ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<frctl>:
  endbr64
  movapd %xmm0,%xmm1
  mov    %esi,%eax
  pxor   %xmm0,%xmm0
  push   %rbp
  cvtsi2sd %rax,%xmm0
  push   %rbx
  mov    %rdi,%rbp
  mov    %rax,%rbx
  sub    $0x18,%rsp
  mulsd  %xmm1,%xmm0
  subsd  D8(%rip),%xmm0
  movsd  %xmm0,0x8(%rsp)
  call   <floor@plt>
  movsd  0x8(%rsp),%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm0,%xmm1
  ja     L14
  lea    -0x1(%rbx),%edx
  pxor   %xmm1,%xmm1
  cvtsi2sd %rdx,%xmm1
  comisd %xmm1,%xmm0
  ja     L15
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  cltq
  subsd  %xmm1,%xmm0
  movsd  D11(%rip),%xmm1
  subsd  %xmm0,%xmm1
  mulsd  0x8(%rbp,%rax,8),%xmm0
  mulsd  0x0(%rbp,%rax,8),%xmm1
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  addsd  %xmm1,%xmm0
  ret
  nopl   0x0(%rax)
L15:
  movsd  0x0(%rbp,%rdx,8),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L14:
  movsd  0x0(%rbp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D8
type_guessed: byte[8]
name: parse_data
arguments: tag: D11
type_guessed: byte[8]

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: {"D11": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double frctl(const double *xx, double p, unsigned int n) {
  double zi, th;
  int i;
  zi = p * n - .5;
  i = floor(zi);
  if (zi < 0)
    return xx[0];
  else if (zi > n - 1)
    return xx[n - 1];
  th = zi - i;
  return ((1 - th) * xx[i] + th * xx[i + 1]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<frctl>:
  endbr64
  movapd %xmm0,%xmm1
  mov    %esi,%eax
  pxor   %xmm0,%xmm0
  push   %rbp
  cvtsi2sd %rax,%xmm0
  push   %rbx
  mov    %rdi,%rbp
  mov    %rax,%rbx
  sub    $0x18,%rsp
  mulsd  %xmm1,%xmm0
  subsd  D9(%rip),%xmm0
  movsd  %xmm0,0x8(%rsp)
  call   <floor@plt>
  movsd  0x8(%rsp),%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm0,%xmm1
  ja     L4
  lea    -0x1(%rbx),%edx
  pxor   %xmm1,%xmm1
  cvtsi2sd %rdx,%xmm1
  comisd %xmm1,%xmm0
  ja     L15
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  cltq
  subsd  %xmm1,%xmm0
  movsd  D8(%rip),%xmm1
  subsd  %xmm0,%xmm1
  mulsd  0x8(%rbp,%rax,8),%xmm0
  mulsd  0x0(%rbp,%rax,8),%xmm1
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  addsd  %xmm1,%xmm0
  ret
  nopl   0x0(%rax)
L15:
  movsd  0x0(%rbp,%rdx,8),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L4:
  movsd  0x0(%rbp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D9
datatype: byte[8]
name: disassemble_data
arguments: data_tag: D8
datatype: byte[8]

tool: {"D9": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double frctl(const double *xx, double p, unsigned int n) {
  double zi, th;
  int i;
  zi = p * n - .5;
  i = floor(zi);
  if (zi < 0)
    return xx[0];
  else if (zi > n - 1)
    return xx[n - 1];
  th = zi - i;
  return ((1 - th) * xx[i] + th * xx[i + 1]);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x150,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x150(%rbp),%rax
  lea    D16(%rip),%rdx
  mov    $0x28,%ecx
  mov    %rax,%rdi
  mov    %rdx,%rsi
  rep movsq %ds:(%rsi),%es:(%rdi)
  mov    %rsi,%rdx
  mov    %rdi,%rax
  mov    (%rdx),%ecx
  mov    %ecx,(%rax)
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x150(%rbp),%rax
  mov    %rax,%rdi
  call   <affichage@plt>
  lea    -0x150(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <estValide@plt>
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x150(%rbp),%rax
  mov    %rax,%rdi
  call   <affichage@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D16
data_type: string
name: read_data
arguments: data_label: D1
data_type: string
name: read_data
arguments: data_label: D11
data_type: string

