main()
{
if ((1||0) && (0||1))
{
printf("OK I am done.");
}
else
{
printf("OK I am gone.");
}
}
a. OK I am done
b. OK I am gone
c. compile error
d. none of the above
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Answer / govind verma
OK I am done. reason bollean expression in if evaluate in such manner first (1||0) will execute and first expresiion of this boolean expression is one so control will not goes to chek any thing i.e whole expression evaluate as (1||0)as 1(true) now expression becom like dis if(1&&(0||1))
now (0||1) expressiion will evaluate in this expression value of left hand side of || operator is 0(false) so control goes to check further and right hand side it recieve 1 then the value of wholl expression becom 1 thn nw orignal expression looks like if(1&&1)
nw (1&&1) is evalute in this case control check both the value of the operator(&&) is one then this return 1 otherwise return 0 then orignal expression like this if(1)
1 is nonzero value this mean condition is true then code of if block will be execute which is OK I am done.
| Is This Answer Correct ? | 3 Yes | 1 No |
Answer / naveen kumar
ok i am gone...because in this case the condition if((1||0)
&&(0||1)
always remain true. hence the answer is ok i am done...i.e.
(a
| Is This Answer Correct ? | 1 Yes | 1 No |
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