ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<reverse_string>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%rsi
  cmp    %rsi,%rbx
  ja     L14
  lea    -0x1(%rbx),%rdi
  mov    %rsi,%rax
  mov    %r12,%rdx
  nop
L8:
  movzbl (%rax),%ecx
  sub    $0x1,%rax
  add    $0x1,%rdx
  mov    %cl,-0x1(%rdx)
  cmp    %rax,%rdi
  jne    L8
  sub    %rbx,%rsi
  lea    0x1(%r12,%rsi,1),%rax
  movb   $0x0,(%rax)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L14:
  mov    %r12,%rax
  movb   $0x0,(%rax)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *reverse_string(char *destination, char *source) {
  char *marker = destination;
  char *ptr = source + strlen(source) - 1;
  while (ptr >= source) {
    *destination++ = *ptr--;
  }
  *destination = '\0';
  return marker;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_memcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L16
L18:
  mov    -0x14(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    -0x14(%rbp),%edx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x14(%rbp)
L16:
  mov    -0x14(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jb     L18
  mov    -0x28(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
void *ft_memcpy(void *dst, const void *src, unsigned int size) {
  unsigned int i;
  int *s;
  int *d;
  s = (int *)src;
  d = (int *)dst;
  i = 0;
  while (i < size) {
    d[i] = s[i];
    i++;
  }
  return (dst);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_memcpy>:
  endbr64
  mov    %rdi,%rax
  test   %edx,%edx
  je     L2
  mov    %edx,%edx
  mov    $0x0,%ecx
L14:
  mov    (%rsi,%rcx,4),%r8d
  mov    %r8d,(%rax,%rcx,4)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L14
L2:
  ret
```

assistant: ```c
void *ft_memcpy(void *dst, const void *src, unsigned int size) {
  unsigned int i;
  int *s;
  int *d;
  s = (int *)src;
  d = (int *)dst;
  i = 0;
  while (i < size) {
    d[i] = s[i];
    i++;
  }
  return (dst);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ft_memcpy>:
  endbr64
  mov    %rdi,%rax
  test   %edx,%edx
  je     L12
  mov    %edx,%edx
  xor    %ecx,%ecx
  nop
L3:
  mov    (%rsi,%rcx,4),%r8d
  mov    %r8d,(%rax,%rcx,4)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L3
L12:
  ret
```

assistant: ```c
void *ft_memcpy(void *dst, const void *src, unsigned int size) {
  unsigned int i;
  int *s;
  int *d;
  s = (int *)src;
  d = (int *)dst;
  i = 0;
  while (i < size) {
    d[i] = s[i];
    i++;
  }
  return (dst);
}
```
user: What is the c source code of the assembly code below:

```asm
<ft_memcpy>:
  endbr64
  mov    %rdi,%r8
  mov    %rsi,%rcx
  test   %edx,%edx
  je     L20
  lea    0x4(%rsi),%rsi
  mov    %rdi,%rax
  sub    %rsi,%rax
  cmp    $0x8,%rax
  jbe    L30
  lea    -0x1(%rdx),%eax
  cmp    $0x2,%eax
  jbe    L30
  mov    %edx,%esi
  xor    %eax,%eax
  shr    $0x2,%esi
  shl    $0x4,%rsi
  nopl   0x0(%rax)
L14:
  movdqu (%rcx,%rax,1),%xmm0
  movups %xmm0,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %rsi,%rax
  jne    L14
  mov    %edx,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%dl
  je     L20
  mov    %eax,%esi
  mov    (%rcx,%rsi,4),%edi
  mov    %edi,(%r8,%rsi,4)
  lea    0x1(%rax),%esi
  cmp    %esi,%edx
  jbe    L20
  mov    (%rcx,%rsi,4),%edi
  add    $0x2,%eax
  mov    %edi,(%r8,%rsi,4)
  cmp    %eax,%edx
  jbe    L20
  mov    (%rcx,%rax,4),%edx
  mov    %edx,(%r8,%rax,4)
L20:
  mov    %r8,%rax
  ret
  xchg   %ax,%ax
L30:
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L44:
  mov    (%rcx,%rax,4),%esi
  mov    %esi,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L44
  mov    %r8,%rax
  ret
```

assistant: ```c
