Printing arrays in C

You have an extra semicolon.

 for (i=0;i<5;i++);
                  ^
                  |
        here -----+

That semicolon means your for loop has an empty body, and you end up printing after the loop completes, equivalently:

printf("%d", array[5]);

Since that value is beyond the end of the array, you get undefined behaviour and some strange output.

Editorial note: You should look into a compiler that gives better warnings. Clang, for example, gives this warning, even with no special flags passed:

example.c:7:20: warning: for loop has empty body [-Wempty-body]
  for (i=0;i<5;i++);
                   ^
example.c:7:20: note: put the semicolon on a separate line to silence this
      warning

Skip the semi-colon in the for-loop:

for (i=0;i<5;i++);

For your logic, you don't need a simi-colon after the for-loop.

Due to semi-colon, the for-loop is evaluated 5 times and by the time it reaches the printf, the value of i is 5. Now, trying to access the index 5 in array is giving you an unknown value since you have initialized the array with 5 values only. This can become clearer, if you were to print the value of i in the printf statement:

  for (i=0;i<5;i++);
  {
     printf("i: %d \t Val: %d",i, array[i]);
  }

Will give you an output such as:

i: 5     Val: 32767

semicolon after for loop was the problem. Before it was printing some random number, since the storage class was automatic.

#include <stdio.h>
#include <math.h>
int main()
{
  int array[]={0,1,2,3,4};
  int i;
  for (i=0;i<5;i++);
  {
      printf("%d",array[i]);
  }
  printf("\n");
}