ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<heapSort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <buildMaxHeap@plt>
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L10
L8:
  mov    -0x10(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <treeReOrderMaxHeap@plt>
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  subl   $0x1,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
L10:
  mov    -0x2c(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jg     L8
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *heapSort(int *array, int n) {
  int temp, i;
  int heapsize = n;
  buildMaxHeap(array, 0, heapsize);
  int *newArray = (int *)malloc(n * sizeof(int));
  for (i = 0; i <= n - 1; i++) {
    treeReOrderMaxHeap(array, 0, heapsize);
    *(newArray + i) = *(array + 0);
    temp = *(array + 0);
    *(array + 0) = *(array + (heapsize - 1));
    *(array + (heapsize - 1)) = temp;
    heapsize--;
  }
  return newArray;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<heapSort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r15d
  mov    %esi,%edx
  mov    $0x0,%esi
  call   <buildMaxHeap@plt>
  lea    0x0(,%r15,4),%edi
  call   <malloc@plt>
  mov    %rax,%r14
  test   %r15d,%r15d
  jle    L7
  movslq %r15d,%rbx
  mov    %rax,%r12
  lea    -0x1(%rbx),%r13
  lea    -0x1(%r15),%eax
  sub    %rax,%r13
L11:
  mov    %ebx,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <treeReOrderMaxHeap@plt>
  mov    0x0(%rbp),%eax
  mov    %eax,(%r12)
  mov    -0x4(%rbp,%rbx,4),%edx
  mov    %edx,0x0(%rbp)
  mov    %eax,-0x4(%rbp,%rbx,4)
  sub    $0x1,%rbx
  add    $0x4,%r12
  cmp    %r13,%rbx
  jne    L11
L7:
  mov    %r14,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int *heapSort(int *array, int n) {
  int temp, i;
  int heapsize = n;
  buildMaxHeap(array, 0, heapsize);
  int *newArray = (int *)malloc(n * sizeof(int));
  for (i = 0; i <= n - 1; i++) {
    treeReOrderMaxHeap(array, 0, heapsize);
    *(newArray + i) = *(array + 0);
    temp = *(array + 0);
    *(array + 0) = *(array + (heapsize - 1));
    *(array + (heapsize - 1)) = temp;
    heapsize--;
  }
  return newArray;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<heapSort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %esi,%rbp
  xor    %esi,%esi
  push   %rbx
  mov    %ebp,%edx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <buildMaxHeap@plt>
  lea    0x0(,%rbp,4),%edi
  call   <malloc@plt>
  mov    %rax,%r15
  test   %ebp,%ebp
  jle    L3
  lea    -0x1(%rbp),%r12
  mov    %rax,%r13
  lea    -0x1(%rbp),%eax
  mov    %r12,%r14
  sub    %rax,%r14
  jmp    L24
  cs nopw 0x0(%rax,%rax,1)
L14:
  sub    $0x1,%r12
L24:
  mov    %ebp,%edx
  xor    %esi,%esi
  mov    %rbx,%rdi
  add    $0x4,%r13
  call   <treeReOrderMaxHeap@plt>
  mov    (%rbx),%eax
  mov    %eax,-0x4(%r13)
  mov    -0x4(%rbx,%rbp,4),%edx
  mov    %edx,(%rbx)
  mov    %eax,-0x4(%rbx,%rbp,4)
  mov    %r12,%rbp
  cmp    %r14,%r12
  jne    L14
L3:
  add    $0x8,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int *heapSort(int *array, int n) {
  int temp, i;
  int heapsize = n;
  buildMaxHeap(array, 0, heapsize);
  int *newArray = (int *)malloc(n * sizeof(int));
  for (i = 0; i <= n - 1; i++) {
    treeReOrderMaxHeap(array, 0, heapsize);
    *(newArray + i) = *(array + 0);
    temp = *(array + 0);
    *(array + 0) = *(array + (heapsize - 1));
    *(array + (heapsize - 1)) = temp;
    heapsize--;
  }
  return newArray;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<heapSort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %esi,%rbp
  xor    %esi,%esi
  push   %rbx
  mov    %ebp,%edx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <buildMaxHeap@plt>
  lea    0x0(,%rbp,4),%edi
  call   <malloc@plt>
  mov    %rax,%r15
  test   %ebp,%ebp
  jle    L24
  lea    -0x1(%rbp),%r12
  mov    %rax,%r13
  lea    -0x1(%rbp),%eax
  mov    %r12,%r14
  sub    %rax,%r14
  jmp    L17
  cs nopw 0x0(%rax,%rax,1)
