Can you reverse the total? (Looping C)

Question

"can you reverse the number" is the number already reverse by using Mod(%). My question, can it be reversed to normal?

For example, if you enter the number "2552" it'll change to "2+5+5+2", which is correct, but, when you enter another number like "4125" it will change to "5+2+1+4" instead of "4+1+2+5"

Ok, I just entered the programming world, a newcomer

With the "if" can it add "+" without exceeding the number like "4+1+2+5+" there are "+" after "5", how can I delete this extra "+"?

#include <stdio.h>
main(){
int a, b, h=0;
    printf("Enter the number : ");
    scanf("%d",&a);

    printf("%d = ", a);

    while(a != 0)
    {
        b=a%10;
        a=a/10;
        printf("%d",b);
        if(b != a)
        {
            printf("+");
        }
        h=h+b;
    }
    printf(" = %d\n", h);

}

Please try to ask a specific question and leave out stuff like "OK", "I mean", "Hm". This reads like you're talking to yourself, being confused. — C-Otto, Jan 16 '18 at 15:41
I don't understand your question. What output do you want? What output are you currently getting. — Martin Bonner supports Monica, Jan 16 '18 at 15:44
not clear what is the question. What I understood is: You want to know how to print the `+` only between two digits, but not after the last one. Is that correct? — 463035818_is_not_an_ai, Jan 16 '18 at 15:44
Well done @François - I have no idea who approved putting in those back-ticks, but they were not an improvement. — Martin Bonner supports Monica, Jan 16 '18 at 15:46
@MartinBonner I did by accident when I tried to remove the c++ tag, sorry. Just cleaning up my own mess. — François Andrieux, Jan 16 '18 at 15:47
Related: https://stackoverflow.com/questions/35372784/idioms-for-for-each-except-the-last-or-between-each-consecutive-pair-of-el/35373017#35373017 — Jarod42, Jan 16 '18 at 15:48

score 1 · Answer 1 · answered Jan 16 '18 at 15:45

1

Instead of scanning for an actual number, you can scan a string from the user. Strings are easy to reverse:

char num[5]; // Has the input
char rev[5]; // Will have the reverse
int len = strlen(num);
rev[len] = 0; // End the string
for(int i = 0; i < len; i++){
    rev[i] = num[len-i-1];
}

answered Jan 16 '18 at 15:45

Felix

2,548
19
48

This code is a little hard to follow, so I wasn't paying attention to the right things. Those buffers are scary small, though. – tadman Jan 16 '18 at 16:19
@tadman Their size can be increased via substituting the '5' with a number that is greater in size ;) – Felix Jan 16 '18 at 16:21
1

Sarcasm aside, there's an extremely high probability someone will type in a number over 5 characters long. Why not 255? Or 1024? Or something someone's less likely to pop by pure accident? – tadman Jan 16 '18 at 16:23
@tadman Yes, that might be appropriate, but is in no way essential to the example I provided. Also, +1 for using powers of two for buffer sizes (although 256 is the *nitpick* right size). – Felix Jan 16 '18 at 16:26
255 has a long tradition in the database world, it's the longest length an 8-bit variable length indicator can handle, and 1024 is a good default for a generic "buffer". This of course glosses over how dangerous C can be if not kept on a very tight leash when it comes to strings in general. – tadman Jan 16 '18 at 16:30
@tadman Cool, but wouldn't the buffer size still be 256? As it is null-terminated, the length would be 255. – Felix Jan 16 '18 at 16:31
In the database context it's effectively a Pascal-style string so there's no null-terminator, and null values are not significant. It takes the form of one byte length field, (up to) 255 bytes of data. – tadman Jan 16 '18 at 16:33
1

@tadman Okay, interesting.. Haven't done a lot with databases yet. But thanks! – Felix Jan 16 '18 at 16:35

score 0 · Answer 2 · answered Jan 16 '18 at 16:12

You can go backwards by taking the highest digits in your number first instead of the lowest, e.g. like so:

int currentLog10;
int powerOfTenForCurrentDigit;

...

while(a != 0)
{
  currentLog10 = (int)log10(a);

  powerOfTenForCurrentDigit = (int)pow(10, currentLog10);

  b = a/powerOfTenForCurrentDigit;
  a = a - b * powerOfTenForCurrentDigit;

  ...
}

How does it work:

log10 of 4125 will give us 3.
b is set to 4125 devided by 1000 (10^3) = 4.
a is set to 4125 - 4 * 1000 = 125
Next step log10 of 125 gives 2, we devide by 10^2 (100) and so on.
Last step 5 gives 0, 5 devided by 10^0 (1) gives 5, a becomes 5 - 5 * 1, which is 0, we are done.

You have to be careful in edge cases, where log10 returns something like (n-1).999999999 instead of n, but it shouldn't happen for small numbers as are entered in your program. Maybe add some sanity check for the input.

Can you reverse the total? (Looping C)

2 Answers2