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

```asm
<technology_get_type_str>:
  endbr64
  cmp    $0x2,%esi
  je     L22
  jg     L14
  test   %esi,%esi
  je     L45
  cmp    $0x1,%esi
  jne    L7
  lea    D16(%rip),%rdx
  mov    $0x80,%esi
  jmp    1050 <snprintf@plt>
  nopl   (%rax)
L14:
  cmp    $0x3,%esi
  jne    L46
  lea    D28(%rip),%rdx
  mov    $0x80,%esi
  jmp    1050 <snprintf@plt>
  xchg   %ax,%ax
L46:
  ret
  nopl   0x0(%rax)
L45:
  lea    D39(%rip),%rdx
  mov    $0x80,%esi
  jmp    1050 <snprintf@plt>
  nopl   0x0(%rax)
L22:
  lea    D17(%rip),%rdx
  mov    $0x80,%esi
  jmp    1050 <snprintf@plt>
L7:
  ret
```

tool_calls: name: read_data
arguments: tag: D16
type_guessed: string
name: read_data
arguments: tag: D28
type_guessed: string
name: read_data
arguments: tag: D39
type_guessed: string
name: read_data
arguments: tag: D17
type_guessed: string

tool: D16: Software

tool: D28: Drugs

tool: D39: Hardware

tool: D17: Ads

assistant: ```c
void technology_get_type_str(char *type_str, int type) {
  switch (type) {
  case 0:
    snprintf(type_str, 128, "Hardware");
    break;
  case 1:
    snprintf(type_str, 128, "Software");
    break;
  case 2:
    snprintf(type_str, 128, "Ads");
    break;
  case 3:
    snprintf(type_str, 128, "Drugs");
    break;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<technology_get_type_str>:
  endbr64
  cmp    $0x2,%esi
  je     L32
  jg     L19
  test   %esi,%esi
  je     L28
  cmp    $0x1,%esi
  jne    L31
  lea    D3(%rip),%rdx
  mov    $0x80,%esi
  jmp    1050 <snprintf@plt>
  nopl   (%rax)
L19:
  cmp    $0x3,%esi
  jne    L22
  lea    D9(%rip),%rdx
  mov    $0x80,%esi
  jmp    1050 <snprintf@plt>
  xchg   %ax,%ax
L22:
  ret
  nopl   0x0(%rax)
L28:
  lea    D17(%rip),%rdx
  mov    $0x80,%esi
  jmp    1050 <snprintf@plt>
  nopl   0x0(%rax)
L32:
  lea    D12(%rip),%rdx
  mov    $0x80,%esi
  jmp    1050 <snprintf@plt>
L31:
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D3
type_guessed: string
name: disassemble_data
arguments: data_label: D9
type_guessed: string
name: disassemble_data
arguments: data_label: D17
type_guessed: string
name: disassemble_data
arguments: data_label: D12
type_guessed: string

tool: D3: Software

tool: D9: Drugs

tool: D17: Hardware

tool: D12: Ads

assistant: ```c
void technology_get_type_str(char *type_str, int type) {
  switch (type) {
  case 0:
    snprintf(type_str, 128, "Hardware");
    break;
  case 1:
    snprintf(type_str, 128, "Software");
    break;
  case 2:
    snprintf(type_str, 128, "Ads");
    break;
  case 3:
    snprintf(type_str, 128, "Drugs");
    break;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<wp_str_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x1505,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L2
L12:
  mov    -0xc(%rbp),%eax
  shl    $0x5,%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  addq   $0x1,-0x8(%rbp)
L2:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L12
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int wp_str_hash(const void *v) {
  const signed char *p;
  unsigned int h = 5381;
  for (p = v; *p != '\0'; p++) {
    h = (h << 5) + h + *p;
  }
  return h;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<wp_str_hash>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L15
  mov    $0x1505,%eax
L8:
  mov    %eax,%ecx
  shl    $0x5,%ecx
  add    %ecx,%eax
  movsbl %dl,%edx
  add    %edx,%eax
  add    $0x1,%rdi
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L8
  ret
L15:
  mov    $0x1505,%eax
  ret
```

assistant: ```c
unsigned int wp_str_hash(const void *v) {
  const signed char *p;
  unsigned int h = 5381;
  for (p = v; *p != '\0'; p++) {
    h = (h << 5) + h + *p;
  }
  return h;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<wp_str_hash>:
  endbr64
  movsbl (%rdi),%edx
  mov    $0x1505,%eax
  test   %dl,%dl
  je     L18
L0:
  mov    %eax,%ecx
  add    $0x1,%rdi
  shl    $0x5,%ecx
  add    %ecx,%eax
  add    %edx,%eax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L0
  ret
  nopl   (%rax)
L18:
  ret
```

