Return type of '?:' (ternary conditional operator)
为什么第一个返回引用？
1 2 3 | int x = 1; int y = 2; (x > y ? x : y) = 100; |
而第二个没有？
1 2 3 | int x = 1; long y = 2; (x > y ? x : y) = 100; |
实际上，第二个根本没有编译-"不是赋值的左值"。
(P)Expressions don't have return type s，they have a type and-a s it's known in the latest C++standard-a value category.(p)(P)a Conditional expression can be an avaloe or an rvalue.这是它的价值分类。(This is some of a simplification，in EDOCX1 universal 4 we have lvalues，xvalues and prvues.)(p)(P)In very broad and simple terms，an a lvalue refers to an object in memory and an rvalue is just a value that may not necessarily be attached to an object in memory.(p)(P)1.An assignment expression assigns a value to an object so the thing being assigned to must be an lvalue.(p)(P)For a conditional expression(EDOCX1)to be an lvaloue(again，in broad and simple terms)，the second and third operations must be lvaloes of the same type.这是因为一种条件表达的类型和价值分类是在竞争的时间决定的，并且必须是适当的，条件是真实的。如果操作人员中有一人必须转为不同类型的操作人员，以杀害其他人，则条件表达方式不能作为这一转换的结果而被拯救。(p)布尔奇1
(P)The type of the ternary EDOCX1 pental 0 common expression is the common type of its second and third argument.如果你有相同的类型，你会得到一个参考回来。如果他们是可以转换成each other，one gets chosen and the other gets converted(promoted in this case).由于你无法返回到一个临时的参考点(转换/升级变量)，你的类型是一个价值类型。(p)
(P)It cannot return a lvaluue since it will have to implicitly promote the type of EDOCX1 original 1 to match the type of EDOCX1 disciplinary 2(since both sides of EDOCX1 disciplinary 3)are not of the same type)，and that it has to create a temporary.(p)What does the standard say？(N1905)(P)Expressions 5.17 assignment and compound assignment operators(p)布尔奇1布尔奇1布尔奇1