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Specified the code segment below and that n is the problem size, answer the following queries:
// . . .
int sum = 0;
if(x > 12){
for(int i = 1; i <= n; i = i * 2) for( intk = 0; k< n * n; k++) sum++; // statement1
}
else{
for(int i = 1; i <=n; i++) sum++; // statement2
for(int i = 1; i <= n; i = i +4) sum++; // statement3
for(int i = 1; i <= n; i++) for( intk = 1; k<= n; k = k * 2) sum++; // statement4
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I think yes...so if 8259 gives interrupt then it will be serviced instantly but since into pin is left hanging the insr bit will never be set...and so whenever any interrupt occurs
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Divide (101110) 2 by (101) 2 Ans. Quotient is = -1001 Remainder is = -001
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