ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<dmp_set_tap_axes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %al,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x0,-0x9(%rbp)
  movzbl -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L1
  movzbl -0x9(%rbp),%eax
  or     $0x30,%eax
  mov    %al,-0x9(%rbp)
L1:
  movzbl -0x14(%rbp),%eax
  and    $0x2,%eax
  test   %eax,%eax
  je     L35
  movzbl -0x9(%rbp),%eax
  or     $0xc,%eax
  mov    %al,-0x9(%rbp)
L35:
  movzbl -0x14(%rbp),%eax
  and    $0x4,%eax
  test   %eax,%eax
  je     L9
  movzbl -0x9(%rbp),%eax
  or     $0x3,%eax
  mov    %al,-0x9(%rbp)
L9:
  lea    -0x9(%rbp),%rax
  mov    %rax,%rdx
  mov    $0x1,%esi
  mov    $0x148,%edi
  call   <mpu_write_mem@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L32
  call   <__stack_chk_fail@plt>
L32:
  leave
  ret
```

assistant: ```c
int dmp_set_tap_axes(unsigned char axis) {
  unsigned char tmp = 0;
  if (axis & (0x01))
    tmp |= 0x30;
  if (axis & (0x02))
    tmp |= 0x0C;
  if (axis & (0x04))
    tmp |= 0x03;
  return mpu_write_mem((256 + 72), 1, &tmp);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<dmp_set_tap_axes>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %edi,%eax
  and    $0x1,%eax
  neg    %eax
  and    $0x30,%eax
  mov    %al,0x7(%rsp)
  test   $0x2,%dil
  je     L24
  orb    $0xc,0x7(%rsp)
L24:
  test   $0x4,%dil
  je     L22
  orb    $0x3,0x7(%rsp)
L22:
  lea    0x7(%rsp),%rdx
  mov    $0x1,%esi
  mov    $0x148,%edi
  call   <mpu_write_mem@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L29
  add    $0x18,%rsp
  ret
L29:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dmp_set_tap_axes(unsigned char axis) {
  unsigned char tmp = 0;
  if (axis & (0x01))
    tmp |= 0x30;
  if (axis & (0x02))
    tmp |= 0x0C;
  if (axis & (0x04))
    tmp |= 0x03;
  return mpu_write_mem((256 + 72), 1, &tmp);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dmp_set_tap_axes>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %edi,%eax
  and    $0x1,%eax
  neg    %eax
  and    $0x30,%eax
  mov    %al,0x7(%rsp)
  test   $0x2,%dil
  je     L3
  orb    $0xc,0x7(%rsp)
L3:
  and    $0x4,%edi
  je     L24
  orb    $0x3,0x7(%rsp)
L24:
  lea    0x7(%rsp),%rdx
  mov    $0x1,%esi
  mov    $0x148,%edi
  call   <mpu_write_mem@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L29
  add    $0x18,%rsp
  ret
L29:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dmp_set_tap_axes(unsigned char axis) {
  unsigned char tmp = 0;
  if (axis & (0x01))
    tmp |= 0x30;
  if (axis & (0x02))
    tmp |= 0x0C;
  if (axis & (0x04))
    tmp |= 0x03;
  return mpu_write_mem((256 + 72), 1, &tmp);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dmp_set_tap_axes>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   $0x1,%dil
  jne    L19
  movb   $0x0,0x7(%rsp)
  mov    $0x3,%eax
  mov    $0xf,%edx
  mov    $0xc,%ecx
L12:
  test   $0x2,%dil
  je     L24
  mov    %cl,0x7(%rsp)
  mov    %edx,%eax
L24:
  and    $0x4,%edi
  je     L5
  mov    %al,0x7(%rsp)
L5:
  lea    0x7(%rsp),%rdx
  mov    $0x1,%esi
  mov    $0x148,%edi
  call   <mpu_write_mem@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L33
  add    $0x18,%rsp
  ret
L19:
  movb   $0x30,0x7(%rsp)
  mov    $0x33,%eax
  mov    $0x3f,%edx
  mov    $0x3c,%ecx
  jmp    L12
L33:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dmp_set_tap_axes(unsigned char axis) {
  unsigned char tmp = 0;
  if (axis & (0x01))
    tmp |= 0x30;
  if (axis & (0x02))
    tmp |= 0x0C;
  if (axis & (0x04))
    tmp |= 0x03;
  return mpu_write_mem((256 + 72), 1, &tmp);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x34(%rbp)
  lea    -0x34(%rbp),%rdi
  lea    -0x12(%rbp),%rcx
  lea    -0x1c(%rbp),%rdx
  lea    -0x26(%rbp),%rsi
  lea    -0x30(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <distrifunc@plt>
  mov    -0x34(%rbp),%ecx
  lea    -0x12(%rbp),%rdx
  lea    -0x1c(%rbp),%rsi
  lea    -0x26(%rbp),%rax
