ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<l2_cpu>:
  endbr64
  test   %edi,%edi
  jle    L13
  movslq %edi,%rdi
  xor    %eax,%eax
  nopl   (%rax)
L6:
  movss  (%rdx,%rax,4),%xmm0
  subss  (%rsi,%rax,4),%xmm0
  movaps %xmm0,%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,(%r8,%rax,4)
  movss  %xmm0,(%rcx,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rdi
  jne    L6
L13:
  ret
```

assistant: ```c
void l2_cpu(int n, float *pred, float *truth, float *delta, float *error) {
  int i;
  for (i = 0; i < n; ++i) {
    float diff = truth[i] - pred[i];
    error[i] = diff * diff;
    delta[i] = diff;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<l2_cpu>:
  endbr64
  movslq %edi,%r9
  mov    %rdx,%rdi
  mov    %rcx,%rdx
  test   %r9d,%r9d
  jle    L67
  lea    0x4(%r8),%r11
  mov    %rdx,%rax
  lea    0x4(%rsi),%rcx
  sub    %r11,%rax
  lea    0x4(%rdi),%r10
  cmp    $0x8,%rax
  seta   %al
  cmp    $0x1,%r9d
  setne  %r11b
  and    %r11d,%eax
  mov    %r8,%r11
  sub    %rcx,%r11
  cmp    $0x8,%r11
  seta   %r11b
  and    %r11d,%eax
  mov    %rdx,%r11
  sub    %rcx,%r11
  cmp    $0x8,%r11
  seta   %cl
  and    %ecx,%eax
  mov    %r8,%rcx
  sub    %r10,%rcx
  cmp    $0x8,%rcx
  seta   %cl
  test   %cl,%al
  je     L70
  mov    %rdx,%rax
  sub    %r10,%rax
  cmp    $0x8,%rax
  jbe    L70
  lea    -0x1(%r9),%eax
  mov    %r9d,%r10d
  cmp    $0x2,%eax
  jbe    L87
  mov    %r9d,%ecx
  xor    %eax,%eax
  shr    $0x2,%ecx
  shl    $0x4,%rcx
  nopl   0x0(%rax,%rax,1)
L44:
  movups (%rdi,%rax,1),%xmm0
  movups (%rsi,%rax,1),%xmm2
  subps  %xmm2,%xmm0
  movaps %xmm0,%xmm1
  mulps  %xmm0,%xmm1
  movups %xmm1,(%r8,%rax,1)
  movups %xmm0,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L44
  mov    %r9d,%ecx
  and    $0xfffffffc,%ecx
  mov    %ecx,%eax
  cmp    %ecx,%r9d
  je     L67
  mov    %r9d,%r10d
  sub    %ecx,%r10d
  cmp    $0x1,%r10d
  je     L32
L95:
  movq   (%rsi,%rcx,4),%xmm1
  movq   (%rdi,%rcx,4),%xmm0
  subps  %xmm1,%xmm0
  movaps %xmm0,%xmm1
  mulps  %xmm0,%xmm1
  movlps %xmm1,(%r8,%rcx,4)
  movlps %xmm0,(%rdx,%rcx,4)
  mov    %r10d,%ecx
  and    $0xfffffffe,%ecx
  add    %ecx,%eax
  cmp    %ecx,%r10d
  je     L67
L32:
  cltq
  movss  (%rdi,%rax,4),%xmm0
  subss  (%rsi,%rax,4),%xmm0
  movaps %xmm0,%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,(%r8,%rax,4)
  movss  %xmm0,(%rdx,%rax,4)
  ret
  nopl   0x0(%rax,%rax,1)
L70:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L42:
  movss  (%rdi,%rax,4),%xmm0
  subss  (%rsi,%rax,4),%xmm0
  movaps %xmm0,%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,(%r8,%rax,4)
  movss  %xmm0,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %r9,%rax
  jne    L42
L67:
  ret
L87:
  xor    %ecx,%ecx
  xor    %eax,%eax
  jmp    L95
```

assistant: ```c
void l2_cpu(int n, float *pred, float *truth, float *delta, float *error) {
  int i;
  for (i = 0; i < n; ++i) {
    float diff = truth[i] - pred[i];
    error[i] = diff * diff;
    delta[i] = diff;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<arrayReverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L10
L15:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
  subl   $0x1,-0x8(%rbp)
L10:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L15
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void arrayReverse(int array[], int size) {
  for (int i = 0, j = size - 1; i < j; i++, j--) {
    int temporary = array[j];
