c – 为什么新的第一次分配1040个额外字节？

@H_618_4@ 我正在创建这个简单的测试程序来演示使用标准new分配内存时对齐的工作方式…

#include <iostream>
#include <iomanip>
#include <cstdint>

//
// Print a reserved block: its asked size,its start address 
//       and the size of the prevIoUs reserved block
//
void print(uint16_t num,uint16_t size_asked,uint8_t* p) {
   static uint8_t* last = nullptr;

   std::cout << "BLOCK " << num << ":   ";
   std::cout << std::setfill('0') << std::setw(2) << size_asked << "b,"; 
   std::cout << std::hex << (void*)p;
   if (last != nullptr) {
      std::cout << "," << std::dec << (uint32_t)(p - last) << "b";
   }
   std::cout << "\n";
   last = p;
}

int main(void) {
   // Sizes of the blocks to be reserved and pointers
   uint16_t s[8] = { 8,8,16,4,6,6 };
   uint8_t* p[8];

   // Reserve some consecutive memory blocks and print
   // pointers (start) and their real sizes
//   std::cout << "          size   start    prev.size\n";
//   std::cout << "-----------------------------------\n";
   for(uint16_t i = 0; i < 8; ++i) {
      p[i] = new uint8_t[s[i]];
      print(i,s[i],p[i]);
   }

   return 0;
}

但是当我执行程序时,我发现了这种奇怪的行为：

[memtest]$g++ -O2 mem.cpp -o mem
[memtest]$./mem 
BLOCK 0:   08b,0xa0ec20
BLOCK 1:   08b,0xa0f050,1072b
BLOCK 2:   16b,0xa0f070,32b
BLOCK 3:   16b,0xa0f090,32b
BLOCK 4:   04b,0xa0f0b0,32b
BLOCK 5:   04b,0xa0f0d0,32b
BLOCK 6:   06b,0xa0f0f0,32b
BLOCK 7:   06b,0xa0f110,32b

如您所见,new分配的第二个块不是在下一个32b内存的对齐地址,而是远在(1040个字节以外).如果这不够奇怪,取消注释打印出表头的2 std :: cout行会产生以下结果：

[memtest]$g++ -O2 mem.cpp -o mem
[memtest]$./mem 
          size   start    prev.size
-----------------------------------
BLOCK 0:   08b,0x1f47030
BLOCK 1:   08b,0x1f47050,32b
BLOCK 2:   16b,0x1f47070,0x1f47090,0x1f470b0,0x1f470d0,0x1f470f0,0x1f47110,32b

这是正常的预期结果.是什么让新手在首次运行中以如此奇怪的方式表现？我正在使用g(GCC)7.1.1 20170516.您可以编译而不进行优化,结果是相同的.

std::cout << "BLOCK " << num << ":   ";