tool: D16: 	

tool: D1: Grille avant


tool: D11: Grille apres


assistant: ```c
int main(void) {
  int grille[9][9] = {{9, 0, 0, 1, 0, 0, 0, 0, 5}, {0, 0, 5, 0, 9, 0, 2, 0, 1}, {8, 0, 0, 0, 4, 0, 0, 0, 0}, {0, 0, 0, 0, 8, 0, 0, 0, 0}, {0, 0, 0, 7, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 2, 6, 0, 0, 9}, {2, 0, 0, 3, 0, 0, 0, 0, 6}, {0, 0, 0, 2, 0, 0, 9, 0, 0}, {0, 0, 1, 9, 0, 4, 5, 7, 0}};
  printf("Grille avant\n");
  affichage(grille);
  estValide(grille, 0);
  printf("Grille apres\n");
  affichage(grille);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x150,%rsp
  mov    $0x28,%ecx
  mov    %fs:(%rcx),%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    D22(%rip),%rsi
  rep movsq %ds:(%rsi),%es:(%rdi)
  mov    (%rsi),%eax
  mov    %eax,(%rdi)
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <affichage@plt>
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <estValide@plt>
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <affichage@plt>
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x150,%rsp
  pop    %rbx
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D22
datatype: string
name: extract_symbolic_data
arguments: data_tag: D13
datatype: string
name: extract_symbolic_data
arguments: data_tag: D3
datatype: string

tool: D22: 	

tool: D13: Grille avant


tool: D3: Grille apres


assistant: ```c
int main(void) {
  int grille[9][9] = {{9, 0, 0, 1, 0, 0, 0, 0, 5}, {0, 0, 5, 0, 9, 0, 2, 0, 1}, {8, 0, 0, 0, 4, 0, 0, 0, 0}, {0, 0, 0, 0, 8, 0, 0, 0, 0}, {0, 0, 0, 7, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 2, 6, 0, 0, 9}, {2, 0, 0, 3, 0, 0, 0, 0, 6}, {0, 0, 0, 2, 0, 0, 9, 0, 0}, {0, 0, 1, 9, 0, 4, 5, 7, 0}};
  printf("Grille avant\n");
  affichage(grille);
  estValide(grille, 0);
  printf("Grille apres\n");
  affichage(grille);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x28,%ecx
  lea    D6(%rip),%rsi
  sub    $0x150,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    %rsp,%rbp
  rep movsq %ds:(%rsi),%es:(%rdi)
  mov    (%rsi),%eax
  mov    %eax,(%rdi)
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <affichage@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <estValide@plt>
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <affichage@plt>
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x150,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D6
datatype: string
name: read_data
arguments: data_tag: D4
datatype: string
name: read_data
arguments: data_tag: D2
datatype: string

tool: D6: 	

tool: D4: Grille avant


tool: D2: Grille apres


assistant: ```c
int main(void) {
  int grille[9][9] = {{9, 0, 0, 1, 0, 0, 0, 0, 5}, {0, 0, 5, 0, 9, 0, 2, 0, 1}, {8, 0, 0, 0, 4, 0, 0, 0, 0}, {0, 0, 0, 0, 8, 0, 0, 0, 0}, {0, 0, 0, 7, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 2, 6, 0, 0, 9}, {2, 0, 0, 3, 0, 0, 0, 0, 6}, {0, 0, 0, 2, 0, 0, 9, 0, 0}, {0, 0, 1, 9, 0, 4, 5, 7, 0}};
  printf("Grille avant\n");
  affichage(grille);
  estValide(grille, 0);
  printf("Grille apres\n");
  affichage(grille);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x28,%ecx
  lea    D5(%rip),%rsi
  sub    $0x150,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    %rsp,%rbp
  rep movsq %ds:(%rsi),%es:(%rdi)
  mov    (%rsi),%eax
  mov    %eax,(%rdi)
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <affichage@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <estValide@plt>
  lea    D21(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <affichage@plt>
  mov    0x148(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x150,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D12
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D21
type_infered: string

tool: D5: 	

