int arr[] = {1, 2, 3, 4, 5}; (int x = 0; x < 5; x++) { printf("%d", [x]); }

Answer 1

printf FORMAT [ARGUMENT]...

Your issue is in your printf usage. By using %d you are requesting to Print an integer as a signed decimal number. See Integer Conversions, for details. ‘%d’ and ‘%i’ are synonymous for output, but are different when used with scanf for input.

https://www.gnu.org/software/libc/manual/html_node/Table-of-Output-Conversions.html

next your argument to use %d is only [x] which in this case would represent nothing.

your int arr = {1,2,3,4,5} set up an array of integers with a variable named arr and indexed 0 - 4 such that:

{[arr[0]] = 1, [arr[1]] = 2, [arr[2]] = 3, [arr[3]] = 4 , [arr[4]] = 5}

So what you need is:

printf("%d", arr[x]);

which will produce:

12345user$

the 12345 comes from the program the user$ is your terminal prompt because of no new line.