ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<length>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movss  (%rax),%xmm1
  mov    -0x18(%rbp),%rax
  movss  (%rax),%xmm0
  mulss  %xmm0,%xmm1
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm2
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mulss  %xmm2,%xmm0
  addss  %xmm0,%xmm1
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm2
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm0
  mulss  %xmm2,%xmm0
  addss  %xmm0,%xmm1
  movd   %xmm1,%eax
  movd   %eax,%xmm0
  call   <sqrt@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  leave
  ret
```

assistant: ```c
float length(float v[3]) {
  float r = sqrt(v[0] * v[0] + v[1] * v[1] + v[2] * v[2]);
  return r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<length>:
  endbr64
  sub    $0x8,%rsp
  movss  (%rdi),%xmm0
  movss  0x4(%rdi),%xmm2
  movss  0x8(%rdi),%xmm1
  mulss  %xmm0,%xmm0
  mulss  %xmm2,%xmm2
  addss  %xmm2,%xmm0
  mulss  %xmm1,%xmm1
  addss  %xmm1,%xmm0
  call   <sqrt@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
float length(float v[3]) {
  float r = sqrt(v[0] * v[0] + v[1] * v[1] + v[2] * v[2]);
  return r;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<length>:
  endbr64
  movss  (%rdi),%xmm0
  movss  0x4(%rdi),%xmm2
  movss  0x8(%rdi),%xmm1
  mulss  %xmm0,%xmm0
  mulss  %xmm2,%xmm2
  mulss  %xmm1,%xmm1
  addss  %xmm2,%xmm0
  addss  %xmm1,%xmm0
  jmp    1050 <sqrt@plt>
```

assistant: ```c
float length(float v[3]) {
  float r = sqrt(v[0] * v[0] + v[1] * v[1] + v[2] * v[2]);
  return r;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<length>:
  endbr64
  movss  (%rdi),%xmm0
  movss  0x4(%rdi),%xmm2
  movss  0x8(%rdi),%xmm1
  mulss  %xmm0,%xmm0
  mulss  %xmm2,%xmm2
  mulss  %xmm1,%xmm1
  addss  %xmm2,%xmm0
  addss  %xmm1,%xmm0
  jmp    1050 <sqrt@plt>
```

assistant: ```c
float length(float v[3]) {
  float r = sqrt(v[0] * v[0] + v[1] * v[1] + v[2] * v[2]);
  return r;
}
```
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
  mov    $0x0,%eax
  call   <FooFloat@plt>
  mov    $0x0,%eax
  call   <FooDouble@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

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

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <FooFloat@plt>
  mov    $0x0,%eax
  call   <FooDouble@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  FooFloat();
  FooDouble();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

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

assistant: ```c
int main(void) {
  FooFloat();
  FooDouble();
}
```
user: What would the C source look like for the assembly code provided here:

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

assistant: ```c
int main(void) {
  FooFloat();
  FooDouble();
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<somatorio_vetor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L10
L3:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  movss  -0x8(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L10:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L3
  movss  -0x8(%rbp),%xmm0
  pop    %rbp
  ret
```

assistant: ```c
float somatorio_vetor(float v[], int n) {
  float sum = 0;
  for (int i = 0; i < n; ++i) {
