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

```asm
<sumGTRCATPROT>:
  endbr64
  push   %rbx
  mov    %edi,%eax
  mov    %rdx,%rdi
  mov    %rcx,%rdx
  mov    0x10(%rsp),%r10
  cmp    $0x1,%eax
  je     L70
  cmp    $0x2,%eax
  je     L125
  test   %eax,%eax
  je     L10
  mov    $0x0,%edi
  call   <assert@plt>
L24:
  pop    %rbx
  ret
L125:
  cmpl   $0x0,0x18(%rsp)
  jle    L24
  mov    0x18(%rsp),%eax
  lea    -0x1(%rax),%eax
  lea    (%rax,%rax,4),%r8
  shl    $0x5,%r8
  add    $0x140,%r8
  mov    $0xa0,%ecx
L29:
  lea    -0xa0(%rcx),%rax
L26:
  movsd  (%rdi,%rax,1),%xmm0
  mulsd  (%rdx,%rax,1),%xmm0
  movsd  %xmm0,(%rsi,%rax,1)
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L26
  add    $0xa0,%rcx
  cmp    %r8,%rcx
  jne    L29
  jmp    L24
L70:
  cmpl   $0x0,0x18(%rsp)
  jle    L24
  mov    0x18(%rsp),%eax
  lea    (%rax,%rax,4),%rbx
  shl    $0x2,%rbx
  mov    $0xa0,%edi
  mov    $0x0,%r11d
  mov    $0x0,%r10d
L142:
  mov    %r10,%rax
  neg    %rax
  movzbl (%r9),%ecx
  lea    (%rcx,%rcx,4),%ecx
  shl    $0x2,%ecx
  movslq %ecx,%rcx
  lea    (%r10,%rcx,8),%rcx
  add    %r8,%rcx
L12:
  movsd  (%rcx,%rax,1),%xmm0
  mulsd  (%rdx,%rax,1),%xmm0
  movsd  %xmm0,(%rsi,%rax,1)
  add    $0x8,%rax
  cmp    %rdi,%rax
  jne    L12
  add    $0x1,%r9
  sub    $0xa0,%r10
  add    $0x14,%r11
  add    $0xa0,%rdi
  cmp    %rbx,%r11
  jne    L142
  jmp    L24
L10:
  cmpl   $0x0,0x18(%rsp)
  jle    L24
  mov    0x18(%rsp),%r11d
  mov    $0x0,%edi
L14:
  movzbl (%r9,%rdi,1),%eax
  lea    (%rax,%rax,4),%eax
  shl    $0x2,%eax
  cltq
  lea    (%r8,%rax,8),%rcx
  movzbl (%r10,%rdi,1),%eax
  lea    (%rax,%rax,4),%eax
  shl    $0x2,%eax
  cltq
  lea    (%r8,%rax,8),%rdx
  mov    $0x0,%eax
L144:
  movsd  (%rcx,%rax,1),%xmm0
  mulsd  (%rdx,%rax,1),%xmm0
  movsd  %xmm0,(%rsi,%rax,1)
  add    $0x8,%rax
  cmp    $0xa0,%rax
  jne    L144
  add    $0x1,%rdi
  add    $0xa0,%rsi
  cmp    %r11,%rdi
  jne    L14
  jmp    L24
```

assistant: ```c
void sumGTRCATPROT(int tipCase, double *sumtable, double *x1, double *x2, double *tipVector, unsigned char *tipX1, unsigned char *tipX2, int n) {
  int i, l;
  double *sum, *left, *right;
  switch (tipCase) {
  case 0:
    for (i = 0; i < n; i++) {
      left = &(tipVector[20 * tipX1[i]]);
      right = &(tipVector[20 * tipX2[i]]);
      sum = &sumtable[20 * i];
      for (l = 0; l < 20; l++)
        sum[l] = left[l] * right[l];
    }
    break;
  case 1:
    for (i = 0; i < n; i++) {
      left = &(tipVector[20 * tipX1[i]]);
      right = &x2[20 * i];
      sum = &sumtable[20 * i];
      for (l = 0; l < 20; l++)
        sum[l] = left[l] * right[l];
    }
    break;
  case 2:
    for (i = 0; i < n; i++) {
      left = &x1[20 * i];
      right = &x2[20 * i];
      sum = &sumtable[20 * i];
      for (l = 0; l < 20; l++)
        sum[l] = left[l] * right[l];
    }
    break;
  default:
    assert(0);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sumGTRCATPROT>:
  endbr64
  mov    %edi,%eax
  push   %rbx
  mov    %rdx,%rdi
  mov    0x10(%rsp),%r10
  movslq 0x18(%rsp),%r11
  mov    %rcx,%rdx
  cmp    $0x1,%eax
  je     L101
  cmp    $0x2,%eax
  je     L86
  test   %eax,%eax
  je     L21
  xor    %edi,%edi
  pop    %rbx
  jmp    1050 <assert@plt>
L101:
  test   %r11d,%r11d
  jle    L4
  lea    (%r11,%r11,4),%rbx
  mov    $0xa0,%edi
  xor    %r11d,%r11d
  xor    %r10d,%r10d
  shl    $0x2,%rbx
  nopl   0x0(%rax)
L119:
  movzbl (%r9),%ecx
  mov    %r10,%rax
  neg    %rax
  lea    (%rcx,%rcx,4),%ecx
  shl    $0x2,%ecx
  movslq %ecx,%rcx
  lea    (%r10,%rcx,8),%rcx
  add    %r8,%rcx
  nopw   0x0(%rax,%rax,1)
L111:
  movsd  (%rcx,%rax,1),%xmm0
  mulsd  (%rdx,%rax,1),%xmm0
  movsd  %xmm0,(%rsi,%rax,1)
  add    $0x8,%rax
  cmp    %rax,%rdi
  jne    L111
  add    $0x14,%r11
  add    $0x1,%r9
  sub    $0xa0,%r10
  add    $0xa0,%rdi
  cmp    %r11,%rbx
  jne    L119
L4:
  pop    %rbx
  ret
L21:
  test   %r11d,%r11d
  jle    L4
  xor    %edi,%edi
  nopl   0x0(%rax)
L115:
  movzbl (%r9,%rdi,1),%eax
  lea    (%rax,%rax,4),%eax
  shl    $0x2,%eax
  cltq
  lea    (%r8,%rax,8),%rcx
  movzbl (%r10,%rdi,1),%eax
  lea    (%rax,%rax,4),%eax
  shl    $0x2,%eax
  cltq
  lea    (%r8,%rax,8),%rdx
  xor    %eax,%eax
  nopl   0x0(%rax)
L11:
  movsd  (%rcx,%rax,1),%xmm0
  mulsd  (%rdx,%rax,1),%xmm0
  movsd  %xmm0,(%rsi,%rax,1)
  add    $0x8,%rax
  cmp    $0xa0,%rax
  jne    L11
  add    $0x1,%rdi
  add    $0xa0,%rsi
  cmp    %rdi,%r11
  jne    L115
  pop    %rbx
  ret
L86:
  test   %r11d,%r11d
  jle    L4
  lea    -0x1(%r11),%eax
  mov    $0xa0,%ecx
  lea    (%rax,%rax,4),%r8
  shl    $0x5,%r8
  add    $0x140,%r8
  nopw   0x0(%rax,%rax,1)
L31:
  lea    -0xa0(%rcx),%rax
  nop
L97:
  movsd  (%rdi,%rax,1),%xmm0
  mulsd  (%rdx,%rax,1),%xmm0
  movsd  %xmm0,(%rsi,%rax,1)
  add    $0x8,%rax
  cmp    %rax,%rcx
  jne    L97
  add    $0xa0,%rcx
  cmp    %rcx,%r8
  jne    L31
  pop    %rbx
  ret
```

assistant: ```c
void sumGTRCATPROT(int tipCase, double *sumtable, double *x1, double *x2, double *tipVector, unsigned char *tipX1, unsigned char *tipX2, int n) {
  int i, l;
  double *sum, *left, *right;
  switch (tipCase) {
  case 0:
    for (i = 0; i < n; i++) {
      left = &(tipVector[20 * tipX1[i]]);
      right = &(tipVector[20 * tipX2[i]]);
      sum = &sumtable[20 * i];
      for (l = 0; l < 20; l++)
        sum[l] = left[l] * right[l];
    }
    break;
  case 1:
    for (i = 0; i < n; i++) {
      left = &(tipVector[20 * tipX1[i]]);
      right = &x2[20 * i];
      sum = &sumtable[20 * i];
      for (l = 0; l < 20; l++)
        sum[l] = left[l] * right[l];
    }
    break;
  case 2:
    for (i = 0; i < n; i++) {
      left = &x1[20 * i];
      right = &x2[20 * i];
      sum = &sumtable[20 * i];
      for (l = 0; l < 20; l++)
        sum[l] = left[l] * right[l];
    }
    break;
  default:
    assert(0);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<readAllWords>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %rcx,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x10(%rbp)
  lea    -0x18(%rbp),%rdx
  lea    -0x20(%rbp),%rcx
  mov    -0x38(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <readInGulp@plt>
  mov    -0x20(%rbp),%rax
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <chopByWhite@plt>
  mov    %eax,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  je     L14
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <needMem@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    -0x24(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <chopByWhite@plt>
L14:
  mov    -0x40(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x48(%rbp),%rax
  mov    -0x24(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rdx
  mov    -0x50(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
void readAllWords(char *fileName, char ***retWords, int *retWordCount, char **retBuf) {
  int wordCount;
  char *buf = NULL;
  char **words = NULL;
  size_t bufSize;
  readInGulp(fileName, &buf, &bufSize);
  wordCount = chopByWhite(buf, NULL, 0);
  if (wordCount != 0) {
    words = needMem(wordCount * sizeof(words[0]));
    chopByWhite(buf, words, wordCount);
  }
  *retWords = words;
  *retWordCount = wordCount;
