsi,%rcx
  add    %rcx,%r12
L108:
  movb   $0x0,(%rax)
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L32:
  sub    %rsi,%r8
  add    %r8,%r12
  jmp    L108
L65:
  xor    %r12d,%r12d
L68:
  mov    %rsi,%rdi
  call   <strlen@plt>
  add    %rax,%r12
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
size_t bsd_strlcat(char *dst, const char *src, size_t siz) {
  char *d = dst;
  const char *s = src;
  size_t n = siz;
  size_t dlen;
  while (n-- != 0 && *d != '\0')
    d++;
  dlen = d - dst;
  n = siz - dlen;
  if (n == 0)
    return (dlen + strlen(s));
  while (*s != '\0') {
    if (n != 1) {
      *d++ = *s;
      n--;
    }
    s++;
  }
  *d = '\0';
  return (dlen + (s - src));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<THDiskFile_reverseMemory>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %rcx,-0x50(%rbp)
  cmpq   $0x1,-0x48(%rbp)
  je     L24
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdx
  shr    $0x3f,%rdx
  add    %rdx,%rax
  sar    %rax
  mov    %rax,-0x8(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L43
L42:
  movq   $0x0,-0x10(%rbp)
  jmp    L35
L36:
  mov    -0x10(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x29(%rbp)
  mov    -0x48(%rbp),%rax
  sub    $0x1,%rax
  sub    -0x10(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x48(%rbp),%rax
  sub    $0x1,%rax
  sub    -0x10(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x29(%rbp),%eax
  mov    %al,(%rdx)
  addq   $0x1,-0x10(%rbp)
L35:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jl     L36
  mov    -0x48(%rbp),%rax
  add    %rax,-0x28(%rbp)
  mov    -0x48(%rbp),%rax
  add    %rax,-0x20(%rbp)
  addq   $0x1,-0x18(%rbp)
L43:
  mov    -0x18(%rbp),%rax
  cmp    -0x50(%rbp),%rax
  jl     L42
L24:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void THDiskFile_reverseMemory(void *dst, const void *src, long blockSize, long numBlocks) {
  if (blockSize != 1) {
    long halfBlockSize = blockSize / 2;
    char *charSrc = (char *)src;
    char *charDst = (char *)dst;
    long b, i;
    for (b = 0; b < numBlocks; b++) {
      for (i = 0; i < halfBlockSize; i++) {
        char z = charSrc[i];
        charDst[i] = charSrc[blockSize - 1 - i];
        charDst[blockSize - 1 - i] = z;
      }
      charSrc += blockSize;
      charDst += blockSize;
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<THDiskFile_reverseMemory>:
  endbr64
  cmp    $0x1,%rdx
  je     L26
  push   %rbx
  mov    %rsi,%r8
  mov    %rcx,%rbx
  mov    %rdx,%r11
  shr    $0x3f,%r11
  add    %rdx,%r11
  sar    %r11
  mov    $0x0,%ecx
  test   %rbx,%rbx
  jg     L43
L54:
  pop    %rbx
  ret
L29:
  movzbl (%r8,%rax,1),%r9d
  movzbl (%r8,%rsi,1),%r10d
  mov    %r10b,(%rdi,%rax,1)
  mov    %r9b,(%rdi,%rsi,1)
  add    $0x1,%rax
  sub    $0x1,%rsi
  cmp    %rax,%r11
  jg     L29
L45:
  add    %rdx,%r8
  add    %rdx,%rdi
  add    $0x1,%rcx
  cmp    %rcx,%rbx
  je     L54
L43:
  lea    -0x1(%rdx),%rsi
  mov    $0x0,%eax
  cmp    $0x1,%rdx
  jg     L29
  jmp    L45
L26:
  ret
```

assistant: ```c
void THDiskFile_reverseMemory(void *dst, const void *src, long blockSize, long numBlocks) {
  if (blockSize != 1) {
    long halfBlockSize = blockSize / 2;
    char *charSrc = (char *)src;
    char *charDst = (char *)dst;
    long b, i;
    for (b = 0; b < numBlocks; b++) {
      for (i = 0; i < halfBlockSize; i++) {
        char z = charSrc[i];
        charDst[i] = charSrc[blockSize - 1 - i];
        charDst[blockSize - 1 - i] = z;
