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
<main>:
  endbr64
  cmp    $0x1,%edi
  jbe    L90
  push   %rbp
  lea    -0x2(%rdi),%edx
  lea    0x8(%rsi),%rax
  xor    %ebp,%ebp
  push   %rbx
  lea    0x10(%rsi,%rdx,8),%rcx
  xor    %edi,%edi
  xor    %ebx,%ebx
  mov    $0x1,%esi
  sub    $0x8,%rsp
  jmp    L112
  nopl   0x0(%rax)
L108:
  add    $0x8,%rax
  cmp    %rcx,%rax
  je     L84
L112:
  mov    (%rax),%rdx
  cmpb   $0x2d,(%rdx)
  jne    L108
  movzbl 0x1(%rdx),%edx
  test   %dl,%dl
  je     L108
  cmp    $0x62,%dl
  je     L129
  cmp    $0x63,%dl
  je     L92
  cmp    $0x61,%dl
  cmove  %esi,%ebx
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L112
  nopw   0x0(%rax,%rax,1)
L84:
  test   %edi,%edi
  je     L89
  lea    D26(%rip),%rdi
  call   <printf@plt>
L89:
  test   %ebx,%ebx
  je     L48
  lea    D20(%rip),%rdi
  call   <printf@plt>
L48:
  test   %ebp,%ebp
  je     L54
  lea    D20(%rip),%rdi
  call   <printf@plt>
L54:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L92:
  mov    $0x1,%ebp
  jmp    L108
L129:
  mov    $0x1,%edi
  jmp    L108
L90:
  xor    %eax,%eax
  ret
```

tool_calls: name: unpack_data
arguments: label: D26
data_type: string
name: unpack_data
arguments: label: D20
data_type: string

tool: D26: You should see this message


tool: D20: You should NOT see this message


assistant: ```c
int main(int argc, char **argv) {
  int flag_a = 0;
  int flag_b = 0;
  int flag_c = 0;
  unsigned iter;
  for (iter = 1; iter < argc; iter++) {
    if (argv[iter][0] == '-' && argv[iter][1]) {
      switch (argv[iter][1]) {
      case 'a':
        flag_a = 1;
        break;
      case 'b':
        flag_b = 1;
        break;
      case 'c':
        flag_c = 1;
        break;
      default:
        break;
      }
    }
  }
  if (flag_b) {
    printf("You should see this message\n");
  }
  if (flag_a) {
    printf("You should NOT see this message\n");
  }
  if (flag_c) {
    printf("You should NOT see this message\n");
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<quit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xc0,%rsp
  mov    %rdi,-0xb8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L4
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L4:
  lea    -0xb8(%rbp),%rax
  add    $0x8,%rax
  mov    -0xb8(%rbp),%rdx
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <vprintf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void quit(char *s, ...) {
  vprintf(s, (void *)((char *)&s + sizeof(s)));
  exit(1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<quit>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  lea    0x10(%rsp),%rsi
  call   <vprintf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
void quit(char *s, ...) {
  vprintf(s, (void *)((char *)&s + sizeof(s)));
  exit(1);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<quit>:
  endbr64
  sub    $0x18,%rsp
  lea    0x10(%rsp),%rsi
  mov    %rdi,0x8(%rsp)
  call   <vprintf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
void quit(char *s, ...) {
  vprintf(s, (void *)((char *)&s + sizeof(s)));
  exit(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<quit>:
  endbr64
  sub    $0x18,%rsp
  lea    0x10(%rsp),%rsi
  mov    %rdi,0x8(%rsp)
  call   <vprintf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
void quit(char *s, ...) {
  vprintf(s, (void *)((char *)&s + sizeof(s)));
  exit(1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hashgen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L28:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L12:
  cmpl   $0x4,-0x4(%rbp)
  jg     L14
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L28
L14:
  mov    -0x8(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x3,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  shl    $0x2,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int hashgen(const char *symbolname) {