L21:
  sub    $0x1,%r12
L17:
  mov    %ebp,%edx
  xor    %esi,%esi
  mov    %rbx,%rdi
  add    $0x4,%r13
  call   <treeReOrderMaxHeap@plt>
  mov    (%rbx),%eax
  mov    %eax,-0x4(%r13)
  mov    -0x4(%rbx,%rbp,4),%edx
  mov    %edx,(%rbx)
  mov    %eax,-0x4(%rbx,%rbp,4)
  mov    %r12,%rbp
  cmp    %r12,%r14
  jne    L21
L24:
  add    $0x8,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int *heapSort(int *array, int n) {
  int temp, i;
  int heapsize = n;
  buildMaxHeap(array, 0, heapsize);
  int *newArray = (int *)malloc(n * sizeof(int));
  for (i = 0; i <= n - 1; i++) {
    treeReOrderMaxHeap(array, 0, heapsize);
    *(newArray + i) = *(array + 0);
    temp = *(array + 0);
    *(array + 0) = *(array + (heapsize - 1));
    *(array + (heapsize - 1)) = temp;
    heapsize--;
  }
  return newArray;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_write_decimal_10000>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x14f8b589,%rax,%rax
  shr    $0x20,%rax
  sar    $0xd,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x186a0,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  movslq %eax,%rdx
  imul   $0x68db8bad,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0xc,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cltq
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L4
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <_write_decimal@plt>
L4:
  nop
  leave
  ret
```

assistant: ```c
void _write_decimal_10000(uint32_t value) {
  uint8_t decimal_10000 = (value % 100000) / 10000;
  if (decimal_10000 != 0) {
    _write_decimal(decimal_10000);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<_write_decimal_10000>:
  endbr64
  movslq %edi,%rax
  imul   $0x14f8b589,%rax,%rax
  sar    $0x2d,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  imul   $0x186a0,%eax,%eax
  sub    %eax,%edi
  movslq %edi,%rax
  imul   $0x68db8bad,%rax,%rax
  sar    $0x2c,%rax
  sar    $0x1f,%edi
  sub    %edi,%eax
  movslq %eax,%rdi
  test   %rdi,%rdi
  jne    L6
  ret
L6:
  sub    $0x8,%rsp
  call   <_write_decimal@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void _write_decimal_10000(uint32_t value) {
  uint8_t decimal_10000 = (value % 100000) / 10000;
  if (decimal_10000 != 0) {
    _write_decimal(decimal_10000);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_write_decimal_10000>:
  endbr64
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x14f8b589,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x2d,%rax
  sub    %edx,%eax
  imul   $0x186a0,%eax,%eax
  sub    %eax,%edi
  movslq %edi,%rax
  sar    $0x1f,%edi
  imul   $0x68db8bad,%rax,%rax
  sar    $0x2c,%rax
  sub    %edi,%eax
  jne    L9
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  movslq %eax,%rdi
  jmp    1050 <_write_decimal@plt>
```

assistant: ```c
void _write_decimal_10000(uint32_t value) {
  uint8_t decimal_10000 = (value % 100000) / 10000;
  if (decimal_10000 != 0) {
    _write_decimal(decimal_10000);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<_write_decimal_10000>:
  endbr64
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x14f8b589,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x2d,%rax
  sub    %edx,%eax
  imul   $0x186a0,%eax,%eax
  sub    %eax,%edi
  movslq %edi,%rax
  sar    $0x1f,%edi
  imul   $0x68db8bad,%rax,%rax
  sar    $0x2c,%rax
  sub    %edi,%eax
  jne    L6
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  movslq %eax,%rdi
  jmp    1050 <_write_decimal@plt>
```

assistant: ```c
void _write_decimal_10000(uint32_t value) {
  uint8_t decimal_10000 = (value % 100000) / 10000;
  if (decimal_10000 != 0) {
    _write_decimal(decimal_10000);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<rot13er>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L112
L63:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L59
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x6d,%al
  jg     L59