  mov    %rax,%rdi
  call   <printfunc@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int main(void) {
  char input[10], ware[10], pro[10], quali[10];
  int a = 0;
  distrifunc(input, ware, pro, quali, &a);
  printfunc(ware, pro, quali, a);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x0,0xc(%rsp)
  lea    0x2e(%rsp),%r12
  lea    0x24(%rsp),%rbp
  lea    0x1a(%rsp),%rbx
  lea    0x10(%rsp),%rdi
  lea    0xc(%rsp),%r8
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  call   <distrifunc@plt>
  mov    0xc(%rsp),%ecx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <printfunc@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  char input[10], ware[10], pro[10], quali[10];
  int a = 0;
  distrifunc(input, ware, pro, quali, &a);
  printfunc(ware, pro, quali, a);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  sub    $0x40,%rsp
  lea    0x2e(%rsp),%r13
  lea    0x24(%rsp),%r12
  lea    0x1a(%rsp),%rbp
  lea    0x10(%rsp),%rdi
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    0xc(%rsp),%r8
  movl   $0x0,0xc(%rsp)
  call   <distrifunc@plt>
  mov    0xc(%rsp),%ecx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <printfunc@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  char input[10], ware[10], pro[10], quali[10];
  int a = 0;
  distrifunc(input, ware, pro, quali, &a);
  printfunc(ware, pro, quali, a);
  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
  sub    $0x40,%rsp
  lea    0x2e(%rsp),%r13
  lea    0x24(%rsp),%r12
  lea    0x1a(%rsp),%rbp
  lea    0x10(%rsp),%rdi
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    0xc(%rsp),%r8
  movl   $0x0,0xc(%rsp)
  call   <distrifunc@plt>
  mov    0xc(%rsp),%ecx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <printfunc@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  char input[10], ware[10], pro[10], quali[10];
  int a = 0;
  distrifunc(input, ware, pro, quali, &a);
  printfunc(ware, pro, quali, a);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<printArr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L11:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0x4(%rbp)
L18:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L11
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

assistant: ```c
void printArr(char arr[], int k) {
  int i;
  for (i = 0; i < k; i++)
    printf("%c", arr[i]);
  printf("\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<printArr>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L3
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x1(%rdi,%rax,1),%rbp
L18:
  movsbl (%rbx),%edi
  call   <putchar@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L18
L3:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void printArr(char arr[], int k) {
  int i;
  for (i = 0; i < k; i++)
    printf("%c", arr[i]);
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<printArr>:
  endbr64
  test   %esi,%esi
  jle    L14
  push   %rbp
  lea    -0x1(%rsi),%eax
  push   %rbx
  lea    0x1(%rdi,%rax,1),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L6:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <putchar@plt>
  cmp    %rbp,%rbx
  jne    L6
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  jmp    1050 <putchar@plt>
  nopl   0x0(%rax)
L14:
  mov    $0xa,%edi
  jmp    1050 <putchar@plt>
```

assistant: ```c
void printArr(char arr[], int k) {
  int i;
  for (i = 0; i < k; i++)
    printf("%c", arr[i]);
  printf("\n");
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<printArr>:
  endbr64
  test   %esi,%esi
  jle    L5
  push   %rbp
  lea    -0x1(%rsi),%eax
  push   %rbx
  lea    0x1(%rdi,%rax,1),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L16:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <putchar@plt>
  cmp    %rbp,%rbx
  jne    L16
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  jmp    1050 <putchar@plt>
  nopl   0x0(%rax)
L5:
  mov    $0xa,%edi
  jmp    1050 <putchar@plt>
