Question

How to declare, initialize and print one-dimensional array of integers in C

Answer 1

#include <stdio.h>

#define LEN 10

int main(void)
{
	int arr[LEN] = { 2, 234, 48, 17, 98, 918, 800, 12237, 100, 28 };
	
	for (int i = 0; i < LEN; i++) 
		printf("arr[%d] = %d\n", i, arr[i]);
    
	return 0;
}
  
    
/*
      
run:

arr[0] = 2
arr[1] = 234
arr[2] = 48
arr[3] = 17
arr[4] = 98
arr[5] = 918
arr[6] = 800
arr[7] = 12237
arr[8] = 100
arr[9] = 28

*/

 

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Answer 2

#include <stdio.h>

int main(void)
{
	int arr[] = { 2, 234, 48, 17, 98, 918, 800, 12237, 100, 28 };
	int arr_size;
	
	arr_size = sizeof(arr) / sizeof(arr[0]);
	
	for (int i = 0; i < arr_size; i++) 
		printf("arr[%d] = %d\n", i, arr[i]);
    
	return 0;
}
  
    
/*
      
run:

arr[0] = 2
arr[1] = 234
arr[2] = 48
arr[3] = 17
arr[4] = 98
arr[5] = 918
arr[6] = 800
arr[7] = 12237
arr[8] = 100
arr[9] = 28

*/

 

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Answer 3

#include <stdio.h>
 
#define LEN 10
 
int main(void)
{
    int arr[LEN];
     
    for (int i = 0; i < LEN; i++) 
		arr[i] = i + 1;
		
    for (int i = 0; i < LEN; i++) 
        printf("arr[%d] = %d\n", i, arr[i]);
     
    return 0;
}
   
     
/*
       
run:
 
arr[0] = 1
arr[1] = 2
arr[2] = 3
arr[3] = 4
arr[4] = 5
arr[5] = 6
arr[6] = 7
arr[7] = 8
arr[8] = 9
arr[9] = 10

*/

 

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