What will be the output of the following code?

What will be the output of the following code and WHY?

Code:

---

#include <stdio.h>

void main()
{
int x,y,z;
x=y=z=1;
z=++x||++y&&++z;
printf("x=%d y=%d z=%d\n",x,y,z);
}

---

Code:

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#include <stdio.h>

void main()
{
float a=0.7
if (0.7>a)
printf("Hi");
else
printf("Hello");
}

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Have you compiled and run it?

int main()

http://faq.cprogramming.com/cgi-bin/...&id=1043284376

Quote:

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yes i have executed it.

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So? Is the output as you expect? Or not? And you don't understand why not?

Quote:

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Originally Posted by vart

Have you compiled and run it?

int main()

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int main() is always not needed. void main() is also correct. The program has output.
yes i have executed it.

Quote:

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Originally Posted by vart

http://faq.cprogramming.com/cgi-bin/...&id=1043284376


So? Is the output as you expect? Or not? And you don't understand why not?

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No, the output is not as expected. I can't understand it.

Perhaps the compiler doesn't allow the increment operator to be applied more than once on the same sentence that uses && and ||. Switching x and y results in y having a value of 2 and x of 1 (as opposed to x = 2 and y = 1) so only the first variable is actually incremented. if we use a fourth variable 'w' to receive the assignment of the operation, 'w' receives the value of 1 and z is not incremented (when we print all four variables). Apparently incrementing more than one variable is not even recognized by the compiler. Then again, that's just my educated guess.

Quote:

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Originally Posted by developersubham

No, the output is not as expected. I can't understand it.

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Look at the following code

Code:

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if(pThing && pThing->bUsed)

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What do you think will happen if pThing is NULL?
Try to figure out why...
Try to figure out what is happening in your case

The other part is a good puzzle. I think only an expert could answer for sure, why the computer says 0.7 is greater than the value of a variable initialized by

Code:

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  float a = 0.7;

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Could it be the internal representation of both numbers (a constant and a variable)? or possibly how the > operator internally in the microarquitecture. I have no clue, but it does lead to think about what lies below the compiler and the executed code.

Quote:

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Originally Posted by jalnewbie

The other part is a good puzzle. I think only an expert could answer for sure, why the computer says 0.7 is greater than the value of a variable initialized by

Code:

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  float a = 0.7;

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Could it be the internal representation of both numbers (a constant and a variable)? or possibly how the > operator internally in the microarquitecture. I have no clue, but it does lead to think about what lies below the compiler and the executed code.

---

Read this http://cboard.cprogramming.com/showt...ight=scientist
for example

Quote:

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Originally Posted by jalnewbie

The other part is a good puzzle. I think only an expert could answer for sure, why the computer says 0.7 is greater than the value of a variable initialized by

Code:

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  float a = 0.7;

---

Could it be the internal representation of both numbers (a constant and a variable)? or possibly how the > operator internally in the microarquitecture. I have no clue, but it does lead to think about what lies below the compiler and the executed code.

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Try this and try to figure out the difference

Code:

---

#include <stdio.h>

int main()
{
    float a=0.7;
    if (0.7f>a)
        printf("Hi");
    else
        printf("Hello");
}

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Kurt

Quote:

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Originally Posted by developersubham

int main() is always not needed. void main() is also correct. The program has output.
yes i have executed it.

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Don't talk such nonsense!:mad:
Read this: http://www.research.att.com/~bs/bs_faq2.html#void-main

Also, you're lying! the second program is missing a semi-colon and CANNOT compile. If you want to know why it does what it does when you add the missing semicolon, you'll need to read this:
http://docs.sun.com/source/806-3568/ncg_goldberg.html

This line has what is called "undefined behaviour":
z=++x||++y&&++z;
You may not modify a variable more than once without intervening sequence points.
See http://msdn.microsoft.com/library/de...m/expre_11.asp.

Quote:

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Originally Posted by vart

Read this http://cboard.cprogramming.com/showt...ight=scientist
for example

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thanks for the help.

Quote:

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Originally Posted by jalnewbie

Perhaps the compiler doesn't allow the increment operator to be applied more than once on the same sentence that uses && and ||. Switching x and y results in y having a value of 2 and x of 1 (as opposed to x = 2 and y = 1) so only the first variable is actually incremented. if we use a fourth variable 'w' to receive the assignment of the operation, 'w' receives the value of 1 and z is not incremented (when we print all four variables). Apparently incrementing more than one variable is not even recognized by the compiler. Then again, that's just my educated guess.

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I also thought like you. But need clarifications. Thanks for giving a try.

Quote:

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Originally Posted by jalnewbie

Perhaps the compiler doesn't allow the increment operator to be applied more than once on the same sentence that uses && and ||. Switching x and y results in y having a value of 2 and x of 1 (as opposed to x = 2 and y = 1) so only the first variable is actually incremented. if we use a fourth variable 'w' to receive the assignment of the operation, 'w' receives the value of 1 and z is not incremented (when we print all four variables). Apparently incrementing more than one variable is not even recognized by the compiler. Then again, that's just my educated guess.

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If a left part of the || is +ve, it is obvious that the output will be 1 therfore compiler doesnot calculate the other side of the ||. It ignores it.
Similarly if left side of && is 0 the result is obviously 0 therfore compiler discards the other part.