program that store integer values into a 2D array - c

Question

It just creates random values
I tried using a separate value for the variable of the array and I also don't know why it starts counting at element 6.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int i;
int array[3][5];
for (i = 0; i<5; i++);
{
printf("Input a whole number for row 1 element %d:\n", i+1);
scanf("%d", &array[0][i]);
}
printf("Row 1 elements:\n");
for(i = 0; i<5; i++)
{
printf("%d\n", array[0][i]);
}
return 0;
}
Ouput:
> Input a whole number for row 1 element 6: 4 Row 1 elements: 0 0 0 0
> 1897665488
>
> Process returned 0 (0x0) execution time : 1.969 s Press any key to
> continue.

Answer 1

It starts counting from 6 because the line for (i = 0; i < 5; i++);, is iterating (incrementing) i 5 times so, i becomes 5, then you print i + 1 to stdout.
So, basically your call to printf() and scanf() functions were never a part of any sort of loop.
NOTE: Adding a semi-colon ;, after any loop means that there is no body for the loop. Basically it's an empty loop. It can be useful for finding the length of a string, and so on.
Some tips:
Also instead to using bare return 0;, use return EXIT_SUCCESS;, which is defined in the header file stdlib.h.
use int main(void) { }, instead of int main() { }
always check whether scanf() input was successful or not
Correct Code
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int i;
int array[3][5];
for (i = 0; i < 5; i++)
{
printf("Input a whole number for row 1 element %d:\n", i + 1);
if (scanf("%d", &array[0][i]) != 1)
{
perror("bad input: only numbers are acceptable\n");
return EXIT_FAILURE;
}
}
printf("Row 1 elements:\n");
for (i = 0; i < 5; i++)
{
printf("%d\n", array[0][i]);
}
return EXIT_SUCCESS;
}
Output:
Input a whole number for row 1 element 1:
1
Input a whole number for row 1 element 2:
2
Input a whole number for row 1 element 3:
3
Input a whole number for row 1 element 4:
5
Input a whole number for row 1 element 5:
7
Row 1 elements:
1
2
3
5
7