问题描述
我有一个像
test(something &)
我正在做
something *ss = new something();
我说
test(ss)
编译器抱怨说初始化了 something& 类型的引用from 表达 something * .
compiler complains saying initialization of reference of type something& from expression something * .
但不是 new 返回地址并且 ss 必须指向该地址！所以如果测试期待一个参考是不是它 ss 代表一个参考?
but isn't that new returns the address and ss must point to that address ! so if test is expecting a reference is not it ss represents a reference ?
推荐答案
你的函数需要一个普通的 something 对象.你不需要在这里使用指针:
Your function expects a normal something object. You don't need to use a pointer here:
something ss;
test(ss);
当您的函数签名看起来像 f(T&) 时，这意味着它接受对 T 对象的引用.当签名为f(T*)时，表示它接受一个指针到一个T对象.
When your function signature looks like f(T&), it means that it accepts a reference to a T object. When the signature is f(T*), it means that it accepts a pointer to a T object.
这篇关于编译器错误，提示对 something& 类型的引用初始化无效来自某物类型的表达*的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！