ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<backfunc>:
  endbr64
  lea    (%rdx,%rsi,1),%eax
  push   %r14
  mov    %edx,%r14d
  push   %r13
  lea    D16(%rip),%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %eax,%ebx
  shr    $0x1f,%ebx
  add    %eax,%ebx
  mov    %edx,%eax
  sub    %esi,%eax
  sar    %ebx
  cmp    $0x1,%eax
  jle    L15
L20:
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    -0x1(%rbx),%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <backfunc>
  lea    0x1(%rbx),%r12d
  lea    (%r12,%r14,1),%eax
  mov    %eax,%ebx
  shr    $0x1f,%ebx
  add    %eax,%ebx
  mov    %r14d,%eax
  sub    %r12d,%eax
  sar    %ebx
  cmp    $0x1,%eax
  jg     L20
L15:
  cmp    %r12d,%r14d
  je     L2
  pop    %rbx
  mov    %r14d,%edx
  pop    %rbp
  mov    %r12d,%esi
  lea    D25(%rip),%rdi
  pop    %r12
  xor    %eax,%eax
  pop    %r13
  pop    %r14
  jmp    1050 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L2:
  pop    %rbx
  mov    %r14d,%esi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  lea    D16(%rip),%rdi
  pop    %r13
  pop    %r14
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D16
type_infered: string
name: extract_symbolic_data
arguments: tag: D25
type_infered: string

tool: D16: %d


tool: D25: %d,%d


assistant: ```c
void backfunc(int *buf, int a, int b) {
  int mid = (a + b) / 2;
  if (b - a < 2) {
    if (a == b)
      printf("%d\n", a);
    else
      printf("%d,%d\n", a, b);
    return;
  } else
    printf("%d\n", mid);
  backfunc(buf, a, mid - 1);
  backfunc(buf, mid + 1, b);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<leadingZero>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    $0x0,%ax
  test   %eax,%eax
  je     L2
  mov    -0x14(%rbp),%eax
  and    $0xff000000,%eax
  test   %eax,%eax
  je     L29
  mov    -0x14(%rbp),%eax
  shr    $0x18,%eax
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D21(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L44
L29:
  mov    -0x14(%rbp),%eax
  shr    $0x10,%eax
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D21(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  add    $0x8,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L44
L2:
  mov    -0x14(%rbp),%eax
  and    $0xff00,%eax
  test   %eax,%eax
  je     L32
  mov    -0x14(%rbp),%eax
  shr    $0x8,%eax
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D21(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  add    $0x10,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L44
L32:
  mov    -0x14(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D21(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  add    $0x18,%eax
  mov    %eax,-0x4(%rbp)
L44:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D21
type: string

tool: D21: 

