looping thru array, increment doesnt work

Question

ARRAY_SIZE[10] = {0};
int i;
for(i = 0; i < ARRAY_SIZE; ++i ) {      
  printf("Before assignment %d\n", array_of_ints[i]);   
}

I was expecting the print out result be 0 thru 9, since ++i increments i. And on the second loop it will print out 1, on the following 2, and so on. However, it prints out 0 ten times. Why is that? Thanks ahead.

Please [read about how to ask good questions](http://stackoverflow.com/help/how-to-ask), and learn how to create a [**Minimal, Complete, and Verifiable Example**](http://stackoverflow.com/help/mcve). The code you have is to minimal, it's not complete, and it's not verifiable. — Some programmer dude, Mar 27 '18 at 05:45
and it does print out 0 to 9. if I add this two lines: `//array_of_ints[i] = i; //printf("After assignment %d\n", array_of_ints[i]);` — pumkinthatcodes, Mar 27 '18 at 05:58
First you have to provide minimal code so that one can compile and check. Regarding your code, you are printing `array_of_ints` and expecting result of `i`. — hafeez, Mar 27 '18 at 06:00
what should `ARRAY_SIZE[10] = {0};` mean? there is no `array_of_ints`defined. We are not clairvoyant. — Kami Kaze, Mar 27 '18 at 06:02

score 0 · Answer 1 · answered Mar 27 '18 at 05:53

0

ARRAY_SIZE will not give you the length of the array. You have to enter it manually.

ARRAY_SIZE[10] = {0}; 
int i; for(i = 0; i < 10; ++i ) 
{        printf("Before assignment %d\n", array_of_ints[i]);    

}

answered Mar 27 '18 at 05:53

AsthaUndefined

1,111
1
11
24

Thanks. but still gives the same output. ten lines of 0. – pumkinthatcodes Mar 27 '18 at 05:57
Yes, because the array is initialized with 0. – AsthaUndefined Mar 27 '18 at 05:59
Got it. i increments thru the index of the array not the value contained at that index. that is why!! – pumkinthatcodes Mar 27 '18 at 06:02
1

What is array_of_ints? – Lakshraj Mar 27 '18 at 06:47

looping thru array, increment doesnt work

1 Answers1