ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<compareOptionString>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%r14
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %eax,0x1c(%rsp)
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%r15d
  test   %eax,%eax
  jle    L64
  mov    %r14,%rbx
  lea    -0x1(%rax),%eax
  lea    0x1(%r14,%rax,1),%r13
L54:
  mov    %rbx,0x8(%rsp)
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L37
  movsbl %dil,%edi
  call   <tolower@plt>
  mov    %rax,%r12
  mov    %rbp,0x10(%rsp)
  movsbl 0x0(%rbp),%edi
  call   <tolower@plt>
  cmp    %rax,%r12
  jne    L55
  add    $0x1,%rbx
  add    $0x1,%rbp
  cmp    %r13,%rbx
  jne    L54
L64:
  mov    $0x0,%eax
  cmp    %r15d,0x1c(%rsp)
  jle    L28
  movslq %r15d,%r15
  movzbl (%r14,%r15,1),%ebx
  movsbl %bl,%edi
  call   <toupper@plt>
  cmp    %al,%bl
  sete   %al
  movzbl %al,%eax
  jmp    L28
L55:
  mov    0x8(%rsp),%rax
  mov    0x10(%rsp),%rcx
  movzbl (%rcx),%ecx
  cmp    %cl,(%rax)
  setg   %al
  movzbl %al,%eax
  lea    -0x1(%rax,%rax,1),%eax
  jmp    L28
L37:
  mov    $0xffffffff,%eax
L28:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int compareOptionString(char *option, char *compareit) {
  int i, l1, l2;
  l1 = strlen(option);
  l2 = strlen(compareit);
  for (i = 0; i < l2; i++) {
    if (option[i] == '\0')
      return -1;
    if (tolower(option[i]) == tolower(compareit[i])) {
    } else {
      if (option[i] > compareit[i])
        return 1;
      else
        return -1;
    }
  }
  if (l1 > l2) {
    if (option[l2] == toupper(option[l2]))
      return 1;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<compareOptionString>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %eax,0xc(%rsp)
  call   <strlen@plt>
  movslq %eax,%r13
  test   %r13d,%r13d
  jle    L62
  lea    -0x1(%r13),%eax
  mov    %r12,%r14
  lea    0x1(%r12,%rax,1),%rbp
  jmp    L60
  nop
L6:
  call   <tolower@plt>
  movsbl (%rbx),%edi
  mov    %rax,%r15
  call   <tolower@plt>
  cmp    %rax,%r15
  jne    L0
  add    $0x1,%r14
  add    $0x1,%rbx
  cmp    %rbp,%r14
  je     L62
L60:
  movsbl (%r14),%edi
  test   %dil,%dil
  jne    L6
L42:
  mov    $0xffffffff,%eax
L31:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L62:
  xor    %eax,%eax
  cmp    %r13d,0xc(%rsp)
  jle    L31
  movsbl (%r12,%r13,1),%edi
  mov    %edi,%ebx
  call   <toupper@plt>
  cmp    %al,%bl
  sete   %al
  add    $0x18,%rsp
  pop    %rbx
  movzbl %al,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L0:
  movzbl (%rbx),%ecx
  mov    $0x1,%eax
  cmp    %cl,(%r14)
  jle    L42
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int compareOptionString(char *option, char *compareit) {
  int i, l1, l2;
  l1 = strlen(option);
  l2 = strlen(compareit);
  for (i = 0; i < l2; i++) {
    if (option[i] == '\0')
      return -1;
    if (tolower(option[i]) == tolower(compareit[i])) {
    } else {
      if (option[i] > compareit[i])
        return 1;
      else
        return -1;
    }
  }
  if (l1 > l2) {
    if (option[l2] == toupper(option[l2]))
      return 1;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<compareOptionString>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %eax,0xc(%rsp)
  call   <strlen@plt>
  movslq %eax,%r13
  test   %r13d,%r13d
  jle    L7
  lea    -0x1(%r13),%eax
  mov    %r12,%r14
  lea    0x1(%r12,%rax,1),%rbp
  jmp    L65
  nop
L44:
  call   <tolower@plt>
  movsbl (%rbx),%edi
  mov    %rax,%r15
  call   <tolower@plt>
  cmp    %rax,%r15
  jne    L41
  add    $0x1,%r14
  add    $0x1,%rbx
  cmp    %rbp,%r14
  je     L7
L65:
  movsbl (%r14),%edi
  test   %dil,%dil
  jne    L44
L59:
  mov    $0xffffffff,%eax
L32:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L7:
  xor    %eax,%eax
  cmp    %r13d,0xc(%rsp)
  jle    L32
  movsbl (%r12,%r13,1),%edi
  mov    %edi,%ebx
  call   <toupper@plt>
  cmp    %al,%bl
  sete   %al
  add    $0x18,%rsp
  pop    %rbx
  movzbl %al,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L41:
  movzbl (%rbx),%ecx
  mov    $0x1,%eax
  cmp    %cl,(%r14)
  jle    L59
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int compareOptionString(char *option, char *compareit) {
  int i, l1, l2;
  l1 = strlen(option);
