为什么输出是9和19 [英] Why the output is 9 and 19

问题描述

include<stdio.h>

int main()
{
    int x=10;
    int a=1,b=2,c=3,d=4;
    x+=a=b*c+d-a;
    printf("%d,%d",a,x);
}

推荐答案

看来你正在学习C / C ++ - 很好，但我建议你先阅读一些基础知识有关它的书籍，所以你不需要询问有关运算符优先级的问题...

It seems you are into learning C/C++ - good, but I would advise you to read some basic books about it so you will not need asking questions about operator precedence...

x+=a=b*c+d-a

根据执行流程和运算符优先级将它分开，你将得到：

By breaking it apart according the execution flow and operator precedence, you will have this:

a=b*c+d-a = > a=((b*c)+d)-a;
x+=a => x=x+a;

从此处必须清楚...

C ++运算符优先级 - cppreference.com [ ^ ]

你有

As you have

int x=10;
   int a=1,b=2,c=3,d=4;
   x+=a=b*c+d-a;
   printf("%d,%d",a,x);

看看你的输出是怎样的，



x + = a& a = b * c + d-a



a = 2 * 3 + 4-1，结果为9

& x + = a表示x = x + a => 10 + 9 = 19

see how your output come,

x+=a & a=b*c+d-a

a= 2*3+4-1 which results as 9
& x+=a means x = x+a => 10+9= 19

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