ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<mem_swap_near_block>:
  endbr64
  test   %edx,%edx
  je     L32
  mov    %rdi,%r8
  test   %ecx,%ecx
  je     L32
  add    %edx,%ecx
  imul   %esi,%ecx
  imul   %esi,%edx
  mov    %rdi,%rax
  or     %rcx,%rax
  test   $0x3,%al
  je     L46
L41:
  movzbl (%r8),%r9d
  mov    $0x0,%edi
  jmp    L44
L46:
  test   $0x3,%dl
  jne    L41
  sub    $0x8,%rsp
  shr    $0x2,%edx
  shr    $0x2,%ecx
  mov    %ecx,%esi
  call   <mem_swap_near_block_32@plt>
  add    $0x8,%rsp
  ret
L45:
  mov    %eax,%edi
L44:
  lea    (%rdi,%rdx,1),%eax
  mov    %eax,%esi
  sub    %ecx,%esi
  cmp    %ecx,%eax
  cmovae %esi,%eax
  mov    %edi,%edi
  add    %r8,%rdi
  mov    %eax,%esi
  movzbl (%r8,%rsi,1),%esi
  mov    %sil,(%rdi)
  test   %eax,%eax
  jne    L45
  mov    %r9b,(%rdi)
  ret
L32:
  ret
```

assistant: ```c
void mem_swap_near_block(void *base, unsigned int elem_size, unsigned int cnt1, unsigned int cnt2) {
  unsigned int sz = elem_size * (cnt1 + cnt2);
  unsigned int shift = elem_size * cnt1;
  if (cnt1 == 0 || cnt2 == 0) {
    return;
  }
  if (((unsigned int)base & 3) || (sz & 3) || (shift & 3)) {
    unsigned int i0 = 0;
    unsigned int i1 = 0;
    char *p = base;
    char tmp = p[i0];
    do {
      i0 = i1;
      i1 = (((shift + i0) >= (sz)) ? ((shift + i0) - (sz)) : (shift + i0));
      p[i0] = p[i1];
    } while (i1 != 0);
    p[i0] = tmp;
  } else {
    mem_swap_near_block_32(base, sz / 4, shift / 4);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mem_swap_near_block>:
  endbr64
  mov    %rdi,%r8
  test   %edx,%edx
  je     L52
  test   %ecx,%ecx
  je     L52
  add    %edx,%ecx
  imul   %esi,%edx
  mov    %rdi,%rax
  imul   %esi,%ecx
  or     %rcx,%rax
  test   $0x3,%al
  je     L15
L12:
  movzbl (%r8),%r9d
  xor    %edi,%edi
  jmp    L55
  nopl   0x0(%rax)
L26:
  mov    %eax,%edi
L55:
  lea    (%rdi,%rdx,1),%eax
  mov    %eax,%esi
  sub    %ecx,%esi
  cmp    %ecx,%eax
  cmovae %esi,%eax
  add    %r8,%rdi
  mov    %eax,%esi
  movzbl (%r8,%rsi,1),%esi
  mov    %sil,(%rdi)
  test   %eax,%eax
  jne    L26
  mov    %r9b,(%rdi)
  ret
  nopl   0x0(%rax,%rax,1)
L52:
  ret
  nopl   0x0(%rax)
L15:
  test   $0x3,%dl
  jne    L12
  shr    $0x2,%ecx
  shr    $0x2,%edx
  mov    %ecx,%esi
  jmp    1050 <mem_swap_near_block_32@plt>
```

