1.
i think if the total no of bits is within no of bits in a int , the
size will b same as int .
here total bits 3 < 32 (bits in int according to gcc, in Turbo c its 16)
so size of structure will be 4(in gcc), 2 (in TC)
if total no of bits is 40 (suppose) than size will be 8(in GCC) , 4(in TC)
2.
By default int is considered as signed int
if in n bits signed int is stored than the range will be -2^(n-1) to 2^(n-1)-1
so for bit1:1 range is -1,0
so for bit2:4 range is -8,7
so for bit3:4 range is -8,7
if the assigned value exceed than it will rotate as per conversion
from unsigned to signed int (hope u know this)
bit1 cant hold 1 .it ll store as -1
others within range
so it ll output as -1 4 4
3.
Yea u can have 3 bitfields in union ...but u can initilize only one at a time
her sizeof union will be always sizeof int bcoz u can use only one
variable at a time in union
and u cant declare bitfield of more than bits of an int (u wil be
compile time error)
On 6/13/10, divya <sweetdivya....@gmail.com> wrote:
> tell the o/p of following with explanations
>
> 1. #include<stdio.h>
> int main()
> {
> struct value
> {
> int bit1:1;
> int bit3:4;
> int bit4:4;
> }bit;
>
> printf("%d\n",sizeof(bit));
> return 0;
> }
>
> 2.
> #include<stdio.h>
> int main()
> {
> struct value
> {
> int bit1: 1;
> int bit3: 4;
> int bit4: 4;
> } bit={1,2,2};
> printf("%d %d %d\n",bit.bit1,bit.bit3,bit.bit4);
> return 0;
> }
>
> 3 can bit field be used in union??
>
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