assistant: ```c
unsigned int wp_str_hash(const void *v) {
  const signed char *p;
  unsigned int h = 5381;
  for (p = v; *p != '\0'; p++) {
    h = (h << 5) + h + *p;
  }
  return h;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<wp_str_hash>:
  endbr64
  movsbl (%rdi),%edx
  mov    $0x1505,%eax
  test   %dl,%dl
  je     L19
L6:
  mov    %eax,%ecx
  add    $0x1,%rdi
  shl    $0x5,%ecx
  add    %ecx,%eax
  add    %edx,%eax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L6
  ret
  nopl   (%rax)
L19:
  ret
```

assistant: ```c
unsigned int wp_str_hash(const void *v) {
  const signed char *p;
  unsigned int h = 5381;
  for (p = v; *p != '\0'; p++) {
    h = (h << 5) + h + *p;
  }
  return h;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  sub    $0xa4d,%ax
  mov    %eax,%edx
  test   $0x1,%al
  je     L27
  lea    (%rax,%rax,1),%edx
L0:
  add    $0x6e28,%dx
  mov    %dx,(%rsi)
  ret
L27:
  test   $0x10,%al
  je     L0
  mov    %al,%dh
  movsbl %ah,%eax
  mov    %al,%dl
  jmp    L0
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  char copy11;
  char copy12;
  char copy13;
  {
    state[0UL] = (input[0UL] + 914778474UL) - (unsigned short)29623;
    if (state[0UL] & (unsigned short)1) {
      if (state[0UL] & (unsigned short)1) {
        if (state[0UL] & (unsigned short)1) {
          state[0UL] += state[0UL];
        } else {
          state[0UL] += state[0UL];
        }
      } else if ((state[0UL] >> (unsigned short)4) & (unsigned short)1) {
        copy11 = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = copy11;
      } else {
        copy12 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy12;
      }
    } else if ((state[0UL] >> (unsigned short)4) & (unsigned short)1) {
      copy13 = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = copy13;
    }
    output[0UL] = state[0UL] + (unsigned short)28200;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%edx
  lea    -0xa4d(%rdx),%eax
  and    $0x1,%edx
  jne    L5
  add    %eax,%eax
  add    $0x6e28,%ax
  mov    %ax,(%rsi)
  ret
  nopl   0x0(%rax)
L5:
  mov    %eax,%edx
  rol    $0x8,%dx
  test   $0x10,%al
  cmovne %edx,%eax
  add    $0x6e28,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  char copy11;
  char copy12;
  char copy13;
  {
    state[0UL] = (input[0UL] + 914778474UL) - (unsigned short)29623;
    if (state[0UL] & (unsigned short)1) {
      if (state[0UL] & (unsigned short)1) {
        if (state[0UL] & (unsigned short)1) {
          state[0UL] += state[0UL];
        } else {
          state[0UL] += state[0UL];
        }
      } else if ((state[0UL] >> (unsigned short)4) & (unsigned short)1) {
        copy11 = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = copy11;
      } else {
        copy12 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy12;
      }
    } else if ((state[0UL] >> (unsigned short)4) & (unsigned short)1) {
      copy13 = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = copy13;
    }
    output[0UL] = state[0UL] + (unsigned short)28200;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%edx
  lea    -0xa4d(%rdx),%eax
  and    $0x1,%edx
  jne    L9
  add    %eax,%eax
  add    $0x6e28,%ax
  mov    %ax,(%rsi)
  ret
  nopl   0x0(%rax)
L9:
  mov    %eax,%edx
  rol    $0x8,%dx
  test   $0x10,%al
  cmovne %edx,%eax
  add    $0x6e28,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  char copy11;
  char copy12;
  char copy13;
  {
    state[0UL] = (input[0UL] + 914778474UL) - (unsigned short)29623;
    if (state[0UL] & (unsigned short)1) {
      if (state[0UL] & (unsigned short)1) {
        if (state[0UL] & (unsigned short)1) {
          state[0UL] += state[0UL];
        } else {
          state[0UL] += state[0UL];
        }
      } else if ((state[0UL] >> (unsigned short)4) & (unsigned short)1) {
        copy11 = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = copy11;
      } else {
        copy12 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy12;
      }
    } else if ((state[0UL] >> (unsigned short)4) & (unsigned short)1) {
      copy13 = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = copy13;
    }
    output[0UL] = state[0UL] + (unsigned short)28200;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<parse_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L27
L36:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <isblank@plt>
  test   %rax,%rax
  je     L56
  addl   $0x1,-0x8(%rbp)
  jmp    L27
L56:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <isalpha@plt>
  test   %rax,%rax
  je     L27
  mov    $0xffffffff,%eax
  jmp    L17
L27:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L36
L17:
  leave
  ret
```

