三元运算符和赋值运算符 [英] ternary operator and assignment operator

问题描述

在执行C/C ++三元运算符实际上与赋值运算符具有相同的优先级吗?

卢希恩·格里戈尔(Luchian Grigore)的回答是说

Luchian Grigore's answer says that cases like

a ? b : c = d

将始终被推断为

a ? b : ( c = d )

因为=和?:左右关联，所以

because both = and ?: associate right to left so

在c ++中

k =  21 > 3 ? j = 12 : j = 10;

和

k = 1 > 3 ? j = 12 : j = 10;

都很好.

在C

k = 21 > 3 ? 12 : j = 10

返回错误

invalid lvalue in assignment.

以上不应该推断为(并且不返回错误)

Shouldn't above be inferred as (and return no error)

k=  21 > 3 ? 12 : ( j = 10 )

我认为现在将其分组为

k = ( 21 > 3 ? 12 : j ) = 10

会产生错误，因为在C中(不在C ++中)三元运算符无法返回左值. 在这种情况下，任何人都可以准确地告诉我如何对运算符进行分组.

which gives error since in C(not in C++) ternary operator cannot return lvalue. Can anyone tell me exactly how operators are grouped in this case.

推荐答案

您的链接问题(

Your linked question's (Does the C/C++ ternary operator actually have the same precedence as assignment operators?) answer by @hvd shows the answer.

?:的C ++和C语法不同.

The C++ and C grammars for ?: are different.

在C ++中，最右边的操作数被允许作为赋值表达式(因此，编译器[贪婪地]将=视为?:的一部分)，而在C语言中，最右边的操作数则是conditional-expression.因此，在C语言中，只要编译器点击=，对?:的分析就完成了，并将其视为k = ( 21 > 3 ? 12 : j ) = 10.

In C++, the rightmost operand is allowed to be an assignment expression (so the compiler [greedily] treats the = are part of the ?:) while in C the rightmost operand is a conditional-expression instead. Thus in C as soon as the compiler hits the = the analysis of ?: is complete and it treats it as k = ( 21 > 3 ? 12 : j ) = 10.

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