  *retBuf = buf;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<readAllWords>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rsi,%r13
  mov    %rdx,%r12
  mov    %rcx,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x8(%rsp)
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  call   <readInGulp@plt>
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    0x8(%rsp),%rdi
  call   <chopByWhite@plt>
  mov    %eax,%ebx
  mov    $0x0,%r14d
  test   %eax,%eax
  jne    L2
L19:
  mov    %r14,0x0(%r13)
  mov    %ebx,(%r12)
  mov    0x8(%rsp),%rax
  mov    %rax,0x0(%rbp)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  lea    0x0(,%rax,8),%edi
  call   <needMem@plt>
  mov    %rax,%r14
  mov    %ebx,%edx
  mov    %rax,%rsi
  mov    0x8(%rsp),%rdi
  call   <chopByWhite@plt>
  jmp    L19
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void readAllWords(char *fileName, char ***retWords, int *retWordCount, char **retBuf) {
  int wordCount;
  char *buf = NULL;
  char **words = NULL;
  size_t bufSize;
  readInGulp(fileName, &buf, &bufSize);
  wordCount = chopByWhite(buf, NULL, 0);
  if (wordCount != 0) {
    words = needMem(wordCount * sizeof(words[0]));
    chopByWhite(buf, words, wordCount);
  }
  *retWords = words;
  *retWordCount = wordCount;
  *retBuf = buf;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<readAllWords>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  movq   $0x0,0x8(%rsp)
  call   <readInGulp@plt>
  mov    0x8(%rsp),%rdi
  xor    %edx,%edx
  xor    %esi,%esi
  call   <chopByWhite@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L14
L10:
  mov    0x8(%rsp),%rax
  mov    %r14,0x0(%r13)
  mov    %ebx,(%r12)
  mov    %rax,0x0(%rbp)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L14:
  lea    0x0(,%rax,8),%edi
  call   <needMem@plt>
  mov    0x8(%rsp),%rdi
  mov    %ebx,%edx
  mov    %rax,%rsi
  mov    %rax,%r14
  call   <chopByWhite@plt>
  jmp    L10
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void readAllWords(char *fileName, char ***retWords, int *retWordCount, char **retBuf) {
  int wordCount;
  char *buf = NULL;
  char **words = NULL;
  size_t bufSize;
  readInGulp(fileName, &buf, &bufSize);
  wordCount = chopByWhite(buf, NULL, 0);
  if (wordCount != 0) {
    words = needMem(wordCount * sizeof(words[0]));
    chopByWhite(buf, words, wordCount);
  }
  *retWords = words;
  *retWordCount = wordCount;
  *retBuf = buf;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<readAllWords>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  movq   $0x0,0x8(%rsp)
  call   <readInGulp@plt>
  mov    0x8(%rsp),%rdi
  xor    %edx,%edx
  xor    %esi,%esi
  call   <chopByWhite@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L22
L19:
  mov    0x8(%rsp),%rax
  mov    %r14,0x0(%r13)
  mov    %ebx,(%r12)
  mov    %rax,0x0(%rbp)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L22:
  lea    0x0(,%rax,8),%edi
  call   <needMem@plt>
  mov    0x8(%rsp),%rdi
  mov    %ebx,%edx
  mov    %rax,%rsi
  mov    %rax,%r14
  call   <chopByWhite@plt>
  jmp    L19
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void readAllWords(char *fileName, char ***retWords, int *retWordCount, char **retBuf) {
  int wordCount;
  char *buf = NULL;
  char **words = NULL;
  size_t bufSize;
  readInGulp(fileName, &buf, &bufSize);
  wordCount = chopByWhite(buf, NULL, 0);
  if (wordCount != 0) {
    words = needMem(wordCount * sizeof(words[0]));
    chopByWhite(buf, words, wordCount);
  }
  *retWords = words;
  *retWordCount = wordCount;
  *retBuf = buf;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_ethernet_speed>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    D20(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x18(%rbp)
