Question

TOPIC-PYTHON
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In the below code why r=0 and please explain r=r*10+d
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s=int (input("Enter any number:"))
r=0
while (s>0):
d=s%10
r=r*10+d
s=s//10
print ("reverse",r)

Answer 1

Explanation:

r=r*10+d shifts the previous value to the left (by multiplying by 10) and adds a new digit.

So imagine r = 123 and you want to append 4 to it.

First r is multiplied by 10, so you have 1230. Then you add 4, and get 1234.

The r variable builds up the reverse of s, so it starts at 0.

Answer 2

This program prints a number with reversed digits. For example, if you input 3564, you'll get 4653.

Suppose you give $s=45$, and let's see what the code does line by line:

We give 45 as input, and define $r=0$.

Then, we enter the while loop. The instruction d = s%10 simply extracts the last digit from s. In this case, $d=5$.

The istruction $r = r\ast 10+d$ adds a 0 at the end of the current value of $r$. Then we add $d$, so now the last digit of $r$ is $d$: we're performing

$0\cdot 10 + 5 = 5$

Finally, the integer division s = s//10 cuts the last digit from $s$. So, after the first loop, we have

$d=5,\quad r=5,\quad s=4$

We enter the loop again. we have

$d = s\%10 = 4\%10 = 4$

The new value of $r$ is

$5\cdot 10 + 4 = 54$

And the division s//10 returns 0, so we exit the loop.

And indeed we have r=54, which is 45 in reverse.