tool: D12: Grille avant


tool: D21: Grille apres


assistant: ```c
int main(void) {
  int grille[9][9] = {{9, 0, 0, 1, 0, 0, 0, 0, 5}, {0, 0, 5, 0, 9, 0, 2, 0, 1}, {8, 0, 0, 0, 4, 0, 0, 0, 0}, {0, 0, 0, 0, 8, 0, 0, 0, 0}, {0, 0, 0, 7, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 2, 6, 0, 0, 9}, {2, 0, 0, 3, 0, 0, 0, 0, 6}, {0, 0, 0, 2, 0, 0, 9, 0, 0}, {0, 0, 1, 9, 0, 4, 5, 7, 0}};
  printf("Grille avant\n");
  affichage(grille);
  estValide(grille, 0);
  printf("Grille apres\n");
  affichage(grille);
}
```
user: Translate the below assembly code into C programming language:

```asm
<rotate>:
  endbr64
  mov    %rdi,%rcx
  mov    %edx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L33
  sub    $0x1,%esi
  test   %esi,%esi
  jle    L18
  movslq %esi,%rdi
  mov    $0x0,%eax
L38:
  mov    (%rcx,%rax,4),%r8d
  mov    (%rcx,%rdi,4),%r9d
  mov    %r9d,(%rcx,%rax,4)
  mov    %r8d,(%rcx,%rdi,4)
  add    $0x1,%rax
  sub    $0x1,%rdi
  cmp    %edi,%eax
  jl     L38
L18:
  lea    -0x1(%rdx),%eax
  test   %eax,%eax
  jle    L25
  cltq
  mov    $0x0,%edi
L14:
  mov    (%rcx,%rdi,4),%r8d
  mov    (%rcx,%rax,4),%r9d
  mov    %r9d,(%rcx,%rdi,4)
  mov    %r8d,(%rcx,%rax,4)
  add    $0x1,%rdi
  sub    $0x1,%rax
  cmp    %eax,%edi
  jl     L14
L25:
  cmp    %esi,%edx
  jge    L33
  movslq %edx,%rdx
  movslq %esi,%rax
L64:
  mov    (%rcx,%rdx,4),%esi
  mov    (%rcx,%rax,4),%edi
  mov    %edi,(%rcx,%rdx,4)
  mov    %esi,(%rcx,%rax,4)
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %eax,%edx
  jl     L64
L33:
  ret
```

assistant: ```c
void rotate(int *nums, int numsSize, int k) {
  int begin = 0;
  int end = numsSize - 1;
  int tmp;
  if (k % numsSize == 0) {
    return;
  }
  k = k % numsSize;
  while (begin < end) {
    tmp = nums[begin];
    nums[begin] = nums[end];
    nums[end] = tmp;
    begin++;
    end--;
  }
  begin = 0;
  end = k - 1;
  while (begin < end) {
    tmp = nums[begin];
    nums[begin] = nums[end];
    nums[end] = tmp;
    begin++;
    end--;
  }
  begin = k;
  end = numsSize - 1;
  while (begin < end) {
    tmp = nums[begin];
    nums[begin] = nums[end];
    nums[end] = tmp;
    begin++;
    end--;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rotate>:
  endbr64
  mov    %edx,%eax
  mov    %rdi,%rcx
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L10
  sub    $0x1,%esi
  test   %esi,%esi
  jle    L60
  movslq %esi,%rdi
  xor    %eax,%eax
L67:
  mov    (%rcx,%rax,4),%r8d
  mov    (%rcx,%rdi,4),%r9d
  mov    %r9d,(%rcx,%rax,4)
  add    $0x1,%rax
  mov    %r8d,(%rcx,%rdi,4)
  sub    $0x1,%rdi
  cmp    %eax,%edi
  jg     L67
L60:
  lea    -0x1(%rdx),%eax
  test   %eax,%eax
  jle    L43
  cltq
  xor    %edi,%edi
  nopw   0x0(%rax,%rax,1)
L17:
  mov    (%rcx,%rdi,4),%r8d
  mov    (%rcx,%rax,4),%r9d
  mov    %r9d,(%rcx,%rdi,4)
  add    $0x1,%rdi
  mov    %r8d,(%rcx,%rax,4)
  sub    $0x1,%rax
  cmp    %edi,%eax
  jg     L17
L43:
  cmp    %esi,%edx
  jge    L10
  movslq %edx,%rdx
  movslq %esi,%rax
  cs nopw 0x0(%rax,%rax,1)
L30:
  mov    (%rcx,%rdx,4),%esi
  mov    (%rcx,%rax,4),%edi
  mov    %edi,(%rcx,%rdx,4)
  add    $0x1,%rdx
  mov    %esi,(%rcx,%rax,4)
  sub    $0x1,%rax
  cmp    %edx,%eax
  jg     L30
L10:
  ret
```

