Some more explanation about an array example I found online - c

Question

I'm just starting out to learn program in C and am trying to get my head around arrays. I found this example online but don't understand how it works. Here is the code:
#include <stdio.h>
int main () {
int n[ 10 ]; /* n is an array of 10 integers */
int i,j;
/* initialize elements of array n to 0 */
for ( i = 0; i < 10; i++ ) {
n[ i ] = i + 100; /* set element at location i to i + 100 */
}
/* output each array element's value */
for (j = 0; j < 10; j++ ) {
printf("Element[%d] = %d\n", j, n[j] );
}
return 0;
}
What I don't understand here is how int j links to int i. In the output, at the printf line, I don't understand why it is not:
printf("Element[%d] = %d\n", i, n[i]);
Why do we need another int j?
And another thing I don't understand is the %d's in that line. Why %d and not %i? d hasn't been declared anywhere right?
Any deeper explanation than on the website where I found this example would be appreciated. Thank you.

Answer 1

Why do we need another int j?
You don't. I'm assuming the author was just demonstrating that you can use any variable to index into an array, as long as the variable has the right type (pretty much any integral type) and holds a value in the correct range.
Array indices can be any integral expression, whether it's an integral constant (a[5]), a variable (a[i], a[j]), or a more complex expression (a[i+j], a[foo(i)], a[1+j*k/2], etc.). All that matters is that the expression has an integral type, and that its result is in the correct range (0 to N-1, where N is total number of elements in the array).
And another thing I don't understand is the %d's in that line. Why %d and not %i? d hasn't been declared anywhere right?
%d is a conversion specifier - it has nothing to do with any variable, and does not need to be declared separately. Conversion specifiers in the format string tell printf the number and types of additional arguments, and how to format those values for display.
In the line
printf("Element[%d] = %d\n", i, n[i]);
the %d specifiers in the format string tell printf that i and n[i] both have type int, and that they are to be formatted as decimal integers (as opposed to hex or octal, for example).
Here's an incomplete list of conversion specifiers:
Specifier Argument Type Output format
--------- ------------- -------------
%d,%i int decimal integer
%u unsigned int decimal integer (non-negative)
%f double decimal floating point
%x,%X unsigned int hexadecimal integer (non-negative)
%o unsigned int octal (non-negative)
%c char single character
%s char * text
Check your handy C reference manual for a complete list. If you don't have a handy C reference manual, I recommend Harbison & Steele's C: A Reference Manual.

Answer 2

Agreeing with # Eugene Sh, that in this case, j is indeed an extra variable.
The coding would be as shown below, and would return the same output.
for (i = 0; i < 10; i++) {
printf("Element[%d] = %d\n", i, n[i]);
}
And another thing I don't understand is the %d's in that line.
The printf() family of functions uses % character as a placeholder. This is a format specifier and does not require initialization. When a % is encountered, printf reads the characters following the % to determine what to do:
%s - Take the next argument and print it as a string
%d - Take the next argument and print it as an int
Sooner or later you might also encounter %s, which indicates that you're expecting a String to be your first print parameter.
Hope this helps.

Answer 3

What I don't understand here is how int j links to int i
They don't "link" and there is actually no connection between the two. The author simply used another integer (j) to avoid confusion.
As for your suggestion:
In the output, at the printf line, I don't understand why it is not:
printf("Element[%d] = %d\n", i, n[i]);
It is perfectly fine to do so if you make sure to use i as the loop variable for the second loop as well like so:
for (i = 0; i < 10; i++ ) {
printf("Element[%d] = %d\n", i, n[i] );
}
And another thing I don't understand is the %d's in that line. Why %d and not %i? d hasn't been declared anywhere right?
%d is a type modifier that tells the printf function to replace it with a decimal number when printing. It has nothing to do with some sort of variable named d. Your confusion might originated from the fact that there is also a %i modifier for printf but it is less commonly used. Refer to this wiki for more details.