ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<RandomFunc>:
  endbr64
  imul   $0xfb16,(%rdi),%ax
  add    $0x1cb1,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  char copy11;
  {
    state[0UL] = (input[0UL] + 914778474UL) * (unsigned short)64278;
    if (state[0UL] & (unsigned short)1) {
      if (state[0UL] & (unsigned short)1) {
        if (state[0UL] & (unsigned short)1) {
          copy11 = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = copy11;
          copy11 = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = copy11;
        } else {
          state[0UL] *= state[0UL];
        }
      } else if (!((state[0UL] >> (unsigned short)4) & (unsigned short)1)) {
        state[0UL] *= state[0UL];
      }
    }
    output[0UL] = state[0UL] + 387129237UL;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<RandomFunc>:
  endbr64
  imul   $0xfb16,(%rdi),%ax
  add    $0x1cb1,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  char copy11;
  {
    state[0UL] = (input[0UL] + 914778474UL) * (unsigned short)64278;
    if (state[0UL] & (unsigned short)1) {
      if (state[0UL] & (unsigned short)1) {
        if (state[0UL] & (unsigned short)1) {
          copy11 = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = copy11;
          copy11 = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = copy11;
        } else {
          state[0UL] *= state[0UL];
        }
      } else if (!((state[0UL] >> (unsigned short)4) & (unsigned short)1)) {
        state[0UL] *= state[0UL];
      }
    }
    output[0UL] = state[0UL] + 387129237UL;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<RandomFunc>:
  endbr64
  imul   $0xfb16,(%rdi),%ax
  add    $0x1cb1,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  char copy11;
  {
    state[0UL] = (input[0UL] + 914778474UL) * (unsigned short)64278;
    if (state[0UL] & (unsigned short)1) {
      if (state[0UL] & (unsigned short)1) {
        if (state[0UL] & (unsigned short)1) {
          copy11 = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = copy11;
          copy11 = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = copy11;
        } else {
          state[0UL] *= state[0UL];
        }
      } else if (!((state[0UL] >> (unsigned short)4) & (unsigned short)1)) {
        state[0UL] *= state[0UL];
      }
    }
    output[0UL] = state[0UL] + 387129237UL;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L5:
  call   <fork@plt>
  jmp    L5
```

assistant: ```c
int main() {
  for (;;)
    fork();
  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
  sub    $0x8,%rsp
L6:
  call   <fork@plt>
  jmp    L6
```

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

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L0:
  call   <fork@plt>
  jmp    L0
```

assistant: ```c
int main() {
  for (;;)
    fork();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L6:
  call   <fork@plt>
  jmp    L6
```

assistant: ```c
int main() {
  for (;;)
    fork();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<vga_mode>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    $0x3ce,%esi
  mov    $0x5,%edi
  call   <outb@plt>
  mov    -0x4(%rbp),%eax
  mov    $0x3cf,%esi
  mov    %eax,%edi
  call   <outb@plt>
  nop
  leave
  ret
```

assistant: ```c
void vga_mode(int mode) {
  outb(0x05, 0x3ce);
  outb(mode, 0x3cf);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<vga_mode>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0x3ce,%esi
  mov    $0x5,%edi
  call   <outb@plt>
  mov    $0x3cf,%esi
  mov    %ebx,%edi
  call   <outb@plt>
  pop    %rbx
  ret
```

