Question: This question looks long but it has a quick answer. working in c++, I have a sentient controlled while loop (see the below picture for
This question looks long but it has a quick answer.
working in c++, I have a sentient controlled while loop (see the below picture for the details), basically the user chooses either a hard or soft copy and whether or not they want a discount. They can then choose to check out another book and the loop starts over, it is cancelled when the user enters something other than "c" when prompted to check out another book and the program then outputs a sale summary. I am have a lot of issues with the math. The problem is that the summary includes "the total before the tax and discount" and also "the total after the tax and discount (6% tax)" Hence the need for both total1 and total2. My code works as long as the first book the user gets applies a discount otherwise it doesn't work... i'll attach the output for my code and the output that my code is supposed to match, any help would be appreciated, thanks!!

MY OUTPUT:

THE OUTPUT THAT THE SUMMARY NEEDS TO MATCH

MY CODE (Non-image form)
while ((action1 == 'c') || (action1 == 'C'))
{
cout
cout
cin >> cost;
cout
cin >> action3;
if (action3 == 'h' || action3 == 'H')
{
total1 += cost + 5;
}
else
{
total1 += cost + 2;
}
cout
cin >> action;
if (action == 'd')
{
total2 += total1 - (total1 * .1);
}
else
{
if (action3 == 'h' || action3 == 'h')
{
total2 = total2 + (cost + 5);
}
else
total2 = total2 + (cost + 2);
}
cout
cin >> action1;
checked++;
}
cout
cout
cout
total7 = (total2 + (total2 * tax));
cout
while ((action! == ) 11 (action! 'C')) = cout action3; total! += cost + 5; else total! += cost + 2; cout action3; total! += cost + 5; else total! += cost + 2; cout
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