why is this double type number getting rounded?

[leehc94]

I have the following very simple code:

#include <iostream>
#include <string>
#include <Windows.h>

void Problem2();

int main()
{
	Problem2();
	return 0;
}


void Problem2()
{
	using namespace std;
	double score;
	char grade, flag = 'n';
	do {
		if (flag == 'y') {
			cout << "Invalid score. Score must be between 0 and 100" <<"\n";
			Sleep(1500);
		}
		cout << "Enter a score: ";
		cin >> score;
		flag = 'y';
	} while (score < 0 || score >100);
	if (90 <= score && score <= 100)
		grade = 'A';
	else if (80 <= score && score < 90)
		grade = 'B';
	else if (70 <= score && score < 80)
		grade = 'C';
	else
		grade = 'F';
	cout << "Your grade is " << grade << "\n";
}

this code checks the score and categorize them into grades;

According to the code, the score of x=89.99999999999999999999

(20 9's after decimal dot)

must award grade of B, but when I run it it actually awards A. What I suspect is that x is rounded to nearest number 90, but why does this happen? As far as I'm aware, doesn't double data type can store data up to far better accuracy?

[Peter87]

I'm afraid double doesn't have better precision than that.

#include <iostream>
#include <iomanip>
#include <string>
#include <cmath>

int main()
{
	std::string inputString = "89.99999999999999999999";
	double inputValue = std::stod(inputString);
	double smallerValue = std::nextafter(inputValue, 0);
	
	std::cout << std::fixed << std::setprecision(20);
	
	std::cout << "        Your input: " << inputString << '\n';
	std::cout << "   Value as double: " << inputValue << '\n';
	std::cout << "Next smaller value: " << smallerValue << '\n';
}

        Your input: 89.99999999999999999999
   Value as double: 90.00000000000000000000
Next smaller value: 89.99999999999998578915