    sum += v[i];
  }
  return sum;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<somatorio_vetor>:
  endbr64
  test   %esi,%esi
  jle    L0
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rdx
  pxor   %xmm0,%xmm0
L12:
  addss  (%rax),%xmm0
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L12
  ret
L0:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
float somatorio_vetor(float v[], int n) {
  float sum = 0;
  for (int i = 0; i < n; ++i) {
    sum += v[i];
  }
  return sum;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<somatorio_vetor>:
  endbr64
  test   %esi,%esi
  jle    L12
  lea    -0x1(%rsi),%eax
  pxor   %xmm0,%xmm0
  lea    0x4(%rdi,%rax,4),%rax
  nopl   0x0(%rax)
L3:
  addss  (%rdi),%xmm0
  add    $0x4,%rdi
  cmp    %rax,%rdi
  jne    L3
  ret
  cs nopw 0x0(%rax,%rax,1)
L12:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
float somatorio_vetor(float v[], int n) {
  float sum = 0;
  for (int i = 0; i < n; ++i) {
    sum += v[i];
  }
  return sum;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<somatorio_vetor>:
  endbr64
  test   %esi,%esi
  jle    L51
  lea    -0x1(%rsi),%eax
  cmp    $0x2,%eax
  jbe    L67
  mov    %esi,%edx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    $0x2,%edx
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
L26:
  movups (%rax),%xmm1
  add    $0x10,%rax
  addss  %xmm1,%xmm0
  movaps %xmm1,%xmm2
  shufps $0x55,%xmm1,%xmm2
  addss  %xmm2,%xmm0
  movaps %xmm1,%xmm2
  unpckhps %xmm1,%xmm2
  shufps $0xff,%xmm1,%xmm1
  addss  %xmm2,%xmm0
  addss  %xmm1,%xmm0
  cmp    %rdx,%rax
  jne    L26
  mov    %esi,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%sil
  je     L35
L15:
  movslq %eax,%rdx
  addss  (%rdi,%rdx,4),%xmm0
  lea    0x0(,%rdx,4),%rcx
  lea    0x1(%rax),%edx
  cmp    %edx,%esi
  jle    L57
  add    $0x2,%eax
  addss  0x4(%rdi,%rcx,1),%xmm0
  cmp    %eax,%esi
  jle    L57
  addss  0x8(%rdi,%rcx,1),%xmm0
  ret
  nopl   0x0(%rax,%rax,1)
L51:
  pxor   %xmm0,%xmm0
L57:
  ret
  nopl   (%rax)
L35:
  ret
L67:
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  jmp    L15
```

assistant: ```c
float somatorio_vetor(float v[], int n) {
  float sum = 0;
  for (int i = 0; i < n; ++i) {
    sum += v[i];
  }
  return sum;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L56
  movl   $0x0,-0x4(%rbp)
  jmp    L57
L37:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x5(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x5(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L57:
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x4(%rbp)
  jl     L37
  jmp    L10
L56:
  movl   $0x0,-0x4(%rbp)
  jmp    L33
L7:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x5(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x5(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L33:
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x4(%rbp)
  jl     L7
L10:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void swap(char a[], int n) {
  int i;
  char tmp;
  if (!(n % 2)) {
    for (i = 0; i < n / 2; i++) {
      tmp = a[i];
      a[i] = a[n / 2 + i];
      a[n / 2 + i] = tmp;
    }
  } else {
    for (i = 0; i < n / 2; i++) {
      tmp = a[i];
      a[i] = a[(n + 1) / 2 + i];