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  lea    0xd(%rax),%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L81
L59:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x6d,%al
  jle    L36
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x7a,%al
  jg     L36
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  lea    -0xd(%rax),%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L81
L36:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L121
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x4d,%al
  jg     L121
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  lea    0xd(%rax),%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L81
L121:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x4d,%al
  jle    L81
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x5a,%al
  jg     L81
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  lea    -0xd(%rax),%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L81:
  addl   $0x1,-0x4(%rbp)
L112:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L63
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rot13er(char *string) {
  int i;
  for (i = 0; string[i] != '\0'; i++) {
    if (string[i] >= 'a' && string[i] <= 'm')
      string[i] += 13;
    else if (string[i] >= 'n' && string[i] <= 'z')
      string[i] -= 13;
    else if (string[i] >= 'A' && string[i] <= 'M')
      string[i] += 13;
    else if (string[i] >= 'N' && string[i] <= 'Z')
      string[i] -= 13;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rot13er>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L30
L56:
  ret
L10:
  sub    $0x6e,%eax
  cmp    $0xc,%al
  ja     L63
  sub    $0xd,%edx
  mov    %dl,(%rdi)
L32:
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  je     L56
L30:
  mov    %eax,%edx
  lea    -0x61(%rax),%ecx
  cmp    $0xc,%cl
  ja     L10
  add    $0xd,%eax
  mov    %al,(%rdi)
  jmp    L32
L63:
  lea    -0x41(%rdx),%eax
  cmp    $0xc,%al
  ja     L51
  add    $0xd,%edx
  mov    %dl,(%rdi)
  jmp    L32
L51:
  lea    -0x4e(%rdx),%eax
  cmp    $0xc,%al
  ja     L32
  sub    $0xd,%edx
  mov    %dl,(%rdi)
  jmp    L32
```

assistant: ```c
void rot13er(char *string) {
  int i;
  for (i = 0; string[i] != '\0'; i++) {
    if (string[i] >= 'a' && string[i] <= 'm')
      string[i] += 13;
    else if (string[i] >= 'n' && string[i] <= 'z')
      string[i] -= 13;
    else if (string[i] >= 'A' && string[i] <= 'M')
      string[i] += 13;
    else if (string[i] >= 'N' && string[i] <= 'Z')
      string[i] -= 13;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<rot13er>:
  endbr64
  jmp    L59
  cs nopw 0x0(%rax,%rax,1)
L32:
  lea    -0x6e(%rax),%edx
  cmp    $0xc,%dl
  jbe    L38
  lea    -0x41(%rax),%edx
  cmp    $0xc,%dl
  jbe    L53
  lea    -0x4e(%rax),%edx
  cmp    $0xc,%dl
  ja     L36
L38:
  sub    $0xd,%eax
  mov    %al,(%rdi)
L36:
  add    $0x1,%rdi
L59:
  movzbl (%rdi),%eax
  test   %al,%al
  je     L51
  lea    -0x61(%rax),%edx
  cmp    $0xc,%dl
  ja     L32
L53:
  add    $0xd,%eax
  mov    %al,(%rdi)
  jmp    L36
  nopw   0x0(%rax,%rax,1)
L51:
  ret
```

assistant: ```c
void rot13er(char *string) {
  int i;
  for (i = 0; string[i] != '\0'; i++) {
    if (string[i] >= 'a' && string[i] <= 'm')
      string[i] += 13;
    else if (string[i] >= 'n' && string[i] <= 'z')
      string[i] -= 13;
    else if (string[i] >= 'A' && string[i] <= 'M')
      string[i] += 13;
    else if (string[i] >= 'N' && string[i] <= 'Z')
      string[i] -= 13;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rot13er>:
  endbr64
  jmp    L41
  cs nopw 0x0(%rax,%rax,1)
L46:
  lea    -0x6e(%rax),%edx
  cmp    $0xc,%dl
  jbe    L38
  lea    -0x41(%rax),%edx
  cmp    $0xc,%dl
  jbe    L1
  lea    -0x4e(%rax),%edx
  cmp    $0xc,%dl
  ja     L36
L38:
  sub    $0xd,%eax
  mov    %al,(%rdi)
L36:
  add    $0x1,%rdi
L41:
  movzbl (%rdi),%eax
  test   %al,%al
  je     L31
  lea    -0x61(%rax),%edx
  cmp    $0xc,%dl
  ja     L46
L1:
  add    $0xd,%eax