    array[j] = array[i];
    array[i] = temporary;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<arrayReverse>:
  endbr64
  sub    $0x1,%esi
  test   %esi,%esi
  jle    L0
  movslq %esi,%rsi
  mov    $0x0,%eax
L2:
  mov    (%rdi,%rsi,4),%edx
  mov    (%rdi,%rax,4),%ecx
  mov    %ecx,(%rdi,%rsi,4)
  mov    %edx,(%rdi,%rax,4)
  add    $0x1,%rax
  sub    $0x1,%rsi
  cmp    %esi,%eax
  jl     L2
L0:
  ret
```

assistant: ```c
void arrayReverse(int array[], int size) {
  for (int i = 0, j = size - 1; i < j; i++, j--) {
    int temporary = array[j];
    array[j] = array[i];
    array[i] = temporary;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<arrayReverse>:
  endbr64
  sub    $0x1,%esi
  test   %esi,%esi
  jle    L7
  movslq %esi,%rsi
  xor    %eax,%eax
L8:
  mov    (%rdi,%rsi,4),%edx
  mov    (%rdi,%rax,4),%ecx
  mov    %ecx,(%rdi,%rsi,4)
  sub    $0x1,%rsi
  mov    %edx,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%esi
  jg     L8
L7:
  ret
```

assistant: ```c
void arrayReverse(int array[], int size) {
  for (int i = 0, j = size - 1; i < j; i++, j--) {
    int temporary = array[j];
    array[j] = array[i];
    array[i] = temporary;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<arrayReverse>:
  endbr64
  lea    -0x1(%rsi),%eax
  test   %eax,%eax
  jle    L73
  mov    %rdi,%rdx
  lea    -0x2(%rsi),%edi
  movslq %esi,%rsi
  mov    %edi,%r11d
  lea    0x0(,%rsi,4),%r8
  shr    %r11d
  mov    %r8,%rsi
  lea    0x1(%r11),%ecx
  mov    %rcx,%r10
  shl    $0x2,%rcx
  sub    %rcx,%rsi
  cmp    %rsi,%rcx
  setle  %cl
  test   %r8,%r8
  setle  %sil
  or     %sil,%cl
  je     L102
  cmp    $0x1,%edi
  jbe    L102
  push   %rbx
  cmp    $0x5,%edi
  jbe    L85
  mov    %r10d,%r9d
  lea    -0x10(%rdx,%r8,1),%rdi
  mov    %rdx,%rcx
  shr    $0x2,%r9d
  lea    -0x10(%rdi),%rsi
  lea    -0x1(%r9),%ebx
  mov    %rsi,%r9
  shl    $0x4,%rbx
  sub    %rbx,%r9
  jmp    L109
  nopl   0x0(%rax,%rax,1)
L76:
  sub    $0x10,%rsi
L109:
  movdqu (%rdi),%xmm0
  movdqu (%rcx),%xmm2
  add    $0x10,%rcx
  pshufd $0x1b,%xmm2,%xmm1
  pshufd $0x1b,%xmm0,%xmm0
  movups %xmm1,(%rdi)
  mov    %rsi,%rdi
  movups %xmm0,-0x10(%rcx)
  cmp    %r9,%rsi
  jne    L76
  mov    %r10d,%esi
  and    $0xfffffffc,%esi
  mov    %esi,%ecx
  sub    %esi,%eax
  cmp    %esi,%r10d
  je     L51
L44:
  sub    %esi,%r10d
  cmp    %esi,%r11d
  je     L18
  neg    %rsi
  shl    $0x2,%rsi
  lea    -0x8(%r8,%rsi,1),%rdi
  mov    %rdx,%r8
  sub    %rsi,%r8
  add    %rdx,%rdi
  mov    %r10d,%esi
  movq   (%rdi),%xmm0
  movq   (%r8),%xmm1
  and    $0xfffffffe,%esi
  add    %esi,%ecx
  sub    %esi,%eax
  pshufd $0xe1,%xmm1,%xmm1
  pshufd $0xe1,%xmm0,%xmm0
  movq   %xmm1,(%rdi)
  movq   %xmm0,(%r8)
  cmp    %esi,%r10d
  je     L51
L18:
  cltq
  lea    (%rdx,%rax,4),%rsi
  movslq %ecx,%rax
  lea    (%rdx,%rax,4),%rax
  mov    (%rsi),%edi
  mov    (%rax),%edx
  mov    %edx,(%rsi)
  mov    %edi,(%rax)
L51:
  pop    %rbx
  ret
  nop
L102:
  cltq
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L17:
  mov    (%rdx,%rax,4),%esi
  mov    (%rdx,%rcx,4),%edi
  mov    %edi,(%rdx,%rax,4)
  sub    $0x1,%rax
  mov    %esi,(%rdx,%rcx,4)
  add    $0x1,%rcx
  cmp    %eax,%ecx
  jl     L17
  ret
  nopl   0x0(%rax)
L73:
  ret
L85:
  xor    %esi,%esi
  xor    %ecx,%ecx
  jmp    L44
```

assistant: ```c
void arrayReverse(int array[], int size) {
  for (int i = 0, j = size - 1; i < j; i++, j--) {
    int temporary = array[j];