      a[(n + 1) / 2 + i] = tmp;
    }
  }
  a[n] = '\0';
}
```
user: What is the c source code of the assembly code below:

```asm
<swap>:
  endbr64
  mov    %rdi,%rdx
  test   $0x1,%sil
  je     L18
  mov    %esi,%r9d
  shr    $0x1f,%r9d
  add    %esi,%r9d
  sar    %r9d
  cmp    $0x1,%esi
  jle    L50
  lea    0x1(%rsi),%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %eax,%ecx
  sar    %ecx
  mov    $0x0,%eax
  movslq %ecx,%rcx
  add    %rdi,%rcx
L3:
  movzbl (%rdx,%rax,1),%edi
  movzbl (%rcx,%rax,1),%r8d
  mov    %r8b,(%rdx,%rax,1)
  mov    %dil,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%r9d
  jg     L3
  jmp    L50
L18:
  mov    %esi,%r9d
  shr    $0x1f,%r9d
  add    %esi,%r9d
  sar    %r9d
  cmp    $0x1,%esi
  jle    L50
  mov    $0x0,%eax
  movslq %r9d,%rcx
  add    %rdi,%rcx
L33:
  movzbl (%rdx,%rax,1),%edi
  movzbl (%rcx,%rax,1),%r8d
  mov    %r8b,(%rdx,%rax,1)
  mov    %dil,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%r9d
  jg     L33
L50:
  movslq %esi,%rsi
  movb   $0x0,(%rdx,%rsi,1)
  ret
```

assistant: ```c
void swap(char a[], int n) {
  int i;
  char tmp;
  if (!(n % 2)) {
    for (i = 0; i < n / 2; i++) {
      tmp = a[i];
      a[i] = a[n / 2 + i];
      a[n / 2 + i] = tmp;
    }
  } else {
    for (i = 0; i < n / 2; i++) {
      tmp = a[i];
      a[i] = a[(n + 1) / 2 + i];
      a[(n + 1) / 2 + i] = tmp;
    }
  }
  a[n] = '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<swap>:
  endbr64
  mov    %rdi,%rdx
  movslq %esi,%rdi
  mov    %edi,%ecx
  shr    $0x1f,%ecx
  add    %edi,%ecx
  sar    %ecx
  test   $0x1,%dil
  je     L3
  cmp    $0x1,%edi
  jle    L18
  lea    0x1(%rdi),%esi
  xor    %eax,%eax
  sar    %esi
  movslq %esi,%rsi
  add    %rdx,%rsi
  nopl   0x0(%rax,%rax,1)
L29:
  movzbl (%rdx,%rax,1),%r8d
  movzbl (%rsi,%rax,1),%r9d
  mov    %r9b,(%rdx,%rax,1)
  mov    %r8b,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%ecx
  jg     L29
L18:
  movb   $0x0,(%rdx,%rdi,1)
  ret
  nop
L3:
  cmp    $0x1,%edi
  jle    L18
  movslq %ecx,%rsi
  xor    %eax,%eax
  add    %rdx,%rsi
  nopl   (%rax)
L49:
  movzbl (%rdx,%rax,1),%r8d
  movzbl (%rsi,%rax,1),%r9d
  mov    %r9b,(%rdx,%rax,1)
  mov    %r8b,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%ecx
  jg     L49
  movb   $0x0,(%rdx,%rdi,1)
  ret
```

assistant: ```c
void swap(char a[], int n) {
  int i;
  char tmp;
  if (!(n % 2)) {
    for (i = 0; i < n / 2; i++) {
      tmp = a[i];
      a[i] = a[n / 2 + i];
      a[n / 2 + i] = tmp;
    }
  } else {
    for (i = 0; i < n / 2; i++) {
      tmp = a[i];
      a[i] = a[(n + 1) / 2 + i];
      a[(n + 1) / 2 + i] = tmp;
    }
  }
  a[n] = '\0';
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x207369206566694c,%rax
  movabs $0x2061202c656b696c,%rdx
  mov    %rax,-0x60(%rbp)
  mov    %rdx,-0x58(%rbp)
  movabs $0x6320666f20786f62,%rax
  movabs $0x6574616c6f636f68,%rdx
  mov    %rax,-0x50(%rbp)
  mov    %rdx,-0x48(%rbp)
  movabs $0x656e20756f79202c,%rax
  movabs $0x6f6e6b202c726576,%rdx
  mov    %rax,-0x40(%rbp)
  mov    %rdx,-0x38(%rbp)
  movabs $0x7920746168772077,%rax
  movabs $0x67202c657227756f,%rdx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  movabs $0x74656720616e6e6f,%rax
  mov    %rax,-0x20(%rbp)
  movb   $0x0,-0x18(%rbp)
