Why does left shift float value work?

Question

As I understand from Left shift Float type one cannot use left shift operator on float values. But when I tried it, it gave the same answer as multiplying by 2ⁿ.

#include <iostream>
#include <cmath>
using namespace std;

int main() {
    // your code goes here
    float a  = 1.1234;
    int b = (int)(a*(1<<10));
    int c = (int)(a*pow(2,10));
    cout << "\n a = " << a << " b = " << b << " c = " << c;
    return 0;
}

It outputs a = 1.1234 b = 1150 c = 1150

In which case will the two outputs (b and c) differ?

Do you know what `1<<10` is? No `float`s involved here. – Maroun Feb 05 '14 at 06:58 — Maroun, Feb 05 '14 at 06:58
@ᴍarounᴍaroun ah! I'm such an idiot. – user13107 Feb 05 '14 at 06:59 — user13107, Feb 05 '14 at 06:59

score 9 · Accepted Answer · answered Feb 05 '14 at 06:59

9

int b = (int)(a*(1<<10));

Here, since both 1 and 10 are integers, you are performing left shift operation on integer instead of on floating-point number.

answered Feb 05 '14 at 06:59

timrau

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score 3 · Answer 2 · answered Feb 05 '14 at 07:01

3

you are multiplying 1024 with value of a(1.1234) in both case. it does not mean you are shifting float value.

answered Feb 05 '14 at 07:01

Amit Chauhan

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Why does left shift float value work?

2 Answers2