      }
      charSrc += blockSize;
      charDst += blockSize;
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<THDiskFile_reverseMemory>:
  endbr64
  cmp    $0x1,%rdx
  je     L18
  mov    %rdx,%r11
  push   %rbx
  mov    %rcx,%rbx
  shr    $0x3f,%r11
  add    %rdx,%r11
  sar    %r11
  test   %rcx,%rcx
  jle    L42
  mov    %rsi,%r8
  xor    %ecx,%ecx
  nopl   (%rax)
L30:
  lea    -0x1(%rdx),%rsi
  xor    %eax,%eax
  cmp    $0x1,%rdx
  jle    L37
  nopl   0x0(%rax)
L33:
  movzbl (%r8,%rsi,1),%r10d
  movzbl (%r8,%rax,1),%r9d
  mov    %r10b,(%rdi,%rax,1)
  add    $0x1,%rax
  mov    %r9b,(%rdi,%rsi,1)
  sub    $0x1,%rsi
  cmp    %rax,%r11
  jg     L33
L37:
  add    $0x1,%rcx
  add    %rdx,%r8
  add    %rdx,%rdi
  cmp    %rcx,%rbx
  jne    L30
L42:
  pop    %rbx
  ret
L18:
  ret
```

assistant: ```c
void THDiskFile_reverseMemory(void *dst, const void *src, long blockSize, long numBlocks) {
  if (blockSize != 1) {
    long halfBlockSize = blockSize / 2;
    char *charSrc = (char *)src;
    char *charDst = (char *)dst;
    long b, i;
    for (b = 0; b < numBlocks; b++) {
      for (i = 0; i < halfBlockSize; i++) {
        char z = charSrc[i];
        charDst[i] = charSrc[blockSize - 1 - i];
        charDst[blockSize - 1 - i] = z;
      }
      charSrc += blockSize;
      charDst += blockSize;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<THDiskFile_reverseMemory>:
  endbr64
  cmp    $0x1,%rdx
  je     L0
  mov    %rdx,%r11
  push   %rbp
  shr    $0x3f,%r11
  push   %rbx
  mov    %rcx,%rbx
  add    %rdx,%r11
  sar    %r11
  test   %rcx,%rcx
  jle    L18
  mov    %rsi,%r8
  xor    %ecx,%ecx
  lea    -0x1(%rdx),%rbp
  cmp    $0x1,%rdx
  jle    L18
L32:
  mov    %rbp,%rsi
  xor    %eax,%eax
  nopl   (%rax)
L7:
  movzbl (%r8,%rsi,1),%r10d
  movzbl (%r8,%rax,1),%r9d
  mov    %r10b,(%rdi,%rax,1)
  add    $0x1,%rax
  mov    %r9b,(%rdi,%rsi,1)
  sub    $0x1,%rsi
  cmp    %rax,%r11
  jg     L7
  add    $0x1,%rcx
  add    %rdx,%r8
  add    %rdx,%rdi
  cmp    %rcx,%rbx
  jne    L32
L18:
  pop    %rbx
  pop    %rbp
  ret
L0:
  ret
```

assistant: ```c
void THDiskFile_reverseMemory(void *dst, const void *src, long blockSize, long numBlocks) {
  if (blockSize != 1) {
    long halfBlockSize = blockSize / 2;
    char *charSrc = (char *)src;
    char *charDst = (char *)dst;
    long b, i;
    for (b = 0; b < numBlocks; b++) {
      for (i = 0; i < halfBlockSize; i++) {
        char z = charSrc[i];
        charDst[i] = charSrc[blockSize - 1 - i];
        charDst[blockSize - 1 - i] = z;
      }
      charSrc += blockSize;
      charDst += blockSize;
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<zig_subw_u8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jge    L13
  mov    -0x18(%rbp),%rax
  sub    -0x10(%rbp),%rax
  sub    -0x8(%rbp),%rax
  add    $0x1,%rax
  jmp    L5
L13:
  mov    -0x8(%rbp),%rax
  sub    -0x10(%rbp),%rax
L5:
  pop    %rbp
  ret
```

assistant: ```c
uint8_t zig_subw_u8(uint8_t lhs, uint8_t rhs, uint8_t max) {
  if (lhs < rhs) {
    return max - rhs - lhs + 1;
  } else {
    return lhs - rhs;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<zig_subw_u8>:
  endbr64
  mov    %rdi,%rax
  sub    %rsi,%rax
  cmp    %rsi,%rdi
  jge    L9
  sub    %rsi,%rdx
  sub    %rdi,%rdx
  lea    0x1(%rdx),%rax
L9:
  ret
```

assistant: ```c
uint8_t zig_subw_u8(uint8_t lhs, uint8_t rhs, uint8_t max) {
  if (lhs < rhs) {
    return max - rhs - lhs + 1;
  } else {