assistant: ```c
unsigned int leadingZero(const unsigned int input) {
  unsigned int l;
  const static unsigned int leadingZero8bit[256] = {8, 7, 6, 6, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  if (input & 0xFFFF0000) {
    if (input & 0xFF000000) {
      l = leadingZero8bit[input >> 24];
    } else {
      l = 8 + leadingZero8bit[input >> 16];
    }
  } else {
    if (input & 0x0000FF00) {
      l = 16 + leadingZero8bit[input >> 8];
    } else {
      l = 24 + leadingZero8bit[input];
    }
  }
  return l;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<leadingZero>:
  endbr64
  test   $0xffff0000,%edi
  je     L1
  test   $0xff000000,%edi
  je     L0
  shr    $0x18,%edi
  mov    %edi,%edi
  lea    D37(%rip),%rax
  mov    (%rax,%rdi,4),%eax
  ret
L0:
  shr    $0x10,%edi
  mov    %edi,%edi
  lea    D37(%rip),%rax
  mov    (%rax,%rdi,4),%eax
  add    $0x8,%eax
  ret
L1:
  test   $0xff00,%edi
  je     L25
  shr    $0x8,%edi
  mov    %edi,%edi
  lea    D37(%rip),%rax
  mov    (%rax,%rdi,4),%eax
  add    $0x10,%eax
  ret
L25:
  mov    %edi,%edi
  lea    D37(%rip),%rax
  mov    (%rax,%rdi,4),%eax
  add    $0x18,%eax
  ret
```

tool_calls: name: unpack_data
arguments: tag: D37
type_infered: string

tool: D37: 

assistant: ```c
unsigned int leadingZero(const unsigned int input) {
  unsigned int l;
  const static unsigned int leadingZero8bit[256] = {8, 7, 6, 6, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  if (input & 0xFFFF0000) {
    if (input & 0xFF000000) {
      l = leadingZero8bit[input >> 24];
    } else {
      l = 8 + leadingZero8bit[input >> 16];
    }
  } else {
    if (input & 0x0000FF00) {
      l = 16 + leadingZero8bit[input >> 8];
    } else {
      l = 24 + leadingZero8bit[input];
    }
  }
  return l;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<leadingZero>:
  endbr64
  test   $0xffff0000,%edi
  je     L27
  test   $0xff000000,%edi
  je     L22
  shr    $0x18,%edi
  lea    D29(%rip),%rax
  mov    (%rax,%rdi,4),%eax
  ret
  nopw   0x0(%rax,%rax,1)
L27:
  test   $0xff00,%edi
  je     L20
  shr    $0x8,%edi
  lea    D29(%rip),%rax
  mov    (%rax,%rdi,4),%eax
  add    $0x10,%eax
  ret
  nopl   0x0(%rax)
L20:
  mov    %edi,%edi
  lea    D29(%rip),%rax
  mov    (%rax,%rdi,4),%eax
  add    $0x18,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L22:
  shr    $0x10,%edi
  lea    D29(%rip),%rax
  mov    (%rax,%rdi,4),%eax
  add    $0x8,%eax
  ret
```

tool_calls: name: parse_data
arguments: label: D29
type_guessed: string

tool: D29: 

assistant: ```c
unsigned int leadingZero(const unsigned int input) {
  unsigned int l;
  const static unsigned int leadingZero8bit[256] = {8, 7, 6, 6, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  if (input & 0xFFFF0000) {
    if (input & 0xFF000000) {
      l = leadingZero8bit[input >> 24];
    } else {
      l = 8 + leadingZero8bit[input >> 16];
    }
  } else {
    if (input & 0x0000FF00) {
      l = 16 + leadingZero8bit[input >> 8];
    } else {
      l = 24 + leadingZero8bit[input];
    }
  }
  return l;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<leadingZero>:
  endbr64
  test   $0xffff0000,%edi
  je     L0
  test   $0xff000000,%edi
  je     L26
  shr    $0x18,%edi
  lea    D23(%rip),%rax
  mov    (%rax,%rdi,4),%eax
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  test   $0xff00,%edi
  je     L21
  shr    $0x8,%edi
  lea    D23(%rip),%rax
  mov    (%rax,%rdi,4),%eax
  add    $0x10,%eax
  ret
  nopl   0x0(%rax)
L21:
  mov    %edi,%edi
  lea    D23(%rip),%rax
  mov    (%rax,%rdi,4),%eax
  add    $0x18,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  shr    $0x10,%edi
  lea    D23(%rip),%rax
  mov    (%rax,%rdi,4),%eax
  add    $0x8,%eax
  ret
```

tool_calls: name: unpack_data
arguments: tag: D23
data_type: string

tool: D23: 

assistant: ```c
unsigned int leadingZero(const unsigned int input) {
  unsigned int l;
  const static unsigned int leadingZero8bit[256] = {8, 7, 6, 6, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  if (input & 0xFFFF0000) {
    if (input & 0xFF000000) {
      l = leadingZero8bit[input >> 24];
    } else {
      l = 8 + leadingZero8bit[input >> 16];
    }
  } else {
    if (input & 0x0000FF00) {
      l = 16 + leadingZero8bit[input >> 8];
    } else {
      l = 24 + leadingZero8bit[input];
    }
  }
  return l;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<remove>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    D2(%rip),%rax
  mov    (%rax),%edx
  mov    D18(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int remove(const char *path) {
  errno = ENOSYS;
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<remove>:
  endbr64
  mov    D14(%rip),%rax
  mov    (%rax),%edx
  mov    D10(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int remove(const char *path) {
  errno = ENOSYS;
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<remove>:
  endbr64
  mov    D7(%rip),%rax
  mov    (%rax),%edx
  mov    D9(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int remove(const char *path) {
  errno = ENOSYS;
  return -1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<remove>:
  endbr64
  mov    D11(%rip),%rax
  mov    (%rax),%edx
  mov    D13(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int remove(const char *path) {
  errno = ENOSYS;
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<mate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L94
  mov    $0x0,%eax
  jmp    L81
L94:
  mov    -0x8(%rbp),%rdx
  mov    -0x20(%rbp),%ecx
  mov    -0x28(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <KMP@plt>
  jmp    L55
L12:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L24
  cmpl   $0xffffffff,-0xc(%rbp)
  jne    L67
L24:
  addl   $0x1,-0x10(%rbp)
  addl   $0x1,-0xc(%rbp)
  jmp    L55
L67:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
L55:
  mov    -0x10(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L15
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L12
L15:
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jne    L56
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  movq   $0x0,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  sub    %rdx,%rax
  mov    %rax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  jmp    L81
L56:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  movq   $0x0,-0x8(%rbp)
  mov    $0x0,%eax
L81:
  leave
  ret
```