    array[j] = array[i];
    array[i] = temporary;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rgb_store>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    D44(%rip),%eax
  lea    0x1(%rax),%edx
  mov    %edx,D29(%rip)
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  cltq
  lea    0x0(,%rax,8),%rcx
  lea    D29(%rip),%rax
  mov    %rdx,(%rcx,%rax,1)
  mov    D29(%rip),%eax
  cmp    $0x3,%eax
  jne    L7
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <RgbToCmyk@plt>
  movl   $0x0,D29(%rip)
L7:
  nop
  leave
  ret
```

assistant: ```c
void rgb_store(int c) {
  static uint8_t rgb_store[(3)];
  static int idx = 0;
  rgb_store[idx++] = (uint8_t)c;
  if (idx == (3)) {
    RgbToCmyk(rgb_store);
    idx = 0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rgb_store>:
  endbr64
  mov    D23(%rip),%eax
  lea    0x1(%rax),%edx
  mov    %edx,D27(%rip)
  cltq
  movslq %edi,%rdi
  lea    D27(%rip),%rcx
  mov    %rdi,(%rcx,%rax,8)
  cmp    $0x3,%edx
  je     L7
  ret
L7:
  sub    $0x8,%rsp
  mov    %rcx,%rdi
  call   <RgbToCmyk@plt>
  movl   $0x0,D27(%rip)
  add    $0x8,%rsp
  ret
```

assistant: ```c
void rgb_store(int c) {
  static uint8_t rgb_store[(3)];
  static int idx = 0;
  rgb_store[idx++] = (uint8_t)c;
  if (idx == (3)) {
    RgbToCmyk(rgb_store);
    idx = 0;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<rgb_store>:
  endbr64
  movslq D36(%rip),%rax
  lea    D8(%rip),%r8
  movslq %edi,%rdi
  lea    0x1(%rax),%edx
  mov    %rdi,(%r8,%rax,8)
  mov    %edx,D15(%rip)
  cmp    $0x3,%edx
  je     L5
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  sub    $0x8,%rsp
  mov    %r8,%rdi
  call   <RgbToCmyk@plt>
  movl   $0x0,D15(%rip)
  add    $0x8,%rsp
  ret
```

assistant: ```c
void rgb_store(int c) {
  static uint8_t rgb_store[(3)];
  static int idx = 0;
  rgb_store[idx++] = (uint8_t)c;
  if (idx == (3)) {
    RgbToCmyk(rgb_store);
    idx = 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rgb_store>:
  endbr64
  movslq D3(%rip),%rax
  lea    D39(%rip),%r8
  movslq %edi,%rdi
  lea    0x1(%rax),%edx
  mov    %rdi,(%r8,%rax,8)
  mov    %edx,D38(%rip)
  cmp    $0x3,%edx
  je     L3
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  sub    $0x8,%rsp
  mov    %r8,%rdi
  call   <RgbToCmyk@plt>
  movl   $0x0,D38(%rip)
  add    $0x8,%rsp
  ret
```

assistant: ```c
void rgb_store(int c) {
  static uint8_t rgb_store[(3)];
  static int idx = 0;
  rgb_store[idx++] = (uint8_t)c;
  if (idx == (3)) {
    RgbToCmyk(rgb_store);
    idx = 0;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<udcReadFloat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <udcMustRead@plt>
  cmpl   $0x0,-0x1c(%rbp)
  je     L12
  mov    -0xc(%rbp),%eax
  movd   %eax,%xmm0
  call   <byteSwapFloat@plt>
  movd   %xmm0,%eax
  mov    %eax,-0xc(%rbp)
L12:
  movss  -0xc(%rbp),%xmm0
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

assistant: ```c
float udcReadFloat(struct udcFile *file, int isSwapped) {
  float val;
  udcMustRead(file, &val, sizeof(val));
  if (isSwapped)
    val = byteSwapFloat(val);
  return val;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<udcReadFloat>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %esi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  mov    $0x4,%edx
  call   <udcMustRead@plt>
  test   %ebx,%ebx
  jne    L24
L15:
  movss  0x4(%rsp),%xmm0
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x10,%rsp