void *ft_memcpy(void *dst, const void *src, unsigned int size) {
  unsigned int i;
  int *s;
  int *d;
  s = (int *)src;
  d = (int *)dst;
  i = 0;
  while (i < size) {
    d[i] = s[i];
    i++;
  }
  return (dst);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<string_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L1
L10:
  mov    -0x10(%rbp),%rax
  shl    $0x4,%rax
  mov    %rax,%rcx
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %rcx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  and    $0xf0000000,%eax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L15
  mov    -0x8(%rbp),%rax
  shr    $0x18,%rax
  xor    %rax,-0x10(%rbp)
L15:
  mov    -0x8(%rbp),%rax
  not    %rax
  and    %rax,-0x10(%rbp)
L1:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L10
  mov    -0x10(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long string_hash(const unsigned char *name) {
  unsigned long h = 0, g;
  while (*name) {
    h = (h << 4) + *name++;
    if ((g = h & 0xf0000000) != 0) {
      h ^= g >> 24;
    }
    h &= ~g;
  }
  return h;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<string_hash>:
  endbr64
  movzbl (%rdi),%ecx
  add    $0x1,%rdi
  movzbl %cl,%edx
  mov    $0x0,%eax
  test   %cl,%cl
  jne    L11
  ret
L0:
  shl    $0x4,%rax
  add    $0x1,%rdi
  movzbl %dl,%edx
  add    %rax,%rdx
  mov    %rdx,%rax
  and    $0xf0000000,%eax
  mov    %rax,%rcx
  shr    $0x18,%rcx
  xor    %rdx,%rcx
  test   %rax,%rax
  cmovne %rcx,%rdx
L11:
  not    %rax
  and    %rdx,%rax
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L0
  ret
```

assistant: ```c
unsigned long string_hash(const unsigned char *name) {
  unsigned long h = 0, g;
  while (*name) {
    h = (h << 4) + *name++;
    if ((g = h & 0xf0000000) != 0) {
      h ^= g >> 24;
    }
    h &= ~g;
  }
  return h;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<string_hash>:
  endbr64
  movzbl (%rdi),%edx
  xor    %eax,%eax
  add    $0x1,%rdi
  test   %dl,%dl
  jne    L11
  jmp    L19
  nopl   0x0(%rax,%rax,1)
L21:
  shl    $0x4,%rax
  add    $0x1,%rdi
  add    %rax,%rdx
  mov    %rdx,%rax
  and    $0xf0000000,%eax
  mov    %rax,%rcx
  shr    $0x18,%rcx
  xor    %rdx,%rcx
  test   %rax,%rax
  cmovne %rcx,%rdx
L11:
  not    %rax
  and    %rdx,%rax
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L21
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  ret
```

assistant: ```c
unsigned long string_hash(const unsigned char *name) {
  unsigned long h = 0, g;
  while (*name) {
    h = (h << 4) + *name++;
    if ((g = h & 0xf0000000) != 0) {
      h ^= g >> 24;
    }
    h &= ~g;
  }
  return h;
}
```
user: Translate the below assembly code into C programming language:

```asm
<string_hash>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L32
  movzbl 0x1(%rdi),%edx
  lea    0x1(%rdi),%rcx
  test   %dl,%dl
  je     L11
  nopw   0x0(%rax,%rax,1)
L25:
  shl    $0x4,%rax
  add    $0x1,%rcx
  add    %rdx,%rax
  mov    %rax,%rdx
  and    $0xf0000000,%edx
  je     L20
  mov    %rdx,%rsi
  not    %rdx
  shr    $0x18,%rsi
  xor    %rsi,%rax
  and    %rdx,%rax
  movzbl (%rcx),%edx
  test   %dl,%dl
  jne    L25
  ret
  xchg   %ax,%ax
L20:
  movzbl (%rcx),%edx
  test   %dl,%dl
  jne    L25
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  ret
  nopl   0x0(%rax)
L32:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned long string_hash(const unsigned char *name) {
  unsigned long h = 0, g;
  while (*name) {
    h = (h << 4) + *name++;
    if ((g = h & 0xf0000000) != 0) {
      h ^= g >> 24;
    }
    h &= ~g;
  }
  return h;
}
```
user: What is the c source code of the assembly code below?

