This non-const reference parameter is not modified意味着你的C++代码中有一个函数,该函数接受一个非const引用参数,但在函数体内没有修改这个参数的值。在C++中,如果你需要传递一个对象到函数中,但不打算在函数内部修改这个对象,最佳实践是将该参数声明为const引用。这样做可以避免意外的修改,同时允许函数
This non-const reference parameter is not modified意味着在C++代码的某个函数中，尽管参数是一个非const引用，但在函数体内部并未对其实际进行修改。在C++中，若要将对象传递至函数而无需在函数内对其进行修改，建议将该参数声明为const引用。这样做可以防止意外修改，并允许函数处理临时对象和常量，从而...
Here,foo()accepts an integer as aconstreference, which clearly communicates that the input reference is not modified byfoo(). From the signature ofbar(), it is not clear whether the input reference is modified or not. To comply with this rule,const-qualify the type of named pointer or re...
回答:匿名不是‘const’的参考不可能一定在non-lvalue 2013-05-23 12:24:58 回答:匿名引用不是以 'const' 不能绑定到非左值 2013-05-23 12:26:38 回答:匿名2013-05-23 12:28:18 回答:匿名
awhat is this what is this [translate] a但是树太高了 But the tree too was high[translate] a他有一头长长的黑头发 He has a long black hair[translate] a而且朋友之间要互相帮助 Moreover between the friend must help mutually [translate] ...
Can I declare a constant array in Java? Yes, in Java, you can declare an array as final to create a constant array. This ensures that the array reference cannot be changed, but the individual elements of the array can still be modified. ...
不是\\ ‘const \\’的参考不可能一定在非lvalue 翻译结果5复制译文编辑译文朗读译文返回顶部 不是\ ‘const \’的参考不可能一定在non-lvalue 相关内容 a检查 维修过程描述 Inspection service process description[translate] aHow does this company stack up to its competitors? 这公司堆由它的竞争者决定?[tr...
Git Source Code Mirror - This is a publish-only repository but pull requests can be turned into patches to the mailing list via GitGitGadget (https://gitgitgadget.github.io/). Please follow Documentation/SubmittingPatches procedure for any of your improv
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The validity, which is one of the __ProjectReferenceOutputValidity values. Returns Int32 If the method succeeds, it returns S_OK. If it fails, it returns an error code. Remarks This function allows the flavored project to override the reference validity check perfo...