Beginner C++ problem

[captpete]

I need a program that prints the "hundreds" digit in a series of integers entered by the user. For instance, if the user enters 2356 and 764 and 894, then the program will display 3, 7, and 8.

[captpete]

I do not want code, just an idea, concept or explanation of how to do it. I want to put the code together myself. I just don't know who to determine the correct digit.

[strangerstill]

((int) n / 100) % 10
or
((int) n % 1000) / 100

Your choice: first probably faster.
%: modulus operator ('remainder on division by')

[captpete]

I see how these expressions would work with an integer of four or more digits, and I greatly appreciate that. But how about three digit integers? I'm not supposed to use if statements yet. And thanks again.

[strangerstill]

OK. Let's try 386:

(int) 386 / 100 = 3
3 % 10 = 3

or

(int) 386 % 1000 = 386
386 / 100 = 3

See? It works.

Note that the (int) cast applies to n, not to (n/100)
This is in case n is a single; of course I'd expect it to be an int already so that's a bit unnecessary: in fact

(i / 100) % 10

will work just as well; because both i and 100 are integers, C++ will use integer division and return an integer.

Oh, and my cryptic reference to speed: thinking about it, both will be equal in speed (one integer division, one integer modulus), but the first looks tidier. On the other hand, if n is of type long, then the following code:

(int)(n % 1000) / 100

will be fastest as it involves one long modulus and an integer division as opposed to

(n / 100) % 10

which would involve a long modulus and a long division.

Hope that makes sense.

[captpete]

Of course you're right - I was confusing the quotient with the modulus thinking the latter would be 0 given 386%1000. When in doubt, read the book.

Thanks once more.

Originally posted by strangerstill
OK. Let's try 386:

(int) 386 / 100 = 3
3 % 10 = 3

or

(int) 386 % 1000 = 386
386 / 100 = 3

See? It works.

Note that the (int) cast applies to n, not to (n/100)
This is in case n is a single; of course I'd expect it to be an int already so that's a bit unnecessary: in fact

(i / 100) % 10

will work just as well; because both i and 100 are integers, C++ will use integer division and return an integer.

Oh, and my cryptic reference to speed: thinking about it, both will be equal in speed (one integer division, one integer modulus), but the first looks tidier. On the other hand, if n is of type long, then the following code:

(int)(n % 1000) / 100

will be fastest as it involves one long modulus and an integer division as opposed to

(n / 100) % 10

which would involve a long modulus and a long division.

Hope that makes sense.

[strangerstill]

Happy to be of help. Hope you get further with C++ than I did.

[most_wanted]

is there source code for this program?? if theres, would you please post it. thank you