  mov    %al,(%rdi)
  jmp    L36
  nopw   0x0(%rax,%rax,1)
L31:
  ret
```

assistant: ```c
void rot13er(char *string) {
  int i;
  for (i = 0; string[i] != '\0'; i++) {
    if (string[i] >= 'a' && string[i] <= 'm')
      string[i] += 13;
    else if (string[i] >= 'n' && string[i] <= 'z')
      string[i] -= 13;
    else if (string[i] >= 'A' && string[i] <= 'M')
      string[i] += 13;
    else if (string[i] >= 'N' && string[i] <= 'Z')
      string[i] -= 13;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<func2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void func2(int var1, int var2) {
  var1 = var2 << 1;
  var1 = var2 << 1;
  var1 = var2 << 1;
  var1 = var2 >> 1;
  var1 = var2 >> 1;
  var1 = var2 >> 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<func2>:
  endbr64
  ret
```

assistant: ```c
void func2(int var1, int var2) {
  var1 = var2 << 1;
  var1 = var2 << 1;
  var1 = var2 << 1;
  var1 = var2 >> 1;
  var1 = var2 >> 1;
  var1 = var2 >> 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<func2>:
  endbr64
  ret
```

assistant: ```c
void func2(int var1, int var2) {
  var1 = var2 << 1;
  var1 = var2 << 1;
  var1 = var2 << 1;
  var1 = var2 >> 1;
  var1 = var2 >> 1;
  var1 = var2 >> 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<func2>:
  endbr64
  ret
```

assistant: ```c
void func2(int var1, int var2) {
  var1 = var2 << 1;
  var1 = var2 << 1;
  var1 = var2 << 1;
  var1 = var2 >> 1;
  var1 = var2 >> 1;
  var1 = var2 >> 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strnpchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  jmp    L24
L39:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x14(%rbp)
  jne    L19
  mov    -0x8(%rbp),%rax
  jmp    L1
L19:
  addq   $0x1,-0x8(%rbp)
L24:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jne    L39
  mov    $0x0,%eax
L1:
  pop    %rbp
  ret
```

assistant: ```c
const char *strnpchr(const char *start, const char *end, int ch) {
  while (start != end) {
    if (*start == ch) {
      return start;
    }
    ++start;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strnpchr>:
  endbr64
  cmp    %rsi,%rdi
  je     L13
L7:
  movsbl (%rdi),%eax
  cmp    %edx,%eax
  je     L10
  add    $0x1,%rdi
  cmp    %rdi,%rsi
  jne    L7
  mov    $0x0,%eax
  ret
L13:
  mov    $0x0,%eax
  ret
L10:
  mov    %rdi,%rax
  ret
```

assistant: ```c
const char *strnpchr(const char *start, const char *end, int ch) {
  while (start != end) {
    if (*start == ch) {
      return start;
    }
    ++start;
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<strnpchr>:
  endbr64
  cmp    %rsi,%rdi
  jne    L24
  jmp    L28
  nopl   0x0(%rax,%rax,1)
L16:
  add    $0x1,%rdi
  cmp    %rdi,%rsi
  je     L28
L24:
  movsbl (%rdi),%eax
  cmp    %edx,%eax
  jne    L16
  mov    %rdi,%rax
  ret
  nopl   0x0(%rax)
L28:
  xor    %eax,%eax
  ret
```

assistant: ```c
const char *strnpchr(const char *start, const char *end, int ch) {
  while (start != end) {
    if (*start == ch) {
      return start;
    }
    ++start;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strnpchr>:
  endbr64
  cmp    %rsi,%rdi
  jne    L7
  jmp    L28
  nopl   0x0(%rax,%rax,1)
L13:
  add    $0x1,%rdi
  cmp    %rdi,%rsi
  je     L28
L7:
  movsbl (%rdi),%eax
  cmp    %edx,%eax
  jne    L13
  mov    %rdi,%rax
  ret
  nopl   0x0(%rax)
L28:
  xor    %eax,%eax
  ret
```

assistant: ```c
const char *strnpchr(const char *start, const char *end, int ch) {
  while (start != end) {
    if (*start == ch) {
      return start;
    }
    ++start;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<matchString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L32
L10:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L19
  mov    $0x0,%eax
  jmp    L27
L19:
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x20(%rbp)
L32:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L10
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    $0x1,%eax
