在C和Java preIncrement和后运营商的行为 [英] Behaviour of PreIncrement and PostIncrement operator in C and Java
本文介绍了在C和Java preIncrement和后运营商的行为的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我跑在Visual C以下为节目++和Java:
I'm running the following programs in Visual C++ and Java:
的Visual C ++
void main()
{
int i = 1, j;
j = i++ + i++ + ++i;
printf("%d\n",j);
}
输出:
6
Java的:
public class Increment {
public static void main(String[] args) {
int i = 1, j;
j = i++ + i++ + ++i;
System.out.println(j);
}
}
输出:
7
为什么这两种语言的输出有什么不同?无论是汉语语言如何看待pre和后运营商不同?
Why the output in these two languages are different? How both the langauges treat pre and postincrement operators differently?
推荐答案
在C / C ++的行为是不确定的,因为在这个前pression I
修改更多的则一旦没有插入顺序点。阅读:什么是我的价值++ + i ++在
In C/C++ behavior is undefined because In this expression i
is modified more then once without an intervening sequence point. read: What's the value of i++ + i++?
中,当然在这种codeS 是很好的界定。
下面是我的回答对Java,一步一步:
Of-course in Java behaviour of this kind of codes is well defined. Below is my answer for Java, step by step:
起初 I
是 1
。
j = i++ + i++ + ++i;
// first step, post increment
j = i++ + i++ + ++i;
// ^^^
j = 1 + i++ + ++i;
// now, i is 2, and another post increment:
j = i++ + i++ + ++i;
// ^^^^^^^^^
j = 1 + 2 + ++i;
// now, i is 3 and we have a pre increment:
j = i++ + i++ + ++i;
// ^^^^^^^^^^^^^^^^
j = 1 + 2 + 4;
j = 7;
这篇关于在C和Java preIncrement和后运营商的行为的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
查看全文