assistant: ```c
void vga_mode(int mode) {
  outb(0x05, 0x3ce);
  outb(mode, 0x3cf);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<vga_mode>:
  endbr64
  push   %rbp
  mov    $0x3ce,%esi
  mov    %edi,%ebp
  mov    $0x5,%edi
  call   <outb@plt>
  mov    %ebp,%edi
  mov    $0x3cf,%esi
  pop    %rbp
  jmp    1050 <outb@plt>
```

assistant: ```c
void vga_mode(int mode) {
  outb(0x05, 0x3ce);
  outb(mode, 0x3cf);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L9
L11:
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  subl   $0x1,-0x4(%rbp)
L9:
  cmpl   $0x0,-0x8(%rbp)
  jg     L11
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int x = __VERIFIER_nondet_int();
  int y = __VERIFIER_nondet_int();
  while (x > 0) {
    x = y;
    y--;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  test   %ebx,%ebx
  jle    L9
L1:
  mov    %eax,%edx
  sub    $0x1,%eax
  test   %edx,%edx
  jg     L1
L9:
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int x = __VERIFIER_nondet_int();
  int y = __VERIFIER_nondet_int();
  while (x > 0) {
    x = y;
    y--;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

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

assistant: ```c
int main() {
  int x = __VERIFIER_nondet_int();
  int y = __VERIFIER_nondet_int();
  while (x > 0) {
    x = y;
    y--;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

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

assistant: ```c
int main() {
  int x = __VERIFIER_nondet_int();
  int y = __VERIFIER_nondet_int();
  while (x > 0) {
    x = y;
    y--;
  }
  return 0;
}
```
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    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x4,%edi
  call   <alloca@plt>
  mov    %rax,-0x40(%rbp)
  mov    $0x4,%edi
  call   <alloca@plt>
  mov    %rax,-0x38(%rbp)
  mov    $0x4,%edi
  call   <alloca@plt>
  mov    %rax,-0x30(%rbp)
  mov    $0x4,%edi
  call   <alloca@plt>
  mov    %rax,-0x28(%rbp)
  mov    $0x4,%edi
  call   <alloca@plt>
  mov    %rax,-0x20(%rbp)
  mov    $0x4,%edi
  call   <alloca@plt>
  mov    %rax,-0x18(%rbp)
  mov    $0x4,%edi
  call   <alloca@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x40(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  movl   $0x0,(%rax)
  jmp    L25
L8:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x1,%eax
  jle    L45
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  add    %eax,%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L25
L45:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
L25:
  mov    -0x38(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jge    L8
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
int main() {
  int *u = alloca(sizeof(int));
  int *x = alloca(sizeof(int));
  int *v = alloca(sizeof(int));
  int *y = alloca(sizeof(int));
  int *w = alloca(sizeof(int));
  int *z = alloca(sizeof(int));
  int *c = alloca(sizeof(int));
  *u = *x;
  *v = *y;
  *w = *z;
  *c = 0;
  while (*x >= *y) {
    *c = *c + 1;
    if (*z > 1) {
      *z = *z - 1;
      *x = *x + *z;
    } else {
      *y = *y + 1;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    $0x4,%edi
  call   <alloca@plt>
  mov    %rax,%r15
  mov    $0x4,%edi
  call   <alloca@plt>
  mov    %rax,%rbx
  mov    $0x4,%edi
  call   <alloca@plt>
  mov    %rax,%r14
  mov    $0x4,%edi
  call   <alloca@plt>
  mov    %rax,%r12
  mov    $0x4,%edi
  call   <alloca@plt>
  mov    %rax,-0x48(%rbp)
  mov    $0x4,%edi
  call   <alloca@plt>
  mov    %rax,%r13
  mov    $0x4,%edi
  call   <alloca@plt>
  mov    (%rbx),%edx
  mov    %edx,(%r15)
  mov    (%r12),%edx
  mov    %edx,(%r14)
  mov    0x0(%r13),%edx
  mov    -0x48(%rbp),%rsi
  mov    %edx,(%rsi)
  movl   $0x0,(%rax)
  mov    (%r12),%esi
  cmp    %esi,(%rbx)
  jge    L42
L37:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L41
  mov    $0x0,%eax
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L5:
  sub    $0x1,%edx
  mov    %edx,0x0(%r13)
  add    %edx,(%rbx)
L1:
  mov    (%r12),%ecx
  cmp    %ecx,(%rbx)
  jl     L37
L42:
  addl   $0x1,(%rax)
  mov    0x0(%r13),%edx
  cmp    $0x1,%edx
  jg     L5
  addl   $0x1,(%r12)
  jmp    L1
L41:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int *u = alloca(sizeof(int));
  int *x = alloca(sizeof(int));
  int *v = alloca(sizeof(int));
  int *y = alloca(sizeof(int));
  int *w = alloca(sizeof(int));
  int *z = alloca(sizeof(int));
  int *c = alloca(sizeof(int));
  *u = *x;
  *v = *y;
  *w = *z;
  *c = 0;
  while (*x >= *y) {
    *c = *c + 1;
    if (*z > 1) {
      *z = *z - 1;
      *x = *x + *z;
    } else {
      *y = *y + 1;
    }
  }
}
```
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    $0x4,%edi
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,-0x48(%rbp)
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,%rbx
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,%r15
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,%r13
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,%r14
