c programming

Question

struct marks{
int p:3; int c:3; int m:2; };
void main(){
struct marks s={2,-6,5};
printf("%d %d %d",s.p,s.c,s.m); }
 

Please explain why result is 221

Comments

int c:3 ----> only 3 bits allocate for c, but you are assigned -6 to it... -6 can't represent by 3bits therefore it is implementation defined...

int m:2 ----> only two bits allocated for m but you are assigning 5 to it... 5 can't represent by 2 bits therefore it is implementation defined...

i hope your compiler implementation is ignore the MSB bits

Answer 1

Such questions should be solved considering the options given, because that way you would be able to judge the way the compiler is implemented.

Here we allocate 3 bits for p, 2 bits for c and 2 bits for m.

Since we are storing 2 which in binary is 10, storing it in 3 bits will give us 010 which still remains to be 2
similarly, we are storing -6 (2's complement of 0110 = 1010), in 2 bits, discarding the MSBs we have 10 which is 2
5 = (101)₂  again discarding MSBs to fit into 2 bits gives 1 which in decimal notation remains to be 1

Hence 221

Comments

@arvin Thank you very much!

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@arvin

if its int x=4 it stores integer value 4 inside x.

in structure declaration, you can't have int x=4

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@shaikh yes brother. here i only meant to differentiate between the = and : .

and yes we cannot use int x=4 inside the structure as its only a declaration.