  l2 = strlen(compareit);
  for (i = 0; i < l2; i++) {
    if (option[i] == '\0')
      return -1;
    if (tolower(option[i]) == tolower(compareit[i])) {
    } else {
      if (option[i] > compareit[i])
        return 1;
      else
        return -1;
    }
  }
  if (l1 > l2) {
    if (option[l2] == toupper(option[l2]))
      return 1;
  }
  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
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L16
  mov    -0x14(%rbp),%eax
  or     $0x5,%eax
  add    $0x8,%eax
  or     $0x17,%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L3
L16:
  mov    -0x14(%rbp),%eax
  add    $0x4,%eax
  and    $0x77,%eax
  mov    %eax,-0x4(%rbp)
L3:
  mov    -0x4(%rbp),%eax
  shl    $0x3,%eax
  cltq
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int x;
  if (argc & 1) {
    x = (((argc | 5) + 8) | 23) + 1;
  } else {
    x = (argc + 4) & 0x77;
  }
  return (*argv) = x << 3;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  test   $0x1,%dil
  je     L17
  or     $0x5,%edi
  lea    0x8(%rdi),%eax
  or     $0x17,%eax
  add    $0x1,%eax
L9:
  shl    $0x3,%eax
  movslq %eax,%rdx
  mov    %rdx,(%rsi)
  ret
L17:
  lea    0x4(%rdi),%eax
  and    $0x77,%eax
  jmp    L9
```

assistant: ```c
int main(int argc, char **argv) {
  int x;
  if (argc & 1) {
    x = (((argc | 5) + 8) | 23) + 1;
  } else {
    x = (argc + 4) & 0x77;
  }
  return (*argv) = x << 3;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  test   $0x1,%dil
  je     L1
  or     $0x5,%edi
  lea    0x8(%rdi),%eax
  or     $0x17,%eax
  add    $0x1,%eax
L19:
  shl    $0x3,%eax
  movslq %eax,%rdx
  mov    %rdx,(%rsi)
  ret
L1:
  lea    0x4(%rdi),%eax
  and    $0x77,%eax
  jmp    L19
```

assistant: ```c
int main(int argc, char **argv) {
  int x;
  if (argc & 1) {
    x = (((argc | 5) + 8) | 23) + 1;
  } else {
    x = (argc + 4) & 0x77;
  }
  return (*argv) = x << 3;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  test   $0x1,%dil
  je     L11
  or     $0x5,%edi
  lea    0x8(%rdi),%eax
  or     $0x17,%eax
  add    $0x1,%eax
L14:
  shl    $0x3,%eax
  movslq %eax,%rdx
  mov    %rdx,(%rsi)
  ret
L11:
  lea    0x4(%rdi),%eax
  and    $0x77,%eax
  jmp    L14
```

assistant: ```c
int main(int argc, char **argv) {
  int x;
  if (argc & 1) {
    x = (((argc | 5) + 8) | 23) + 1;
  } else {
    x = (argc + 4) & 0x77;
  }
  return (*argv) = x << 3;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<random_2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1,-0x4(%rbp)
  jmp    L26
L14:
  mov    -0x4(%rbp),%eax
  mov    $0x1f4,%edx
  mov    %eax,%esi
  mov    $0x5,%edi
  call   <random_1@plt>
  addl   $0x1,-0x4(%rbp)
L26:
  cmpl   $0x13,-0x4(%rbp)
  jle    L14
  movl   $0x14,-0x4(%rbp)
  jmp    L33
L2:
  mov    -0x4(%rbp),%eax
  mov    $0x1f4,%edx
  mov    %eax,%esi
  mov    $0x5,%edi
  call   <random_1@plt>
  subl   $0x1,-0x4(%rbp)
L33:
  cmpl   $0x0,-0x4(%rbp)
  jg     L2
  nop
  nop
  leave
  ret
```

assistant: ```c
void random_2(void) {
  int i;
  for (i = 1; i < 20; i++) {
    random_1(5, i, 500);
  }
  for (i = 20; i >= 1; i--) {
    random_1(5, i, 500);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<random_2>:
  endbr64
  push   %rbx
  mov    $0x1,%ebx
L4:
  mov    $0x1f4,%edx
  mov    %ebx,%esi
  mov    $0x5,%edi
  call   <random_1@plt>
  add    $0x1,%ebx
  cmp    $0x14,%ebx
  jne    L4
L5:
  mov    $0x1f4,%edx
  mov    %ebx,%esi
  mov    $0x5,%edi
  call   <random_1@plt>
  sub    $0x1,%ebx
  jne    L5
  pop    %rbx
  ret
```

assistant: ```c
void random_2(void) {
  int i;
  for (i = 1; i < 20; i++) {
    random_1(5, i, 500);
  }
  for (i = 20; i >= 1; i--) {
    random_1(5, i, 500);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<random_2>:
  endbr64
  push   %rbx
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L9:
  mov    %ebx,%esi
  mov    $0x1f4,%edx
  mov    $0x5,%edi
  add    $0x1,%ebx
  call   <random_1@plt>
  cmp    $0x14,%ebx
  jne    L9
  nopl   0x0(%rax)
L11:
  mov    %ebx,%esi
  mov    $0x1f4,%edx
  mov    $0x5,%edi
  call   <random_1@plt>
  sub    $0x1,%ebx
  jne    L11
  pop    %rbx
  ret
```

assistant: ```c
void random_2(void) {
  int i;
  for (i = 1; i < 20; i++) {
    random_1(5, i, 500);
  }
  for (i = 20; i >= 1; i--) {
