What does 6 >> 1 evaluate to in C?

3. Right shift moves bits toward the low end and each position halves the value: binary 110 becomes 011. One shift right is integer division by 2.

Is right shift the same as dividing by 2 in C?

For non-negative integers yes: x >> n is x / 2ⁿ truncated. For negative signed values the behavior is implementation-defined and can differ from division — on gcc and clang, -7 >> 1 is -4 while -7 / 2 is -3.

What is the difference between arithmetic and logical right shift?

Arithmetic shift copies the sign bit in from the left, keeping negatives negative; logical shift feeds in zeros. C leaves the choice to the implementation for signed types — gcc and clang use arithmetic shift. Unsigned types always shift in zeros.

Right Shift Operator in C - Why 6 >> 1 Is 3

The right shift operator >> is left shift’s mirror: bits slide toward the low end, and each position shifted halves the value. This question from our C Programming Quiz App is quick if you know that rule — and the wrong options map exactly onto the usual confusions.

The Quiz Question

int a = 6;
printf("%d", a >> 1);

What is printed by this code?

The Correct Answer: 3

6 in binary is 110. Shift every bit one place right and the lowest bit falls off the end: 011, which is 3. One shift right divides by 2. Verified on gcc 13.3 and Apple clang 21, warning-free:

$ gcc -Wall -Wextra rshift.c && ./a.out
3

  6      =  1 1 0
  6 >> 1 =  0 1 1  =  3   (low bit discarded)

Why Each Wrong Answer Is Wrong

Why not 12?

12 is the left shift: 6 << 1 doubles to 12. Mixing up the arrow direction is the most common slip — the arrows literally point the direction the bits move: >> pushes them right (smaller), << pushes them left (bigger).

Why not 2?

2 would be 6 ÷ 3 — reading the shift count as a divisor. Like left shift, the count is an exponent: x >> n divides by 2ⁿ, so >> 1 divides by 2, not by the count’s value… and definitely not by 3.

Why not 6?

A shift by 1 always moves something (unless the value is 0). Only x >> 0 leaves the number unchanged.

Two Subtleties Worth Knowing

Odd numbers truncate. The dropped bit is simply gone: 7 >> 1 is 3 (not 3.5, not 4) — we verified this in the same run. Right shift is integer division by two, flooring toward negative infinity for the compilers you’re likely using.

Negative numbers are implementation-defined. The C standard doesn’t say whether >> on a negative signed value copies the sign bit in (arithmetic shift) or feeds in zeros (logical shift). Both gcc 13.3 and clang 21 do the arithmetic version:

printf("%d\n", -7 >> 1);   /* -4 on gcc and clang (sign bit copied in) */

Note it’s -4, not -3: arithmetic shift floors, while C’s division truncates toward zero (-7 / 2 is -3). So >> and / 2 genuinely disagree on negative odd numbers — one more reason to do bit tricks on unsigned types, where the behavior is fully defined. For the full family of shift pitfalls, see the left shift post and the 1 << 31 signed-overflow trap.

Frequently Asked Questions

Is right shift the same as dividing by 2?

For non-negative integers, yes: x >> n equals x / 2ⁿ, truncating. For negative signed values the two can differ (-7 >> 1 is -4 on gcc/clang but -7 / 2 is -3), and the shift’s behavior is implementation-defined.

What happens to the bits that fall off in a right shift?

They’re discarded. 7 >> 1 drops the lowest 1-bit and yields 3 — there’s no rounding and no carry.

What fills in from the left when right-shifting?

For unsigned types: zeros, always. For signed negative values it’s implementation-defined — gcc and clang copy the sign bit (arithmetic shift), so negative numbers stay negative.

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 #61 in the C Programming Quiz App — 155 questions with explanations covering operators, pointers, memory, and more.
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Right Shift Operator in C – Why 6 >> 1 Is 3