assistant: ```c
void rotate(int *nums, int numsSize, int k) {
  int begin = 0;
  int end = numsSize - 1;
  int tmp;
  if (k % numsSize == 0) {
    return;
  }
  k = k % numsSize;
  while (begin < end) {
    tmp = nums[begin];
    nums[begin] = nums[end];
    nums[end] = tmp;
    begin++;
    end--;
  }
  begin = 0;
  end = k - 1;
  while (begin < end) {
    tmp = nums[begin];
    nums[begin] = nums[end];
    nums[end] = tmp;
    begin++;
    end--;
  }
  begin = k;
  end = numsSize - 1;
  while (begin < end) {
    tmp = nums[begin];
    nums[begin] = nums[end];
    nums[end] = tmp;
    begin++;
    end--;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<taa_h264_inverse_hadamard_4x4>:
  endbr64
  push   %rbx
  mov    %rsi,%r11
  mov    %rdi,%rcx
  mov    %rsi,%rax
  add    $0x40,%rdi
L6:
  mov    (%rcx),%edx
  mov    %edx,%r9d
  add    0x8(%rcx),%r9d
  sub    0x8(%rcx),%edx
  mov    0x4(%rcx),%r8d
  mov    %r8d,%r10d
  sub    0xc(%rcx),%r10d
  add    0xc(%rcx),%r8d
  lea    (%r9,%r8,1),%ebx
  mov    %ebx,(%rsi)
  lea    (%rdx,%r10,1),%ebx
  mov    %ebx,0x4(%rsi)
  sub    %r10d,%edx
  mov    %edx,0x8(%rsi)
  sub    %r8d,%r9d
  mov    %r9d,0xc(%rsi)
  add    $0x10,%rcx
  add    $0x10,%rsi
  cmp    %rdi,%rcx
  jne    L6
  lea    0x10(%r11),%r8
L11:
  mov    (%rax),%edx
  mov    %edx,%esi
  add    0x20(%rax),%esi
  sub    0x20(%rax),%edx
  mov    0x10(%rax),%ecx
  mov    %ecx,%edi
  sub    0x30(%rax),%edi
  add    0x30(%rax),%ecx
  lea    (%rsi,%rcx,1),%r9d
  mov    %r9d,(%rax)
  lea    (%rdx,%rdi,1),%r9d
  mov    %r9d,0x10(%rax)
  sub    %edi,%edx
  mov    %edx,0x20(%rax)
  sub    %ecx,%esi
  mov    %esi,0x30(%rax)
  add    $0x4,%rax
  cmp    %r8,%rax
  jne    L11
  pop    %rbx
  ret
```

assistant: ```c
void taa_h264_inverse_hadamard_4x4(int16_t *coeff, int16_t *block) {
  int tmp0, tmp1, tmp2, tmp3;
  for (int i = 0; i < 4; i++) {
    tmp0 = coeff[i * 4 + 0] + coeff[i * 4 + 2];
    tmp1 = coeff[i * 4 + 0] - coeff[i * 4 + 2];
    tmp2 = coeff[i * 4 + 1] - coeff[i * 4 + 3];
    tmp3 = coeff[i * 4 + 1] + coeff[i * 4 + 3];
    block[i * 4 + 0] = (int16_t)(tmp0 + tmp3);
    block[i * 4 + 1] = (int16_t)(tmp1 + tmp2);
    block[i * 4 + 2] = (int16_t)(tmp1 - tmp2);
    block[i * 4 + 3] = (int16_t)(tmp0 - tmp3);
  }
  for (int i = 0; i < 4; i++) {
    tmp0 = block[0 * 4 + i] + block[2 * 4 + i];
    tmp1 = block[0 * 4 + i] - block[2 * 4 + i];
    tmp2 = block[1 * 4 + i] - block[3 * 4 + i];
    tmp3 = block[1 * 4 + i] + block[3 * 4 + i];
    block[0 * 4 + i] = (int16_t)(tmp0 + tmp3);
    block[1 * 4 + i] = (int16_t)(tmp1 + tmp2);
    block[2 * 4 + i] = (int16_t)(tmp1 - tmp2);
    block[3 * 4 + i] = (int16_t)(tmp0 - tmp3);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<taa_h264_inverse_hadamard_4x4>:
  endbr64
  push   %rbx
  mov    %rsi,%r10
  mov    %rsi,%rax
  lea    0x40(%rdi),%rbx
  mov    %rsi,%rcx
L15:
  mov    (%rdi),%edx
  mov    0x4(%rdi),%esi
  add    $0x10,%rdi
  add    $0x10,%rcx
  mov    -0x8(%rdi),%r8d
  mov    %esi,%r9d
  add    -0x4(%rdi),%esi
  sub    -0x4(%rdi),%r9d
  add    %edx,%r8d
  sub    -0x8(%rdi),%edx
  lea    (%r8,%rsi,1),%r11d
  sub    %esi,%r8d
  mov    %r11d,-0x10(%rcx)
  lea    (%rdx,%r9,1),%r11d
  sub    %r9d,%edx
  mov    %r11d,-0xc(%rcx)
  mov    %edx,-0x8(%rcx)
  mov    %r8d,-0x4(%rcx)
  cmp    %rdi,%rbx
  jne    L15
  lea    0x10(%r10),%r8
L13:
  mov    (%rax),%edx
  mov    0x10(%rax),%ecx
  add    $0x4,%rax
  mov    0x1c(%rax),%esi
  mov    %ecx,%edi
  add    0x2c(%rax),%ecx
  sub    0x2c(%rax),%edi
  add    %edx,%esi
  sub    0x1c(%rax),%edx
  lea    (%rsi,%rcx,1),%r9d
  sub    %ecx,%esi
  mov    %r9d,-0x4(%rax)
  lea    (%rdx,%rdi,1),%r9d
  sub    %edi,%edx
  mov    %r9d,0xc(%rax)
  mov    %edx,0x1c(%rax)
  mov    %esi,0x2c(%rax)
  cmp    %rax,%r8
  jne    L13
  pop    %rbx
  ret
```