  int sum = 0;
  for (int i = 0; (i < 5 && symbolname[i] != '\0'); i++)
    sum += symbolname[i];
  return sum % 20;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<hashgen>:
  endbr64
  lea    0x5(%rdi),%rcx
  mov    $0x0,%edx
L8:
  movzbl (%rdi),%eax
  test   %al,%al
  je     L2
  movsbl %al,%eax
  add    %eax,%edx
  add    $0x1,%rdi
  cmp    %rcx,%rdi
  jne    L8
L2:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x23,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  shl    $0x2,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  ret
```

assistant: ```c
int hashgen(const char *symbolname) {
  int sum = 0;
  for (int i = 0; (i < 5 && symbolname[i] != '\0'); i++)
    sum += symbolname[i];
  return sum % 20;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<hashgen>:
  endbr64
  lea    0x5(%rdi),%rcx
  xor    %edx,%edx
L11:
  movsbl (%rdi),%eax
  test   %al,%al
  je     L6
  add    $0x1,%rdi
  add    %eax,%edx
  cmp    %rcx,%rdi
  jne    L11
L6:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x23,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %edx,%eax
  shl    $0x2,%ecx
  sub    %ecx,%eax
  ret
```

assistant: ```c
int hashgen(const char *symbolname) {
  int sum = 0;
  for (int i = 0; (i < 5 && symbolname[i] != '\0'); i++)
    sum += symbolname[i];
  return sum % 20;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<hashgen>:
  endbr64
  movsbl (%rdi),%edx
  test   %dl,%dl
  je     L29
  movsbl 0x1(%rdi),%eax
  test   %al,%al
  je     L49
  add    %eax,%edx
  movsbl 0x2(%rdi),%eax
  test   %al,%al
  je     L49
  add    %eax,%edx
  movsbl 0x3(%rdi),%eax
  test   %al,%al
  je     L49
  add    %eax,%edx
  movsbl 0x4(%rdi),%eax
  mov    %eax,%ecx
  add    %edx,%eax
  test   %cl,%cl
  cmovne %eax,%edx
L49:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x23,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %edx,%eax
  shl    $0x2,%ecx
  sub    %ecx,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L29:
  xor    %edx,%edx
  jmp    L49
```

assistant: ```c
int hashgen(const char *symbolname) {
  int sum = 0;
  for (int i = 0; (i < 5 && symbolname[i] != '\0'); i++)
    sum += symbolname[i];
  return sum % 20;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getMax>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movss  %xmm0,-0x4(%rbp)
  movss  %xmm1,-0x8(%rbp)
  movss  -0x4(%rbp),%xmm0
  comiss -0x8(%rbp),%xmm0
  jbe    L7
  movss  -0x4(%rbp),%xmm0
  jmp    L0
L7:
  movss  -0x8(%rbp),%xmm0
L0:
  pop    %rbp
  ret
```

assistant: ```c
float getMax(float a, float b) {
  if (a > b)
    return a;
  else
    return b;
}
```
user: What is the c source code of the assembly code below?

```asm
<getMax>:
  endbr64
  maxss  %xmm1,%xmm0
  ret
```

assistant: ```c
float getMax(float a, float b) {
  if (a > b)
    return a;
  else
    return b;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getMax>:
  endbr64
  maxss  %xmm1,%xmm0
  ret
```

assistant: ```c
float getMax(float a, float b) {
  if (a > b)
    return a;
  else
    return b;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<getMax>:
  endbr64
  maxss  %xmm1,%xmm0
  ret
```

assistant: ```c
float getMax(float a, float b) {
  if (a > b)
    return a;
  else
    return b;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<next>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L0
L4:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x14(%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,-0x14(%rbp)
  mov    -0x4(%rbp),%eax
  imul   %eax,%eax
  add    %eax,-0x8(%rbp)
L0:
  cmpl   $0x0,-0x14(%rbp)
  jne    L4
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int next(int number) {
  int sum = 0;
  while (number != 0) {
    int digit = number % 10;
    number /= 10;
    sum += digit * digit;
  }
  return sum;
}
```
user: What is the c source code of the assembly code below?