```

assistant: ```c
void printArr(char arr[], int k) {
  int i;
  for (i = 0; i < k; i++)
    printf("%c", arr[i]);
  printf("\n");
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<cal_ex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %rdi,-0x68(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %rdx,-0x78(%rbp)
  mov    %ecx,-0x7c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,-0x30(%rbp)
  movl   $0x0,-0x54(%rbp)
  jmp    L18
L21:
  mov    -0x54(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x78(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x7c(%rbp),%eax
  sub    -0x54(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x70(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rcx
  lea    -0x50(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <mm_mul@plt>
  movsd  -0x30(%rbp),%xmm1
  movsd  -0x50(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x30(%rbp)
  movsd  -0x28(%rbp),%xmm1
  movsd  -0x48(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movsd  -0x20(%rbp),%xmm1
  movsd  -0x40(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x18(%rbp),%xmm1
  movsd  -0x38(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  addl   $0x1,-0x54(%rbp)
L18:
  mov    -0x54(%rbp),%eax
  cmp    -0x7c(%rbp),%eax
  jl     L21
  movsd  -0x30(%rbp),%xmm0
  mov    -0x68(%rbp),%rax
  movsd  %xmm0,(%rax)
  mov    -0x68(%rbp),%rax
  add    $0x8,%rax
  movsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x68(%rbp),%rax
  add    $0x10,%rax
  movsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x68(%rbp),%rax
  add    $0x18,%rax
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

assistant: ```c
void cal_ex(double *ex, double **r, double **x, const int i) {
  int j;
  double t[4], s[4];
  s[0] = s[1] = s[2] = s[3] = 0.;
  for (j = 0; j < i; j++) {
    mm_mul(t, r[i - j], x[j]);
    s[0] += t[0];
    s[1] += t[1];
    s[2] += t[2];
    s[3] += t[3];
  }
  ex[0] = s[0];
  ex[1] = s[1];
  ex[2] = s[2];
  ex[3] = s[3];
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cal_ex>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdi,%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  test   %ecx,%ecx
  jle    L15
  mov    %rdx,%rbx
  movslq %ecx,%rax
  lea    (%rsi,%rax,8),%rbp
  lea    -0x1(%rcx),%eax
  lea    0x8(%rdx,%rax,8),%r14
  mov    D17(%rip),%r12
  mov    %r12,0x8(%rsp)
  mov    %r12,0x10(%rsp)
  mov    %r12,0x18(%rsp)
  lea    0x20(%rsp),%r13
L2:
  mov    (%rbx),%rdx
  mov    0x0(%rbp),%rsi
  mov    %r13,%rdi
  call   <mm_mul@plt>
  movsd  0x8(%rsp),%xmm0
  addsd  0x20(%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
  movsd  0x10(%rsp),%xmm1
  addsd  0x28(%rsp),%xmm1
  movsd  %xmm1,0x10(%rsp)
  movsd  0x18(%rsp),%xmm2
  addsd  0x30(%rsp),%xmm2
  movsd  %xmm2,0x18(%rsp)
  movq   %r12,%xmm3
  addsd  0x38(%rsp),%xmm3
  movq   %xmm3,%r12
  add    $0x8,%rbx
  sub    $0x8,%rbp
  cmp    %r14,%rbx
  jne    L2
L39:
  movsd  0x8(%rsp),%xmm4
  movsd  %xmm4,(%r15)
  movsd  0x10(%rsp),%xmm5
  movsd  %xmm5,0x8(%r15)
  movsd  0x18(%rsp),%xmm6
  movsd  %xmm6,0x10(%r15)
  mov    %r12,0x18(%r15)
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L15:
  mov    D17(%rip),%r12
  mov    %r12,0x8(%rsp)
  mov    %r12,0x10(%rsp)
  mov    %r12,0x18(%rsp)
  jmp    L39
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D17
datatype: f64

tool: D17: 0

assistant: ```c
void cal_ex(double *ex, double **r, double **x, const int i) {
  int j;
  double t[4], s[4];
  s[0] = s[1] = s[2] = s[3] = 0.;
  for (j = 0; j < i; j++) {
    mm_mul(t, r[i - j], x[j]);
    s[0] += t[0];
    s[1] += t[1];
    s[2] += t[2];
    s[3] += t[3];
  }
  ex[0] = s[0];
  ex[1] = s[1];
  ex[2] = s[2];
  ex[3] = s[3];
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<cal_ex>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  test   %ecx,%ecx
  jle    L10
  movslq %ecx,%rax
  pxor   %xmm0,%xmm0
  lea    0x20(%rsp),%r12
  mov    %rdx,%rbx