L27:
  pop    %rbp
  ret
```

assistant: ```c
int matchString(const char **strptr, const char *against) {
  const char *str = *strptr;
  while (*against != 0) {
    if (*str != *against)
      return 0;
    str++;
    against++;
  }
  *strptr = str;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<matchString>:
  endbr64
  mov    (%rdi),%rdx
  movzbl (%rsi),%eax
  test   %al,%al
  je     L2
L14:
  cmp    %al,(%rdx)
  jne    L27
  add    $0x1,%rdx
  add    $0x1,%rsi
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L14
L2:
  mov    %rdx,(%rdi)
  mov    $0x1,%eax
  ret
L27:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int matchString(const char **strptr, const char *against) {
  const char *str = *strptr;
  while (*against != 0) {
    if (*str != *against)
      return 0;
    str++;
    against++;
  }
  *strptr = str;
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<matchString>:
  endbr64
  movzbl (%rsi),%eax
  mov    (%rdi),%rdx
  test   %al,%al
  jne    L20
  jmp    L37
L16:
  movzbl 0x1(%rsi),%eax
  add    $0x1,%rsi
  add    $0x1,%rdx
  test   %al,%al
  je     L37
L20:
  cmp    %al,(%rdx)
  je     L16
  xor    %eax,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L37:
  mov    %rdx,(%rdi)
  mov    $0x1,%eax
  ret
```

assistant: ```c
int matchString(const char **strptr, const char *against) {
  const char *str = *strptr;
  while (*against != 0) {
    if (*str != *against)
      return 0;
    str++;
    against++;
  }
  *strptr = str;
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<matchString>:
  endbr64
  movzbl (%rsi),%eax
  mov    (%rdi),%rdx
  test   %al,%al
  jne    L22
  jmp    L39
L6:
  movzbl 0x1(%rsi),%eax
  add    $0x1,%rsi
  add    $0x1,%rdx
  test   %al,%al
  je     L39
L22:
  cmp    %al,(%rdx)
  je     L6
  xor    %eax,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L39:
  mov    %rdx,(%rdi)
  mov    $0x1,%eax
  ret
```

assistant: ```c
int matchString(const char **strptr, const char *against) {
  const char *str = *strptr;
  while (*against != 0) {
    if (*str != *against)
      return 0;
    str++;
    against++;
  }
  *strptr = str;
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<generate_public_key>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L21
  cmpq   $0x0,-0x20(%rbp)
  je     L21
  mov    $0x1,%eax
  jmp    L13
L21:
  mov    $0x0,%eax
L13:
  mov    %eax,%edi
  call   <assert@plt>
  movq   $0x0,-0x8(%rbp)
  jmp    L24
L1:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  imul   -0x28(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x38(%rbp),%rdx
  add    %rdx,%rcx
  mov    $0x0,%edx
  divq   -0x30(%rbp)
  mov    %rdx,%rax
  mov    %rax,(%rcx)
  addq   $0x1,-0x8(%rbp)
L24:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L1
  nop
  nop
  leave
  ret
```

assistant: ```c
void generate_public_key(const size_t *private_key, const size_t key_size, const size_t n, const size_t m, size_t *public_key) {
  assert(private_key != NULL && key_size > 0);
  size_t i;
  for (i = 0; i < key_size; i++) {
    public_key[i] = (private_key[i] * n) % m;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<generate_public_key>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %rdx,%r13
  mov    %rcx,%rbx
  mov    %r8,%r14
  test   %rdi,%rdi
  setne  %al
  test   %rsi,%rsi
  setne  %dil
  movzbl %dil,%edi
  and    %eax,%edi
  call   <assert@plt>
  test   %rbp,%rbp
  je     L6
  mov    $0x0,%ecx
L16:
  mov    %r13,%rax
  imul   (%r12,%rcx,8),%rax
  mov    $0x0,%edx
  div    %rbx
  mov    %rdx,(%r14,%rcx,8)
  add    $0x1,%rcx
  cmp    %rcx,%rbp
  jne    L16
L6:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void generate_public_key(const size_t *private_key, const size_t key_size, const size_t n, const size_t m, size_t *public_key) {
  assert(private_key != NULL && key_size > 0);
  size_t i;
  for (i = 0; i < key_size; i++) {
    public_key[i] = (private_key[i] * n) % m;
  }
}
```