what will be the output of this programme?

Question

1.00/5 (2 votes)

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C++

#include <stdio.h>
    void main()
    {
        int x = 0;
        if (x = 0)
            printf("Its zero\n");
        else
            printf("Its not zero\n");
    }

Posted 7-Oct-15 21:37pm

Member 12042196

Updated 7-Oct-15 21:47pm

Afzaal Ahmad Zeeshan

v2

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Comments

Richard MacCutchan 8-Oct-15 3:48am

Not what you expect. Go back to your study notes and read again assignments and expressions. Then run the code to see exactly what happens. Then correct it so it works correctly.

Afzaal Ahmad Zeeshan 8-Oct-15 4:00am

Thanks for that correction that you commented on my answer. :-) That was something new to me.

Richard MacCutchan 8-Oct-15 4:06am

What happened to my comment?

Afzaal Ahmad Zeeshan 8-Oct-15 4:07am

Sorry, I had the answer deleted and then read your comment in the email notification. :-) So, I replied here.

F-ES Sitecore 8-Oct-15 3:57am

I'll give you a clue...if you look at really really old code, rather than comparing their variable with a value (if x is 10), they used to compare the value with their variable (if 10 is x). It might make no sense but it was done to avoid the problem you have here. So give that way a go and see if you can work it out for yourself :)

Richard MacCutchan 8-Oct-15 4:07am

see if you can work it out for yourself
Lol, that's what CodeProject is for. :)

CPallini 8-Oct-15 4:07am

Basically the compiler is telling you: "It's C, not Visual Basic".

Afzaal Ahmad Zeeshan 8-Oct-15 4:41am

Or compiler is telling him, "Did you miss C programming guide somewhere? You may want to read it again". ;-)

2 solutions

Solution 3

Since you didn't specify the executable there would be an error while compiling your program.
For fine working of your program the declared variable(x) in your question may not be initialized.Because initialized variables have a defined value.In order to execute the if else statement in your question the input should be given by you and it can be succeeded by an input function(scanf). In-order to keep the stability use the header file conio.h and getch(); function.

kindly use the below code for your reference:

XML

#include<stdio.h>
#include<conio.h>
void main()
{
 int x;
 printf("please enter your number");
 scanf("%d",&x);
 if(x==0)
 printf("It is zero");
 else
 printf("It is non zero");
getch();
}

Posted 7-Oct-15 22:22pm

VISWESWARAN1998

Updated 8-Oct-15 5:10am

v4

Comments

Andreas Gieriet 8-Oct-15 5:43am

This is crap, sorry.
Why the heck should the variable *not* be initialized?!
You are not answering the OP's question.
And conio.h is not needed at all, or am I missing something?
The obvious problem in the OP's question is that he does not know the difference between = and ==.
Regards
Andi

[no name] 8-Oct-15 6:11am

Pardon me Sir who is OP.Sir, conio.h header file is necessary to display the output till user enters any key.Else the result will be disappeared within a fraction of second and I have mentioned correctly that the code has been given for his reference purpose only.Excuse me since English is not my first language.I misspelled may be to should be.

phil.o 8-Oct-15 6:29am

OP is an abbreviation for Original Poster => this is the person who asked the question.

[no name] 8-Oct-15 9:20am

Thank you Sir for your kind information.

Andreas Gieriet 8-Oct-15 7:05am

getch() comes from stdio.h (see stdio.h).
conio.h is not a standard C header file and might come with some compilers on some systems (see conio.h.
The odd thing is that both header files declare a getch() function and it's a matter of the linker command arguments sequence on which implementation is finally linked into the program.
Regards
Andi
PS: If you start your program in a console, it will not disappear when completed. If you run it during development in Visual Studio, then it will/will not disappear, depending on if you run it in/ouitside the Debugger (F5 versus CTRL-F5).

[no name] 8-Oct-15 9:19am

Sorry Sir because I am unable to read the complete article in Wikipedia.Being a good learner and a discipline student I was supposed to ask to you whether my solution regarding conio.h is correct or wrong.

Andreas Gieriet 8-Oct-15 10:13am

1) What hinders you to read the Wikipedia article?
2) If using conio.h or not is up to you - it's not right or wrong. But *you* should have a clear understanding *yourself* why you think you need to use a proprietary version of e.g. getch() and why you think that the standard functions (like getch() from stdio.h) are not sufficient. I personally don't use conio.h since the standards (stdio.h) all deliver what I need - like getch() - and I never had the need to clear the screen and alike.
3) Your advise to *not* initialize the variable is at least strange in my eyes - or is it a language issue? Your reasoning is very odd.
4) You do not answer the OP's question. Therefore, your solution is not worth anything with respect to the OP's question.