```asm
<roweliminate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  movsbq -0x1c(%rbp),%rax
  imul   $0x5a,%rax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  movsbl -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L35
  mov    $0x0,%eax
  jmp    L33
L35:
  movb   $0x0,-0x1(%rbp)
  movb   $0x0,-0x2(%rbp)
  jmp    L29
L15:
  movzbl -0x2(%rbp),%eax
  cmp    -0x20(%rbp),%al
  je     L6
  movsbq -0x1c(%rbp),%rax
  imul   $0x5a,%rax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rdi
  movsbl -0x2(%rbp),%ecx
  movsbq -0x1c(%rbp),%rax
  imul   $0x5a,%rax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  movsbl -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rsi,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  movslq %eax,%rsi
  movslq %ecx,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rdi,%rax
  add    %rsi,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L6
  movsbq -0x1c(%rbp),%rax
  imul   $0x5a,%rax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rdi
  movsbl -0x2(%rbp),%ecx
  movsbq -0x1c(%rbp),%rax
  imul   $0x5a,%rax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  movsbl -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rsi,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  movslq %eax,%rsi
  movslq %ecx,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rdi,%rax
  add    %rsi,%rax
  movb   $0x0,(%rax)
  movb   $0x1,-0x1(%rbp)
L6:
  movzbl -0x2(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x2(%rbp)
L29:
  cmpb   $0x8,-0x2(%rbp)
  jle    L15
  movsbl -0x1(%rbp),%eax
L33:
  pop    %rbp
  ret
```

assistant: ```c
int roweliminate(char grid[9][9][10], char m, char n) {
  if (!grid[m][n][0])
    return 0;
  char j, f = 0;
  for (j = 0; j < 9; ++j)
    if (j != n)
      if (grid[m][j][grid[m][n][0]])
        grid[m][j][grid[m][n][0]] = 0, f = 1;
  return f;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<roweliminate>:
  endbr64
  mov    %edx,%r9d
  movsbq %sil,%rsi
  imul   $0x5a,%rsi,%rsi
  add    %rsi,%rdi
  movsbq %dl,%rdx
  lea    (%rdx,%rdx,4),%rax
  lea    (%rdi,%rax,2),%r10
  mov    $0x0,%eax
  cmpb   $0x0,(%r10)
  je     L4
  mov    $0x0,%edx
  mov    $0x1,%r11d
  jmp    L32
L53:
  add    $0x1,%edx
  cmp    $0x9,%edx
  je     L25
L32:
  cmp    %dl,%r9b
  je     L53
  movsbl (%r10),%ecx
  movslq %ecx,%r8
  movslq %edx,%rsi
  lea    (%rsi,%rsi,4),%rsi
  lea    (%rdi,%rsi,2),%rsi
  cmpb   $0x0,(%rsi,%r8,1)
  je     L53
  movb   $0x0,(%rsi,%r8,1)
  mov    %r11d,%eax
  jmp    L53
L25:
  movsbl %al,%eax
L4:
  ret
```

assistant: ```c
int roweliminate(char grid[9][9][10], char m, char n) {
  if (!grid[m][n][0])
    return 0;
  char j, f = 0;
  for (j = 0; j < 9; ++j)
    if (j != n)
      if (grid[m][j][grid[m][n][0]])
        grid[m][j][grid[m][n][0]] = 0, f = 1;
  return f;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<roweliminate>:
  endbr64
  movsbq %sil,%rsi
  mov    %edx,%r8d
  movsbq %dl,%rdx
  imul   $0x5a,%rsi,%rsi
  lea    (%rdx,%rdx,4),%rax
  add    %rsi,%rdi
  lea    (%rdi,%rax,2),%r9
  xor    %eax,%eax
  cmpb   $0x0,(%r9)
  je     L2
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L21:
  cmp    %dl,%r8b
  je     L36
  movslq %edx,%rcx
  movsbq (%r9),%rsi
  lea    (%rcx,%rcx,4),%rcx
  lea    (%rdi,%rcx,2),%rcx