  pop    %rbx
  ret
L24:
  movss  0x4(%rsp),%xmm0
  call   <byteSwapFloat@plt>
  movss  %xmm0,0x4(%rsp)
  jmp    L15
L29:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
float udcReadFloat(struct udcFile *file, int isSwapped) {
  float val;
  udcMustRead(file, &val, sizeof(val));
  if (isSwapped)
    val = byteSwapFloat(val);
  return val;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<udcReadFloat>:
  endbr64
  push   %rbx
  mov    $0x4,%edx
  mov    %esi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <udcMustRead@plt>
  movss  0x4(%rsp),%xmm0
  test   %ebx,%ebx
  jne    L26
L14:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L26:
  call   <byteSwapFloat@plt>
  jmp    L14
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
float udcReadFloat(struct udcFile *file, int isSwapped) {
  float val;
  udcMustRead(file, &val, sizeof(val));
  if (isSwapped)
    val = byteSwapFloat(val);
  return val;
}
```
user: What is the c source code of the assembly code below?

```asm
<udcReadFloat>:
  endbr64
  push   %rbx
  mov    $0x4,%edx
  mov    %esi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <udcMustRead@plt>
  movss  0x4(%rsp),%xmm0
  test   %ebx,%ebx
  jne    L9
L23:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x10,%rsp
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  call   <byteSwapFloat@plt>
  jmp    L23
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
float udcReadFloat(struct udcFile *file, int isSwapped) {
  float val;
  udcMustRead(file, &val, sizeof(val));
  if (isSwapped)
    val = byteSwapFloat(val);
  return val;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strspn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L16
L21:
  addq   $0x1,-0x8(%rbp)
L16:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L19
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L21
L19:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
size_t strspn(const char *str1, const char *str2) {
  size_t index = 0;
  while (*str1 && strchr(str2, *str1++))
    index++;
  return index;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<strspn>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  movzbl (%rdi),%esi
  mov    $0x0,%ebx
  test   %sil,%sil
  je     L0
L3:
  movsbl %sil,%esi
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L0
  add    $0x1,%rbx
  movzbl 0x0(%rbp,%rbx,1),%esi
  test   %sil,%sil
  jne    L3
L0:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
size_t strspn(const char *str1, const char *str2) {
  size_t index = 0;
  while (*str1 && strchr(str2, *str1++))
    index++;
  return index;
}
```
user: What is the c source code of the assembly code below:

```asm
<strspn>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  movsbl (%rdi),%esi
  mov    %rdi,%rbx
  test   %sil,%sil
  jne    L24
  jmp    L15
  nopl   0x0(%rax,%rax,1)
L30:
  add    $0x1,%r12
  movsbl (%rbx,%r12,1),%esi
  test   %sil,%sil
  je     L15
L24:
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L30
L15:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
size_t strspn(const char *str1, const char *str2) {
  size_t index = 0;
  while (*str1 && strchr(str2, *str1++))
    index++;
  return index;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strspn>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  movsbl (%rdi),%esi
  mov    %rdi,%rbx
  test   %sil,%sil
  jne    L11
  jmp    L30
  nopl   0x0(%rax,%rax,1)
L4:
  add    $0x1,%r12