```asm
<next>:
  endbr64
  test   %edi,%edi
  je     L3
  mov    $0x0,%esi
L7:
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,%ecx
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  mov    %edi,%eax
  sub    %edx,%eax
  mov    %ecx,%edi
  imul   %eax,%eax
  add    %eax,%esi
  test   %ecx,%ecx
  jne    L7
L9:
  mov    %esi,%eax
  ret
L3:
  mov    %edi,%esi
  jmp    L9
```

assistant: ```c
int next(int number) {
  int sum = 0;
  while (number != 0) {
    int digit = number % 10;
    number /= 10;
    sum += digit * digit;
  }
  return sum;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<next>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L6
  nopl   0x0(%rax,%rax,1)
L8:
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %eax,%edx
  mov    %edi,%eax
  add    %ecx,%ecx
  mov    %edx,%edi
  sub    %ecx,%eax
  imul   %eax,%eax
  add    %eax,%r8d
  test   %edx,%edx
  jne    L8
L6:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int next(int number) {
  int sum = 0;
  while (number != 0) {
    int digit = number % 10;
    number /= 10;
    sum += digit * digit;
  }
  return sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<next>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L8
  nopl   0x0(%rax,%rax,1)
L7:
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%ecx
  mov    %eax,%edx
  mov    %edi,%eax
  add    %ecx,%ecx
  mov    %edx,%edi
  sub    %ecx,%eax
  imul   %eax,%eax
  add    %eax,%r8d
  test   %edx,%edx
  jne    L7
L8:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int next(int number) {
  int sum = 0;
  while (number != 0) {
    int digit = number % 10;
    number /= 10;
    sum += digit * digit;
  }
  return sum;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <f@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_guessed: string

tool: D1: %d

assistant: ```c
int main(void) {
  int i;
  i = f();
  printf("%d", i);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <f@plt>
  mov    %eax,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
data_type: string

tool: D2: %d

assistant: ```c
int main(void) {
  int i;
  i = f();
  printf("%d", i);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <f@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D6
type_guessed: string

tool: D6: %d

assistant: ```c
int main(void) {
  int i;
  i = f();
  printf("%d", i);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <f@plt>
  lea    D1(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D1
type_infered: string

tool: D1: %d

assistant: ```c
int main(void) {
  int i;
  i = f();
  printf("%d", i);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strAdd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L13
L18:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L13:
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x10(%rbp)
  jl     L18
  nop
  nop
  leave
  ret
```

assistant: ```c
void strAdd(char *a, char *b) {
  int lenA = strlen(a);
  int lenB = strlen(b);
  int i, j;
  for (j = 0, i = lenA; i < lenA + lenB; j++, i++) {
    a[i] = b[j];
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strAdd>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %eax,%r12d
  mov    %rbp,%rdi
  call   <strlen@plt>
  add    %r12d,%eax
  cmp    %eax,%r12d
  jge    L17
  movslq %r12d,%r12
  mov    %r12,%rdx
  mov    %rbp,%rsi
  sub    %r12,%rsi
L3:
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rbx,%rdx,1)
  add    $0x1,%rdx
  cmp    %edx,%eax
  jg     L3