  add    %rsi,%rcx
  cmpb   $0x0,(%rcx)
  je     L36
  movb   $0x0,(%rcx)
  mov    $0x1,%eax
L36:
  add    $0x1,%edx
  cmp    $0x9,%edx
  jne    L21
  movsbl %al,%eax
L2:
  ret
```

assistant: ```c
int roweliminate(char grid[9][9][10], char m, char n) {
  if (!grid[m][n][0])
    return 0;
  char j, f = 0;
  for (j = 0; j < 9; ++j)
    if (j != n)
      if (grid[m][j][grid[m][n][0]])
        grid[m][j][grid[m][n][0]] = 0, f = 1;
  return f;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<roweliminate>:
  endbr64
  movsbq %sil,%rsi
  mov    %edx,%r8d
  imul   $0x5a,%rsi,%rsi
  movsbq %r8b,%rax
  lea    (%rax,%rax,4),%rax
  lea    (%rdi,%rsi,1),%rdx
  lea    (%rdx,%rax,2),%rsi
  xor    %eax,%eax
  movsbq (%rsi),%rcx
  test   %cl,%cl
  je     L159
  test   %r8b,%r8b
  je     L167
  movsbl (%rdx,%rcx,1),%eax
  test   %al,%al
  je     L24
  movb   $0x0,(%rdx,%rcx,1)
  mov    $0x1,%eax
L24:
  cmp    $0x1,%r8b
  je     L153
  movsbq (%rsi),%rdi
  cmpb   $0x0,0xa(%rdx,%rdi,1)
  mov    %rdi,%rcx
  je     L92
L60:
  movb   $0x0,0xa(%rdx,%rdi,1)
  movsbq (%rsi),%rcx
  mov    $0x1,%eax
L92:
  cmp    $0x2,%r8b
  je     L45
  movslq %ecx,%rdi
  cmpb   $0x0,0x14(%rdx,%rdi,1)
  je     L184
L99:
  movb   $0x0,0x14(%rdx,%rdi,1)
  movsbq (%rsi),%rcx
  mov    $0x1,%eax
L184:
  cmp    $0x3,%r8b
  je     L204
L45:
  movslq %ecx,%rdi
  cmpb   $0x0,0x1e(%rdx,%rdi,1)
  je     L112
  movb   $0x0,0x1e(%rdx,%rdi,1)
  movsbq (%rsi),%rcx
  mov    $0x1,%eax
L112:
  cmp    $0x4,%r8b
  je     L94
L204:
  movslq %ecx,%rdi
  cmpb   $0x0,0x28(%rdx,%rdi,1)
  je     L155
  movb   $0x0,0x28(%rdx,%rdi,1)
  movsbq (%rsi),%rcx
  mov    $0x1,%eax
L155:
  cmp    $0x5,%r8b
  je     L141
L94:
  movslq %ecx,%rdi
  cmpb   $0x0,0x32(%rdx,%rdi,1)
  je     L110
  movb   $0x0,0x32(%rdx,%rdi,1)
  movsbq (%rsi),%rcx
  mov    $0x1,%eax
L110:
  cmp    $0x6,%r8b
  je     L81
L141:
  movslq %ecx,%rdi
  cmpb   $0x0,0x3c(%rdx,%rdi,1)
  je     L76
  movb   $0x0,0x3c(%rdx,%rdi,1)
  movsbq (%rsi),%rcx
  mov    $0x1,%eax
L76:
  cmp    $0x7,%r8b
  je     L104
L81:
  cmpb   $0x0,0x46(%rdx,%rcx,1)
  je     L128
  movb   $0x0,0x46(%rdx,%rcx,1)
  mov    $0x1,%eax
L128:
  cmp    $0x8,%r8b
  je     L159
  movsbq (%rsi),%rcx
L104:
  cmpb   $0x0,0x50(%rdx,%rcx,1)
  je     L159
  movb   $0x0,0x50(%rdx,%rcx,1)
  mov    $0x1,%eax
L159:
  ret
  cs nopw 0x0(%rax,%rax,1)
L153:
  movsbq (%rsi),%rcx
  movslq %ecx,%rdi
  cmpb   $0x0,0x14(%rdx,%rdi,1)
  je     L184
  jmp    L99
  nopw   0x0(%rax,%rax,1)
L167:
  movsbq (%rsi),%rdi
  xor    %eax,%eax
  cmpb   $0x0,0xa(%rdx,%rdi,1)
  mov    %rdi,%rcx
  je     L92
  jmp    L60
```

assistant: ```c
int roweliminate(char grid[9][9][10], char m, char n) {
  if (!grid[m][n][0])
    return 0;
  char j, f = 0;
  for (j = 0; j < 9; ++j)
    if (j != n)
      if (grid[m][j][grid[m][n][0]])
        grid[m][j][grid[m][n][0]] = 0, f = 1;
  return f;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<signed_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movb   $0x65,-0x5(%rbp)
  movl   $0x3,-0x4(%rbp)
  movsbl -0x5(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  movsbl %al,%edx
  movsbl -0x5(%rbp),%esi
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  movsbl %al,%eax
  mov    %edx,%ecx
  mov    %eax,%edx
