Thread: I really need some help here

Originally Posted by ಠ_ಠ

EDIT: unless you mean "...keep track if you've read an even or odd number of times..."

Yep.


Quzah.

Originally Posted by quzah

Yep.


Quzah.

Then what you said would be best, I tried reading it in as a formatted integer (%2d) but the EOF flag is vital in this case


Edit:
Unless you're guaranteed a certain number of numbers, then the problem becomes much easier

edit2:
examples 2 and 3 seem to kill that possibility

Originally Posted by newbie1973

I've tried this

Code:

long integer l = getlongnumber()
while (l >= 10) {
print ascii(l % 100)
l /= 100;
}
and for right to left (using recursion for fun)

print_ascii_code(getlonginteger())

print_ascii_code(long integer l)
if (l >= 10) {
print_ascii_code(l / 100)
print ascii(l % 100)
}

can't get it to work

Ok. Why dont you get it converted to C code. Just input a long unsigned integer from the user using scanf. Then make a while loop and do what you're doing in the colored line(taking the last two digits using % operator and obtaining the remianing number using / operator). For printing the ASCII code you dont have to make a separate function just use %c format specifier in your printf statement. And also do include the condition when the number becomes zero.

Originally Posted by BEN10

Ok. Why dont you get it converted to C code. Just input a long unsigned integer from the user using scanf.

I hope you don't want anything bigger than a four letter word.


Quzah.

Originally Posted by quzah

I hope you don't want anything bigger than a four letter word.


Quzah.

This is what the question says

Take an unsigned long variable, ask the user for input.

That's why I said to input a long unsigned integer.