为什么这个volatile变量的地址总是为1？ [英] Why is the address of this volatile variable always at 1?

问题描述

我想检查变量的地址

volatile int clock;
cout << &clock;

但它总是说x在地址1.我做错了什么？

But it always says that x is at address 1. Am i doing something wrong??

推荐答案

iostreams将投射大多数指针到 void * 为 volatile 指针。因此，C ++回落到 bool 的隐式转换。如果您要打印地址，则显式地转换为 void * ：

iostreams will cast most pointers to void * for display - but no conversion exists for volatile pointers. As such C++ falls back to the implicit cast to bool. Cast to void* explicitly if you want to print the address:

std::cout << (void*)&clock;

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