- Option : C
- Explanation : ((strlen(s) – strlen(t)) > c) ? strlen (s) : strlen (t)

= (3 – 5 > 0)

= (-2 > 0)

Important point here is while comparing -2 with c, result will be a positive number as c is unsigned. So, out of these two, strlen (s) will be printed. Therefore, option c is correct

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- Option : A
- Explanation :

Union of context free language is also context free language.

L_{1}= { a^{n}b^{n}c^{m}| m >= 0 and n >= 0 } and

L_{2}= { a^{m}b^{n}c^{n}| n >= 0 and m >= 0 }

L_{3}= L_{1}∪ L_{2}= { a^{n}b^{n}c^{m}∪ a^{m}b^{n}c^{n}| n >= 0, m >= 0 } is also context free.

L_{1}says number of a’s should be equal to number of b’s and L_{2}says number of b’s should be equal to number of c’s. Their union says either of two conditions to be true. So it is also context free language.

Intersection of CFG may or may not be CFG.

L_{3}= L_{1}∩ L_{2}= { a^{n}b^{n}c^{n}| n >= 0 } need not be context free

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- Option : C
- Explanation :

For 2 “if statements”, 22 = 4 control flow paths are possible:

So for 10 “If statements”, control flow paths will be there.

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- Option : A
- Explanation :

Given,

if TS(T_{2}) <TS(T_{1}) then

T_{1}is killed

else T_{2}waits.

- T_{1}holds a lock on the resource R

- T_{2}has requested a conflicting lock on the same resource R

According to algo, TS(T_{2}) <TS(T_{1}) then T1 is killed else T2 will wait. So in both cases neither deadlock will happen nor starvation.

Therefore, option A is correct

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