  movw   $0x2c,-0x62(%rbp)
  lea    -0x62(%rbp),%rdx
  lea    -0x60(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x70(%rbp)
  jmp    L3
L8:
  mov    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x62(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x70(%rbp)
L3:
  cmpq   $0x0,-0x70(%rbp)
  jne    L8
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D0
datatype: string

tool: D0: %s 


assistant: ```c
int main() {
  char str[] = "Life is like, a box of chocolate, you never, know what you're, gonna get";
  char delims[] = ",";
  char *result;
  result = strtok(str, delims);
  while (result != NULL) {
    printf("%s \n", result);
    result = strtok(NULL, delims);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movabs $0x207369206566694c,%rax
  movabs $0x2061202c656b696c,%rdx
  mov    %rax,0x10(%rsp)
  mov    %rdx,0x18(%rsp)
  movabs $0x6320666f20786f62,%rax
  movabs $0x6574616c6f636f68,%rdx
  mov    %rax,0x20(%rsp)
  mov    %rdx,0x28(%rsp)
  movabs $0x656e20756f79202c,%rax
  movabs $0x6f6e6b202c726576,%rdx
  mov    %rax,0x30(%rsp)
  mov    %rdx,0x38(%rsp)
  movabs $0x7920746168772077,%rax
  movabs $0x67202c657227756f,%rdx
  mov    %rax,0x40(%rsp)
  mov    %rdx,0x48(%rsp)
  movabs $0x74656720616e6e6f,%rax
  mov    %rax,0x50(%rsp)
  movb   $0x0,0x58(%rsp)
  movw   $0x2c,0xe(%rsp)
  lea    0xe(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L22
  mov    %rax,%rsi
  lea    D5(%rip),%rbx
L27:
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    0xe(%rsp),%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L27
L22:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  mov    $0x0,%eax
  add    $0x70,%rsp
  pop    %rbx
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D5
data_type: string

tool: D5: %s 


assistant: ```c
int main() {
  char str[] = "Life is like, a box of chocolate, you never, know what you're, gonna get";
  char delims[] = ",";
  char *result;
  result = strtok(str, delims);
  while (result != NULL) {
    printf("%s \n", result);
    result = strtok(NULL, delims);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  movdqa D27(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rbx
  lea    0x10(%rsp),%rdi
  movabs $0x74656720616e6e6f,%rax
  movaps %xmm0,0x10(%rsp)
  movdqa D30(%rip),%xmm0
  mov    %rbx,%rsi
  mov    %rax,0x50(%rsp)
  mov    $0x2c,%eax
  movaps %xmm0,0x20(%rsp)
  movdqa D39(%rip),%xmm0
  movb   $0x0,0x58(%rsp)
  movaps %xmm0,0x30(%rsp)
  movdqa D1(%rip),%xmm0
  mov    %ax,0xe(%rsp)
  movaps %xmm0,0x40(%rsp)
  call   <strtok@plt>
  test   %rax,%rax
  je     L2
  mov    %rax,%rsi
  lea    D48(%rip),%rbp
  nopl   (%rax)
L7:
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbx,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L7
L2:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D27
datatype: byte[16]
name: unpack_data
arguments: data_tag: D30
datatype: byte[16]
name: unpack_data
arguments: data_tag: D39
datatype: byte[16]
name: unpack_data
arguments: data_tag: D1
datatype: byte[16]
name: unpack_data
arguments: data_tag: D48
datatype: string

tool: {"D27": [0x4c,0x69,0x66,0x65,0x20,0x69,0x73,0x20,0x6c,0x69,0x6b,0x65,0x2c,0x20,0x61,0x20]}

tool: {"D30": [0x62,0x6f,0x78,0x20,0x6f,0x66,0x20,0x63,0x68,0x6f,0x63,0x6f,0x6c,0x61,0x74,0x65]}

