ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<util_getdate>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,0x4(%rsp)
  lea    0x4(%rsp),%rdi
  call   <localtime@plt>
  mov    %eax,%ecx
  lea    D1(%rip),%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <strftime@plt>
  test   %rbx,%rbx
  je     L3
  test   %rax,%rax
  je     L1
L3:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  movb   $0x0,0x0(%rbp)
  jmp    L3
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D1
data_type: string

tool: D1: %c

assistant: ```c
void util_getdate(char *s, size_t max) {
  time_t t = time(NULL);
  if (strftime(s, max, "%c", localtime(&t)) == 0 && max > 0)
    *s = '\0';
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<util_getdate>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <time@plt>
  lea    0x4(%rsp),%rdi
  mov    %eax,0x4(%rsp)
  call   <localtime@plt>
  lea    D7(%rip),%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  mov    %eax,%ecx
  call   <strftime@plt>
  test   %rbx,%rbx
  je     L17
  test   %rax,%rax
  je     L20
L17:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L20:
  movb   $0x0,0x0(%rbp)
  jmp    L17
L36:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D7
datatype: string

tool: D7: %c

assistant: ```c
void util_getdate(char *s, size_t max) {
  time_t t = time(NULL);
  if (strftime(s, max, "%c", localtime(&t)) == 0 && max > 0)
    *s = '\0';
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<util_getdate>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <time@plt>
  lea    0x4(%rsp),%rdi
  mov    %eax,0x4(%rsp)
  call   <localtime@plt>
  lea    D7(%rip),%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  mov    %eax,%ecx
  call   <strftime@plt>
  test   %rbx,%rbx
  je     L28
  test   %rax,%rax
  je     L5
L28:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  movb   $0x0,0x0(%rbp)
  jmp    L28
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
type: string

tool: D7: %c

assistant: ```c
void util_getdate(char *s, size_t max) {
  time_t t = time(NULL);
  if (strftime(s, max, "%c", localtime(&t)) == 0 && max > 0)
    *s = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<qsort>:
  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    -0x38(%rbp),%rax
  mov    %eax,-0x24(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L58
L33:
  mov    -0x24(%rbp),%edx
  mov    -0x2c(%rbp),%eax
  cltq
  imul   -0x48(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x2c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x28(%rbp)
  jmp    L32
L39:
  mov    -0x24(%rbp),%edx
  mov    -0x28(%rbp),%eax
  cltq
  imul   -0x48(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    -0x50(%rbp),%rcx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   *%rcx
  test   %eax,%eax
  jns    L2
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x20(%rbp)
L2:
  addl   $0x1,-0x28(%rbp)
L32:
  mov    -0x28(%rbp),%eax
  cltq
  cmp    %rax,-0x40(%rbp)
  ja     L39
  mov    -0x20(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  je     L31
  mov    -0x48(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x48(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x48(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L31:
  addl   $0x1,-0x2c(%rbp)
L58:
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x40(%rbp),%rdx
  sub    $0x1,%rdx
  cmp    %rdx,%rax
  jb     L33
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