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,%r12
  call   <alloca@plt>
  mov    -0x48(%rbp),%rdx
  mov    %rax,%rcx
  mov    (%rbx),%eax
  mov    %eax,(%rdx)
  mov    0x0(%r13),%eax
  mov    %eax,(%r15)
  mov    (%r12),%eax
  mov    %eax,(%r14)
  movl   $0x0,(%rcx)
  mov    (%rbx),%eax
  cmp    %eax,0x0(%r13)
  jg     L15
  xor    %eax,%eax
  jmp    L13
  nopl   0x0(%rax,%rax,1)
L42:
  sub    $0x1,%eax
  mov    %eax,(%r12)
  add    (%rbx),%eax
  mov    %eax,(%rbx)
  mov    0x0(%r13),%edx
  cmp    %edx,%eax
  jl     L15
L47:
  mov    (%rcx),%eax
L13:
  add    $0x1,%eax
  mov    %eax,(%rcx)
  mov    (%r12),%eax
  cmp    $0x1,%eax
  jg     L42
  mov    0x0(%r13),%eax
  lea    0x1(%rax),%edx
  mov    %edx,0x0(%r13)
  mov    (%rbx),%eax
  cmp    %edx,%eax
  jge    L47
L15:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int *u = alloca(sizeof(int));
  int *x = alloca(sizeof(int));
  int *v = alloca(sizeof(int));
  int *y = alloca(sizeof(int));
  int *w = alloca(sizeof(int));
  int *z = alloca(sizeof(int));
  int *c = alloca(sizeof(int));
  *u = *x;
  *v = *y;
  *w = *z;
  *c = 0;
  while (*x >= *y) {
    *c = *c + 1;
    if (*z > 1) {
      *z = *z - 1;
      *x = *x + *z;
    } else {
      *y = *y + 1;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x4,%edi
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,-0x48(%rbp)
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,%rbx
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,%r15
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,%r13
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,%r14
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,%r12
  call   <alloca@plt>
  mov    -0x48(%rbp),%rcx
  mov    %rax,%rdx
  mov    (%rbx),%eax
  mov    %eax,(%rcx)
  mov    0x0(%r13),%eax
  mov    %eax,(%r15)
  mov    (%r12),%eax
  mov    %eax,(%r14)
  movl   $0x0,(%rdx)
  mov    (%rbx),%eax
  cmp    %eax,0x0(%r13)
  jg     L50
  xor    %eax,%eax
  add    $0x1,%eax
  mov    %eax,(%rdx)
  mov    (%r12),%eax
  cmp    $0x1,%eax
  jle    L59
  nopl   0x0(%rax)
L40:
  sub    $0x1,%eax
  mov    %eax,(%r12)
  add    (%rbx),%eax
  mov    %eax,(%rbx)
  cmp    0x0(%r13),%eax
  jl     L50
L13:
  mov    (%rdx),%eax
  add    $0x1,%eax
  mov    %eax,(%rdx)
  mov    (%r12),%eax
  cmp    $0x1,%eax
  jg     L40
L59:
  mov    0x0(%r13),%eax
  add    $0x1,%eax
  mov    %eax,0x0(%r13)
  cmp    (%rbx),%eax
  jle    L13
L50:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L58
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L58:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int *u = alloca(sizeof(int));
  int *x = alloca(sizeof(int));
  int *v = alloca(sizeof(int));
  int *y = alloca(sizeof(int));
  int *w = alloca(sizeof(int));
  int *z = alloca(sizeof(int));
  int *c = alloca(sizeof(int));
  *u = *x;
  *v = *y;
  *w = *z;
  *c = 0;
  while (*x >= *y) {
    *c = *c + 1;
    if (*z > 1) {
      *z = *z - 1;
      *x = *x + *z;
    } else {
      *y = *y + 1;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mon_getenvp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <monLock@plt>
  mov    $0x0,%eax
  call   <_getenvp@plt>
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <monUnlock@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *mon_getenvp(void) {
  char *ret;
  monLock();
  ret = _getenvp();
  monUnlock();
  return (ret);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<mon_getenvp>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <monLock@plt>
  mov    $0x0,%eax
  call   <_getenvp@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  call   <monUnlock@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *mon_getenvp(void) {
  char *ret;
  monLock();
  ret = _getenvp();
  monUnlock();
  return (ret);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mon_getenvp>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <monLock@plt>
  xor    %eax,%eax
  call   <_getenvp@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <monUnlock@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *mon_getenvp(void) {
  char *ret;
  monLock();
  ret = _getenvp();
  monUnlock();
  return (ret);
}
```
user: Translate the below assembly code into C programming language:

```asm
<mon_getenvp>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <monLock@plt>
  xor    %eax,%eax
  call   <_getenvp@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <monUnlock@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *mon_getenvp(void) {
  char *ret;
  monLock();
  ret = _getenvp();
  monUnlock();
  return (ret);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strrchrc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L30
L37:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L20
  mov    -0x8(%rbp),%rax
  jmp    L28
L20:
  subq   $0x1,-0x8(%rbp)
L30:
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jne    L37
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L15
  mov    -0x8(%rbp),%rax
  jmp    L28
L15:
  mov    $0x0,%eax
L28:
  leave
  ret
```