assistant: ```c
int parse_line(char *line) {
  int i = 0;
  int c = 0;
  while ((c = *line + i) != '\0') {
    if (isblank(c)) {
      i++;
    } else {
      if (isalpha(c)) {
        return -1;
      } else {
      }
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<parse_line>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  movsbl (%rdi),%ebx
  test   %ebx,%ebx
  je     L18
  mov    %rdi,%r12
  mov    $0x0,%ebp
  jmp    L55
L28:
  mov    %ebx,%edi
  call   <isalpha@plt>
  test   %rax,%rax
  jne    L56
L48:
  movsbl (%r12),%eax
  add    %ebp,%eax
  mov    %eax,%ebx
  je     L18
L55:
  mov    %ebx,%edi
  call   <isblank@plt>
  test   %rax,%rax
  je     L28
  add    $0x1,%ebp
  jmp    L48
L56:
  mov    $0xffffffff,%eax
L18:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int parse_line(char *line) {
  int i = 0;
  int c = 0;
  while ((c = *line + i) != '\0') {
    if (isblank(c)) {
      i++;
    } else {
      if (isalpha(c)) {
        return -1;
      } else {
      }
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<parse_line>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  movsbl (%rdi),%ebp
  test   %ebp,%ebp
  je     L38
  mov    %rdi,%r12
  xor    %ebx,%ebx
  jmp    L36
  cs nopw 0x0(%rax,%rax,1)
L11:
  add    $0x1,%ebx
L37:
  movsbl (%r12),%eax
  add    %ebx,%eax
  mov    %eax,%ebp
  je     L38
L36:
  mov    %ebp,%edi
  call   <isblank@plt>
  test   %rax,%rax
  jne    L11
  mov    %ebp,%edi
  call   <isalpha@plt>
  test   %rax,%rax
  je     L37
  mov    $0xffffffff,%eax
L38:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int parse_line(char *line) {
  int i = 0;
  int c = 0;
  while ((c = *line + i) != '\0') {
    if (isblank(c)) {
      i++;
    } else {
      if (isalpha(c)) {
        return -1;
      } else {
      }
    }
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<parse_line>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  movsbl (%rdi),%ebp
  test   %ebp,%ebp
  je     L16
  mov    %rdi,%r12
  xor    %ebx,%ebx
  jmp    L5
  cs nopw 0x0(%rax,%rax,1)
L21:
  add    $0x1,%ebx
L19:
  movsbl (%r12),%eax
  add    %ebx,%eax
  mov    %eax,%ebp
  je     L16
L5:
  mov    %ebp,%edi
  call   <isblank@plt>
  test   %rax,%rax
  jne    L21
  mov    %ebp,%edi
  call   <isalpha@plt>
  test   %rax,%rax
  je     L19
  mov    $0xffffffff,%eax
L16:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int parse_line(char *line) {
  int i = 0;
  int c = 0;
  while ((c = *line + i) != '\0') {
    if (isblank(c)) {
      i++;
    } else {
      if (isalpha(c)) {
        return -1;
      } else {
      }
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void swap(int *array, int index1, int index2) {
  int temp = 0;
  temp = *(array + index1);
  array[index1] = array[index2];
  array[index2] = temp;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<swap>:
  endbr64
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rcx
  mov    (%rcx),%esi
  movslq %edx,%rdx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
  ret
```

assistant: ```c
void swap(int *array, int index1, int index2) {
  int temp = 0;
  temp = *(array + index1);
  array[index1] = array[index2];
  array[index2] = temp;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<swap>:
  endbr64
  movslq %esi,%rsi
  movslq %edx,%rdx
  lea    (%rdi,%rsi,4),%rcx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rcx),%esi
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
  ret
```