assistant: ```c
void qsort(void *base, size_t n, size_t size, int (*cmp)(const void *, const void *)) {
  void *a, *b, *swap;
  int i, j;
  unsigned int pos = (unsigned int)base;
  void *tmp = malloc(size);
  for (i = 0; i < (n - 1); i++) {
    a = (void *)(pos + i * size);
    swap = a;
    for (j = i + 1; j < n; j++) {
      b = (void *)(pos + j * size);
      if (cmp(b, swap) < 0)
        swap = b;
    }
    if (swap != a) {
      memcpy(tmp, a, size);
      memcpy(a, swap, size);
      memcpy(swap, tmp, size);
    }
  }
  free(tmp);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<qsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r13
  mov    %rdx,0x8(%rsp)
  mov    %rcx,%r14
  mov    %rdx,%rdi
  call   <malloc@plt>
  mov    %rax,0x20(%rsp)
  cmp    $0x1,%r13
  je     L41
  mov    %ebx,%eax
  mov    %rax,0x10(%rsp)
  mov    $0x1,%r15d
  mov    %rax,0x28(%rsp)
  jmp    L68
L40:
  add    $0x1,%r15
  mov    0x8(%rsp),%rcx
  add    %rcx,0x10(%rsp)
  cmp    %r15,%r13
  je     L41
L68:
  mov    0x10(%rsp),%rbp
  mov    %rbp,0x18(%rsp)
  cmp    %r15,%r13
  jbe    L40
  movslq %r15d,%r12
  mov    0x8(%rsp),%rbx
  imul   %r12,%rbx
  add    0x28(%rsp),%rbx
L44:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   *%r14
  test   %eax,%eax
  cmovs  %rbx,%rbp
  add    $0x1,%r12
  add    0x8(%rsp),%rbx
  cmp    %r13,%r12
  jb     L44
  mov    0x18(%rsp),%rbx
  cmp    %rbx,%rbp
  je     L40
  mov    0x8(%rsp),%r12
  mov    %r12,%rdx
  mov    %rbx,%rsi
  mov    0x20(%rsp),%rdi
  call   <memcpy@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    %r12,%rdx
  mov    0x20(%rsp),%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  jmp    L40
L41:
  mov    0x20(%rsp),%rdi
  call   <free@plt>
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void qsort(void *base, size_t n, size_t size, int (*cmp)(const void *, const void *)) {
  void *a, *b, *swap;
  int i, j;
  unsigned int pos = (unsigned int)base;
  void *tmp = malloc(size);
  for (i = 0; i < (n - 1); i++) {
    a = (void *)(pos + i * size);
    swap = a;
    for (j = i + 1; j < n; j++) {
      b = (void *)(pos + j * size);
      if (cmp(b, swap) < 0)
        swap = b;
    }
    if (swap != a) {
      memcpy(tmp, a, size);
      memcpy(a, swap, size);
      memcpy(swap, tmp, size);
    }
  }
  free(tmp);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<qsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  mov    %rdx,%rdi
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x28,%rsp
  call   <malloc@plt>
  movq   $0x1,(%rsp)
  mov    %rax,0x18(%rsp)
  mov    %r13d,%eax
  mov    %rax,0x10(%rsp)
  mov    %rax,0x8(%rsp)
  cmp    $0x1,%rbp
  je     L27
  nopw   0x0(%rax,%rax,1)
L7:
  mov    (%rsp),%r13
  cmp    %r13,%rbp
  jbe    L28
  mov    %r13,%rdx
  mov    0x10(%rsp),%rax
  mov    0x8(%rsp),%r15
  imul   %rbx,%rdx
  lea    (%rdx,%rax,1),%r14
  xchg   %ax,%ax
L48:
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   *%r12
  test   %eax,%eax
  cmovs  %r14,%r15
  add    $0x1,%r13
  add    %rbx,%r14
  cmp    %r13,%rbp
  ja     L48
  mov    0x8(%rsp),%r14
  cmp    %r14,%r15
  je     L28
  mov    0x18(%rsp),%r13
  mov    %rbx,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rdx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <memcpy@plt>
L28:
  addq   $0x1,(%rsp)
  mov    (%rsp),%rax
  add    %rbx,0x8(%rsp)
  cmp    %rax,%rbp
  jne    L7
L27:
  mov    0x18(%rsp),%rdi