    return lhs - rhs;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<zig_subw_u8>:
  endbr64
  mov    %rdi,%rax
  sub    %rsi,%rax
  cmp    %rsi,%rdi
  jge    L8
  sub    %rsi,%rdx
  sub    %rdi,%rdx
  lea    0x1(%rdx),%rax
L8:
  ret
```

assistant: ```c
uint8_t zig_subw_u8(uint8_t lhs, uint8_t rhs, uint8_t max) {
  if (lhs < rhs) {
    return max - rhs - lhs + 1;
  } else {
    return lhs - rhs;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<zig_subw_u8>:
  endbr64
  mov    %rdi,%rax
  sub    %rsi,%rax
  cmp    %rsi,%rdi
  jge    L3
  sub    %rsi,%rdx
  sub    %rdi,%rdx
  lea    0x1(%rdx),%rax
L3:
  ret
```

assistant: ```c
uint8_t zig_subw_u8(uint8_t lhs, uint8_t rhs, uint8_t max) {
  if (lhs < rhs) {
    return max - rhs - lhs + 1;
  } else {
    return lhs - rhs;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<FaultISR>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L3:
  jmp    L3
```

assistant: ```c
void FaultISR(void) {
  while (1) {
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<FaultISR>:
  endbr64
L6:
  jmp    L6
```

assistant: ```c
void FaultISR(void) {
  while (1) {
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<FaultISR>:
  endbr64
  nopl   0x0(%rax)
L5:
  jmp    L5
```

assistant: ```c
void FaultISR(void) {
  while (1) {
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<count_1bits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    $0x55555555,%edi
  call   <UINT64_C@plt>
  mov    %eax,-0x10(%rbp)
  mov    $0x33333333,%edi
  call   <UINT64_C@plt>
  mov    %eax,-0xc(%rbp)
  mov    $0xf0f0f0f,%edi
  call   <UINT64_C@plt>
  mov    %eax,-0x8(%rbp)
  mov    $0x1010101,%edi
  call   <UINT64_C@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  sar    %eax
  and    -0x10(%rbp),%eax
  sub    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  and    -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  sar    $0x2,%eax
  and    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  sar    $0x4,%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  and    -0x8(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  imul   -0x4(%rbp),%eax
  mov    $0x38,%edx
  mov    %edx,%ecx
  sar    %cl,%eax
  leave
  ret
```

assistant: ```c
unsigned int count_1bits(uint64_t x) {
  const uint64_t h55 = UINT64_C(0x5555555555555555);
  const uint64_t h33 = UINT64_C(0x3333333333333333);
  const uint64_t h0F = UINT64_C(0x0F0F0F0F0F0F0F0F);
  const uint64_t h01 = UINT64_C(0x0101010101010101);
  x -= (x >> 1) & h55;
  x = (x & h33) + ((x >> 2) & h33);
  x = (x + (x >> 4)) & h0F;
  return (x * h01) >> 56;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<count_1bits>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    $0x55555555,%edi
  call   <UINT64_C@plt>
  mov    %eax,%r13d
  mov    $0x33333333,%edi
  call   <UINT64_C@plt>
  mov    %eax,%r12d
  mov    $0xf0f0f0f,%edi
  call   <UINT64_C@plt>
  mov    %eax,%ebp
  mov    $0x1010101,%edi
  call   <UINT64_C@plt>
  mov    %eax,%ecx
  mov    %ebx,%eax
  sar    %eax
  and    %r13d,%eax
  sub    %eax,%ebx
  mov    %r12d,%edx
  and    %ebx,%edx
  sar    $0x2,%ebx
  and    %r12d,%ebx
  add    %ebx,%edx
  mov    %edx,%eax
  sar    $0x4,%eax
  add    %edx,%eax
  and    %ebp,%eax
  imul   %ecx,%eax
  mov    $0x38,%ecx
  sar    %cl,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
unsigned int count_1bits(uint64_t x) {
  const uint64_t h55 = UINT64_C(0x5555555555555555);
