if statement inside of for loop not being executed - c++

Question

Writing a program to solve problem four of project euler: Find the largest palindrome made from the product of two 2-digit numbers. Heres my reprex:
#include <iostream>
int reverseNumber(int testNum)
{
int reversedNum, remainder = 0;
int temp = testNum;
while(temp != 0)
{
remainder = temp % 10;
reversedNum = reversedNum * 10 + remainder;
temp /= 10;
}
return reversedNum;
}
int main()
{
const int MIN = 100;
int numOne = 99;
int product = 0;
for(int numTwo = 10; numTwo < 100; numTwo++)
{
product = numOne * numTwo;
if (reverseNumber(product) == product)
{
int solution = product;
std::cout << solution << '\n';
return 0;
}
}
return 0;
}
My main thought process behind this is that the for loop will go through every number from 10 to 99 and multiply it by 99. My intended outcome is for it to print 9009 which is the largest palindrome with 2 factors of 2 digits. So what I think should happen here is the for loop will go from 10 to 99, and each loop it should go through the parameters of the if statement which reverses the number and sees if it equals itself.
I've made sure it wasn't a compiler issue, as this is recurring between different compilers. The reverseNumber() function returns the proper number every time I've tested it, so that shouldn't be the problem, however this problem only occurs when the function is involved in the logical comparison. By this I mean if that even I set it equal to a variable and put the variable in the if parameters, the issue still occurs. I'm pretty much stumped. I just hope it's not some silly mistake as I've been on this for a couple days now.

Answer 1

int reversedNum, remainder = 0;
You should be aware that this gives you (in an automatic variable context) a zero remainder but an arbitrary reversedNum. This is actually one of the reasons some development shops have the "one variable per declaration" rule.
In other words, it should probably be:
int reversedNum = 0, remainder;
or even:
int reversedNum = 0;
int remainder;
One other thing that often helps out is to limit the scope of variable to as small an area as possible, only bringing them into existence when needed. An example of that would be:
int reverseNumber(int testNum) {
int reversedNum = 0;
while (testNum != 0) {
int remainder = testNum % 10;
reversedNum = reversedNum * 10 + remainder;
testNum /= 10;
}
return reversedNum;
}
In fact, I'd probably go further and eliminate remainder altogether since you only use it once:
reversedNum = reversedNum * 10 + testNum % 10;
You'll notice I've gotten rid of temp there as well. There's little to gain by putting testNum into a temporary variable since it's already a copy of the original (as it was passed in by value).
And one other note, more to do with the problem rather than the code. You seem to be assuming that there is a palindrome formed that is a multiple of 99. That may be the case but a cautious programmer wouldn't rely on it - if you're allowed to assume things like that, you could just replace your entire program with:
print 9009
Hence you should probably check all possibilities.
You also get the first one you find which is not necessarily the highest one (for example, let's assume that 99 * 17 and 99 * 29 are both palindromic - you don't want the first one.
And, since you're checking all possibilities, you probably don't want to stop at the first one, even if the nested loops are decrementing instead of incrementing. That's because, if 99 * 3 and 97 * 97 are both palindromic, you want the highest, not the first.
So a better approach may be to start high and do an exhaustive search, while also ensuring you ignore the palindrome check of candidates that are smaller that your current maximum, something like (pseudo-code)
# Current highest palindrome.
high = -1
# Check in reverse order, to quickly get a relatively high one.
for num1 in 99 .. 0 inclusive:
# Only need to check num2 values <= num1: if there was a
# better palindrome at (num2 * num1), we would have
# already found in with (num1 * num2).
for num2 in num1 .. 0 inclusive:
mult = num1 * num2
# Don't waste time doing palindrome check if it's
# not greater than current maximum - we can't use
# it then anyway. Also, if we find one, it's the
# highest possible for THIS num1 value (since num2
# is decreasing), so we can exit the num2 loop
# right away.
if mult > high:
if mult == reversed(mult):
high = mult
break
if high >= 0:
print "Solution is ", high
else:
print "No solution"

Answer 2

In addition to properly initializing your variables, if you want the largest palindrome, you should switch the direction of your for loop -- like:
for(int numTwo = 100; numTwo > 10; numTwo--) {
...
}
or else you are just printing the first palindrome within your specified range