in second problem a:1 and b:3 means default initialization but not in c it is D programming language but u may b trying to compile it with c complier that's why it is showing wrong result..........if u will compile it with D language compiler than it will print 6 and 2.....and if u will not initialize them again with 6 and 2 then it will print 1 and 2
On Thu, Jul 14, 2011 at 8:27 AM, John Hayes <[email protected]>wrote: > Hey Guys, plz help me in getting these 2 C output problems > > 1st problem > > > *#*include<stdio.h> > int main() > { > short int a,b,c; > scanf("%d%d",&a,&b); > c=a+b; > printf("%d",c); > return 0; > } > INPUT- > 1 1 > > OUTPUT > 1 > > i am not getting why 1 is coming in the output.....what difference is > using short making in the code ??? > > Problem No. 2 > * > *#include<stdio.h> > main() > { > struct > { > int a:1; > int b:2; > }t; > t.b=6; > t.a=2; > printf("%d %d",t.a,t.b); > } > > OUTPUT > 0 -2 > > What does the statement a:1 and b:1 mean and what are they doing.....i am > seeing them first time ever...hence not able to get the output....if someone > has any idea plz help !! > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- shilpa gupta b tech 2nd year computer science and engineering mnnit allahabad -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