    random_1(5, i, 500);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<random_2>:
  endbr64
  push   %rbx
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L2:
  mov    %ebx,%esi
  mov    $0x1f4,%edx
  mov    $0x5,%edi
  add    $0x1,%ebx
  call   <random_1@plt>
  cmp    $0x14,%ebx
  jne    L2
  nopl   0x0(%rax)
L8:
  mov    %ebx,%esi
  mov    $0x1f4,%edx
  mov    $0x5,%edi
  call   <random_1@plt>
  sub    $0x1,%ebx
  jne    L8
  pop    %rbx
  ret
```

assistant: ```c
void random_2(void) {
  int i;
  for (i = 1; i < 20; i++) {
    random_1(5, i, 500);
  }
  for (i = 20; i >= 1; i--) {
    random_1(5, i, 500);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L20
L18:
  mov    -0x10(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  add    %edx,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x10(%rbp)
L20:
  mov    -0x10(%rbp),%eax
  test   %eax,%eax
  jne    L18
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D27
data_type: string
name: parse_data
arguments: data_tag: D11
data_type: string
name: parse_data
arguments: data_tag: D13
data_type: string

tool: D27: Enter a number: 

tool: D11: %d

tool: D13: The sum is %d


assistant: ```c
int main(void) {
  int num, sum = 0, di;
  printf("%s\n", "Enter a number: ");
  scanf("%d", &num);
  while (num) {
    sum += num % 10;
    num /= 10;
  }
  printf("The sum is %d\n", sum);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D25(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D8(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  je     L6
  mov    $0x0,%esi
L36:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  add    %edx,%esi
  mov    %eax,%edx
  test   %eax,%eax
  jne    L36
  movl   $0x0,0x4(%rsp)
L3:
  lea    D21(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L6:
  mov    %edx,%esi
  jmp    L3
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D25
type: string
name: read_data
arguments: data_tag: D8
type: string
name: read_data
arguments: data_tag: D21
type: string

tool: D25: Enter a number: 

tool: D8: %d

tool: D21: The sum is %d


assistant: ```c
int main(void) {
  int num, sum = 0, di;
  printf("%s\n", "Enter a number: ");
  scanf("%d", &num);
  while (num) {
    sum += num % 10;
    num /= 10;
  }
  printf("The sum is %d\n", sum);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D23(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  xor    %esi,%esi
  test   %edx,%edx
  je     L14
  nop
L2:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  add    %edx,%esi
  mov    %eax,%edx
  test   %eax,%eax
  jne    L2
  movl   $0x0,0x4(%rsp)
L14:
  xor    %eax,%eax
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D23
datatype: string
name: read_data
arguments: data_label: D25
datatype: string
name: read_data
arguments: data_label: D18
datatype: string

tool: D23: Enter a number: 

tool: D25: %d

tool: D18: The sum is %d


assistant: ```c
int main(void) {
  int num, sum = 0, di;
  printf("%s\n", "Enter a number: ");
  scanf("%d", &num);
  while (num) {
    sum += num % 10;
    num /= 10;
  }
  printf("The sum is %d\n", sum);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D18(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D22(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  xor    %esi,%esi
  test   %edx,%edx
  je     L29
  nop
L25:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  add    %edx,%esi
  mov    %eax,%edx
  test   %eax,%eax
  jne    L25
  movl   $0x0,0x4(%rsp)
L29:
  xor    %eax,%eax
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D18
data_type: string
name: retrieve_data
arguments: data_label: D22
data_type: string
name: retrieve_data
arguments: data_label: D5
data_type: string

tool: D18: Enter a number: 

tool: D22: %d

tool: D5: The sum is %d


assistant: ```c
int main(void) {
  int num, sum = 0, di;
  printf("%s\n", "Enter a number: ");
  scanf("%d", &num);
  while (num) {
    sum += num % 10;
    num /= 10;
  }
  printf("The sum is %d\n", sum);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<DoCodeBccB>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <cctrue@plt>
  test   %rax,%rax
  je     L9
  mov    $0x0,%eax
  call   <DoCodeBraB@plt>