assistant: ```c
void swap(int *array, int index1, int index2) {
  int temp = 0;
  temp = *(array + index1);
  array[index1] = array[index2];
  array[index2] = temp;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<swap>:
  endbr64
  movslq %esi,%rsi
  movslq %edx,%rdx
  lea    (%rdi,%rsi,4),%rcx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rcx),%esi
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
  ret
```

assistant: ```c
void swap(int *array, int index1, int index2) {
  int temp = 0;
  temp = *(array + index1);
  array[index1] = array[index2];
  array[index2] = temp;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<merge>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jle    L18
  cmpl   $0x0,-0x10(%rbp)
  jle    L18
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <merge_internal@plt>
  jmp    L17
L18:
  cmpl   $0x0,-0xc(%rbp)
  jns    L50
  cmpl   $0x0,-0x10(%rbp)
  jns    L50
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <merge_leaf@plt>
  jmp    L17
L50:
  mov    $0x0,%eax
  call   <abort@plt>
L17:
  leave
  ret
```

assistant: ```c
int merge(struct bplus *tree, int leftid, int rightid) {
  if (leftid > 0 && rightid > 0)
    return merge_internal(tree, leftid, rightid);
  else if (leftid < 0 && rightid < 0)
    return merge_leaf(tree, leftid, rightid);
  else
    abort();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<merge>:
  endbr64
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L3
  test   %edx,%edx
  jg     L15
L3:
  test   %edx,%esi
  js     L25
  mov    $0x0,%eax
  call   <abort@plt>
L45:
  add    $0x8,%rsp
  ret
L15:
  call   <merge_internal@plt>
  jmp    L45
L25:
  call   <merge_leaf@plt>
  jmp    L45
```

assistant: ```c
int merge(struct bplus *tree, int leftid, int rightid) {
  if (leftid > 0 && rightid > 0)
    return merge_internal(tree, leftid, rightid);
  else if (leftid < 0 && rightid < 0)
    return merge_leaf(tree, leftid, rightid);
  else
    abort();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<merge>:
  endbr64
  test   %esi,%esi
  jle    L26
  test   %edx,%edx
  jg     L29
L26:
  test   %edx,%esi
  js     L18
  xor    %eax,%eax
  jmp    1070 <abort@plt>
  nopw   0x0(%rax,%rax,1)
L29:
  jmp    1090 <merge_internal@plt>
  nopl   (%rax)
L18:
  jmp    1080 <merge_leaf@plt>
```

assistant: ```c
int merge(struct bplus *tree, int leftid, int rightid) {
  if (leftid > 0 && rightid > 0)
    return merge_internal(tree, leftid, rightid);
  else if (leftid < 0 && rightid < 0)
    return merge_leaf(tree, leftid, rightid);
  else
    abort();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<merge>:
  endbr64
  test   %esi,%esi
  jle    L10
  test   %edx,%edx
  jg     L24
L10:
  test   %edx,%esi
  js     L7
  xor    %eax,%eax
  jmp    1070 <abort@plt>
  nopw   0x0(%rax,%rax,1)
L24:
  jmp    1090 <merge_internal@plt>
  nopl   (%rax)
L7:
  jmp    1080 <merge_leaf@plt>
```

assistant: ```c
int merge(struct bplus *tree, int leftid, int rightid) {
  if (leftid > 0 && rightid > 0)
    return merge_internal(tree, leftid, rightid);
  else if (leftid < 0 && rightid < 0)
    return merge_leaf(tree, leftid, rightid);
  else
    abort();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<skippathname>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L14
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  jmp    L19
L14:
  mov    -0x18(%rbp),%rax
L19:
  leave
  ret
```