  movsbl (%rbx,%r12,1),%esi
  test   %sil,%sil
  je     L30
L11:
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L4
L30:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
size_t strspn(const char *str1, const char *str2) {
  size_t index = 0;
  while (*str1 && strchr(str2, *str1++))
    index++;
  return index;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<copy_recursive_dirs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movq   $0x0,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L12
  mov    $0xffffffff,%eax
  jmp    L39
L12:
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L36
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0xffffffff,%eax
  jmp    L39
L36:
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <copy_recursive_exec@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  cmpl   $0xfffffffe,-0x14(%rbp)
  je     L48
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <copy_dir_entries@plt>
L48:
  mov    -0x14(%rbp),%eax
L39:
  leave
  ret
```

assistant: ```c
int copy_recursive_dirs(const char *oldpath, const char *cachepath) {
  char *dup_path = NULL;
  char *dup_cachepath = NULL;
  int ret = 0;
  dup_path = strdup(oldpath);
  if (!dup_path)
    return -1;
  dup_cachepath = strdup(cachepath);
  if (!dup_cachepath) {
    free(dup_path);
    return -1;
  }
  ret = copy_recursive_exec(dup_path, dup_cachepath);
  free(dup_path);
  free(dup_cachepath);
  if (ret != -2)
    copy_dir_entries(oldpath, cachepath);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<copy_recursive_dirs>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %rsi,%r14
  call   <strdup@plt>
  test   %rax,%rax
  je     L45
  mov    %rax,%rbx
  mov    %r14,%rdi
  call   <strdup@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L52
  mov    %rax,%rsi
  mov    %rbx,%rdi
  call   <copy_recursive_exec@plt>
  mov    %eax,%r12d
  mov    %rbx,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  cmp    $0xfffffffe,%r12d
  jne    L34
L8:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L52:
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0xffffffff,%r12d
  jmp    L8
L34:
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <copy_dir_entries@plt>
  jmp    L8
L45:
  mov    $0xffffffff,%r12d
  jmp    L8
```

assistant: ```c
int copy_recursive_dirs(const char *oldpath, const char *cachepath) {
  char *dup_path = NULL;
  char *dup_cachepath = NULL;
  int ret = 0;
  dup_path = strdup(oldpath);
  if (!dup_path)
    return -1;
  dup_cachepath = strdup(cachepath);
  if (!dup_cachepath) {
    free(dup_path);
    return -1;
  }
  ret = copy_recursive_exec(dup_path, dup_cachepath);
  free(dup_path);
  free(dup_cachepath);
  if (ret != -2)
    copy_dir_entries(oldpath, cachepath);
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<copy_recursive_dirs>:
  endbr64
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  call   <strdup@plt>
  test   %rax,%rax
  je     L28
  mov    %r14,%rdi
  mov    %rax,%rbp
  call   <strdup@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L20
  mov    %rax,%rsi
  mov    %rbp,%rdi
  call   <copy_recursive_exec@plt>
  mov    %rbp,%rdi
  mov    %eax,%r15d
  call   <free@plt>
  mov    %r12,%rdi
  call   <free@plt>
  cmp    $0xfffffffe,%r15d
  je     L36
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <copy_dir_entries@plt>
L36:
  pop    %rbp
  mov    %r15d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  mov    %rbp,%rdi