L9:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void DoCodeBccB(void) {
  if (cctrue()) {
    DoCodeBraB();
  } else {
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<DoCodeBccB>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <cctrue@plt>
  test   %rax,%rax
  jne    L6
L11:
  add    $0x8,%rsp
  ret
L6:
  mov    $0x0,%eax
  call   <DoCodeBraB@plt>
  jmp    L11
```

assistant: ```c
void DoCodeBccB(void) {
  if (cctrue()) {
    DoCodeBraB();
  } else {
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<DoCodeBccB>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <cctrue@plt>
  test   %rax,%rax
  jne    L0
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax)
L0:
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <DoCodeBraB@plt>
```

assistant: ```c
void DoCodeBccB(void) {
  if (cctrue()) {
    DoCodeBraB();
  } else {
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<DoCodeBccB>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <cctrue@plt>
  test   %rax,%rax
  jne    L5
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax)
L5:
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <DoCodeBraB@plt>
```

assistant: ```c
void DoCodeBccB(void) {
  if (cctrue()) {
    DoCodeBraB();
  } else {
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<wildmatch>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  lea    D4(%rip),%rdx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <dowild@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int wildmatch(const char *pattern, const char *text) {
  static const unsigned char *nomore[1];
  return dowild((const unsigned char *)pattern, (const unsigned char *)text, nomore) == 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<wildmatch>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdx
  call   <dowild@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int wildmatch(const char *pattern, const char *text) {
  static const unsigned char *nomore[1];
  return dowild((const unsigned char *)pattern, (const unsigned char *)text, nomore) == 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<wildmatch>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdx
  call   <dowild@plt>
  cmp    $0x1,%eax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int wildmatch(const char *pattern, const char *text) {
  static const unsigned char *nomore[1];
  return dowild((const unsigned char *)pattern, (const unsigned char *)text, nomore) == 1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<wildmatch>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdx
  call   <dowild@plt>
  cmp    $0x1,%eax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int wildmatch(const char *pattern, const char *text) {
  static const unsigned char *nomore[1];
  return dowild((const unsigned char *)pattern, (const unsigned char *)text, nomore) == 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
<opa_unicode_last_utf8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L13
  cmpl   $0x0,-0x20(%rbp)
  jne    L97
L13:
  mov    $0xffffffff,%eax
  jmp    L87
L97:
  mov    -0x20(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x5(%rbp)
  movzbl -0x5(%rbp),%eax
  test   %al,%al
  js     L52
  mov    -0x4(%rbp),%eax
  jmp    L87
L52:
  subl   $0x1,-0x4(%rbp)
  jmp    L68
L64:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x5(%rbp)
  movzbl -0x5(%rbp),%eax
  and    $0xc0,%eax
  cmp    $0x80,%eax
  jne    L8
  subl   $0x1,-0x4(%rbp)
L68:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L6
  mov    -0x20(%rbp),%eax
  sub    $0x4,%eax
  cmp    %eax,-0x4(%rbp)
  jge    L64
  jmp    L6
L8:
  nop
L6:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L96
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L96:
  mov    -0x4(%rbp),%eax
L87:
  pop    %rbp
  ret
```

assistant: ```c
int opa_unicode_last_utf8(const char *in, int start, int end) {
  if (start >= end || end == 0) {
    return -1;
  }
  int i = end - 1;
  unsigned char c = in[i];
  if (c < 0x80) {
    return i;
  }
  for (i = i - 1; i >= start && i >= end - 4; i--) {
    c = in[i];
    if ((c & 0xc0) != 0x80) {
      break;
    }
  }
  if (i < start) {
    i = start;
  }
  return i;
}
```
user: What is the c source code of the assembly code below?