assistant: ```c
char *mate(char *str, int str_len, char *pattern, int len) {
  int i = 0, j = 0;
  int *arr = (int *)malloc(sizeof(int) * len);
  if (NULL == arr) {
    return NULL;
  }
  KMP(pattern, len, arr);
  while (i < str_len && j < len) {
    if (str[i] == pattern[j] || -1 == j) {
      i++;
      j++;
    } else {
      j = arr[j];
    }
  }
  if (j == len) {
    free(arr);
    arr = NULL;
    return str + i - j;
  }
  free(arr);
  arr = NULL;
  return NULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mate>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r15
  mov    %esi,%r13d
  mov    %rdx,%r14
  mov    %ecx,%r12d
  lea    0x0(,%rcx,4),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,0x8(%rsp)
  mov    $0x0,%eax
  test   %rdi,%rdi
  je     L98
  mov    %rdi,%rdx
  mov    %r12d,%esi
  mov    %r14,%rdi
  call   <KMP@plt>
  test   %r13d,%r13d
  jle    L33
  test   %r12d,%r12d
  jle    L33
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  jmp    L15
L80:
  add    $0x1,%ebp
  add    $0x1,%ebx
L27:
  cmp    %r13d,%ebp
  jge    L13
  cmp    %r12d,%ebx
  jge    L13
L15:
  movslq %ebx,%rax
  movslq %ebp,%rdx
  movzbl (%r14,%rax,1),%ecx
  cmp    %cl,(%r15,%rdx,1)
  je     L80
  cmp    $0xffffffff,%ebx
  je     L80
  mov    0x8(%rsp),%rsi
  mov    (%rsi,%rax,4),%ebx
  jmp    L27
L33:
  mov    $0x0,%ebx
  mov    $0x0,%ebp
L13:
  cmp    %ebx,%r12d
  je     L103
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    $0x0,%eax
L98:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L103:
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  movslq %ebp,%rax
  movslq %ebx,%rbx
  sub    %rbx,%rax
  lea    (%r15,%rax,1),%rax
  jmp    L98
```

assistant: ```c
char *mate(char *str, int str_len, char *pattern, int len) {
  int i = 0, j = 0;
  int *arr = (int *)malloc(sizeof(int) * len);
  if (NULL == arr) {
    return NULL;
  }
  KMP(pattern, len, arr);
  while (i < str_len && j < len) {
    if (str[i] == pattern[j] || -1 == j) {
      i++;
      j++;
    } else {
      j = arr[j];
    }
  }
  if (j == len) {
    free(arr);
    arr = NULL;
    return str + i - j;
  }
  free(arr);
  arr = NULL;
  return NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<mate>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  movslq %ecx,%rbp
  push   %rbx
  lea    0x0(,%rbp,4),%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r15
  xor    %eax,%eax
  test   %r15,%r15
  je     L38
  mov    %r15,%rdx
  mov    %ebp,%esi
  mov    %r13,%rdi
  xor    %ebx,%ebx
  call   <KMP@plt>
  xor    %eax,%eax
  test   %ebp,%ebp
  jle    L75
  test   %r14d,%r14d
  jg     L14
  jmp    L75
L87:
  cmp    $0xffffffff,%eax
  je     L34
  mov    (%r15,%rdx,4),%eax
  cmp    %r14d,%ebx
  jge    L75
L48:
  cmp    %ebp,%eax
  jge    L75
L14:
  movslq %eax,%rdx
  movslq %ebx,%rcx
  movzbl 0x0(%r13,%rdx,1),%esi
  cmp    %sil,(%r12,%rcx,1)
