What is a output?. How will Come?.

kharade Sagar M · #21 (**permalink**) 07-11-2009

0 0 1 2 -1

kharade Sagar M · #22 (**permalink**) 07-11-2009

Quote:

Originally Posted by Golda

main()
{
int i=5;
printf("%d%d%d%d%d",i++, i--, ++i, --i,i);
}
I got the output 45545. How it will come?...

the output is 5 6 6 5 5

harini_sree · #23 (**permalink**) 07-11-2009

0,0,1,3,0
is this correct

TANMOYPALIT · #24 (**permalink**) 07-29-2009

the output will be 00131
All the effect will be occur at next statement because of post increment
the || operator is the last element and the value of l is 2 so that any value of the previous generation will be ommited and the value of m will be true i.e. 1
m=0||1 is 1
i=-1 --> 0
j=-1 --> 0
k=0 --> 1
l=2 --> 3
m = 1

zaster · #25 (**permalink**) 08-13-2009

Quote:

Originally Posted by kharade Sagar M

the output is 5 6 6 5 5

The output for this expression is ambiguous. The ANSI C standard does not define the order of evaluation of arguments to a function. Hence, the output may vary depending on the compiler you use.

zaster · #26 (**permalink**) 08-13-2009

Quote:

Originally Posted by Golda

what about this?

void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u ",gotit=-giveit);
}

"giveit" has a bit pattern

Code:

11111111 11111111 11111111 11111111

Incrementing it by 1 gives you the following bit pattern

Code:

1 00000000 00000000 00000000 00000000

This 1 results from overflow, and is ignored. So, you effectively get a zero value on return

rsmadhu2008 · #27 (**permalink**) 09-07-2009

Quote:

Originally Posted by Golda

main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}

Output : 0 0 1 3 1
EXP : As the expression have 3 && operators at the begin it will check for all the case s ,& will get 0 as resultant wch is || with 1 (i.e.,2) So that m=0||1=1.

lilliput2009 · #28 (**permalink**) 10-03-2009

i=0,j=0,k=1,l=3,m=0

mahammuda.begum · #29 (**permalink**) 10-03-2009

Quote:

Originally Posted by Golda

what about this?

void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u ",gotit=-giveit);
}

++giveit=2 n -gotit= -2

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What is a output?. How will Come?.

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