assistant: ```c
const char *strrchrc(const char *str, char v) {
  const char *s = str + strlen(str) - 1;
  while (s != str) {
    if (*s == v)
      return s;
    s--;
  }
  if (*s == v)
    return s;
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<strrchrc>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%r12d
  mov    %esi,%ebp
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%rax
  cmp    %rax,%rbx
  je     L17
L15:
  cmp    %bpl,(%rax)
  je     L8
  sub    $0x1,%rax
  cmp    %rax,%rbx
  jne    L15
L17:
  cmp    %r12b,(%rax)
  mov    $0x0,%edx
  cmovne %rdx,%rax
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
const char *strrchrc(const char *str, char v) {
  const char *s = str + strlen(str) - 1;
  while (s != str) {
    if (*s == v)
      return s;
    s--;
  }
  if (*s == v)
    return s;
  return NULL;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strrchrc>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%rax
  cmp    %rax,%rbx
  jne    L8
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L19:
  sub    $0x1,%rax
  cmp    %rax,%rbx
  je     L33
L8:
  cmp    %bpl,(%rax)
  jne    L19
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L33:
  xor    %edx,%edx
  cmp    %r12b,(%rax)
  pop    %rbx
  cmovne %rdx,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
const char *strrchrc(const char *str, char v) {
  const char *s = str + strlen(str) - 1;
  while (s != str) {
    if (*s == v)
      return s;
    s--;
  }
  if (*s == v)
    return s;
  return NULL;
}
```
user: What is the c source code of the assembly code below:

```asm
<strrchrc>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%rax
  cmp    %rax,%rbx
  jne    L13
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L22:
  sub    $0x1,%rax
  cmp    %rax,%rbx
  je     L25
L13:
  cmp    %bpl,(%rax)
  jne    L22
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  xor    %edx,%edx
  cmp    %r12b,(%rax)
  pop    %rbx
  cmovne %rdx,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
const char *strrchrc(const char *str, char v) {
  const char *s = str + strlen(str) - 1;
  while (s != str) {
    if (*s == v)
      return s;
    s--;
  }
  if (*s == v)
    return s;
  return NULL;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<task_13>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <OS_TaskRegister@plt>
L8:
  jmp    L8
```

assistant: ```c
void task_13(void) {
  OS_TaskRegister();
  while (1)
    ;
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<task_13>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <OS_TaskRegister@plt>
L9:
  jmp    L9
```

assistant: ```c
void task_13(void) {
  OS_TaskRegister();
  while (1)
    ;
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<task_13>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <OS_TaskRegister@plt>
L5:
  jmp    L5
```

assistant: ```c
void task_13(void) {
  OS_TaskRegister();
  while (1)
    ;
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<task_13>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <OS_TaskRegister@plt>
L1:
  jmp    L1
```

assistant: ```c
void task_13(void) {
  OS_TaskRegister();
  while (1)
    ;
  return;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<count>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,%eax
  mov    %al,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L3
  mov    $0x0,%eax
  jmp    L38
L3:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0xc(%rbp)
  jne    L13
  movsbl -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  add    $0x1,%rdx
  mov    %eax,%esi
  mov    %rdx,%rdi
  call   <count>
  add    $0x1,%eax
  jmp    L38
L13:
  movsbl -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  add    $0x1,%rdx
  mov    %eax,%esi
  mov    %rdx,%rdi
  call   <count>
L38:
  leave
  ret
```

assistant: ```c
int count(char s[], char c) {
  if (strlen(s) == 0) {
    return 0;
  } else if (s[0] == c) {
    return 1 + count((s + 1), c);
  } else {
    return count((s + 1), c);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<count>:
  endbr64
  movzbl (%rdi),%edx
  mov    $0x0,%eax
  test   %dl,%dl
  je     L0
  sub    $0x8,%rsp
  cmp    %sil,%dl
  je     L27
  movsbl %sil,%esi
  add    $0x1,%rdi
  call   <count>
L21:
  add    $0x8,%rsp
  ret
L27:
  movsbl %sil,%esi
  add    $0x1,%rdi
  call   <count>
  add    $0x1,%eax
  jmp    L21
L0:
  ret
```

assistant: ```c
int count(char s[], char c) {
  if (strlen(s) == 0) {
    return 0;
  } else if (s[0] == c) {
    return 1 + count((s + 1), c);
  } else {
    return count((s + 1), c);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<count>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L37
  sub    $0x8,%rsp
  movsbl %sil,%r8d
  movzbl 0x1(%rdi),%ecx
  cmp    %sil,%dl
  je     L50
  test   %cl,%cl
  je     L52
  cmp    %cl,%sil
  je     L65
  movzbl 0x2(%rdi),%eax
  test   %al,%al
  je     L52
  add    $0x3,%rdi
  cmp    %al,%sil
  mov    %r8d,%esi
  je     L88
  add    $0x8,%rsp
  jmp    1100 <count>
  nop
L50:
  mov    $0x1,%eax
  test   %cl,%cl
  je     L62
  add    $0x2,%rdi
  mov    %r8d,%esi
  cmp    %cl,%dl
  je     L73
L88:
  call   <count>
  add    $0x1,%eax
L62:
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax)
L37:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L52:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L65:
  add    $0x2,%rdi
  mov    %r8d,%esi
  call   <count>
  add    $0x1,%eax
  jmp    L62
  nopl   0x0(%rax)
L73:
  call   <count>
  add    $0x2,%eax
  jmp    L62
```

assistant: ```c
int count(char s[], char c) {
  if (strlen(s) == 0) {
    return 0;
  } else if (s[0] == c) {
    return 1 + count((s + 1), c);
  } else {
    return count((s + 1), c);
  }
}
```