Rainfall Week

A variable that outlives one pass of the loop

There is no new syntax in this exercise. A std::vector, a range-based for, an if, std::setw, + — you have met every piece of it. That is the point. Knowing five things separately is not the same as getting them to work together in one program, and the gap between those two is where most people stall.

The idea that joins them up is small, but it matters: a variable declared before a loop is still standing on every pass, and still standing after the loop ends, so whatever one pass leaves in it is what the next pass finds. Declare that same variable inside the loop instead, and the declaration runs again every time round, discarding whatever the pass before it left there — and in C++, a variable declared inside a loop's braces stops existing the moment that pass finishes, so trying to read it afterwards is not even legal. Anything the loop must carry from one day to the next has to live above it.

Two things are worth carrying that way here.

A running total. Declare a variable at zero above the loop, and inside the loop write total = total + mm;. C++ works out the right-hand side first — whatever total was a moment ago, plus this pass's value — and stores the answer back under the same name. After seven passes it holds the sum of all seven days.

The best so far. Declare a second variable at zero above the loop, then inside the loop ask whether this pass beats it: if (mm > wettest_mm) {. When it does, store the new number. That variable then holds the largest value seen up to that moment — and after the final pass, the largest so far is simply the largest there is.

Remembering which day was wettest is one more line, not a new idea: inside the same if, store the day alongside the amount. They change together because they are one fact in two halves.

A worked example

Four coins from a jar, with the running total printed on every pass so you can watch it climb:

#include <iostream>
#include <vector>

int main() {
    std::vector<int> coins = {2, 5, 1, 5};
    int total = 0;
    int biggest = 0;
    for (int coin : coins) {
        total = total + coin;
        if (coin > biggest) {
            biggest = coin;
        }
        std::cout << "coin " << coin << "  total " << total << std::endl;
    }
    std::cout << "biggest coin: " << biggest << std::endl;
}

coin 2  total 2
coin 5  total 7
coin 1  total 8
coin 5  total 13
biggest coin: 5

total never drops back to zero, because the line that sets it to zero sits above the loop and runs once, before any counting starts. biggest ends on 5 — and notice it does not change on the last pass, even though that coin is a 5 too: 5 > 5 is false, so a tie leaves the first record standing.

Your turn

The editor holds a week of rainfall in rain, a std::vector<int> of seven measurements in millimetres, and the loop already prints a row for each one: a day number, a bar of one # per millimetre, and the measurement itself. Day two was dry, so its bar is empty — that is correct, not broken, since zero millimetres draws zero # characters.

Press Run first. The chart is correct; the two lines beneath it are not. The total reads 0 mm, and the wettest day comes out as day 0, which is not one of the seven.

Both of your additions go inside the loop, after the std::cout that is already there:

1. Add this day's rain to total.
2. If this day beats wettest_mm, store both the new amount and the day number.

total, wettest_mm and wettest_day already exist above the loop, all holding zero — you are not declaring them, only changing them as the loop goes round. The counter day = day + 1; already shows the shape your first new line needs.

If something goes wrong

If the compiler stops on use of undeclared identifier 'mm', a line meant for inside the loop has ended up after its closing brace — mm belongs to the loop's own scope and stops existing the moment that brace is reached.

If the millimetres come out right but the wettest day stays at 0, the if is updating wettest_mm and forgetting wettest_day. Both lines belong inside the if, indented one step further than the if itself.

Nothing here can break. Once the week adds up, change one of the numbers in rain and run it again: the bar, the total and the wettest day should all move on their own.