Left shift and right shift behavior in Embedded C - Step-by-Step Execution

Concept Flow - Left shift and right shift behavior

Start with number

↓

Apply left shift <<

↓

Bits move left, zeros fill right

↓

Result is number * 2^shift_count

↓

Apply right shift >>

↓

Bits move right, fill depends on sign

↓

Result is number / 2^shift_count (floor)

↓

End

Left shift moves bits left, multiplying by powers of two; right shift moves bits right, dividing by powers of two, with fill depending on sign.

Execution Sample

Embedded C

int x = 4;  // binary 00000100
int y = x << 1; // left shift by 1
int z = x >> 1; // right shift by 1

This code shifts bits of x left and right by 1, showing multiplication and division by 2.

Execution Table

Step	Variable	Value (decimal)	Value (binary 8-bit)	Operation	Result Explanation
1	x	4	00000100	Initial value	Starting number in binary
2	y	8	00001000	x << 1	Bits shifted left by 1, multiply by 2
3	z	2	00000010	x >> 1	Bits shifted right by 1, divide by 2
4	x	-8	11111000	x = -8 (two's complement)	Negative number binary representation
5	y	-16	11110000	x << 1	Left shift negative number, bits shifted left
6	z	-4	11111100	x >> 1	Right shift negative number, sign bit fills left

💡 All shifts done; left shift multiplies by 2, right shift divides by 2 with sign extension for negatives.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6
x	4	4	4	-8	-8	-8
y	undefined	8	8	8	-16	-16
z	undefined	undefined	2	2	2	-4

Key Moments - 3 Insights

Why does right shifting a negative number fill with 1s on the left?

Because in embedded C, right shift of signed negative numbers usually performs arithmetic shift, preserving the sign by filling left bits with 1s (see step 6 in execution_table).

Does left shifting always multiply the number by 2?

Left shift multiplies by 2 only if no bits are lost (no overflow). For example, shifting -8 left by 1 gives -16 (step 5), but if bits shift out, the value changes unexpectedly.

What happens to bits shifted out of the variable?

Bits shifted out are lost and do not wrap around; this can cause data loss or change in sign (see step 5 and 6).

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the binary value of z?

A 00001000

B 00000100

C 00000010

D 11111100

💡 Hint

Check the 'Value (binary 8-bit)' column for step 3 in execution_table.

At which step does the variable x become negative?

A Step 2

B Step 4

C Step 5

D Step 6

💡 Hint

Look at the 'Value (decimal)' column for x in execution_table rows.

If we right shift x = -8 by 1, what is the expected decimal result according to the table?

A -4

B -2

C 4

D 8

💡 Hint

See step 6 in execution_table for x >> 1 result.

Concept Snapshot

Left shift (<<) moves bits left, filling right with zeros, multiplying by 2 per shift.
Right shift (>>) moves bits right, filling left with sign bit for signed numbers, dividing by 2 per shift.
Shifting can cause data loss if bits shift out.
Right shift of negative numbers usually keeps sign (arithmetic shift).
Use shifts carefully to avoid overflow or unexpected sign changes.

Full Transcript

This lesson shows how left and right bit shifts work in embedded C. Starting with a number, left shift moves bits left, adding zeros on the right, effectively multiplying the number by powers of two. Right shift moves bits right, filling the left side depending on the sign: zeros for positive, ones for negative numbers, effectively dividing by powers of two. The execution table traces values step-by-step, including positive and negative examples. Key points include how negative numbers keep their sign when right shifted and how bits shifted out are lost. The quiz tests understanding of binary values and sign behavior during shifts.