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

I am just learning C and I have a little knowledge of Objective-C due to dabbling in iOS development, however, in Objective-C I was using NSLog(@"%i", x); to print the variable x to the console however I have been reading a few C tutorials and they are saying to use %d instead of %i.

printf("%d", x); and printf("%i", x); both print x to the console correctly.

These both seem to get me to the same place so I am asking the experienced developers which is preferred? Is one more semantically correct or is right?

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.

both %d and %i can be used to print an integer

%d stands for "decimal", and %i for "integer." You can use %x to print in hexadecimal, and %o to print in octal.

You can use %i as a synonym for %d, if you prefer to indicate "integer" instead of "decimal."

On input, using scanf(), you can use use both %i and %d as well. %i means parse it as an integer in any base (octal, hexadecimal, or decimal, as indicated by a 0 or 0x prefix), while %d means parse it as a decimal integer.

check here for more explanation

https://stackoverflow.com/questions/13409014/why-does-d-stand-for-integer

As others said, they produce identical output on printf, but behave differently on scanf. I would prefer %d over %i for this reason. A number that is printed with %d can be read in with %d and you will get the same number. That is not always true with %i, if you ever choose to use zero padding. Because it is common to copy printf format strings into scanf format strings, I would avoid %i, since it could give you a surprising bug introduction:

I write fprintf("%i ...", ...);

You copy and write fscanf(%i ...", ...);

I decide I want to align columns more nicely and make alphabetization behave the same as sorting: fprintf("%03i ...", ...); (or %04d)

Now when you read my numbers, anything between 10 and 99 is interpreted in octal. Oops.

If you want decimal formatting, just say so.