assistant: ```c
void mem_swap_near_block(void *base, unsigned int elem_size, unsigned int cnt1, unsigned int cnt2) {
  unsigned int sz = elem_size * (cnt1 + cnt2);
  unsigned int shift = elem_size * cnt1;
  if (cnt1 == 0 || cnt2 == 0) {
    return;
  }
  if (((unsigned int)base & 3) || (sz & 3) || (shift & 3)) {
    unsigned int i0 = 0;
    unsigned int i1 = 0;
    char *p = base;
    char tmp = p[i0];
    do {
      i0 = i1;
      i1 = (((shift + i0) >= (sz)) ? ((shift + i0) - (sz)) : (shift + i0));
      p[i0] = p[i1];
    } while (i1 != 0);
    p[i0] = tmp;
  } else {
    mem_swap_near_block_32(base, sz / 4, shift / 4);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<mem_swap_near_block>:
  endbr64
  mov    %rdi,%r8
  test   %edx,%edx
  je     L29
  test   %ecx,%ecx
  je     L29
  add    %edx,%ecx
  imul   %esi,%edx
  mov    %rdi,%rax
  imul   %esi,%ecx
  or     %rcx,%rax
  test   $0x3,%al
  je     L2
L20:
  movzbl (%r8),%r9d
  xor    %edi,%edi
  jmp    L25
  nopl   0x0(%rax)
L46:
  mov    %eax,%edi
L25:
  lea    (%rdx,%rdi,1),%eax
  mov    %eax,%esi
  sub    %ecx,%esi
  cmp    %eax,%ecx
  cmovbe %esi,%eax
  add    %r8,%rdi
  mov    %eax,%esi
  movzbl (%r8,%rsi,1),%esi
  mov    %sil,(%rdi)
  test   %eax,%eax
  jne    L46
  mov    %r9b,(%rdi)
  ret
  nopl   0x0(%rax,%rax,1)
L29:
  ret
  nopl   0x0(%rax)
L2:
  test   $0x3,%dl
  jne    L20
  shr    $0x2,%ecx
  shr    $0x2,%edx
  mov    %ecx,%esi
  jmp    1050 <mem_swap_near_block_32@plt>
```

assistant: ```c
void mem_swap_near_block(void *base, unsigned int elem_size, unsigned int cnt1, unsigned int cnt2) {
  unsigned int sz = elem_size * (cnt1 + cnt2);
  unsigned int shift = elem_size * cnt1;
  if (cnt1 == 0 || cnt2 == 0) {
    return;
  }
  if (((unsigned int)base & 3) || (sz & 3) || (shift & 3)) {
    unsigned int i0 = 0;
    unsigned int i1 = 0;
    char *p = base;
    char tmp = p[i0];
    do {
      i0 = i1;
      i1 = (((shift + i0) >= (sz)) ? ((shift + i0) - (sz)) : (shift + i0));
      p[i0] = p[i1];
    } while (i1 != 0);
    p[i0] = tmp;
  } else {
    mem_swap_near_block_32(base, sz / 4, shift / 4);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L13
  mov    $0x0,%eax
  jmp    L24
L13:
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  js     L2
L47:
  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
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%eax
  add    $0x30,%eax
  subq   $0x1,-0x8(%rbp)
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  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)
  cmpl   $0x0,-0x14(%rbp)
  jne    L47
  mov    -0x8(%rbp),%rax
  jmp    L24
L2:
  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
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    $0x30,%eax
  sub    %edx,%eax
  subq   $0x1,-0x8(%rbp)
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  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)
  cmpl   $0x0,-0x14(%rbp)
  jne    L2
  subq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movb   $0x2d,(%rax)
  mov    -0x8(%rbp),%rax
L24:
  pop    %rbp
  ret
```

