具有“cout” myint“。为什么它工作？ [英] Typo with "cout < myint". Why does it work?

问题描述

我有这个代码，我搜索小时为什么它不能打印我的收入

I have this code and I searched for hours why it fails to print my income

int const income = 0;
std::cout << "I'm sorry, your income is: " < income;

直到我发现我错过写<< 但写了< 。为什么编译器没有检测到这个错误？我不知道为什么比较 cout 有意义？

Until I found I missed to write << but wrote <. Why doesn't the compiler detect this and error out? I'm not sure why comparing cout makes sense?

推荐答案

常量0也是一个空指针常量 - 它可以与 ostream '运算符void * 的结果进行比较。请注意，如果常数的值为0，则它​​将失败。

integral constant 0 is also a null pointer constant - it can be compared to the result of ostream's operator void *. Note that it'll fail if the constant has any value but 0.

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