  movq   $0x0,-0x28(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %eax,-0x24(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x1,%rax
  cmp    $0x4,%rax
  ja     L63
  mov    -0x30(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x28(%rbp),%eax
  mov    -0x40(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <memcpy@plt>
  mov    D11(%rip),%rax
  mov    (%rax),%ecx
  lea    -0x28(%rbp),%rdx
  mov    -0x34(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <ioctl@plt>
  test   %rax,%rax
  js     L1
  mov    -0x20(%rbp),%rdx
  mov    D6(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L40
  mov    -0x20(%rbp),%rax
  jmp    L11
L63:
  nop
  jmp    L27
L1:
  nop
  jmp    L27
L40:
  nop
L27:
  mov    $0x0,%eax
L11:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
int get_ethernet_speed(int sockfd, const char *iface) {
  struct ethtool_cmd ecmd = {.cmd = ETHTOOL_GSET};
  struct ifreq ifr = {.ifr_data = (caddr_t)&ecmd};
  size_t niface = strlen(iface);
  if (niface + 1 > sizeof(ifr.ifr_name))
    goto fail;
  memcpy(ifr.ifr_name, iface, niface + 1);
  if (ioctl(sockfd, SIOCETHTOOL, &ifr) < 0)
    goto fail;
  if (ecmd.speed == USHRT_MAX)
    goto fail;
  return ecmd.speed;
fail:
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<get_ethernet_speed>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,%ebp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x18(%rsp)
  movl   $0x0,0x8(%rsp)
  lea    0x10(%rsp),%rax
  mov    %eax,0xc(%rsp)
  mov    %rsi,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdx
  mov    $0x0,%eax
  cmp    $0x4,%rdx
  jbe    L30
L16:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L30:
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <memcpy@plt>
  lea    0x8(%rsp),%rdx
  mov    D6(%rip),%rax
  mov    (%rax),%esi
  mov    %ebp,%edi
  call   <ioctl@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  js     L16
  mov    0x10(%rsp),%rax
  mov    D2(%rip),%rdx
  cmp    (%rdx),%rax
  mov    $0x0,%edx
  cmove  %edx,%eax
  jmp    L16
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_ethernet_speed(int sockfd, const char *iface) {
  struct ethtool_cmd ecmd = {.cmd = ETHTOOL_GSET};
  struct ifreq ifr = {.ifr_data = (caddr_t)&ecmd};
  size_t niface = strlen(iface);
  if (niface + 1 > sizeof(ifr.ifr_name))
    goto fail;
  memcpy(ifr.ifr_name, iface, niface + 1);
  if (ioctl(sockfd, SIOCETHTOOL, &ifr) < 0)
    goto fail;
  if (ecmd.speed == USHRT_MAX)
    goto fail;
  return ecmd.speed;
fail:
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<get_ethernet_speed>:
  endbr64
  push   %r12
  pxor   %xmm0,%xmm0
  mov    %edi,%r12d
  mov    %rsi,%rdi
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D14(%rip),%rax
  movaps %xmm0,0x10(%rsp)
  movl   $0x0,0x8(%rsp)
  mov    (%rax),%eax
  mov    %eax,0x18(%rsp)
  lea    0x10(%rsp),%rax
  mov    %eax,0xc(%rsp)
  call   <strlen@plt>
  lea    0x1(%rax),%rdx
  cmp    $0x4,%rdx
  jbe    L47
L0:
  xor    %eax,%eax
L39:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x38,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L47:
  xor    %edi,%edi
  mov    %rbp,%rsi
  call   <memcpy@plt>
  mov    D19(%rip),%rax
  lea    0x8(%rsp),%rdx
  mov    %r12d,%edi
  mov    (%rax),%esi
  call   <ioctl@plt>
  test   %rax,%rax
  js     L0
  mov    D13(%rip),%rdx
  mov    0x10(%rsp),%rax
  cmp    (%rdx),%rax
  jne    L39
  jmp    L0
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_ethernet_speed(int sockfd, const char *iface) {
  struct ethtool_cmd ecmd = {.cmd = ETHTOOL_GSET};