L17:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void strAdd(char *a, char *b) {
  int lenA = strlen(a);
  int lenB = strlen(b);
  int i, j;
  for (j = 0, i = lenA; i < lenA + lenB; j++, i++) {
    a[i] = b[j];
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<strAdd>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  add    %ebx,%eax
  cmp    %eax,%ebx
  jge    L15
  movslq %ebx,%rdx
  sub    %rdx,%rbp
  nopl   0x0(%rax)
L8:
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    %cl,(%r12,%rdx,1)
  add    $0x1,%rdx
  cmp    %edx,%eax
  jg     L8
L15:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void strAdd(char *a, char *b) {
  int lenA = strlen(a);
  int lenB = strlen(b);
  int i, j;
  for (j = 0, i = lenA; i < lenA + lenB; j++, i++) {
    a[i] = b[j];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strAdd>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <strlen@plt>
  mov    %rbx,%rdi
  mov    %eax,%ebp
  call   <strlen@plt>
  lea    0x0(%rbp,%rax,1),%esi
  cmp    %esi,%ebp
  jge    L47
  mov    %eax,%edx
  lea    -0x1(%rax),%ecx
  lea    0x1(%rbx),%r10
  mov    %eax,%r8d
  movslq %ebp,%rax
  lea    (%r12,%rax,1),%rdi
  mov    %rdi,%r9
  sub    %r10,%r9
  cmp    $0xe,%r9
  jbe    L61
  cmp    $0x6,%ecx
  jbe    L61
  cmp    $0xe,%ecx
  jbe    L16
  mov    %edx,%ecx
  xor    %eax,%eax
  shr    $0x4,%ecx
  shl    $0x4,%rcx
  nopl   0x0(%rax)
L152:
  movdqu (%rbx,%rax,1),%xmm0
  movups %xmm0,(%rdi,%rax,1)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L152
  mov    %edx,%eax
  and    $0xfffffff0,%eax
  mov    %eax,%ecx
  add    %eax,%ebp
  cmp    %eax,%edx
  je     L47
  sub    %eax,%edx
  mov    %edx,%r8d
  lea    -0x1(%rdx),%edx
  cmp    $0x6,%edx
  jbe    L75
L104:
  mov    (%rbx,%rax,1),%rdx
  mov    %rdx,(%rdi,%rax,1)
  mov    %r8d,%eax
  and    $0xfffffff8,%eax
  add    %eax,%ebp
  add    %eax,%ecx
  cmp    %eax,%r8d
  je     L47
L75:
  movslq %ecx,%rax
  movzbl (%rbx,%rax,1),%edx
  movslq %ebp,%rax
  mov    %dl,(%r12,%rax,1)
  lea    0x1(%rbp),%eax
  lea    0x1(%rcx),%edx
  cmp    %eax,%esi
  jle    L47
  movslq %edx,%rdx
  cltq
  movzbl (%rbx,%rdx,1),%edx
  mov    %dl,(%r12,%rax,1)
  lea    0x2(%rbp),%eax
  lea    0x2(%rcx),%edx
  cmp    %eax,%esi
  jle    L47
  movslq %edx,%rdx
  cltq
  movzbl (%rbx,%rdx,1),%edx
  mov    %dl,(%r12,%rax,1)
  lea    0x3(%rbp),%eax
  lea    0x3(%rcx),%edx
  cmp    %eax,%esi
  jle    L47
  movslq %edx,%rdx
  cltq
  movzbl (%rbx,%rdx,1),%edx
  mov    %dl,(%r12,%rax,1)
  lea    0x4(%rbp),%eax
  lea    0x4(%rcx),%edx
  cmp    %eax,%esi
  jle    L47
  movslq %edx,%rdx
  cltq
  movzbl (%rbx,%rdx,1),%edx
  mov    %dl,(%r12,%rax,1)
  lea    0x5(%rbp),%eax
  lea    0x5(%rcx),%edx
  cmp    %eax,%esi
  jle    L47
  movslq %edx,%rdx
  cltq
  add    $0x6,%ebp
  add    $0x6,%ecx
  movzbl (%rbx,%rdx,1),%edx
  mov    %dl,(%r12,%rax,1)
  cmp    %ebp,%esi
  jle    L47
  movslq %ecx,%rcx
  movslq %ebp,%rbp
  movzbl (%rbx,%rcx,1),%eax
  mov    %al,(%r12,%rbp,1)
L47:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L61:
  sub    %rax,%rbx
  nopl   0x0(%rax,%rax,1)
L83:
  movzbl (%rbx,%rax,1),%edx
  mov    %dl,(%r12,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%esi
  jg     L83
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  xor    %eax,%eax
  xor    %ecx,%ecx
  jmp    L104
```

assistant: ```c
void strAdd(char *a, char *b) {
  int lenA = strlen(a);
  int lenB = strlen(b);
  int i, j;
  for (j = 0, i = lenA; i < lenA + lenB; j++, i++) {
    a[i] = b[j];