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <free@plt>
```

assistant: ```c
void qsort(void *base, size_t n, size_t size, int (*cmp)(const void *, const void *)) {
  void *a, *b, *swap;
  int i, j;
  unsigned int pos = (unsigned int)base;
  void *tmp = malloc(size);
  for (i = 0; i < (n - 1); i++) {
    a = (void *)(pos + i * size);
    swap = a;
    for (j = i + 1; j < n; j++) {
      b = (void *)(pos + j * size);
      if (cmp(b, swap) < 0)
        swap = b;
    }
    if (swap != a) {
      memcpy(tmp, a, size);
      memcpy(a, swap, size);
      memcpy(swap, tmp, size);
    }
  }
  free(tmp);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<qsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  mov    %rdx,%rdi
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x28,%rsp
  call   <malloc@plt>
  movq   $0x1,(%rsp)
  mov    %rax,0x18(%rsp)
  mov    %r13d,%eax
  mov    %rax,0x10(%rsp)
  mov    %rax,0x8(%rsp)
  cmp    $0x1,%rbp
  je     L0
  nopw   0x0(%rax,%rax,1)
L22:
  mov    (%rsp),%r13
  cmp    %rbp,%r13
  jae    L19
  mov    %r13,%rdx
  mov    0x10(%rsp),%rax
  mov    0x8(%rsp),%r15
  imul   %rbx,%rdx
  lea    (%rdx,%rax,1),%r14
  xchg   %ax,%ax
L18:
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   *%r12
  test   %eax,%eax
  cmovs  %r14,%r15
  add    $0x1,%r13
  add    %rbx,%r14
  cmp    %r13,%rbp
  ja     L18
  mov    0x8(%rsp),%r14
  cmp    %r14,%r15
  je     L19
  mov    0x18(%rsp),%r13
  mov    %rbx,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rdx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <memcpy@plt>
L19:
  addq   $0x1,(%rsp)
  mov    (%rsp),%rax
  add    %rbx,0x8(%rsp)
  cmp    %rbp,%rax
  jne    L22
L0:
  mov    0x18(%rsp),%rdi
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <free@plt>
```

assistant: ```c
void qsort(void *base, size_t n, size_t size, int (*cmp)(const void *, const void *)) {
  void *a, *b, *swap;
  int i, j;
  unsigned int pos = (unsigned int)base;
  void *tmp = malloc(size);
  for (i = 0; i < (n - 1); i++) {
    a = (void *)(pos + i * size);
    swap = a;
    for (j = i + 1; j < n; j++) {
      b = (void *)(pos + j * size);
      if (cmp(b, swap) < 0)
        swap = b;
    }
    if (swap != a) {
      memcpy(tmp, a, size);
      memcpy(a, swap, size);
      memcpy(swap, tmp, size);
    }
  }
  free(tmp);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<HexToDec>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r13
  mov    %rdx,%r14
  mov    %rcx,%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x8(%rsp)
  movl   $0x0,0xc(%rsp)
  movl   $0x0,0x10(%rsp)
  mov    $0x0,%edi
  lea    0x16(%rsp),%r12
  lea    D8(%rip),%rbx
  jmp    L16
  mov    $0xa,%r11d
  jmp    L32
  movsbl %r10b,%r10d
  lea    -0x30(%r10),%r11d
L32:
  test   %eax,%eax
  je     L74
  imul   %r11d,%ecx
  shl    $0x4,%ecx
  add    %ecx,%esi
  cmp    $0x1,%r8d
  jle    L91
  movslq %edx,%rdx
  mov    %esi,0x8(%rsp,%rdx,4)
  add    $0x2,%rdi
  cmp    $0x6,%rdi
  je     L61
L16:
  movzbl 0x1(%rbp,%rdi,1),%eax
  mov    %al,0x15(%rsp)
  movzbl 0x2(%rbp,%rdi,1),%eax
  mov    %al,0x16(%rsp)
  movb   $0x0,0x17(%rsp)
  mov    %edi,%edx
  shr    $0x1f,%edx
  add    %edi,%edx
  sar    %edx
  movslq %edx,%rax
  mov    0x8(%rsp,%rax,4),%esi
  mov    $0x1,%r8d
  mov    $0x0,%eax
  jmp    L60
  mov    $0xb,%r11d
  jmp    L32
  mov    $0xc,%r11d
  jmp    L32
  mov    $0xd,%r11d
  jmp    L32
  mov    $0xe,%r11d
  jmp    L32
  mov    $0xf,%r11d
  jmp    L32
