本文介绍了#include< stdio.h> int main(){int a =(1,2,3); int b =(3,2,1); for(; a> 0; a--)for(; b< 3; b ++); printf("%d",a * b);返回0; }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
我已经尝试了很多,但我没有得到解决你会帮我解决这个问题 谢谢你...... !!
I have tried a lot but I didn't get solved will you help me out with this thank you...!!
#include <stdio.h> int main() { int a=(1, 2, 3); int b=(3, 2, 1); for(; a>0; a--) for(; b<3; b++); printf("%d ", a*b); return 0; }我的尝试: 想要使用合适的数字进行expiation
推荐答案查看C的逗号运算符。当你知道它是如何工作的时候,你就会知道a和b将被初始化为什么。 Look into C's comma operator. When you know how that works, you will know what a and b will be initialized to.
逗号运算符如下所示: The Comma operator looks like this: x = a, b;
它计算表达式 a ,丢弃结果,计算 b 并将其返回。 因此 a 和 b 的代码执行,并将x设置为 b 的值。 您的代码只是该代码的扩展名:有效
It evaluates the expression a, discards the result, evaluates b and returns it. So the code for a and b both get executed, and x is set to the value of b. Your code is just an extension of that: effectively
x = ((a, b) , c);所以它评估 a 和 b ,然后评估 c 并设置 x 达到那个价值。 即您的代码是:
So it evaluates a and b, then evaluates c and sets x to that value. I.e. Your code is:
int main() { int a = 3; int b = 1; for(; a>0; a--) for(; b<3; b++); printf("%d ", a*b); return 0; }
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#include< stdio.h> int main(){int a =(1,2,3); int b =(3,2,1); for(
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