  jne    L87
L34:
  add    $0x1,%ebx
  add    $0x1,%eax
  cmp    %r14d,%ebx
  jl     L48
L75:
  mov    %r15,%rdi
  cmp    %eax,%ebp
  je     L37
  call   <free@plt>
  xor    %eax,%eax
L38:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L37:
  call   <free@plt>
  movslq %ebx,%rax
  add    $0x8,%rsp
  sub    %rbp,%rax
  pop    %rbx
  pop    %rbp
  lea    (%r12,%rax,1),%rax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *mate(char *str, int str_len, char *pattern, int len) {
  int i = 0, j = 0;
  int *arr = (int *)malloc(sizeof(int) * len);
  if (NULL == arr) {
    return NULL;
  }
  KMP(pattern, len, arr);
  while (i < str_len && j < len) {
    if (str[i] == pattern[j] || -1 == j) {
      i++;
      j++;
    } else {
      j = arr[j];
    }
  }
  if (j == len) {
    free(arr);
    arr = NULL;
    return str + i - j;
  }
  free(arr);
  arr = NULL;
  return NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mate>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  movslq %ecx,%rbp
  push   %rbx
  lea    0x0(,%rbp,4),%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r15
  xor    %eax,%eax
  test   %r15,%r15
  je     L25
  mov    %r15,%rdx
  mov    %ebp,%esi
  mov    %r13,%rdi
  xor    %ebx,%ebx
  call   <KMP@plt>
  xor    %eax,%eax
  test   %ebp,%ebp
  jle    L48
  test   %r14d,%r14d
  jg     L1
  jmp    L48
L97:
  cmp    $0xffffffff,%eax
  je     L66
  mov    (%r15,%rdx,4),%eax
  cmp    %r14d,%ebx
  jge    L48
L38:
  cmp    %ebp,%eax
  jge    L48
L1:
  movslq %eax,%rdx
  movslq %ebx,%rcx
  movzbl 0x0(%r13,%rdx,1),%esi
  cmp    %sil,(%r12,%rcx,1)
  jne    L97
L66:
  add    $0x1,%ebx
  add    $0x1,%eax
  cmp    %r14d,%ebx
  jl     L38
L48:
  mov    %r15,%rdi
  cmp    %eax,%ebp
  je     L85
  call   <free@plt>
  xor    %eax,%eax
L25:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L85:
  call   <free@plt>
  movslq %ebx,%rax
  add    $0x8,%rsp
  sub    %rbp,%rax
  pop    %rbx
  pop    %rbp
  lea    (%r12,%rax,1),%rax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *mate(char *str, int str_len, char *pattern, int len) {
  int i = 0, j = 0;
  int *arr = (int *)malloc(sizeof(int) * len);
  if (NULL == arr) {
    return NULL;
  }
  KMP(pattern, len, arr);
  while (i < str_len && j < len) {
    if (str[i] == pattern[j] || -1 == j) {
      i++;
      j++;
    } else {
      j = arr[j];
    }
  }
  if (j == len) {
    free(arr);
    arr = NULL;
    return str + i - j;
  }
  free(arr);
  arr = NULL;
  return NULL;
}
```
user: What is the c source code of the assembly code below:

```asm
<test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L3
L16:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <test2@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <test2@plt>
  addl   $0x1,-0x4(%rbp)
L3:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L16
  nop
  leave
  ret
```