L74:
  add    %r11d,%esi
L91:
  add    $0x1,%rax
  add    $0x1,%r8d
L60:
  mov    %eax,%ecx
  mov    %r12,%r9
  sub    %rax,%r9
  movzbl (%r9),%r10d
  lea    -0x30(%r10),%r9d
  cmp    $0x36,%r9b
  ja     L32
  movzbl %r9b,%r9d
  movslq (%rbx,%r9,4),%r9
  add    %rbx,%r9
  notrack jmp *%r9
L61:
  mov    0x8(%rsp),%eax
  mov    %eax,0x0(%r13)
  mov    0xc(%rsp),%eax
  mov    %eax,(%r14)
  mov    0x10(%rsp),%eax
  mov    %eax,(%r15)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L124
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L124:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void HexToDec(char *hRGB, int *dR, int *dG, int *dB) {
  char tempCh[3];
  int i, j, rgb[3], tempI;
  for (i = 0; i < 3; i++) {
    rgb[i] = 0;
  }
  for (i = 1; i < 7; i += 2) {
    for (j = 0; j < 2; j++)
      tempCh[j] = hRGB[i + j];
    tempCh[2] = '\0';
    for (j = 0; j < 2; j++) {
      switch (tempCh[1 - j]) {
      case '0':
      case '1':
      case '2':
      case '3':
      case '4':
      case '5':
      case '6':
      case '7':
      case '8':
      case '9':
        tempI = tempCh[1 - j] - '0';
        break;
      case 'A':
      case 'a':
        tempI = 10;
        break;
      case 'B':
      case 'b':
        tempI = 11;
        break;
      case 'C':
      case 'c':
        tempI = 12;
        break;
      case 'D':
      case 'd':
        tempI = 13;
        break;
      case 'E':
      case 'e':
        tempI = 14;
        break;
      case 'F':
      case 'f':
        tempI = 15;
        break;
      }
      if (j == 0)
        rgb[(i - 1) / 2] += tempI;
      else
        rgb[(i - 1) / 2] += tempI * 16 * j;
    }
  }
  *dR = rgb[0];
  *dG = rgb[1];
  *dB = rgb[2];
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<HexToDec>:
  endbr64
  push   %r14
  lea    D8(%rip),%r11
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %rcx,%rbp
  xor    %ecx,%ecx
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x10(%rsp)
  lea    0x15(%rsp),%r14
  movq   $0x0,0x8(%rsp)
L109:
  movzbl 0x2(%rbx,%rcx,1),%edi
  movzbl 0x1(%rbx,%rcx,1),%eax
  mov    %ecx,%r9d
  mov    %r14,%r8
  sar    %r9d
  mov    %dil,0x16(%rsp)
  mov    %al,0x15(%rsp)
  xor    %eax,%eax
L99:
  lea    -0x30(%rdi),%edx
  cmp    $0x36,%dl
  ja     L21
  movzbl %dl,%edx
  movslq (%r11,%rdx,4),%rdx
  add    %r11,%rdx
  notrack jmp *%rdx
  sub    $0x30,%edi
  movsbl %dil,%r10d
  nopw   0x0(%rax,%rax,1)
L21:
  add    $0x1,%eax
  cmp    $0x1,%eax
  je     L11
L83:
  mov    %r10d,%edx
  shl    $0x4,%edx
  add    %edx,%esi
  cmp    $0x2,%eax
  jne    L24
  movslq %r9d,%r9
  add    $0x2,%rcx
  mov    %esi,0x8(%rsp,%r9,4)
  cmp    $0x6,%rcx
  je     L39
  mov    %ecx,%eax
  sar    %eax
  cltq
  mov    0x8(%rsp,%rax,4),%esi
  jmp    L109
  nopl   0x0(%rax,%rax,1)
  add    $0x1,%eax
  mov    $0xf,%r10d
  cmp    $0x1,%eax
  jne    L83
  cs nopw 0x0(%rax,%rax,1)
L11:
  add    %r10d,%esi
L24:
  movzbl (%r8),%edi
  sub    $0x1,%r8
  jmp    L99
  mov    $0xe,%r10d
  jmp    L21
  nopl   0x0(%rax,%rax,1)
  mov    $0xd,%r10d
  jmp    L21
  nopl   0x0(%rax,%rax,1)
  mov    $0xc,%r10d
  jmp    L21
  nopl   0x0(%rax,%rax,1)
  mov    $0xb,%r10d
  jmp    L21
  nopl   0x0(%rax,%rax,1)
L39:
  mov    0x8(%rsp),%eax
  mov    %eax,(%r12)
  mov    0xc(%rsp),%eax
  mov    %eax,0x0(%r13)
  mov    0x10(%rsp),%eax
  mov    %eax,0x0(%rbp)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
  mov    $0xa,%r10d
  jmp    L21