```asm
<opa_unicode_last_utf8>:
  endbr64
  cmp    %edx,%esi
  jge    L3
  test   %edx,%edx
  je     L3
  lea    -0x1(%rdx),%eax
  movslq %eax,%rcx
  cmpb   $0x0,(%rdi,%rcx,1)
  jns    L57
  lea    -0x2(%rdx),%eax
  cmp    %eax,%esi
  jg     L63
  sub    $0x4,%edx
  movslq %eax,%rcx
  add    %rcx,%rdi
L5:
  movzbl (%rdi),%ecx
  and    $0xffffffc0,%ecx
  cmp    $0x80,%cl
  jne    L63
  sub    $0x1,%eax
  cmp    %eax,%esi
  jg     L63
  sub    $0x1,%rdi
  cmp    %edx,%eax
  jge    L5
L63:
  cmp    %eax,%esi
  cmovge %esi,%eax
  ret
L3:
  mov    $0xffffffff,%eax
L57:
  ret
```

assistant: ```c
int opa_unicode_last_utf8(const char *in, int start, int end) {
  if (start >= end || end == 0) {
    return -1;
  }
  int i = end - 1;
  unsigned char c = in[i];
  if (c < 0x80) {
    return i;
  }
  for (i = i - 1; i >= start && i >= end - 4; i--) {
    c = in[i];
    if ((c & 0xc0) != 0x80) {
      break;
    }
  }
  if (i < start) {
    i = start;
  }
  return i;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<opa_unicode_last_utf8>:
  endbr64
  cmp    %edx,%esi
  jge    L51
  test   %edx,%edx
  je     L51
  lea    -0x1(%rdx),%eax
  movslq %eax,%rcx
  cmpb   $0x0,(%rdi,%rcx,1)
  jns    L76
  lea    -0x2(%rdx),%eax
  cmp    %eax,%esi
  jg     L21
  movslq %eax,%rcx
  sub    $0x4,%edx
  add    %rcx,%rdi
  jmp    L62
  nopw   0x0(%rax,%rax,1)
L11:
  sub    $0x1,%eax
  cmp    %eax,%esi
  jg     L21
  sub    $0x1,%rdi
  cmp    %edx,%eax
  jl     L21
L62:
  movzbl (%rdi),%ecx
  and    $0xffffffc0,%ecx
  cmp    $0x80,%cl
  je     L11
L21:
  cmp    %eax,%esi
  cmovge %esi,%eax
  ret
L51:
  mov    $0xffffffff,%eax
L76:
  ret
```

assistant: ```c
int opa_unicode_last_utf8(const char *in, int start, int end) {
  if (start >= end || end == 0) {
    return -1;
  }
  int i = end - 1;
  unsigned char c = in[i];
  if (c < 0x80) {
    return i;
  }
  for (i = i - 1; i >= start && i >= end - 4; i--) {
    c = in[i];
    if ((c & 0xc0) != 0x80) {
      break;
    }
  }
  if (i < start) {
    i = start;
  }
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<opa_unicode_last_utf8>:
  endbr64
  cmp    %edx,%esi
  jge    L31
  test   %edx,%edx
  je     L31
  lea    -0x1(%rdx),%eax
  movslq %eax,%rcx
  cmpb   $0x0,(%rdi,%rcx,1)
  jns    L56
  lea    -0x2(%rdx),%eax
  cmp    %eax,%esi
  jg     L62
  movslq %eax,%rcx
  sub    $0x4,%edx
  add    %rcx,%rdi
  jmp    L64
  nopw   0x0(%rax,%rax,1)
L6:
  sub    $0x1,%eax
  cmp    %eax,%esi
  jg     L62
  sub    $0x1,%rdi
  cmp    %edx,%eax
  jl     L62
L64:
  movzbl (%rdi),%ecx
  and    $0xffffffc0,%ecx
  cmp    $0x80,%cl
  je     L6
L62:
  cmp    %eax,%esi
  cmovge %esi,%eax
  ret
L31:
  mov    $0xffffffff,%eax
L56:
  ret
```

assistant: ```c
int opa_unicode_last_utf8(const char *in, int start, int end) {
  if (start >= end || end == 0) {
    return -1;
  }
  int i = end - 1;
  unsigned char c = in[i];
  if (c < 0x80) {
    return i;
  }
  for (i = i - 1; i >= start && i >= end - 4; i--) {
    c = in[i];
    if ((c & 0xc0) != 0x80) {
      break;
    }
  }
  if (i < start) {
    i = start;
  }
  return i;
}
```