CODEWITHSUNDEEP
c
mie.ppa
mie.ppa

Reputation: 125

Program to find trailing number of zeroes is giving wrong output

My C program to find trailing number of zeroes is giving wrong output. The loop gets terminated after 1 test case and returns garbage outputs. Here is my code:

 #include <stdio.h>
 #include <math.h>
 int main()
 {
      int testcase, no_of_zeroes, i, c, number;
      scanf(" %d \n", &testcase);

      for(i = 0; i<testcase; i++)
      {  
           no_of_zeroes = 0;
           printf(" %d \n", &number);
           c = 5;

           while((number/c) > 0)
           {
                no_of_zeroes += (number/5);
                c *= 5;
           }
           printf(" %d \n", no_of_zeroes);
      }
      return 0;
 }

Upvotes: 0

Views: 84

Answers (2)

Yu Jiaao
Yu Jiaao

Reputation: 4714

warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int *’

 printf(" %d \n", number);

you must initialize it, int number =0;

or maybe you mean(?):

 scanf(" %d \n", &number);

Upvotes: 2

Charles Srstka
Charles Srstka

Reputation: 17040

You're never initializing number, and then you're printing a pointer to it instead of the number itself. Of course you're going to print garbage results.

Also, I don't understand how your algorithm is supposed to work. Dividing by five and adding that to the number of zeroes? If I did that with the number 100, that would add 20, but 100 doesn't have 20 trailing zeroes.

Upvotes: 3

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