L45:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void HexToDec(char *hRGB, int *dR, int *dG, int *dB) {
  char tempCh[3];
  int i, j, rgb[3], tempI;
  for (i = 0; i < 3; i++) {
    rgb[i] = 0;
  }
  for (i = 1; i < 7; i += 2) {
    for (j = 0; j < 2; j++)
      tempCh[j] = hRGB[i + j];
    tempCh[2] = '\0';
    for (j = 0; j < 2; j++) {
      switch (tempCh[1 - j]) {
      case '0':
      case '1':
      case '2':
      case '3':
      case '4':
      case '5':
      case '6':
      case '7':
      case '8':
      case '9':
        tempI = tempCh[1 - j] - '0';
        break;
      case 'A':
      case 'a':
        tempI = 10;
        break;
      case 'B':
      case 'b':
        tempI = 11;
        break;
      case 'C':
      case 'c':
        tempI = 12;
        break;
      case 'D':
      case 'd':
        tempI = 13;
        break;
      case 'E':
      case 'e':
        tempI = 14;
        break;
      case 'F':
      case 'f':
        tempI = 15;
        break;
      }
      if (j == 0)
        rgb[(i - 1) / 2] += tempI;
      else
        rgb[(i - 1) / 2] += tempI * 16 * j;
    }
  }
  *dR = rgb[0];
  *dG = rgb[1];
  *dB = rgb[2];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strrev_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L29
L14:
  addq   $0x1,-0x8(%rbp)
L29:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L14
  subq   $0x1,-0x8(%rbp)
  jmp    L22
L15:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x11(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  movzbl -0x11(%rbp),%edx
  mov    %dl,(%rax)
  addq   $0x1,-0x10(%rbp)
  subq   $0x1,-0x8(%rbp)
L22:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L15
  mov    -0x28(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *strrev_(char *pStr) {
  assert(pStr != NULL);
  char *head = pStr;
  char *tail = pStr;
  char tmp;
  while (*tail != '\0')
    ++tail;
  for (--tail; head < tail; ++head, --tail) {
    tmp = *head;
    *head = *tail;
    *tail = tmp;
  }
  return pStr;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strrev_>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmpb   $0x0,(%rbx)
  je     L7
  mov    %rbx,%rax
L18:
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L18
L8:
  sub    $0x1,%rax
  cmp    %rax,%rbx
  jae    L9
  mov    %rbx,%rdx
L30:
  movzbl (%rdx),%ecx
  movzbl (%rax),%esi
  mov    %sil,(%rdx)
  mov    %cl,(%rax)
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %rax,%rdx
  jb     L30
L9:
  mov    %rbx,%rax
  pop    %rbx
  ret
L7:
  mov    %rbx,%rax
  jmp    L8
```

assistant: ```c
char *strrev_(char *pStr) {
  assert(pStr != NULL);
  char *head = pStr;
  char *tail = pStr;
  char tmp;
  while (*tail != '\0')
    ++tail;
  for (--tail; head < tail; ++head, --tail) {
    tmp = *head;
    *head = *tail;
    *tail = tmp;
  }
  return pStr;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strrev_>:
  endbr64
  push   %r12
  test   %rdi,%rdi
  mov    %rdi,%r12
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  movzbl (%r12),%ecx
  test   %cl,%cl
  je     L43
  mov    %r12,%rdx
  nopl   (%rax)
L15:
  mov    %rdx,%rax
  add    $0x1,%rdx
  cmpb   $0x0,(%rdx)
  jne    L15
L31:
  cmp    %r12,%rax
  jbe    L11
  mov    %r12,%rdx
  jmp    L51
  xchg   %ax,%ax
L47:
  movzbl (%rdx),%ecx
L51:
  movzbl (%rax),%esi
  add    $0x1,%rdx
  sub    $0x1,%rax
  mov    %sil,-0x1(%rdx)
  mov    %cl,0x1(%rax)
  cmp    %rax,%rdx
  jb     L47
L11:
  mov    %r12,%rax
  pop    %r12
  ret
L43:
  lea    -0x1(%r12),%rax
  jmp    L31
```

assistant: ```c
char *strrev_(char *pStr) {
  assert(pStr != NULL);
  char *head = pStr;
  char *tail = pStr;
  char tmp;
  while (*tail != '\0')
    ++tail;
  for (--tail; head < tail; ++head, --tail) {
    tmp = *head;
    *head = *tail;
    *tail = tmp;
  }
  return pStr;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strrev_>:
  endbr64
  test   %rdi,%rdi
  push   %r12
  mov    %rdi,%r12
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmpb   $0x0,(%r12)
  je     L46
  mov    %r12,%rdx
  nopl   0x0(%rax,%rax,1)
L12:
  mov    %rdx,%rax
  add    $0x1,%rdx
  cmpb   $0x0,(%rdx)
  jne    L12
L24:
  mov    %r12,%rdx
  cmp    %r12,%rax
  jbe    L32
  nopl   0x0(%rax)
L45:
  movzbl (%rdx),%ecx
  movzbl (%rax),%esi
  add    $0x1,%rdx
  sub    $0x1,%rax
  mov    %sil,-0x1(%rdx)
  mov    %cl,0x1(%rax)
  cmp    %rax,%rdx
  jb     L45
L32:
  mov    %r12,%rax
  pop    %r12
  ret
L46:
  lea    -0x1(%r12),%rax
  jmp    L24
```

assistant: ```c
char *strrev_(char *pStr) {
  assert(pStr != NULL);
  char *head = pStr;
  char *tail = pStr;
  char tmp;
  while (*tail != '\0')
    ++tail;
  for (--tail; head < tail; ++head, --tail) {
    tmp = *head;
    *head = *tail;
    *tail = tmp;
  }
  return pStr;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <runalltests_behaviour_skip_list@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  runalltests_behaviour_skip_list();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

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

assistant: ```c
int main(void) {
  runalltests_behaviour_skip_list();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

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

assistant: ```c
int main(void) {
  runalltests_behaviour_skip_list();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int main(void) {
  runalltests_behaviour_skip_list();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<__fdim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x18(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <fpclassify@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  call   <fpclassify@plt>
  mov    %eax,-0xc(%rbp)
  mov    D52(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x10(%rbp)
  je     L62
  mov    D52(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0xc(%rbp)
  jne    L23
L62:
  movsd  -0x18(%rbp),%xmm0
  subsd  -0x20(%rbp),%xmm0
  jmp    L39
L23:
  movsd  -0x20(%rbp),%xmm0
  comisd -0x18(%rbp),%xmm0
  jb     L1
  pxor   %xmm0,%xmm0
  jmp    L39
L1:
  movsd  -0x18(%rbp),%xmm0
  subsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <fpclassify@plt>
  mov    D38(%rip),%rdx
  mov    (%rdx),%edx
  cmp    %edx,%eax
  jne    L46
  mov    D38(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x10(%rbp)
  je     L46
  mov    D38(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0xc(%rbp)
  je     L46
  mov    D36(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <__set_errno@plt>
L46:
  movsd  -0x8(%rbp),%xmm0
L39:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double __fdim(double x, double y) {
  int clsx = fpclassify(x);
  int clsy = fpclassify(y);
  if (clsx == FP_NAN || clsy == FP_NAN)
    return x - y;
  if (x <= y)
    return 0.0;
  double r = x - y;
  if (fpclassify(r) == FP_INFINITE && clsx != FP_INFINITE && clsy != FP_INFINITE)
    __set_errno(ERANGE);
  return r;
}
```