  struct ifreq ifr = {.ifr_data = (caddr_t)&ecmd};
  size_t niface = strlen(iface);
  if (niface + 1 > sizeof(ifr.ifr_name))
    goto fail;
  memcpy(ifr.ifr_name, iface, niface + 1);
  if (ioctl(sockfd, SIOCETHTOOL, &ifr) < 0)
    goto fail;
  if (ecmd.speed == USHRT_MAX)
    goto fail;
  return ecmd.speed;
fail:
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_ethernet_speed>:
  endbr64
  push   %r12
  pxor   %xmm0,%xmm0
  mov    %edi,%r12d
  mov    %rsi,%rdi
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D9(%rip),%rax
  movaps %xmm0,0x10(%rsp)
  movl   $0x0,0x8(%rsp)
  mov    (%rax),%eax
  mov    %eax,0x18(%rsp)
  lea    0x10(%rsp),%rax
  mov    %eax,0xc(%rsp)
  call   <strlen@plt>
  lea    0x1(%rax),%rdx
  cmp    $0x4,%rdx
  jbe    L3
L13:
  xor    %eax,%eax
L26:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L27
  add    $0x38,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L3:
  xor    %edi,%edi
  mov    %rbp,%rsi
  call   <memcpy@plt>
  mov    D19(%rip),%rax
  lea    0x8(%rsp),%rdx
  mov    %r12d,%edi
  mov    (%rax),%esi
  call   <ioctl@plt>
  test   %rax,%rax
  js     L13
  mov    D21(%rip),%rdx
  mov    0x10(%rsp),%rax
  cmp    (%rdx),%rax
  jne    L26
  jmp    L13
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_ethernet_speed(int sockfd, const char *iface) {
  struct ethtool_cmd ecmd = {.cmd = ETHTOOL_GSET};
  struct ifreq ifr = {.ifr_data = (caddr_t)&ecmd};
  size_t niface = strlen(iface);
  if (niface + 1 > sizeof(ifr.ifr_name))
    goto fail;
  memcpy(ifr.ifr_name, iface, niface + 1);
  if (ioctl(sockfd, SIOCETHTOOL, &ifr) < 0)
    goto fail;
  if (ecmd.speed == USHRT_MAX)
    goto fail;
  return ecmd.speed;
fail:
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<write_to_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %rsi,-0x10(%rbp)
  mov    %al,-0x4(%rbp)
  movzbl -0x4(%rbp),%eax
  cmp    $0xf,%eax
  ja     L0
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D0(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D0(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    -0x10(%rbp),%rax
  movb   $0x30,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x31,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x32,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x33,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x34,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x35,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x36,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x37,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x38,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x39,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x41,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x42,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x43,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x44,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x45,(%rax)
  jmp    L106
  mov    -0x10(%rbp),%rax
  movb   $0x46,(%rax)
  jmp    L106
L0:
  mov    -0x10(%rbp),%rax
  movb   $0x3f,(%rax)
  nop
L106:
  mov    -0x10(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *write_to_str(unsigned char hex, char *dest) {
  switch (hex) {
  case 0x00:
    *dest = '0';
    break;
  case 0x01:
    *dest = '1';
    break;
  case 0x02:
    *dest = '2';
    break;
  case 0x03:
    *dest = '3';
    break;
  case 0x04:
    *dest = '4';
    break;
  case 0x05:
    *dest = '5';
    break;
  case 0x06:
    *dest = '6';
    break;
  case 0x07:
    *dest = '7';
    break;
  case 0x08:
    *dest = '8';
    break;
  case 0x09:
    *dest = '9';
    break;
  case 0x0a:
    *dest = 'A';
    break;
  case 0x0b:
