找到给定代码的输出并给出适当的解释 [英] Find the output of the given code with proper explanation

问题描述

#include<conio.h>
#include<stdio.h>
using namespace std;
int main()
{   int i=4,j=-1,k=0,w,x,y,z;
  w=i||j||k;
  x=i&&j&&k;
  y=i||j&&k;
  z=i&&j||k;
	printf("w=%d x=%d y=%d z=%d\n",w,x,y,z);
	getch();
	return 0;
}

我的尝试：



我有输出但我无法理解.... PLzz给我一些解释

What I have tried:

I have got the output but i can't understand that ....PLzz provide me some explanation

推荐答案

对不起，这是你的作业，不是我们的 - 所以你必须在这里工作。



首先记住C没有真或假 - 它认为任何数字不为零为真，任何数字为零为假。

然后记住||和&&是逻辑运算符，而不是二进制：所以输出|| b或&& b为true（非零）或false（零）。

最后，查看逻辑AND和OR运算符的作用：它非常简单，结果很明显。

Sorry, but it's your homework, not ours - so you will have to do the work here.

Start by remembering that C does not have a "true" or "false" - it considers any number that is not zero to be "true", and any number that is zero to be "false".
Then remember that || and && are logical operators, not binary: so the output of a || b or a && b with be "true" (non-zero) or "false" (zero).
Finally, look up what logical AND and OR operators do: it pretty simple, and the results become obvious.

首先让我们把它写成一个正确的 C 程序（因为 C ++ 不是 C ）：

First let's write it as a proper C program (as C++ is not C):

#include <stdio.h>
int main()
{
  int i=4, j=-1, k=0, w, x, y, z;
  w = i || j || k;
  x = i && j && k;
  y = i || j && k;
  z = i && j || k;
  printf("w=%d x=%d y=%d z=%d\n", w, x, y, z);
  getchar();
  return 0;
}

然后，考虑

Then, consider

w = i || j || k = (4) || (-1) || (0) = true AND true AND true = true = 1;

其中带有'true'，'AND'等的表达式是'imaginary'，只是为了说明评估是如何发生的。

where the expressions with 'true', 'AND', etc. are 'imaginary', just there to illustrate how evaluation happens.

x = i && j && k = .. = true AND true AND false = false = 0;

y = i || j && k = .. = true OR (true AND false) = true OR false = true = 1;

z = i && j || k = .. = (true AND true) OR false = true OR false = true = 1;

在您看到的最新两个表达式中 C 运营商优先规则 [ ^ ]正在行动（&的优先级;& 运算符高于 || 运算符的优先级。

in the latest two expressions you see the C operator precedence rules[^] in action (the precedence of && operator is higher than the precedence of the || operator).

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