What does 4 | 2 evaluate to in C?

6. Bitwise OR merges the bit patterns: 100 | 010 = 110 binary = 6. Because 4 and 2 share no bits, the result here equals their sum — but 4 | 4 is 4, not 8, since OR never carries.

How are flags combined with bitwise OR in C?

Each option gets its own bit and OR packs them into one integer, as in open(path, O_RDWR | O_CREAT). Setting an already-set flag is harmless because OR is idempotent: 4 | 4 stays 4.

Bitwise OR in C - Why 4 | 2 Is 6 but 4

Q: What is the difference between | and || in C?

| is bitwise OR: it merges bits (4 | 2 = 6) and evaluates both sides. || is logical OR: it yields exactly 0 or 1 (4 || 2 = 1) and skips the right side when the left is nonzero.

Bitwise OR merges two bit patterns: a result bit is 1 if either operand has it. This question from our C Programming Quiz App uses two numbers whose bits don’t overlap — which makes OR behave exactly like addition, and that’s precisely what makes it a good test of whether you’re thinking in bits.

The Quiz Question

int a = 4, b = 2;
printf("%d", a | b);

What is printed by this code?

The Correct Answer: 6

Write the operands in binary and OR each column:

  4  =  1 0 0
  2  =  0 1 0
  ─────────── |
       1 1 0  =  6

Each number contributes its own bit; the union is 110 = 6. Verified on gcc 13.3 and Apple clang 21, warning-free:

$ gcc -Wall -Wextra or.c && ./a.out
6

Because 4 and 2 share no bits, 4 | 2 equals 4 + 2 — but that’s a coincidence of non-overlapping bits, not a rule. The proof is what happens when bits do overlap (from the same verified run):

printf("%d\n", 4 | 4);   /* 4 — not 8! a set bit stays set, no carrying */

Why Each Wrong Answer Is Wrong

Why not 4?

4 would mean b contributed nothing — true only if its bits were already present in a (like 4 | 4 or 6 | 2). But 2’s bit is new, so the result grows.

Why not 2?

Same reasoning mirrored — OR never discards bits from either side. The result always has at least every bit of both operands: a | b >= a and >= b, always.

Why not 0?

0 is what bitwise AND gives for these operands: 4 & 2 = 0, since they share no bits. OR and AND are duals — OR unions the bits, AND intersects them. Picking 0 means applying the intersection where the union was asked.

The Flags Idiom — Where | Lives in Real Code

Bitwise OR is how C combines options into one integer. Every flag gets its own bit (built with left shift), and OR packs them:

fd = open(path, O_RDWR | O_CREAT | O_APPEND);   /* three options, one int */
flags |= FLAG_VISIBLE;                          /* turn a bit on          */

The 4 | 4 = 4 behavior is a feature here: OR-ing a flag that’s already set changes nothing — setting is idempotent, which is exactly what you want from “turn this option on”. The other members of the family: & tests bits, &~ clears them, ^ toggles them.

Don’t confuse | with ||: the double-pipe is logical OR, which collapses everything to a 1-or-0 verdict — 4 || 2 is 1, not 6 (verified in the same run; see the logical OR post).

Frequently Asked Questions

Is bitwise OR the same as addition?

Only when the operands share no bits: 4 | 2 = 6 = 4 + 2, but 4 | 4 = 4 while 4 + 4 = 8. OR has no carrying — a set bit simply stays set.

What is the difference between | and || in C?

| is bitwise: it merges bit patterns (4 | 2 = 6) and always evaluates both operands. || is logical: it yields 0 or 1 (4 || 2 = 1) and short-circuits.

How do I set, clear, and test a flag bit in C?

Set with OR: flags |= F. Clear with AND-NOT: flags &= ~F. Test with AND: if (flags & F). Toggle with XOR: flags ^= F.

Recommended Books

The C Programming Language – Kernighan & Ritchie (India) | Amazon.com
C Programming: A Modern Approach – K.N. King (India) | Amazon.com

This question is #153 in the C Programming Quiz App — 155 questions with explanations covering operators, pointers, memory, and more.
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Bitwise OR in C – Why 4 | 2 Is 6 but 4 | 4 Is 4