  lea    (%rsi,%rax,8),%rbp
  lea    -0x1(%rcx),%eax
  movapd %xmm0,%xmm3
  lea    0x8(%rdx,%rax,8),%r13
  movapd %xmm0,%xmm2
  movapd %xmm0,%xmm1
  nopw   0x0(%rax,%rax,1)
L12:
  mov    (%rbx),%rdx
  mov    0x0(%rbp),%rsi
  mov    %r12,%rdi
  add    $0x8,%rbx
  movsd  %xmm0,0x18(%rsp)
  sub    $0x8,%rbp
  movsd  %xmm1,0x10(%rsp)
  movsd  %xmm2,0x8(%rsp)
  movsd  %xmm3,(%rsp)
  call   <mm_mul@plt>
  movsd  (%rsp),%xmm3
  movsd  0x8(%rsp),%xmm2
  movsd  0x10(%rsp),%xmm1
  movsd  0x18(%rsp),%xmm0
  addsd  0x20(%rsp),%xmm3
  addsd  0x28(%rsp),%xmm2
  addsd  0x30(%rsp),%xmm1
  addsd  0x38(%rsp),%xmm0
  cmp    %r13,%rbx
  jne    L12
L14:
  movsd  %xmm3,(%r14)
  movsd  %xmm2,0x8(%r14)
  movsd  %xmm1,0x10(%r14)
  movsd  %xmm0,0x18(%r14)
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x50,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  xchg   %ax,%ax
L10:
  pxor   %xmm0,%xmm0
  movapd %xmm0,%xmm3
  movapd %xmm0,%xmm2
  movapd %xmm0,%xmm1
  jmp    L14
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void cal_ex(double *ex, double **r, double **x, const int i) {
  int j;
  double t[4], s[4];
  s[0] = s[1] = s[2] = s[3] = 0.;
  for (j = 0; j < i; j++) {
    mm_mul(t, r[i - j], x[j]);
    s[0] += t[0];
    s[1] += t[1];
    s[2] += t[2];
    s[3] += t[3];
  }
  ex[0] = s[0];
  ex[1] = s[1];
  ex[2] = s[2];
  ex[3] = s[3];
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cal_ex>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  test   %ecx,%ecx
  jle    L6
  movslq %ecx,%rax
  pxor   %xmm0,%xmm0
  lea    0x20(%rsp),%r12
  mov    %rdx,%rbx
  lea    (%rsi,%rax,8),%rbp
  lea    -0x1(%rcx),%eax
  movapd %xmm0,%xmm1
  lea    0x8(%rdx,%rax,8),%r14
  nopw   0x0(%rax,%rax,1)
L38:
  mov    (%rbx),%rdx
  mov    0x0(%rbp),%rsi
  mov    %r12,%rdi
  add    $0x8,%rbx
  movaps %xmm1,0x10(%rsp)
  sub    $0x8,%rbp
  movaps %xmm0,(%rsp)
  call   <mm_mul@plt>
  movapd 0x10(%rsp),%xmm1
  movapd (%rsp),%xmm0
  addpd  0x20(%rsp),%xmm1
  addpd  0x30(%rsp),%xmm0
  cmp    %r14,%rbx
  jne    L38
L32:
  movups %xmm1,0x0(%r13)
  movups %xmm0,0x10(%r13)
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x50,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  pxor   %xmm0,%xmm0
  movapd %xmm0,%xmm1
  jmp    L32
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void cal_ex(double *ex, double **r, double **x, const int i) {
  int j;
  double t[4], s[4];
  s[0] = s[1] = s[2] = s[3] = 0.;
  for (j = 0; j < i; j++) {
    mm_mul(t, r[i - j], x[j]);
    s[0] += t[0];
    s[1] += t[1];
    s[2] += t[2];
    s[3] += t[3];
  }
  ex[0] = s[0];
  ex[1] = s[1];
  ex[2] = s[2];
  ex[3] = s[3];
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<strpbrk>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  jmp    L22
L13:
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L37
L42:
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jne    L37
  mov    -0x18(%rbp),%rax
  sub    $0x1,%rax
  jmp    L34
L37:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jne    L42
L22:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x10(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  jne    L13
  mov    $0x0,%eax
L34:
  pop    %rbp
  ret
```

assistant: ```c
char *strpbrk(const char *s1, const char *s2) {
  const char *scanp;
  int c, sc;
  while ((c = *s1++) != 0) {
    for (scanp = s2; (sc = *scanp++) != '\0';)
      if (sc == c) {
        return ((char *)(s1 - 1));
      }
  }
  return (NULL);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strpbrk>:
  endbr64
  movsbl (%rdi),%ecx
  test   %ecx,%ecx
  je     L53
L40:
  mov    %rdi,%r8
  mov    %rsi,%rax
L58:
  add    $0x1,%rax
  movsbl -0x1(%rax),%edx
  test   %edx,%edx
  je     L26
  cmp    %ecx,%edx
  jne    L58
L9:
  mov    %r8,%rax
  ret
L26:
  movsbl 0x1(%r8),%ecx
  add    $0x1,%rdi
  test   %ecx,%ecx
  jne    L40
  mov    $0x0,%r8d
  jmp    L9
L53:
  mov    $0x0,%r8d
  jmp    L9
```

assistant: ```c