assistant: ```c
int test(int a) {
  int i;
  for (i = 0; i < a; i++) {
    test2(a);
    test2(a);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test>:
  endbr64
  test   %edi,%edi
  jle    L8
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    $0x0,%ebp
L16:
  mov    %ebx,%edi
  call   <test2@plt>
  mov    %ebx,%edi
  call   <test2@plt>
  add    $0x1,%ebp
  cmp    %ebp,%ebx
  jne    L16
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  ret
```

assistant: ```c
int test(int a) {
  int i;
  for (i = 0; i < a; i++) {
    test2(a);
    test2(a);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test>:
  endbr64
  test   %edi,%edi
  jle    L11
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L9:
  mov    %ebx,%edi
  add    $0x1,%ebp
  call   <test2@plt>
  mov    %ebx,%edi
  call   <test2@plt>
  cmp    %ebp,%ebx
  jne    L9
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L11:
  ret
```

assistant: ```c
int test(int a) {
  int i;
  for (i = 0; i < a; i++) {
    test2(a);
    test2(a);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test>:
  endbr64
  test   %edi,%edi
  jle    L2
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L17:
  mov    %ebx,%edi
  add    $0x1,%ebp
  call   <test2@plt>
  mov    %ebx,%edi
  call   <test2@plt>
  cmp    %ebp,%ebx
  jne    L17
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L2:
  ret
```

assistant: ```c
int test(int a) {
  int i;
  for (i = 0; i < a; i++) {
    test2(a);
    test2(a);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<mat2_rotation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movss  %xmm0,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  movd   %eax,%xmm0
  call   <cosf@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  movd   %eax,%xmm0
  call   <sinf@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  movss  -0x8(%rbp),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  movss  -0x4(%rbp),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movss  -0x4(%rbp),%xmm0
  movss  D6(%rip),%xmm1
  xorps  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0xc,%rax
  movss  -0x8(%rbp),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D6
type: byte[4]

tool: {"D6": [0x00,0x00,0x00,0x80]}

assistant: ```c
float *mat2_rotation(float *result, float angle) {
  float c = cosf(angle);
  float s = sinf(angle);
  result[0] = c;
  result[1] = s;
  result[2] = -s;
  result[3] = c;
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<mat2_rotation>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  lea    0xc(%rsp),%rdi
  lea    0x8(%rsp),%rsi
  call   <sincosf@plt>
  movss  0x8(%rsp),%xmm1
  movss  0xc(%rsp),%xmm0
  movss  %xmm1,(%rbx)
  movss  %xmm0,0x4(%rbx)
  xorps  D1(%rip),%xmm0
  movss  %xmm0,0x8(%rbx)
  movss  %xmm1,0xc(%rbx)
  mov    %rbx,%rax
  add    $0x10,%rsp
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D1
datatype: byte[16]

tool: {"D1": [0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
float *mat2_rotation(float *result, float angle) {
  float c = cosf(angle);
  float s = sinf(angle);
  result[0] = c;
  result[1] = s;
  result[2] = -s;
  result[3] = c;
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<mat2_rotation>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  sub    $0x10,%rsp
  lea    0xc(%rsp),%rdi
  lea    0x8(%rsp),%rsi
  call   <sincosf@plt>
  movss  0xc(%rsp),%xmm0
  movss  0x8(%rsp),%xmm1
  mov    %r12,%rax
  movss  %xmm0,0x4(%r12)
  xorps  D0(%rip),%xmm0
  movss  %xmm1,(%r12)
  movss  %xmm0,0x8(%r12)
  movss  %xmm1,0xc(%r12)
  add    $0x10,%rsp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D0
datatype: byte[16]

tool: {"D0": [0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
float *mat2_rotation(float *result, float angle) {
  float c = cosf(angle);
  float s = sinf(angle);
  result[0] = c;
  result[1] = s;
  result[2] = -s;
  result[3] = c;
  return result;
}
```