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I am trying to reverse a two digit number, and I understand there may be better ways of doing this, but I am curious now why the way I chose does not work.

If I input 48, it produces 84 (a successful reversal).

If I input 84, it produces 38. If I input 47, it produces 64. These are just some examples of unsuccessful reversals.

    int digit_one, digit_two, input;
    float a, b;

    printf("Enter a two-digit number: ");
    scanf("%d", &input);

    a = input * 0.1; // turns the two digit input figure into a float with a digit after the decimal point
    digit_one = a; // turns the float into an integer, eliminating the digit after the decimal point

    b = a - digit_one; // produces a float that has a 0 before the decimal point, and a digit after the decimal point
    digit_two = b * 10; // moves the digit that was after the decimal point, to before the decimal point

    printf("The reversal is: %d%d\n", digit_two, digit_one);

Thank you!

Redtunic
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    Floating point arithmetic will not work as you seem to expect (see [Is floating point math broken?](https://stackoverflow.com/questions/588004/is-floating-point-math-broken)). Use integer operations like modulo and division instead, – Some programmer dude Jul 20 '22 at 06:58
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    Also, the solution might not work for two digit numbers ending with 0 , for e.g. 10,20,30 etc ! – Harsh Jul 20 '22 at 06:59
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    Just print the results of your calculation `... *0.1` and you will see why this happens: https://godbolt.org/z/4c1ocTqee – mch Jul 20 '22 at 07:07
  • This would be a great time to try out your debugger. It'll let you step through each line and watch your variables change with each operation. – Caleb Jul 20 '22 at 07:11

3 Answers3

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a - digit_one is a fractional number. If it's slightly less than the exact result (because for example 0.7 cannot be represented exactly as a float), then (a - digit_one) * 10 will be slightly less than the desired integer, and so digit_two will be one less than you expect.

You can avoid floating point, and write int digit_one, digit_two = input / 10, input % 10;

Paul Hankin
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Working with floats is not the best approach here. With the input "84", b * 10 = 3.999996 which is 3 when you convert it to an integer.

This is a classic computer science problem with floats. Here are some links where this has been explained very well:

Is floating point math broken?

https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html

Your problem can be solved differently:

int digit_one, digit_two, input, a, b;

printf("Enter a two-digit number: ");
scanf("%d", &input);

digit_one = input % 10;
digit_two = (input / 10) % 10;

printf("The reversal is: %d%d\n", digit_one, digit_two);
nilskch
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  • `digit_two = (input / 10) % 10`; you don't need %10 at the end. Just `digit_two = (input / 10);` is enough – ThongDT Jul 20 '22 at 08:21
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int result =0;
        
do{
    result = (result * 10) + (input % 10);
    input /= 10;
} while(input)
printf("The reversal is: %d\n", result);

You can print the 'result' variable to get any integer value irrespective of 2 digit or 3 digit

Arun Varghese
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