    *dest = 'B';
    break;
  case 0x0c:
    *dest = 'C';
    break;
  case 0x0d:
    *dest = 'D';
    break;
  case 0x0e:
    *dest = 'E';
    break;
  case 0x0f:
    *dest = 'F';
    break;
  default:
    *dest = '?';
    break;
  }
  return dest;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<write_to_str>:
  endbr64
  mov    %rsi,%rax
  cmp    $0xf,%dil
  ja     L56
  movzbl %dil,%edi
  lea    D1(%rip),%rcx
  movslq (%rcx,%rdi,4),%rdx
  add    %rcx,%rdx
  notrack jmp *%rdx
  mov    $0x30,%edx
L129:
  mov    %dl,(%rax)
  ret
  mov    $0x32,%edx
  jmp    L129
  mov    $0x33,%edx
  jmp    L129
  mov    $0x34,%edx
  jmp    L129
  mov    $0x35,%edx
  jmp    L129
  mov    $0x36,%edx
  jmp    L129
  mov    $0x37,%edx
  jmp    L129
  mov    $0x38,%edx
  jmp    L129
  mov    $0x39,%edx
  jmp    L129
  mov    $0x41,%edx
  jmp    L129
  mov    $0x42,%edx
  jmp    L129
  mov    $0x43,%edx
  jmp    L129
  mov    $0x44,%edx
  jmp    L129
  mov    $0x45,%edx
  jmp    L129
  mov    $0x46,%edx
  jmp    L129
L56:
  mov    $0x3f,%edx
  jmp    L129
  mov    $0x31,%edx
  jmp    L129
```

assistant: ```c
char *write_to_str(unsigned char hex, char *dest) {
  switch (hex) {
  case 0x00:
    *dest = '0';
    break;
  case 0x01:
    *dest = '1';
    break;
  case 0x02:
    *dest = '2';
    break;
  case 0x03:
    *dest = '3';
    break;
  case 0x04:
    *dest = '4';
    break;
  case 0x05:
    *dest = '5';
    break;
  case 0x06:
    *dest = '6';
    break;
  case 0x07:
    *dest = '7';
    break;
  case 0x08:
    *dest = '8';
    break;
  case 0x09:
    *dest = '9';
    break;
  case 0x0a:
    *dest = 'A';
    break;
  case 0x0b:
    *dest = 'B';
    break;
  case 0x0c:
    *dest = 'C';
    break;
  case 0x0d:
    *dest = 'D';
    break;
  case 0x0e:
    *dest = 'E';
    break;
  case 0x0f:
    *dest = 'F';
    break;
  default:
    *dest = '?';
    break;
  }
  return dest;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<bar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0x10(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,%r10
  mov    $0x0,%edi
  call   <foo.0>
  cmp    $0x2,%eax
  je     L34
  call   <abort@plt>
L34:
  lea    -0x20(%rbp),%rax
  mov    %rax,%r10
  mov    $0x0,%edi
  call   <foo.0>
  cmp    $0x7,%eax
  je     L9
  call   <abort@plt>
L9:
  mov    -0x20(%rbp),%eax
  cmp    $0xb,%eax
  je     L1
  call   <abort@plt>
L1:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
void bar(int N) {
  int foo(char a[2][++N]) {
    N += 4;
    return sizeof(a[0]);
  }
  if (foo(0) != 2)
    abort();
  if (foo(0) != 7)
    abort();
  if (N != 11)
    abort();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bar>:
  endbr64
  cmp    $0x1,%edi
  jne    L7
  ret
L7:
  sub    $0x8,%rsp
  call   <abort@plt>
```

assistant: ```c
void bar(int N) {
  int foo(char a[2][++N]) {
    N += 4;
    return sizeof(a[0]);
  }
  if (foo(0) != 2)
    abort();
  if (foo(0) != 7)
    abort();
  if (N != 11)
    abort();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bar>:
  endbr64
  cmp    $0x1,%edi
  jne    1060 <bar.cold>
  ret
```

assistant: ```c
void bar(int N) {
  int foo(char a[2][++N]) {
    N += 4;
    return sizeof(a[0]);
  }
  if (foo(0) != 2)
    abort();
  if (foo(0) != 7)
    abort();
  if (N != 11)
    abort();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<bar>:
  endbr64
  cmp    $0x1,%edi
  jne    1060 <bar.cold>
  ret
```

assistant: ```c
void bar(int N) {
  int foo(char a[2][++N]) {
    N += 4;
    return sizeof(a[0]);
  }
  if (foo(0) != 2)
    abort();
  if (foo(0) != 7)
    abort();
  if (N != 11)
    abort();
}
```