c++ - 在 C++ 中将 float 转换为 int 时丢失数据

Question

除了分数值的明显预期损失外，我不明白为什么会发生这种情况......

考虑家庭作业中用于将数字转换为英语短语的以下片段:

int main() {
    float dollars;

    cout << "Please supply a dollar amount between $0.01 and $9999.99 (without the dollar sign): ";
    while (true) {
        cin >> dollars;

        if (cin.fail() || dollars < 0.01 || dollars > 9999.99) {
            cin.clear();
            cin.ignore();
            cout << "You provided an invalid number. Please try again: ";
        } else {
            // Test code:
            cout << "You entered: " << dollars << endl;
            cout << "Times 100 without cast: " << dollars * 100 << endl;
            cout << "Times 100 with cast: " << (int)(dollars * 100) << endl;
            cout << "Cents: " << ((int)(dollars * 100)) % 100 << endl;
            break;
        }
    }

    printDollarsAsString(dollars);

    return 0;
}

我注意到当提供值 9999.21 时，else block 中第二个和第三个 cout 语句的输出不同1. 例如，这是我运行这段代码时的输出:

Please supply a dollar amount between $0.01 and $9999.99 (without the dollar sign): 9999.21
You entered: 9999.21
Times 100 without cast: 999921
Times 100 with cast: 999920
Cents: 20

我该如何解决这个问题，以便正确检索美分值？我可以采用其他方法吗？

Answer 1

        cout << "Times 100 without cast: " << dollars * 100 << endl;
        cout << "Times 100 with cast: " << (int)(dollars * 100) << endl;

当您转换为 int 时，您是在截断数字，而不是四舍五入。在这种情况下，.21 不能完全按二进制 float 存储，因为它实际上是一个重复数。重复数略小于.21。因此，当您乘以 100 时，您会得到 999920.9...，它会通过转换截断为 999920。