If you write a batch application (I deliberately say "batch", not "Console" since this is the basic motivation for me to write such programs at all), the point of it is that you can automate something. So, the application should take some input and process it to produce an output. If you want to feed some output of some program into your program, you should be able to "pipe" it from STDOUT of the predecessor to the STDIN of you program. Likewise with your output.
If you want to program a batch application that shows "menu" like UI on the terminal window (Console), then you might use console focused I/O - but you can perfectly live without, too. I consider this kind of usage a tiny minority of batch programs. Consider also this: conio is not portable - it was made for DOS. If you go in your professional life to other systems (e.g. embedded toys like Arduino and alike), then you have no such thing like conio.h. There you have terminal emulations at best. But even for these, try to live without - there is more hassle than benefit if your are not going over STDIN/OUT/ERR.

Regards
Andi

[no name] 8-Oct-15 11:08am

1) What hinders you to read the Wikipedia article?
Sir, I have read the article which related to my solution and your comments only.

3) Your advise to *not* initialize the variable is at least strange in my eyes - or is it a language issue? Your reasoning is very odd.

Sir, my advice is not to neglect it completely. But in his case it is much more easier for him to understand it.

4) You do not answer the OP's question. Therefore, your solution is not worth anything with respect to the OP's question

OP's question: what will be the output of this programme?
I am sure that there will be an error in output because the executable is not specified.But I did not mention it in solution and so you are correct.I will improve my solution.

Thank you for your kind advice Sir.

Andreas Gieriet 8-Oct-15 16:11pm

I have no clue how you get to your replies to my items 3) and 4)...
You do not give any evidence that allows to follow your reasoning.
I cannot follow your thought - too weird, sorry.

The OP's program is very well defined - no room for interpretation. Just a flaw the in the comparison. If using =, the else-branch is taken, if using ==, the if-branch is taken.
Or am I missing something?
Regards
Andi

[no name] 9-Oct-15 7:33am

Yes Sir you are right for the programmers part but he has given x=0 and in the if part he has given x=0 so the program usually goes for the else part and the output says "it is not zero" so even his program is correct the output is mathematically wrong.This was the message I try to convey.Sorry Sir if any wrong.

Member 12042196 8-Oct-15 6:51am

andi pandi i know the difference between = and == the i got confused at the comparing part and yes there is need of conio.h it is used display the output.
regards
OP

Andreas Gieriet 8-Oct-15 7:10am

Regarding conio.h: see my comment above.
Regarding = versus ==: all what we see is the code in your question, and that code has the obvious flaw of having an assignment in the condition. Now you say differently, so, why not fixing it in the question and tell that that was a typo only which should not distract from the real question, etc. etc.
Regards
Andi

CHill60 8-Oct-15 12:55pm

A small word of advice ... His name is "Andi" NOT "Andi Pandi" - where I come from calling someone that would have probably got you a punch in the jaw :-)

Andreas Gieriet 8-Oct-15 16:23pm

Thanks for your support, CHill60!
I saw it too and decided to not start any arguments about it.
Why bother on immature kids who play from the anonymity of the internet.
If he feels better like this, let him be happy. Seems to me paltry behavior.
Cheers
Andi
PS: Reported as Troll - I could not resist.

CHill60 8-Oct-15 20:16pm

:thumbup:

Mohibur Rashid 8-Oct-15 18:56pm

What conpiler do you use?

[no name] 8-Oct-15 21:18pm

Can you please say me to whom you are asking Sir

Member 12042196 9-Oct-15 1:49am

chill60 take a chill pill baby. u better keep your ass off from :) here. over the internet neither i can touch you, nor you can.if you were here till now u would have been bleeding from every possible part of your body. p.s. i wish you were here.

[no name] 9-Oct-15 7:34am

Please do not insult an experienced person.

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Afzaal Ahmad Zeeshan · Accepted Answer · 2015-10-07T21:59:00

The program doesn't work because you are using assignment operator where you need to use the equality check operator. That is why the results are opposite and not what you had expected, as Richard said above.

What happens is that if(x = 0) would resolve the expression to zero. In C/C++, integers are directly resolved to boolean, everything non-zero is a true value and zero is the false. Since this resolves to zero, the expression is false and control is transferred to the else block.

Re-write it in this manner,

C++

#include <stdio.h>
void main()
{
    int x = 0;
    if (!x) { // If zero, then true, otherwise false. 
        printf("Its zero\n");
    }
    else {
        printf("Its not zero\n");
    }
}

Also remember to add the { } around the statements, it is not a good practice as you may confuse yourself or add a semicolon by mistake, compiler will not complain but the logic would be wrong.

For more on C operators please refer to this tutorial: http://www.tutorialspoint.com/cprogramming/c_operators.htm[^]