ASan报new_delete_type_mismatch：对齐new需配对对齐delete吗？

问题重现 #

以下代码用g++ 14.2.1加-Wall -pedantic编译无警告，运行正常，但启用-fsanitize=address后，ASan抛出类型不匹配错误：

#include <new>
int main() {
    delete new (std::align_val_t(16)) char;
}

ASan报错内容：

==17707==ERROR: AddressSanitizer: new-delete-type-mismatch on 0x503000000040 in thread T0:
  object passed to delete has wrong type:
  size of the allocated type:   1 bytes;
  size of the deallocated type: 1 bytes.
  alignment of the allocated type:   16 bytes;
  alignment of the deallocated type: default-aligned.
    #0 0x70912beff4f2 in operator delete(void*, unsigned long) /usr/src/debug/gcc/gcc/libsanitizer/asan/asan_new_delete.cpp:164
    #1 0x6493bc2dd18d in main (/tmp/a.out+0x118d) (BuildId: e8f06d0aee8e301d7f1522801237096f3006c790)
    #2 0x70912b835487  (/usr/lib/libc.so.6+0x27487) (BuildId: 695cfc6aac7d0f77bb7caba0ef01b2e868762b02)
    #3 0x70912b83554b in __libc_start_main (/usr/lib/libc.so.6+0x2754b) (BuildId: 695cfc6aac7d0f77bb7caba0ef01b2e868762b02)
    #4 0x6493bc2dd094 in _start (/tmp/a.out+0x1094) (BuildId: e8f06d0aee8e301d7f1522801237096f3006c790)

0x503000000040 is located 0 bytes inside of 1-byte region [0x503000000040,0x503000000041)
allocated by thread T0 here:
    #0 0x70912befe97a in operator new(unsigned long, std::align_val_t) /usr/src/debug/gcc/gcc/libsanitizer/asan/asan_new_delete.cpp:107
    #1 0x6493bc2dd17b in main (/tmp/a.out+0x117b) (BuildId: e8f06d0aee8e301d7f1522801237096f3006c790)
    #2 0x70912b835487  (/usr/lib/libc.so.6+0x27487) (BuildId: 695cfc6aac7d0f77bb7caba0ef01b2e868762b02)
    #3 0x70912b83554b in __libc_start_main (/usr/lib/libc.so.6+0x2754b) (BuildId: 695cfc6aac7d0f77bb7caba0ef01b2e868762b02)
    #4 0x6493bc2dd094 in _start (/tmp/a.out+0x1094) (BuildId: e8f06d0aee8e301d7f1522801237096f3006c790)

SUMMARY: AddressSanitizer: new-delete-type-mismatch /usr/src/debug/gcc/gcc/libsanitizer/asan/asan_new_delete.cpp:164 in operator delete(void*, unsigned long)

核心问题解答 #

1. 这是不是ASan的误报？
   不是误报。ASan准确检测到了分配和释放时的对齐属性不匹配——用16字节对齐的new分配内存，却用默认对齐的delete释放，这属于标准定义的问题场景。

2. 必须将对齐new和对齐delete配对吗？
   是的，必须配对。根据C++标准[new.delete]章节的规定：用带std::align_val_t参数的operator new分配的内存，必须用带对应std::align_val_t参数的operator delete来释放。

3. 不配对会导致未定义行为吗？
   会。标准明确指出，这种不匹配的释放操作属于未定义行为。虽然某些编译器/平台下可能“看起来正常”，但这完全依赖实现细节，移植性和稳定性无法保证，极端情况下可能导致内存泄露、堆损坏或程序崩溃。

过度对齐内存释放的最佳实践 #

对于SIMD场景中32字节对齐的double数组这类过度对齐内存，最佳实践如下：

• 显式调用带对齐参数的operator delete：分配时用new (std::align_val_t(32)) double[N]，释放时必须对应调用operator delete[](ptr, std::align_val_t(32))，数组版的对齐new要对应数组版的对齐delete。
• 避免裸指针管理：优先使用智能指针，并自定义删除器来确保对齐释放。示例：

  #include <memory>
  #include <new>
  
  int main() {
      constexpr std::size_t align = 32;
      auto deleter = [](double* p) {
          operator delete[](p, std::align_val_t(align));
      };
      std::unique_ptr<double, decltype(deleter)> ptr(
          new (std::align_val_t(align)) double[8], deleter
      );
  }
  

• 用std::aligned_alloc替代（C风格分配）：如果用C风格的std::aligned_alloc分配，必须用std::free释放，但要注意std::aligned_alloc的大小必须是对齐值的整数倍，这点和C++的对齐new不同。

内容的提问来源于stack exchange，提问作者YiFei