assistant: ```c
void taa_h264_inverse_hadamard_4x4(int16_t *coeff, int16_t *block) {
  int tmp0, tmp1, tmp2, tmp3;
  for (int i = 0; i < 4; i++) {
    tmp0 = coeff[i * 4 + 0] + coeff[i * 4 + 2];
    tmp1 = coeff[i * 4 + 0] - coeff[i * 4 + 2];
    tmp2 = coeff[i * 4 + 1] - coeff[i * 4 + 3];
    tmp3 = coeff[i * 4 + 1] + coeff[i * 4 + 3];
    block[i * 4 + 0] = (int16_t)(tmp0 + tmp3);
    block[i * 4 + 1] = (int16_t)(tmp1 + tmp2);
    block[i * 4 + 2] = (int16_t)(tmp1 - tmp2);
    block[i * 4 + 3] = (int16_t)(tmp0 - tmp3);
  }
  for (int i = 0; i < 4; i++) {
    tmp0 = block[0 * 4 + i] + block[2 * 4 + i];
    tmp1 = block[0 * 4 + i] - block[2 * 4 + i];
    tmp2 = block[1 * 4 + i] - block[3 * 4 + i];
    tmp3 = block[1 * 4 + i] + block[3 * 4 + i];
    block[0 * 4 + i] = (int16_t)(tmp0 + tmp3);
    block[1 * 4 + i] = (int16_t)(tmp1 + tmp2);
    block[2 * 4 + i] = (int16_t)(tmp1 - tmp2);
    block[3 * 4 + i] = (int16_t)(tmp0 - tmp3);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<clar__assert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x8(%rbp)
  mov    %rcx,-0x18(%rbp)
  mov    %r8,-0x20(%rbp)
  mov    %r9d,-0x24(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L16
  cmpl   $0x0,-0x24(%rbp)
  je     L16
  mov    $0x0,%eax
  call   <__coverity_panic__@plt>
L16:
  nop
  leave
  ret
```

assistant: ```c
void clar__assert(int condition, const char *file, int line, const char *error, const char *description, int should_abort) {
  if (!condition && should_abort)
    __coverity_panic__();
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<clar__assert>:
  endbr64
  test   %edi,%edi
  jne    L2
  test   %r9d,%r9d
  jne    L13
L2:
  ret
L13:
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <__coverity_panic__@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void clar__assert(int condition, const char *file, int line, const char *error, const char *description, int should_abort) {
  if (!condition && should_abort)
    __coverity_panic__();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<clar__assert>:
  endbr64
  test   %edi,%edi
  jne    L17
  test   %r9d,%r9d
  jne    L18
L17:
  ret
  xchg   %ax,%ax
L18:
  xor    %eax,%eax
  jmp    1050 <__coverity_panic__@plt>
```