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<test_strnchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movl   $0x5,-0x3c(%rbp)
  mov    -0x3c(%rbp),%eax
  cltq
  sub    $0x1,%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x3c(%rbp),%eax
  cltq
  mov    %rax,%r8
  mov    $0x0,%r9d
  mov    -0x3c(%rbp),%eax
  cltq
  mov    %rax,%rsi
  mov    $0x0,%edi
  mov    -0x3c(%rbp),%eax
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%edi
  mov    $0x0,%edx
  div    %rdi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L25:
  cmp    %rdx,%rsp
  je     L14
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L25
L14:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L4
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L4:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x3c(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <init_buff@plt>
  mov    $0x0,%eax
  call   <mksym_i32@plt>
  mov    %eax,-0x38(%rbp)
  movl   $0x4,-0x34(%rbp)
  mov    -0x38(%rbp),%edx
  mov    -0x34(%rbp),%ecx
  mov    -0x28(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <strnchr@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rcx
  mov    -0x38(%rbp),%edx
  mov    -0x34(%rbp),%esi
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <gen_strnchr@plt>
  mov    %rbx,%rsp
  nop
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void test_strnchr() {
  const int len = 5;
  char s[len];
  init_buff(s, len);
  int c = mksym_i32();
  int n = 4;
  char *res = strnchr(s, n, c);
  gen_strnchr(s, n, c, res);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<test_strnchr>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rbp
  mov    $0x5,%esi
  mov    %rbp,%rdi
  call   <init_buff@plt>
  mov    $0x0,%eax
  call   <mksym_i32@plt>
  mov    %eax,%ebx
  mov    %eax,%edx
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <strnchr@plt>
  mov    %rax,%rcx
  mov    %ebx,%edx
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <gen_strnchr@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test_strnchr() {
  const int len = 5;
  char s[len];
  init_buff(s, len);
  int c = mksym_i32();
  int n = 4;
  char *res = strnchr(s, n, c);
  gen_strnchr(s, n, c, res);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<test_strnchr>:
  endbr64
  push   %r12
  mov    $0x5,%esi
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rbp
  mov    %rbp,%rdi
  call   <init_buff@plt>
  xor    %eax,%eax
  call   <mksym_i32@plt>
  mov    $0x4,%esi
  mov    %rbp,%rdi
  mov    %eax,%edx
  mov    %eax,%r12d
  call   <strnchr@plt>
  mov    %r12d,%edx
  mov    $0x4,%esi
  mov    %rbp,%rdi
  mov    %rax,%rcx
  call   <gen_strnchr@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test_strnchr() {
  const int len = 5;
  char s[len];
  init_buff(s, len);
  int c = mksym_i32();
  int n = 4;
  char *res = strnchr(s, n, c);
  gen_strnchr(s, n, c, res);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<test_strnchr>:
  endbr64
  push   %r12
  mov    $0x5,%esi
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rbp
  mov    %rbp,%rdi
  call   <init_buff@plt>
  xor    %eax,%eax
  call   <mksym_i32@plt>
  mov    $0x4,%esi
  mov    %rbp,%rdi
  mov    %eax,%edx
  mov    %eax,%r12d
  call   <strnchr@plt>
  mov    %r12d,%edx
  mov    $0x4,%esi
  mov    %rbp,%rdi
  mov    %rax,%rcx
  call   <gen_strnchr@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test_strnchr() {
  const int len = 5;
  char s[len];
  init_buff(s, len);
  int c = mksym_i32();
  int n = 4;
  char *res = strnchr(s, n, c);
  gen_strnchr(s, n, c, res);
}
```