在条件运算符中返回值 [英] Returning value in conditional operator

问题描述

我试图通过使用条件运算符根据条件返回值true或false，但出现错误。这是我的代码，

I was trying to return value true or false depending upon the condition by using a conditional operator but I got an error. Here is my code,

bool isEmpty()
{
    int listSize = Node::size();
    listSize > 0 ? return (true) : return (false);
    return false;
}

这是错误，

error C2107: illegal index, indirection not allowed

<现在我被困在这里。我不明白这一点。从逻辑上讲，我认为应该是正确的。请指导我。谢谢

Now I am stuck here. I don't get the point.Logically I think it should be correct. Please guide me about it . Thanks

推荐答案

您只能使用表达式*作为三元条件运算符，而不是语句。通常的说法是：

You can only have expressions* as the operands of the ternary conditional, not statements. The usual way to say this is:

return listSize > 0 ? true : false;

甚至更好，

return listSize > 0;

甚至更好，

bool isEmpty() { return Node::size() > 0; }

*）由于您同时将其标记为C和C ++，因此请知道存在细微差异在两种语言的允许表达之间。

*) Since you tagged this as both C and C++, know that there is a subtle difference between the admissible expressions in the two languages.

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