Find the output of the given code with proper explanation

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#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;
}

What I have tried:

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

Posted 9-Mar-18 21:06pm

Member 13593396

Updated 9-Mar-18 22:30pm

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Richard MacCutchan 10-Mar-18 4:17am

Read your course notes, or get a copy of a C reference guide, which explains all the logical operators. Also learn to use spacing in your code to make it readable.

2 solutions

Solution 1

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.

Posted 9-Mar-18 21:34pm

OriginalGriff

Comments

Member 13593396 10-Mar-18 4:01am

sir i got the output as w=1 x=0 y=1 z=2 but i cannot understand why????

OriginalGriff 10-Mar-18 4:09am

a) No, you didn't. Look again.
b) Which bit of what you did get did you not understand?

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CPallini · Accepted Answer · 2018-03-09T22:31:00

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

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;
}

Then, consider

C

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

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

C

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

C

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

C

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

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).