What will be the output of the following program? int a = 5; cout << "FIRST "…

Concept: In C++, the operator << is overloaded. When its left operand is a stream object such as cout, << means STREAM INSERTION — it prints the right operand as text and returns the stream so the next << can chain onto it. This is different from the BITWISE LEFT-SHIFT <<, which applies only when BOTH operands are plain integers on their own (for example, a standalone 5 << 2 equals 20).

Chained << calls associate left to right, and whitespace never changes tokenization, so a<<2 and a << 2 are exactly the same two tokens (<< then 2). Once the chain already starts from cout, every later << in that same chain is stream insertion, not bit-shift.

Application: Trace cout << "FIRST " << a << 2 << "SECOND"; left to right, with a = 5:

1. cout << "FIRST " prints FIRST (with its trailing space) and returns cout.

2. << a prints the value of a, i.e. 5 → output so far: FIRST 5

3. << 2 is stream insertion (the left operand is the stream returned above, not a bare int), so it prints the literal 2, not a shifted value → output so far: FIRST 52

4. << "SECOND" appends SECOND with no separating space → final output: "FIRST 52SECOND"

Cross-check against the options:

• "FIRST 52 SECOND" inserts an extra space before SECOND that the code never produces.

• "FIRST 20 SECOND" would only hold if a << 2 were read as the bitwise shift 5 << 2 = 20 — but the left operand here is a stream, not a bare int, so that reading does not apply.

• "SECOND 25 FIRST" reverses both the print order and the digit order, contradicting left-to-right stream evaluation.

• Since none of the three specific strings equal "FIRST 52SECOND", "more than one of the above" cannot hold either.

Correct Answer: None of the above