How to print string first n bytes when the string's length is greater than n?

Question

So I have a string that has a certain amount of bytes (or length). I say bytes because there is no NULL terminator at the end of the string. Though, I know how long the string is. Normally, as we all know, when you printf("%s", str);, it will keep printing every byte until it gets to a NULL character. I know there is no C string that is not NULL terminated, but I have a weird situation where I'm storing stuff (Not specifically strings) and I don't store the NULL, but the length of the "thing".

Here is a little sample:

char* str = "Hello_World"; //Let's use our imagination and pretend this doesn't have a NULL terminator after the 'd' in World
long len = 5;

//Print the first 'len' bytes (or char's) of 'str'

I know you are allowed to do something like this:

printf("%.5s", str);

But with that situation, I'm hard coding the 5 in, though with my situation, the 5 is in a variable. I would do something like this:

printf("%.(%l)s", len, str);

But I know you can't do that. But gives you an idea of what I'm trying to accomplish.

Answer 1

printf("%.*s", len, str);

and also, there is no C string that is not NULL terminated.

Answer 2

You can do this:

for (int i =0; i<len; i++)
{
    printf("%c", str[i]);
}

Which will print them in the same line, looping for whatever lenght you need to print.