  mov    $0x4,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D4
data_type: string

tool: D4: %lu, %d, %d


assistant: ```c
int signed_char(void) {
  signed char a = 0x65;
  int b = 3;
  return printf("%lu, %d, %d\n", sizeof(a << b), a << b, a >> b);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<signed_char>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xc,%ecx
  mov    $0x28,%edx
  mov    $0x4,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D4
data_type: string

tool: D4: %lu, %d, %d


assistant: ```c
int signed_char(void) {
  signed char a = 0x65;
  int b = 3;
  return printf("%lu, %d, %d\n", sizeof(a << b), a << b, a >> b);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<signed_char>:
  endbr64
  mov    $0xc,%ecx
  mov    $0x28,%edx
  mov    $0x4,%esi
  lea    D7(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: data_tag: D7
type: string

tool: D7: %lu, %d, %d


assistant: ```c
int signed_char(void) {
  signed char a = 0x65;
  int b = 3;
  return printf("%lu, %d, %d\n", sizeof(a << b), a << b, a >> b);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<signed_char>:
  endbr64
  mov    $0xc,%ecx
  mov    $0x28,%edx
  mov    $0x4,%esi
  lea    D8(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: tag: D8
type_infered: string

tool: D8: %lu, %d, %d


assistant: ```c
int signed_char(void) {
  signed char a = 0x65;
  int b = 3;
  return printf("%lu, %d, %d\n", sizeof(a << b), a << b, a >> b);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<plusClairNB>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    %rcx,-0x38(%rbp)
  mov    %r8,-0x40(%rbp)
  mov    -0x2c(%rbp),%eax
  imul   -0x30(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x2c(%rbp),%eax
  imul   -0x30(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L13
L29:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x14(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  shr    %al
  movzbl %al,%eax
  add    %edx,%eax
  mov    %eax,-0x10(%rbp)
  cmpl   $0xff,-0x10(%rbp)
  jbe    L25
  movl   $0xff,-0x10(%rbp)
L25:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x14(%rbp)
L13:
  mov    -0x14(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L29
  mov    -0x38(%rbp),%rax
  mov    -0x2c(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x40(%rbp),%rax
  mov    -0x30(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
unsigned char *plusClairNB(unsigned char *img, int xs, int ys, int *nc, int *nl) {
  unsigned char *res = (unsigned char *)malloc(xs * ys);
  int nbPixels = xs * ys;
  int i;
  for (i = 0; i < nbPixels; i++) {
    unsigned int v = img[i] + img[i] / 2;
    if (v > 255) {
      v = 255;
    }
    res[i] = v;
  }
  *nc = xs;
  *nl = ys;
  return res;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<plusClairNB>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r14
  mov    %esi,%ebp
  mov    %edx,%ebx
  mov    %rcx,%r13
  mov    %r8,%r12
  mov    %esi,%r15d
  imul   %edx,%r15d
  mov    %r15d,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  test   %r15d,%r15d
  jle    L4
  mov    %r15d,%esi
  add    %rax,%rsi
  mov    %rax,%rdx
  sub    %rax,%r14
L1:
  movzbl (%r14,%rdx,1),%ecx
  mov    %ecx,%eax
  shr    %al
  movzbl %al,%eax
  movzbl %cl,%ecx
  add    %ecx,%eax
  mov    $0xff,%ecx