assistant: ```c
const char *itoa(int i, char *str, unsigned long length) {
  if (length <= 0)
    return 0;
  str[length - 1] = '\0';
  char *p = str + length - 1;
  if (i >= 0) {
    do {
      *--p = '0' + (i % 10);
      i /= 10;
    } while (i != 0);
    return p;
  } else {
    do {
      *--p = '0' - (i % 10);
      i /= 10;
    } while (i != 0);
    *--p = '-';
  }
  return p;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<itoa>:
  endbr64
  mov    %edi,%ecx
  test   %rdx,%rdx
  je     L43
  lea    -0x1(%rsi,%rdx,1),%rdx
  movb   $0x0,(%rdx)
  mov    $0x30,%edi
  test   %ecx,%ecx
  jns    L22
L21:
  mov    %rdx,%r8
  sub    $0x1,%rdx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  lea    (%rdi,%rsi,2),%esi
  sub    %ecx,%esi
  mov    %sil,(%rdx)
  mov    %eax,%ecx
  test   %eax,%eax
  jne    L21
  movb   $0x2d,-0x1(%rdx)
  lea    -0x2(%r8),%rdx
L31:
  mov    %rdx,%rax
  ret
L22:
  sub    $0x1,%rdx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%ecx
  add    $0x30,%ecx
  mov    %cl,(%rdx)
  mov    %eax,%ecx
  test   %eax,%eax
  jne    L22
  jmp    L31
L43:
  mov    $0x0,%edx
  jmp    L31
```

assistant: ```c
const char *itoa(int i, char *str, unsigned long length) {
  if (length <= 0)
    return 0;
  str[length - 1] = '\0';
  char *p = str + length - 1;
  if (i >= 0) {
    do {
      *--p = '0' + (i % 10);
      i /= 10;
    } while (i != 0);
    return p;
  } else {
    do {
      *--p = '0' - (i % 10);
      i /= 10;
    } while (i != 0);
    *--p = '-';
  }
  return p;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<itoa>:
  endbr64
  mov    %edi,%eax
  test   %rdx,%rdx
  je     L31
  lea    -0x1(%rsi,%rdx,1),%r8
  mov    $0x30,%esi
  movb   $0x0,(%r8)
  test   %edi,%edi
  jns    L22
  nopl   0x0(%rax)
L29:
  movslq %eax,%rdx
  mov    %eax,%ecx
  mov    %r8,%rdi
  sub    $0x1,%r8
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  lea    (%rsi,%rcx,2),%ecx
  sub    %eax,%ecx
  mov    %edx,%eax
  mov    %cl,(%r8)
  test   %edx,%edx
  jne    L29
  movb   $0x2d,-0x1(%r8)
  lea    -0x2(%rdi),%r8
  mov    %r8,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L22:
  mov    $0xcccccccd,%esi
  nopl   (%rax)
L11:
  mov    %eax,%edx
  sub    $0x1,%r8
  imul   %rsi,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  add    $0x30,%eax
  mov    %al,(%r8)
  mov    %edx,%eax
  test   %edx,%edx
  jne    L11
  mov    %r8,%rax
  ret
  nopl   (%rax)
L31:
  xor    %r8d,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
const char *itoa(int i, char *str, unsigned long length) {
  if (length <= 0)
    return 0;
  str[length - 1] = '\0';
  char *p = str + length - 1;
  if (i >= 0) {
    do {
      *--p = '0' + (i % 10);
      i /= 10;
    } while (i != 0);
    return p;
  } else {
    do {
      *--p = '0' - (i % 10);
      i /= 10;
    } while (i != 0);
    *--p = '-';
  }
  return p;
}
```
user: What is the c source code of the assembly code below?

```asm
<itoa>:
  endbr64
  mov    %edi,%eax
  test   %rdx,%rdx
  je     L37
  lea    -0x1(%rsi,%rdx,1),%r8
  mov    $0x30,%esi
  movb   $0x0,(%r8)
  test   %edi,%edi
  jns    L14
  nopl   0x0(%rax)
L38:
  movslq %eax,%rdx
  mov    %eax,%ecx
  mov    %r8,%rdi
  sub    $0x1,%r8
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  lea    (%rsi,%rcx,2),%ecx
  sub    %eax,%ecx
  mov    %edx,%eax
  mov    %cl,(%r8)
  test   %edx,%edx
  jne    L38
  movb   $0x2d,-0x1(%r8)
  lea    -0x2(%rdi),%r8
  mov    %r8,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  mov    $0xcccccccd,%esi
  nopl   (%rax)
L18:
  mov    %eax,%edx
  sub    $0x1,%r8
  imul   %rsi,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  add    $0x30,%eax
  mov    %al,(%r8)
  mov    %edx,%eax
  test   %edx,%edx
  jne    L18
  mov    %r8,%rax
  ret
  nopl   (%rax)
L37:
  xor    %r8d,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
const char *itoa(int i, char *str, unsigned long length) {
  if (length <= 0)
    return 0;
  str[length - 1] = '\0';
  char *p = str + length - 1;
  if (i >= 0) {
    do {
      *--p = '0' + (i % 10);
      i /= 10;
    } while (i != 0);
    return p;
  } else {
    do {
      *--p = '0' - (i % 10);
      i /= 10;
    } while (i != 0);
    *--p = '-';
  }
  return p;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<scout_path_noscheme>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x3a,%esi
  mov    %rax,%rdi
  call   <scout_any_segment@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L7