assistant: ```c
char *skippathname(char const *name) {
  char const *p;
  p = strrchr(name, '/');
  return (char *)(p ? p + 1 : name);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<skippathname>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x2f,%esi
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  cmovne %rdx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *skippathname(char const *name) {
  char const *p;
  p = strrchr(name, '/');
  return (char *)(p ? p + 1 : name);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<skippathname>:
  endbr64
  push   %rbx
  mov    $0x2f,%esi
  mov    %rdi,%rbx
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  pop    %rbx
  cmovne %rdx,%rax
  ret
```

assistant: ```c
char *skippathname(char const *name) {
  char const *p;
  p = strrchr(name, '/');
  return (char *)(p ? p + 1 : name);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<skippathname>:
  endbr64
  push   %rbx
  mov    $0x2f,%esi
  mov    %rdi,%rbx
  call   <strrchr@plt>
  lea    0x1(%rax),%rdx
  test   %rax,%rax
  mov    %rbx,%rax
  pop    %rbx
  cmovne %rdx,%rax
  ret
```

assistant: ```c
char *skippathname(char const *name) {
  char const *p;
  p = strrchr(name, '/');
  return (char *)(p ? p + 1 : name);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<xora64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cld
  mov    -0x18(%rbp),%rsi
  mov    -0x20(%rbp),%rcx
  xor    %rdx,%rdx
  lods   %ds:(%rsi),%rax
  xor    %rax,%rdx
  loop   1115 <xora64+0x1c>
  mov    %rdx,%rdi
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
uint64_t xora64(void *ptr, size_t size64) {
  uint64_t result;
  asm volatile("cld;\n"
               "mov %1, %%rsi;\n"
               "mov %2, %%rcx;\n"
               "xor %%rdx, %%rdx\n"
               "1: lodsq;\n"
               "xor %%rax, %%rdx;\n"
               "loop 1b;\n"
               "mov %%rdx, %0;\n"
               : "=g"(result)
               : "g"(ptr), "g"(size64)
               : "rax", "rcx", "rdx", "rsi");
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xora64>:
  endbr64
  mov    %rsi,%r8
  cld
  mov    %rdi,%rsi
  mov    %r8,%rcx
  xor    %rdx,%rdx
  lods   %ds:(%rsi),%rax
  xor    %rax,%rdx
  loop   110a <xora64+0x11>
  mov    %rdx,%rdi
  mov    %rdi,%rax
  ret
```

assistant: ```c
uint64_t xora64(void *ptr, size_t size64) {
  uint64_t result;
  asm volatile("cld;\n"
               "mov %1, %%rsi;\n"
               "mov %2, %%rcx;\n"
               "xor %%rdx, %%rdx\n"
               "1: lodsq;\n"
               "xor %%rax, %%rdx;\n"
               "loop 1b;\n"
               "mov %%rdx, %0;\n"
               : "=g"(result)
               : "g"(ptr), "g"(size64)
               : "rax", "rcx", "rdx", "rsi");
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xora64>:
  endbr64
  mov    %rsi,%r8
  cld
  mov    %rdi,%rsi
  mov    %r8,%rcx
  xor    %rdx,%rdx
  lods   %ds:(%rsi),%rax
  xor    %rax,%rdx
  loop   1111 <xora64+0x11>
  mov    %rdx,%r8
  mov    %r8,%rax
  ret
```

assistant: ```c
uint64_t xora64(void *ptr, size_t size64) {
  uint64_t result;
  asm volatile("cld;\n"
               "mov %1, %%rsi;\n"
               "mov %2, %%rcx;\n"
               "xor %%rdx, %%rdx\n"
               "1: lodsq;\n"
               "xor %%rax, %%rdx;\n"
               "loop 1b;\n"
               "mov %%rdx, %0;\n"
               : "=g"(result)
               : "g"(ptr), "g"(size64)
               : "rax", "rcx", "rdx", "rsi");
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<xora64>:
  endbr64
  mov    %rsi,%r8
  cld
  mov    %rdi,%rsi
  mov    %r8,%rcx
  xor    %rdx,%rdx
  lods   %ds:(%rsi),%rax
  xor    %rax,%rdx
  loop   1111 <xora64+0x11>
  mov    %rdx,%r8
  mov    %r8,%rax
  ret
```

assistant: ```c
uint64_t xora64(void *ptr, size_t size64) {
  uint64_t result;
  asm volatile("cld;\n"
               "mov %1, %%rsi;\n"
               "mov %2, %%rcx;\n"
               "xor %%rdx, %%rdx\n"
               "1: lodsq;\n"
               "xor %%rax, %%rdx;\n"
               "loop 1b;\n"
               "mov %%rdx, %0;\n"
               : "=g"(result)
               : "g"(ptr), "g"(size64)
               : "rax", "rcx", "rdx", "rsi");
  return result;
}
```