  cmp    %ecx,%eax
  cmova  %ecx,%eax
  mov    %al,(%rdx)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L1
L4:
  mov    %ebp,0x0(%r13)
  mov    %ebx,(%r12)
  mov    %rdi,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
unsigned char *plusClairNB(unsigned char *img, int xs, int ys, int *nc, int *nl) {
  unsigned char *res = (unsigned char *)malloc(xs * ys);
  int nbPixels = xs * ys;
  int i;
  for (i = 0; i < nbPixels; i++) {
    unsigned int v = img[i] + img[i] / 2;
    if (v > 255) {
      v = 255;
    }
    res[i] = v;
  }
  *nc = xs;
  *nl = ys;
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<plusClairNB>:
  endbr64
  push   %r15
  mov    %esi,%r15d
  push   %r14
  imul   %edx,%r15d
  mov    %r8,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    %r15d,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r8
  test   %r15d,%r15d
  jle    L6
  movslq %r15d,%rsi
  mov    %rax,%rdx
  sub    %rax,%rbx
  add    %rax,%rsi
  nopl   0x0(%rax,%rax,1)
L5:
  movzbl (%rbx,%rdx,1),%ecx
  mov    %ecx,%eax
  shr    %al
  movzbl %al,%eax
  add    %ecx,%eax
  mov    $0xff,%ecx
  cmp    %ecx,%eax
  cmova  %ecx,%eax
  add    $0x1,%rdx
  mov    %al,-0x1(%rdx)
  cmp    %rdx,%rsi
  jne    L5
L6:
  mov    %ebp,0x0(%r13)
  mov    %r8,%rax
  mov    %r12d,(%r14)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
unsigned char *plusClairNB(unsigned char *img, int xs, int ys, int *nc, int *nl) {
  unsigned char *res = (unsigned char *)malloc(xs * ys);
  int nbPixels = xs * ys;
  int i;
  for (i = 0; i < nbPixels; i++) {
    unsigned int v = img[i] + img[i] / 2;
    if (v > 255) {
      v = 255;
    }
    res[i] = v;
  }
  *nc = xs;
  *nl = ys;
  return res;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<strncpy_empty>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x333231,-0xc(%rbp)
  movb   $0x0,-0xd(%rbp)
  lea    -0xd(%rbp),%rcx
  lea    -0xc(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
void strncpy_empty() {
  char dst[] = "123";
  char src[] = "";
  strncpy(dst, src, 4);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<strncpy_empty>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x333231,0x4(%rsp)
  movb   $0x0,0x3(%rsp)
  lea    0x3(%rsp),%rsi
  lea    0x4(%rsp),%rdi
  mov    $0x4,%edx
  call   <strncpy@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void strncpy_empty() {
  char dst[] = "123";
  char src[] = "";
  strncpy(dst, src, 4);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strncpy_empty>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x4,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rsi
  lea    0x4(%rsp),%rdi
  movl   $0x333231,0x4(%rsp)
  movb   $0x0,0x3(%rsp)
  call   <strncpy@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void strncpy_empty() {
  char dst[] = "123";
  char src[] = "";
  strncpy(dst, src, 4);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strncpy_empty>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x4,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rsi
  lea    0x4(%rsp),%rdi
  movl   $0x333231,0x4(%rsp)
  movb   $0x0,0x3(%rsp)
  call   <strncpy@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void strncpy_empty() {
  char dst[] = "123";
  char src[] = "";
  strncpy(dst, src, 4);
}
```