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    %rax,-0x8(%rbp)
L29:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <scout_path_abempty@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L13
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    %rax,-0x8(%rbp)
L13:
  cmpq   $0x0,-0x18(%rbp)
  jne    L29
L7:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
const char *scout_path_noscheme(const char *c) {
  const char *p = NULL;
  c = scout_any_segment(c, ':');
  if (c != NULL) {
    p = c++;
    do {
      c = scout_path_abempty(c);
      if (c != NULL)
        p = c++;
    } while (c != NULL);
  }
  return p;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<scout_path_noscheme>:
  endbr64
  push   %rbx
  mov    $0x3a,%esi
  call   <scout_any_segment@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L4
  lea    0x1(%rax),%rdi
  jmp    L27
L11:
  lea    0x1(%rax),%rdi
  mov    %rax,%rbx
L27:
  call   <scout_path_abempty@plt>
  test   %rax,%rax
  jne    L11
L4:
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
const char *scout_path_noscheme(const char *c) {
  const char *p = NULL;
  c = scout_any_segment(c, ':');
  if (c != NULL) {
    p = c++;
    do {
      c = scout_path_abempty(c);
      if (c != NULL)
        p = c++;
    } while (c != NULL);
  }
  return p;
}
```
user: Translate the below assembly code into C programming language:

```asm
<scout_path_noscheme>:
  endbr64
  push   %r12
  mov    $0x3a,%esi
  call   <scout_any_segment@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L19
  lea    0x1(%rax),%rdi
  call   <scout_path_abempty@plt>
  test   %rax,%rax
  je     L19
  cs nopw 0x0(%rax,%rax,1)
L26:
  lea    0x1(%rax),%rdi
  mov    %rax,%r12
  call   <scout_path_abempty@plt>
  test   %rax,%rax
  jne    L26
L19:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
const char *scout_path_noscheme(const char *c) {
  const char *p = NULL;
  c = scout_any_segment(c, ':');
  if (c != NULL) {
    p = c++;
    do {
      c = scout_path_abempty(c);
      if (c != NULL)
        p = c++;
    } while (c != NULL);
  }
  return p;
}
```
user: What is the c source code of the assembly code below:

```asm
<scout_path_noscheme>:
  endbr64
  push   %r12
  mov    $0x3a,%esi
  call   <scout_any_segment@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L2
  lea    0x1(%rax),%rdi
  call   <scout_path_abempty@plt>
  test   %rax,%rax
  je     L2
  cs nopw 0x0(%rax,%rax,1)
L24:
  lea    0x1(%rax),%rdi
  mov    %rax,%r12
  call   <scout_path_abempty@plt>
  test   %rax,%rax
  jne    L24
L2:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
const char *scout_path_noscheme(const char *c) {
  const char *p = NULL;
  c = scout_any_segment(c, ':');
  if (c != NULL) {
    p = c++;
    do {
      c = scout_path_abempty(c);
      if (c != NULL)
        p = c++;
    } while (c != NULL);
  }
  return p;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<func_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  mov    $0x1,%edi
  call   <step_hash@plt>
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int func_1(void) {
  int l_2 = 0L;
  step_hash(1);
  return l_2;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<func_1>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <step_hash@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int func_1(void) {
  int l_2 = 0L;
  step_hash(1);
  return l_2;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<func_1>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <step_hash@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int func_1(void) {
  int l_2 = 0L;
  step_hash(1);
  return l_2;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<func_1>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <step_hash@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int func_1(void) {
  int l_2 = 0L;
  step_hash(1);
  return l_2;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mstrncmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L54
L12:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L10
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jne    L16
L10:
  mov    $0x0,%eax
  jmp    L31
L16:
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
L54:
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  cmp    %eax,-0x24(%rbp)
  jle    L24
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L12
L24:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
L31:
  pop    %rbp
  ret
```