tool: {"D39": [0x2c,0x20,0x79,0x6f,0x75,0x20,0x6e,0x65,0x76,0x65,0x72,0x2c,0x20,0x6b,0x6e,0x6f]}

tool: {"D1": [0x77,0x20,0x77,0x68,0x61,0x74,0x20,0x79,0x6f,0x75,0x27,0x72,0x65,0x2c,0x20,0x67]}

tool: D48: %s 


assistant: ```c
int main() {
  char str[] = "Life is like, a box of chocolate, you never, know what you're, gonna get";
  char delims[] = ",";
  char *result;
  result = strtok(str, delims);
  while (result != NULL) {
    printf("%s \n", result);
    result = strtok(NULL, delims);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  movdqa D34(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rbx
  lea    0x10(%rsp),%rdi
  movabs $0x74656720616e6e6f,%rax
  movaps %xmm0,0x10(%rsp)
  movdqa D39(%rip),%xmm0
  mov    %rbx,%rsi
  mov    %rax,0x50(%rsp)
  mov    $0x2c,%eax
  movaps %xmm0,0x20(%rsp)
  movdqa D37(%rip),%xmm0
  movb   $0x0,0x58(%rsp)
  movaps %xmm0,0x30(%rsp)
  movdqa D41(%rip),%xmm0
  mov    %ax,0xe(%rsp)
  movaps %xmm0,0x40(%rsp)
  call   <strtok@plt>
  test   %rax,%rax
  je     L19
  mov    %rax,%rsi
  lea    D42(%rip),%rbp
  nopl   (%rax)
L14:
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbx,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L14
L19:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D34
type: byte[16]
name: disassemble_data
arguments: tag: D39
type: byte[16]
name: disassemble_data
arguments: tag: D37
type: byte[16]
name: disassemble_data
arguments: tag: D41
type: byte[16]
name: disassemble_data
arguments: tag: D42
type: string

tool: {"D34": [0x4c,0x69,0x66,0x65,0x20,0x69,0x73,0x20,0x6c,0x69,0x6b,0x65,0x2c,0x20,0x61,0x20]}

tool: {"D39": [0x62,0x6f,0x78,0x20,0x6f,0x66,0x20,0x63,0x68,0x6f,0x63,0x6f,0x6c,0x61,0x74,0x65]}

tool: {"D37": [0x2c,0x20,0x79,0x6f,0x75,0x20,0x6e,0x65,0x76,0x65,0x72,0x2c,0x20,0x6b,0x6e,0x6f]}

tool: {"D41": [0x77,0x20,0x77,0x68,0x61,0x74,0x20,0x79,0x6f,0x75,0x27,0x72,0x65,0x2c,0x20,0x67]}

tool: D42: %s 


assistant: ```c
int main() {
  char str[] = "Life is like, a box of chocolate, you never, know what you're, gonna get";
  char delims[] = ",";
  char *result;
  result = strtok(str, delims);
  while (result != NULL) {
    printf("%s \n", result);
    result = strtok(NULL, delims);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<af_sa_to_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x80,%eax
  je     L8
  cmp    $0x81,%eax
  jne    L45
  mov    D13(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x8(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x10(%rbp),%rdx
  mov    %eax,%edi
  call   <inet_ntop@plt>
  jmp    L22
L8:
  mov    D13(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x8(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x10(%rbp),%rdx
  mov    %eax,%edi
  call   <inet_ntop@plt>
  jmp    L22
L45:
  mov    $0x0,%eax
  jmp    L21
L22:
  mov    -0x10(%rbp),%rax
L21:
  leave
  ret
```

assistant: ```c
char *af_sa_to_str(const struct sockaddr *sa, char dst[INET6_ADDRSTRLEN]) {
  dst[0] = 0;
  switch (sa->sa_family) {
  case PF_INET:
    inet_ntop(sa->sa_family, &((struct sockaddr_in *)(sa))->sin_addr, dst, INET6_ADDRSTRLEN);
    break;
  case PF_INET6:
    inet_ntop(sa->sa_family, &((struct sockaddr_in6 *)(sa))->sin6_addr, dst, INET6_ADDRSTRLEN);
