如何在 C++ 中将指针转换/转换为引用
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How to cast/convert pointer to reference in C++
提问by Dewsworld
How can I pass a pointer (Object *ob) to a function which prototype is void foo(Object &)?
如何将指针 ( Object *ob)传递给原型是的函数void foo(Object &)？
回答by David Heffernan
Call it like this:
像这样调用它：
foo(*ob);
Note that there is no casting going on here, as suggested in your question title. All we have done is de-referenced the pointer to the object which we then pass to the function.
请注意，正如您的问题标题中所建议的那样，此处没有进行转换。我们所做的只是取消引用指向对象的指针，然后将其传递给函数。
回答by Roee Gavirel
foo(*ob);
You don't need to cast it because it's the same Object type, you just need to dereference it.
您不需要强制转换它，因为它是相同的 Object 类型，您只需要取消引用它。