assistant: ```c
int mstrncmp(char *s, char *t, int n) {
  int i = 0;
  for (; (i++ < n) && (*s == *t); s++, t++) {
    if ((*s == '\0') || (i == n))
      return 0;
  }
  return *s - *t;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mstrncmp>:
  endbr64
  test   %edx,%edx
  jle    L16
  lea    0x1(%rdx),%r8d
  mov    $0x1,%eax
L10:
  movzbl (%rdi),%ecx
  cmp    (%rsi),%cl
  jne    L16
  cmp    %eax,%edx
  je     L28
  test   %cl,%cl
  je     L28
  add    $0x1,%rdi
  add    $0x1,%rsi
  add    $0x1,%eax
  cmp    %r8d,%eax
  jne    L10
L16:
  movsbl (%rdi),%eax
  movsbl (%rsi),%edx
  sub    %edx,%eax
  ret
L28:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int mstrncmp(char *s, char *t, int n) {
  int i = 0;
  for (; (i++ < n) && (*s == *t); s++, t++) {
    if ((*s == '\0') || (i == n))
      return 0;
  }
  return *s - *t;
}
```
user: Translate the below assembly code into C programming language:

```asm
<mstrncmp>:
  endbr64
  lea    0x1(%rdx),%r9d
  mov    $0x1,%eax
  test   %edx,%edx
  jg     L49
  jmp    L8
  nopl   0x0(%rax,%rax,1)
L10:
  cmp    %eax,%edx
  je     L38
  test   %cl,%cl
  je     L38
  add    $0x1,%eax
  lea    0x1(%rdi),%rcx
  lea    0x1(%rsi),%r8
  cmp    %r9d,%eax
  je     L41
  mov    %r8,%rsi
  mov    %rcx,%rdi
L49:
  movzbl (%rdi),%ecx
  movsbl (%rsi),%r8d
  cmp    %r8b,%cl
  je     L10
L27:
  movsbl %cl,%eax
  sub    %r8d,%eax
  ret
  nopl   0x0(%rax)
L38:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L41:
  movzbl 0x1(%rdi),%ecx
  movsbl 0x1(%rsi),%r8d
  movsbl %cl,%eax
  sub    %r8d,%eax
  ret
L8:
  movzbl (%rdi),%ecx
  movsbl (%rsi),%r8d
  jmp    L27
```

assistant: ```c
int mstrncmp(char *s, char *t, int n) {
  int i = 0;
  for (; (i++ < n) && (*s == *t); s++, t++) {
    if ((*s == '\0') || (i == n))
      return 0;
  }
  return *s - *t;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<mstrncmp>:
  endbr64
  lea    0x1(%rdx),%r9d
  mov    $0x1,%eax
  test   %edx,%edx
  jg     L60
  jmp    L34
  nopl   0x0(%rax,%rax,1)
L20:
  cmp    %eax,%edx
  je     L38
  test   %cl,%cl
  je     L38
  add    $0x1,%eax
  lea    0x1(%rdi),%rcx
  lea    0x1(%rsi),%r8
  cmp    %r9d,%eax
  je     L22
  mov    %r8,%rsi
  mov    %rcx,%rdi
L60:
  movzbl (%rdi),%ecx
  movsbl (%rsi),%r8d
  cmp    %r8b,%cl
  je     L20
L40:
  movsbl %cl,%eax
  sub    %r8d,%eax
  ret
  nopl   0x0(%rax)
L38:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L22:
  movzbl 0x1(%rdi),%ecx
  movsbl 0x1(%rsi),%r8d
  movsbl %cl,%eax
  sub    %r8d,%eax
  ret
L34:
  movzbl (%rdi),%ecx
  movsbl (%rsi),%r8d
  jmp    L40
```

assistant: ```c
int mstrncmp(char *s, char *t, int n) {
  int i = 0;
  for (; (i++ < n) && (*s == *t); s++, t++) {
    if ((*s == '\0') || (i == n))
      return 0;
  }
  return *s - *t;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x18(%rbp),%rsi
  lea    -0x20(%rbp),%rcx
  lea    -0x28(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D70(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D89(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D77(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movsd  -0x10(%rbp),%xmm0
  movsd  -0x30(%rbp),%xmm1
  comisd %xmm1,%xmm0
  jb     L28
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L4
L28:
  movsd  -0x10(%rbp),%xmm0
  movsd  -0x28(%rbp),%xmm1
  comisd %xmm1,%xmm0
  jb     L3
  lea    D75(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L4
L3:
  movsd  -0x10(%rbp),%xmm0
  movsd  -0x20(%rbp),%xmm1
  comisd %xmm1,%xmm0
  jb     L57
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L4
L57:
  movsd  -0x10(%rbp),%xmm0
  movsd  -0x18(%rbp),%xmm1
  comisd %xmm1,%xmm0
  jb     L7
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L4
L7:
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L4:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L22
  call   <__stack_chk_fail@plt>
L22:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
data_type: string
name: retrieve_data
arguments: data_tag: D70
data_type: string
name: retrieve_data
arguments: data_tag: D89
data_type: string
name: retrieve_data
arguments: data_tag: D77
data_type: string
name: retrieve_data
arguments: data_tag: D59
data_type: string
name: retrieve_data
arguments: data_tag: D75
data_type: string
name: retrieve_data
arguments: data_tag: D22
data_type: string
name: retrieve_data
arguments: data_tag: D24
data_type: string
name: retrieve_data
arguments: data_tag: D54
data_type: string

tool: D3: Enter thresholds for A, B, C, D
in that order, decreasing percentages > 

tool: D70: %lf %lf %lf %lf

tool: D89: Thank you. Now enter student score (percent) >

tool: D77: %lf

tool: D59: Student has an A grade


tool: D75: Student has an B grade


tool: D22: Student has an C grade


tool: D24: Student has an D grade


tool: D54: Student has failed the course


assistant: ```c
int main() {
  double a, b, c, d;
  double score;
  printf("Enter thresholds for A, B, C, D\nin that order, decreasing percentages > ");
  scanf("%lf %lf %lf %lf", &a, &b, &c, &d);
  printf("Thank you. Now enter student score (percent) >");
  scanf("%lf", &score);
  if (score >= a)
    printf("Student has an A grade\n");
  else if (score >= b)
    printf("Student has an B grade\n");
  else if (score >= c)
    printf("Student has an C grade\n");
  else if (score >= d)
    printf("Student has an D grade\n");
  else
    printf("Student has failed the course\n");
  return 0;
}
```