    break;
  default:
    return (NULL);
  }
  return (dst);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<af_sa_to_str>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  movb   $0x0,(%rsi)
  mov    (%rdi),%eax
  cmp    $0x80,%eax
  je     L15
  mov    $0x0,%edx
  cmp    $0x81,%eax
  jne    L12
  mov    D5(%rip),%rax
  mov    (%rax),%ecx
  mov    %rsi,%rdx
  mov    %rdi,%rsi
  mov    $0x81,%edi
  call   <inet_ntop@plt>
  mov    %rbx,%rdx
L12:
  mov    %rdx,%rax
  pop    %rbx
  ret
L15:
  mov    D5(%rip),%rax
  mov    (%rax),%ecx
  mov    %rsi,%rdx
  mov    %rdi,%rsi
  mov    $0x80,%edi
  call   <inet_ntop@plt>
  mov    %rbx,%rdx
  jmp    L12
```

assistant: ```c
char *af_sa_to_str(const struct sockaddr *sa, char dst[INET6_ADDRSTRLEN]) {
  dst[0] = 0;
  switch (sa->sa_family) {
  case PF_INET:
    inet_ntop(sa->sa_family, &((struct sockaddr_in *)(sa))->sin_addr, dst, INET6_ADDRSTRLEN);
    break;
  case PF_INET6:
    inet_ntop(sa->sa_family, &((struct sockaddr_in6 *)(sa))->sin6_addr, dst, INET6_ADDRSTRLEN);
    break;
  default:
    return (NULL);
  }
  return (dst);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<af_sa_to_str>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  movb   $0x0,(%rsi)
  mov    (%rdi),%eax
  cmp    $0x80,%eax
  je     L9
  xor    %r9d,%r9d
  cmp    $0x81,%eax
  jne    L6
  mov    D1(%rip),%rax
  mov    %rsi,%rdx
  mov    %rdi,%rsi
  mov    $0x81,%edi
  mov    (%rax),%ecx
  call   <inet_ntop@plt>
  mov    %rbx,%r9
L6:
  mov    %r9,%rax
  pop    %rbx
  ret
  nop
L9:
  mov    D1(%rip),%rax
  mov    %rsi,%rdx
  mov    %rdi,%rsi
  mov    $0x80,%edi
  mov    (%rax),%ecx
  call   <inet_ntop@plt>
  mov    %rbx,%r9
  pop    %rbx
  mov    %r9,%rax
  ret
```

assistant: ```c
char *af_sa_to_str(const struct sockaddr *sa, char dst[INET6_ADDRSTRLEN]) {
  dst[0] = 0;
  switch (sa->sa_family) {
  case PF_INET:
    inet_ntop(sa->sa_family, &((struct sockaddr_in *)(sa))->sin_addr, dst, INET6_ADDRSTRLEN);
    break;
  case PF_INET6:
    inet_ntop(sa->sa_family, &((struct sockaddr_in6 *)(sa))->sin6_addr, dst, INET6_ADDRSTRLEN);
    break;
  default:
    return (NULL);
  }
  return (dst);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<af_sa_to_str>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  movb   $0x0,(%rsi)
  mov    (%rdi),%eax
  cmp    $0x80,%eax
  je     L17
  xor    %r9d,%r9d
  cmp    $0x81,%eax
  jne    L15
  mov    D17(%rip),%rax
  mov    %rsi,%rdx
  mov    %rdi,%rsi
  mov    $0x81,%edi
  mov    (%rax),%ecx
  call   <inet_ntop@plt>
  mov    %rbx,%r9
L15:
  mov    %r9,%rax
  pop    %rbx
  ret
  nop
L17:
  mov    D17(%rip),%rax
  mov    %rsi,%rdx
  mov    %rdi,%rsi
  mov    $0x80,%edi
  mov    (%rax),%ecx
  call   <inet_ntop@plt>
  mov    %rbx,%r9
  pop    %rbx
  mov    %r9,%rax
  ret
```

assistant: ```c
char *af_sa_to_str(const struct sockaddr *sa, char dst[INET6_ADDRSTRLEN]) {
  dst[0] = 0;
  switch (sa->sa_family) {
  case PF_INET:
    inet_ntop(sa->sa_family, &((struct sockaddr_in *)(sa))->sin_addr, dst, INET6_ADDRSTRLEN);
    break;
  case PF_INET6:
    inet_ntop(sa->sa_family, &((struct sockaddr_in6 *)(sa))->sin6_addr, dst, INET6_ADDRSTRLEN);
    break;
  default:
    return (NULL);
  }
  return (dst);
}
```