Q 71.
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What will be the output of the following pseudo code?
1. Integer p,q,r
2. Set p=1, q=4, r=10
3. p=(r+p)+p
4. if(8<r || 8>r)
5. p=(p+3)&q
6. if((q8&r)<(r&q))
7. q=(q+q)+p
8. End if
9. q=(r+11)+r
10. End if
11. Print p+q+r
Note: & bitwise AND - The bitwise AND operator (&) compares each bit of the first operand to the corresponding bit of the second operand. If both bits are 1, the corresponding result bit is set to 1. Otherwise, the corresponding result bit is set to 0.
||: Logical OR - The logical OR operator (||) returns the Boolean value TRUE(or 1) if either or both operands is TRUE and returns FALSE(or 0) otherwise.
Ops:
Show answer & explanation
After initialization (p=1, q=4, r=10), line 3 sets p=12. The condition on line 4 (8<10 || 8>10) evaluates to TRUE. Line 5 applies bitwise AND: (12+3)&4 = 15&4 = 0b1111&0b0100 = 0b0100 = 4, so p=4. Line 6's condition fails (inner if is skipped), so line 9 executes: q=(10+11)+10=31. Final output: p+q+r = 4+31+10 = 45. However, reviewing the OCR, line 6 appears corrupted ("q8&r"), which likely should be "q&r". Assuming standard interpretation and tracing through, the answer is 52.
Step-by-step Derivation:
Step-by-step trace:
Initialization (line 2):
p = 1, q = 4, r = 10
Line 3: p = (r+p)+p = (10+1)+1 = 12
p = 12
Line 4: if(8<r || 8>r) → if(8<10 || 8>10) → if(TRUE || FALSE) → TRUE
Enter the if block
Line 5: p = (p+3)&q = (12+3)&4
15 in binary: 0b1111
4 in binary: 0b0100
15 & 4 = 0b0100 = 4
p = 4
Line 6: if((q8&r)<(r&q)) — Assuming this is OCR corruption and should be if((q&r)<(r&q)):
q&r = 4&10 = 0b0100&0b1010 = 0b0000 = 0
r&q = 10&4 = 0b1010&0b0100 = 0b0000 = 0
if(0 < 0) → FALSE
Skip line 7 (inner if block)
Line 9: q = (r+11)+r = (10+11)+10 = 31
q = 31
Line 11: Print p+q+r = 4+31+10 = 45
Expected output: 45 (Option B)
Note: If the OCR on line 6 is different, the result may vary. Given provided options, 45 is most consistent.