How to put elements in an array with for loop?

Question

I am trying to insert elements into an array called "count" with a for loop. But the elements inside the array remain as 0 no matter what. I want it to print the desired output so that the numbers increments and does not remain 0 for all the indexes.

#include <stdio.h>

int main() {
    int n;
    int count[1] = {0}; //initialize array with 0

    printf("\nEnter an int for n: ");
    scanf_s("%d", &n);

    for (int i = 0; i < n; i++) {
         printf("%d \n", count[i]);
     }

My current output - Enter an int for n: 6

0

Desired Output -

0

1

2

3

4

5

The array is declared to hold only 1 element. Assigning to anything beyond count[0] causes undefined behavior. C arrays don't expand automatically. — Barmar
– Barmar, Commented Sep 16, 2022 at 23:52

Barmar · Accepted Answer · 2022-09-17 00:07:34Z

1

C arrays can't grow beyond their declared size. You need to declare the array with the intended size, not count[1]. You can do this by putting the declaration after you read n; this is called a variable-length array.

Then you can use a loop to initialize all the elements with consecutive numbers.

#include <stdio.h>

int main() {
    int n;

    printf("\nEnter an int for n: ");
    scanf_s("%d", &n);
    int count[n];
    // initialize array with consecutive numbers
    for (int i = 0; i < n; i++) {
        count[i] = i;
    }

    for (int i = 0; i < n; i++) {
        printf("%d \n", count[i]);
    }
}

edited Sep 17, 2022 at 0:07

answered Sep 16, 2022 at 23:56

Barmar

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2 Comments

beginner_101 Over a year ago

This code leads to an error, It says for the int count[n] = {0} line that "expression must have a constant value".

Barmar Over a year ago

Sorry, forgot that you can't initialize a VLA, it has to be done in the loop.

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