  const uint64_t h33 = UINT64_C(0x3333333333333333);
  const uint64_t h0F = UINT64_C(0x0F0F0F0F0F0F0F0F);
  const uint64_t h01 = UINT64_C(0x0101010101010101);
  x -= (x >> 1) & h55;
  x = (x & h33) + ((x >> 2) & h33);
  x = (x + (x >> 4)) & h0F;
  return (x * h01) >> 56;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<count_1bits>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    $0x55555555,%edi
  sub    $0x8,%rsp
  call   <UINT64_C@plt>
  mov    $0x33333333,%edi
  mov    %eax,%r13d
  call   <UINT64_C@plt>
  mov    $0xf0f0f0f,%edi
  mov    %eax,%r12d
  call   <UINT64_C@plt>
  mov    $0x1010101,%edi
  mov    %eax,%ebp
  call   <UINT64_C@plt>
  mov    %r12d,%edx
  add    $0x8,%rsp
  mov    $0x38,%ecx
  mov    %eax,%r8d
  mov    %ebx,%eax
  sar    %eax
  and    %r13d,%eax
  sub    %eax,%ebx
  and    %ebx,%edx
  sar    $0x2,%ebx
  and    %r12d,%ebx
  add    %ebx,%edx
  pop    %rbx
  mov    %edx,%eax
  sar    $0x4,%eax
  add    %edx,%eax
  and    %ebp,%eax
  pop    %rbp
  pop    %r12
  imul   %r8d,%eax
  pop    %r13
  sar    %cl,%eax
  ret
```

assistant: ```c
unsigned int count_1bits(uint64_t x) {
  const uint64_t h55 = UINT64_C(0x5555555555555555);
  const uint64_t h33 = UINT64_C(0x3333333333333333);
  const uint64_t h0F = UINT64_C(0x0F0F0F0F0F0F0F0F);
  const uint64_t h01 = UINT64_C(0x0101010101010101);
  x -= (x >> 1) & h55;
  x = (x & h33) + ((x >> 2) & h33);
  x = (x + (x >> 4)) & h0F;
  return (x * h01) >> 56;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<count_1bits>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    $0x55555555,%edi
  sub    $0x8,%rsp
  call   <UINT64_C@plt>
  mov    $0x33333333,%edi
  mov    %eax,%r13d
  call   <UINT64_C@plt>
  mov    $0xf0f0f0f,%edi
  mov    %eax,%r12d
  call   <UINT64_C@plt>
  mov    $0x1010101,%edi
  mov    %eax,%ebp
  call   <UINT64_C@plt>
  mov    %r12d,%edx
  add    $0x8,%rsp
  mov    $0x38,%ecx
  mov    %eax,%r8d
  mov    %ebx,%eax
  sar    %eax
  and    %r13d,%eax
  sub    %eax,%ebx
  and    %ebx,%edx
  sar    $0x2,%ebx
  and    %r12d,%ebx
  add    %ebx,%edx
  pop    %rbx
  mov    %edx,%eax
  sar    $0x4,%eax
  add    %edx,%eax
  and    %ebp,%eax
  pop    %rbp
  pop    %r12
  imul   %r8d,%eax
  pop    %r13
  sar    %cl,%eax
  ret
```

assistant: ```c
unsigned int count_1bits(uint64_t x) {
  const uint64_t h55 = UINT64_C(0x5555555555555555);
  const uint64_t h33 = UINT64_C(0x3333333333333333);
  const uint64_t h0F = UINT64_C(0x0F0F0F0F0F0F0F0F);
  const uint64_t h01 = UINT64_C(0x0101010101010101);
  x -= (x >> 1) & h55;
  x = (x & h33) + ((x >> 2) & h33);
  x = (x + (x >> 4)) & h0F;
  return (x * h01) >> 56;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<sys_open>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    D18(%rip),%rax
  mov    (%rax),%edx
  mov    D14(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int sys_open(const char *name, int flags, int mode) {
  errno = ENOMEM;
  return -1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<sys_open>:
  endbr64
  mov    D9(%rip),%rax
  mov    (%rax),%edx
  mov    D3(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int sys_open(const char *name, int flags, int mode) {
  errno = ENOMEM;
  return -1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sys_open>:
  endbr64
  mov    D4(%rip),%rax
  mov    (%rax),%edx
  mov    D12(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int sys_open(const char *name, int flags, int mode) {
  errno = ENOMEM;
  return -1;
}
```