char *strpbrk(const char *s1, const char *s2) {
  const char *scanp;
  int c, sc;
  while ((c = *s1++) != 0) {
    for (scanp = s2; (sc = *scanp++) != '\0';)
      if (sc == c) {
        return ((char *)(s1 - 1));
      }
  }
  return (NULL);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strpbrk>:
  endbr64
  movsbl (%rdi),%ecx
  test   %ecx,%ecx
  je     L23
L47:
  mov    %rdi,%r8
  mov    %rsi,%rax
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L29:
  cmp    %ecx,%edx
  je     L15
L33:
  movsbl (%rax),%edx
  add    $0x1,%rax
  test   %edx,%edx
  jne    L29
  movsbl 0x1(%rdi),%ecx
  add    $0x1,%rdi
  test   %ecx,%ecx
  jne    L47
L23:
  xor    %r8d,%r8d
L15:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *strpbrk(const char *s1, const char *s2) {
  const char *scanp;
  int c, sc;
  while ((c = *s1++) != 0) {
    for (scanp = s2; (sc = *scanp++) != '\0';)
      if (sc == c) {
        return ((char *)(s1 - 1));
      }
  }
  return (NULL);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strpbrk>:
  endbr64
  movsbl (%rdi),%ecx
  test   %ecx,%ecx
  je     L31
L34:
  mov    %rdi,%r8
  mov    %rsi,%rax
  jmp    L37
  nopl   0x0(%rax,%rax,1)
L41:
  cmp    %ecx,%edx
  je     L47
L37:
  movsbl (%rax),%edx
  add    $0x1,%rax
  test   %edx,%edx
  jne    L41
  movsbl 0x1(%rdi),%ecx
  add    $0x1,%rdi
  test   %ecx,%ecx
  jne    L34
L31:
  xor    %r8d,%r8d
L47:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *strpbrk(const char *s1, const char *s2) {
  const char *scanp;
  int c, sc;
  while ((c = *s1++) != 0) {
    for (scanp = s2; (sc = *scanp++) != '\0';)
      if (sc == c) {
        return ((char *)(s1 - 1));
      }
  }
  return (NULL);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<checksum_path>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x20(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    $0x8,%esi
  mov    %eax,%edi
  call   <roof@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shr    $0x3,%rax
  mov    %eax,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <alloca@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x28(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L20
L18:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  xor    %rax,-0x18(%rbp)
  addq   $0x8,-0x10(%rbp)
  subl   $0x1,-0x20(%rbp)
L20:
  cmpl   $0x0,-0x20(%rbp)
  jne    L18
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

assistant: ```c
unsigned long long checksum_path(const char *path) {
  unsigned long long csum = 0;
  unsigned long long *nums = 0;
  int len = 0;
  int cnt = 0;
  len = roof(strlen(path), sizeof(csum));
  cnt = len / sizeof(csum);
  nums = alloca(len);
  memset(nums, 0, len);
  strcpy((char *)nums, path);
  while (cnt) {
    csum ^= *nums;
    nums++;
    cnt--;
  }
  return csum;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<checksum_path>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%r13
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,%edi
  mov    $0x8,%esi
  call   <roof@plt>
  mov    %eax,%r12d
  movslq %eax,%r14
  shr    $0x3,%r14
  mov    %eax,%edi
  call   <alloca@plt>
  mov    %rax,%rbx
  mov    %r12d,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  test   %r14d,%r14d
  je     L29
  lea    -0x1(%r14),%eax
  lea    0x8(%rbx,%rax,8),%rdx
  mov    $0x0,%eax
L21:
  xor    (%rbx),%rax
  add    $0x8,%rbx
  cmp    %rdx,%rbx
  jne    L21
L18:
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L36
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L29:
  mov    $0x0,%eax
  jmp    L18
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long checksum_path(const char *path) {
  unsigned long long csum = 0;
  unsigned long long *nums = 0;
  int len = 0;
  int cnt = 0;
  len = roof(strlen(path), sizeof(csum));
  cnt = len / sizeof(csum);
  nums = alloca(len);
  memset(nums, 0, len);
  strcpy((char *)nums, path);
  while (cnt) {
    csum ^= *nums;
    nums++;
    cnt--;
  }
  return csum;
}
```