assistant: ```c
void clar__assert(int condition, const char *file, int line, const char *error, const char *description, int should_abort) {
  if (!condition && should_abort)
    __coverity_panic__();
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<clar__assert>:
  endbr64
  test   %edi,%edi
  jne    L9
  test   %r9d,%r9d
  jne    L15
L9:
  ret
  xchg   %ax,%ax
L15:
  xor    %eax,%eax
  jmp    1050 <__coverity_panic__@plt>
```

assistant: ```c
void clar__assert(int condition, const char *file, int line, const char *error, const char *description, int should_abort) {
  if (!condition && should_abort)
    __coverity_panic__();
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x18(%rbp)
  movl   $0x1,-0x14(%rbp)
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x10(%rbp)
  jmp    L25
L26:
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0x10(%rbp)
L25:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x10(%rbp)
  jl     L26
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D3
type: string
name: parse_data
arguments: data_label: D6
type: string
name: parse_data
arguments: data_label: D14
type: string

tool: D3: %d

tool: D6: 0 

tool: D14: %d 

assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  int penultimate = 0, last = 1;
  int current;
  int i;
  printf("0 ");
  for (i = 0; i < n - 1; i++) {
    current = penultimate + last;
    printf("%d ", current);
    penultimate = last;
    last = current;
  }
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  lea    D15(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L5
  mov    $0x0,%ebp
  mov    $0x1,%ebx
  mov    $0x0,%edx
  lea    D5(%rip),%r13
L22:
  mov    %ebx,%r12d
  add    %edx,%ebx
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebp
  mov    %r12d,%edx
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  cmp    %ebp,%eax
  jg     L22
L5:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D9
datatype: string
name: disassemble_data
arguments: data_tag: D15
datatype: string
name: disassemble_data
arguments: data_tag: D5
datatype: string

tool: D9: %d

tool: D15: 0 

tool: D5: %d 

assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  int penultimate = 0, last = 1;
  int current;
  int i;
  printf("0 ");
  for (i = 0; i < n - 1; i++) {
    current = penultimate + last;
    printf("%d ", current);
    penultimate = last;
    last = current;
  }
  printf("\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D12(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  xor    %eax,%eax
  lea    D20(%rip),%rdi
  call   <printf@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L12
  xor    %ebp,%ebp
  mov    $0x1,%ebx
  lea    D10(%rip),%r13
  xor    %edx,%edx
  nopl   0x0(%rax)
L9:
  mov    %ebx,%r12d
  add    %edx,%ebx
  mov    %r13,%rdi
  xor    %eax,%eax
  mov    %ebx,%esi
  add    $0x1,%ebp
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  mov    %r12d,%edx
  sub    $0x1,%eax
  cmp    %ebp,%eax
  jg     L9
L12:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D12
type_infered: string
name: retrieve_data
arguments: data_tag: D20
type_infered: string
name: retrieve_data
arguments: data_tag: D10
type_infered: string

tool: D12: %d

tool: D20: 0 

tool: D10: %d 

assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  int penultimate = 0, last = 1;
  int current;
  int i;
  printf("0 ");
  for (i = 0; i < n - 1; i++) {
    current = penultimate + last;
    printf("%d ", current);
    penultimate = last;
    last = current;
  }
  printf("\n");
  return 0;
}
```