  or     $0xffffffff,%r15d
  call   <free@plt>
  jmp    L36
L28:
  or     $0xffffffff,%r15d
  jmp    L36
```

assistant: ```c
int copy_recursive_dirs(const char *oldpath, const char *cachepath) {
  char *dup_path = NULL;
  char *dup_cachepath = NULL;
  int ret = 0;
  dup_path = strdup(oldpath);
  if (!dup_path)
    return -1;
  dup_cachepath = strdup(cachepath);
  if (!dup_cachepath) {
    free(dup_path);
    return -1;
  }
  ret = copy_recursive_exec(dup_path, dup_cachepath);
  free(dup_path);
  free(dup_cachepath);
  if (ret != -2)
    copy_dir_entries(oldpath, cachepath);
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<copy_recursive_dirs>:
  endbr64
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  call   <strdup@plt>
  test   %rax,%rax
  je     L14
  mov    %r14,%rdi
  mov    %rax,%rbp
  call   <strdup@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L33
  mov    %rax,%rsi
  mov    %rbp,%rdi
  call   <copy_recursive_exec@plt>
  mov    %rbp,%rdi
  mov    %eax,%r15d
  call   <free@plt>
  mov    %r12,%rdi
  call   <free@plt>
  cmp    $0xfffffffe,%r15d
  je     L12
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <copy_dir_entries@plt>
L12:
  pop    %rbp
  mov    %r15d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L33:
  mov    %rbp,%rdi
  or     $0xffffffff,%r15d
  call   <free@plt>
  jmp    L12
L14:
  or     $0xffffffff,%r15d
  jmp    L12
```

assistant: ```c
int copy_recursive_dirs(const char *oldpath, const char *cachepath) {
  char *dup_path = NULL;
  char *dup_cachepath = NULL;
  int ret = 0;
  dup_path = strdup(oldpath);
  if (!dup_path)
    return -1;
  dup_cachepath = strdup(cachepath);
  if (!dup_cachepath) {
    free(dup_path);
    return -1;
  }
  ret = copy_recursive_exec(dup_path, dup_cachepath);
  free(dup_path);
  free(dup_cachepath);
  if (ret != -2)
    copy_dir_entries(oldpath, cachepath);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<gettime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  lea    -0xc(%rbp),%rsi
  lea    -0x10(%rbp),%rcx
  lea    -0x14(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rsi,%r8
  lea    D9(%rip),%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  sub    $0x1,%eax
  cmp    $0x2,%eax
  ja     L44
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  js     L2
  mov    -0x14(%rbp),%eax
  cmp    $0x18,%eax
  jg     L2
  mov    -0x10(%rbp),%eax
  test   %eax,%eax
  js     L2
  mov    -0x10(%rbp),%eax
  cmp    $0x3c,%eax
  jg     L2
  mov    -0xc(%rbp),%eax
  test   %eax,%eax
  js     L2
  mov    -0xc(%rbp),%eax
  cmp    $0x3c,%eax
  jle    L29
L2:
  mov    $0xffffffff,%eax
  jmp    L96
L29:
  mov    -0x14(%rbp),%eax
  imul   $0xe10,%eax,%edx
  mov    -0x10(%rbp),%eax
  imul   $0x3c,%eax,%eax
  add    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  jmp    L96
L44:
  nop
  mov    $0xffffffff,%eax
L96:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L93
  call   <__stack_chk_fail@plt>
L93:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D9
type_infered: string

tool: D9: %2d:%2d:%2d

assistant: ```c
int gettime(const char *digit) {
  int hour, min, sec;
  hour = min = sec = 0;
  switch (sscanf(digit, "%2d:%2d:%2d", &hour, &min, &sec)) {
  case 1:
  case 2:
  case 3:
    if (hour < 0 || hour > 24 || min < 0 || min > 60 || sec < 0 || sec > 60)
      return -1;
    return hour * 3600 + min * 60 + sec;
  default:
    break;
  }
  return -1;
}
```