%i or %d to print integer in C using printf()?

Question

They are completely equivalent when used with printf(). Personally, I prefer %d, it's used more often (should I say "it's the idiomatic conversion specifier for int"?).

(One difference between %i and %d is that when used with scanf(), then %d always expects a decimal integer, whereas %i recognizes the 0 and 0x prefixes as octal and hexadecimal, but no sane programmer uses scanf() anyway so this should not be a concern.)

I am just adding example here because I think examples make it easier to understand.

In printf() they behave identically so you can use any either %d or %i. But they behave differently in scanf().

For example:

int main()
{
    int num,num2;
    scanf("%d%i",&num,&num2);// reading num using %d and num2 using %i

    printf("%d\t%d",num,num2);
    return 0;
}

Output:

You can see the different results for identical inputs.

num:

We are reading num using %d so when we enter 010 it ignores the first 0 and treats it as decimal 10.

num2:

We are reading num2 using %i.

That means it will treat decimals, octals, and hexadecimals differently.

When it give num2 010 it sees the leading 0 and parses it as octal.

When we print it using %d it prints the decimal equivalent of octal 010 which is 8.

d and i conversion specifiers behave the same with fprintf but behave differently for fscanf.

As some other wrote in their answer, the idiomatic way to print an int is using d conversion specifier.

Regarding i specifier and fprintf, C99 Rationale says that:

The %i conversion specifier was added in C89 for programmer convenience to provide symmetry with fscanf’s %i conversion specifier, even though it has exactly the same meaning as the %d conversion specifier when used with fprintf.

%d seems to be the norm for printing integers, I never figured out why, they behave identically.