编程原理和实践使用C++第 4 章演练，步骤 7

Programming Principles and Practice Using C++ Chapter 4 Drill, Step 7

本文关键字：演练 步骤 C++ 编程      更新时间：2023-10-16

我正在学习编程原理和使用C++的实践，但我被困在第 7 章练习的第 4 步。我在这里发现了类似的问题，但是转换单位/值并查看哪个更大/更小时有些东西不起作用。程序运行良好，但由于某种原因，某些转换返回不正确，例如，如果我输入 2 m，然后 2 ft. 2 ft 作为较大的值返回。

我知道代码可能看起来有点丑陋，如果我能让它工作，我会将转换放在一个函数中。提前谢谢。

int main() {
double doubNum = 0;
double smallestNum = ' ';
double largestNum = 0;
string unitOfDistance = " ";
double testNum = 0;
cout << "Enter a distance with a unit of measure (ft, in, cm, m): ";
while (cin >> doubNum >> unitOfDistance) { //while tests to see if the input is a double and unit is legal
    //check the unitOfDistance and convert all values to cm and hold in testNum for comparison
    if (unitOfDistance == "in") { //in to cm
        testNum = doubNum * 2.54;
    }
    else if (unitOfDistance == "ft") { //ft to cm
        testNum = (doubNum * 12) * 2.54;
    }
    else if (unitOfDistance == "cm") { //cm
        testNum = doubNum;
    }
    else if (unitOfDistance == "m") { //m to cm
        testNum = doubNum * 100;
    }
    else {
        cout << "I don't know that unit.n";
        return 0;
    }
    //check to see if testNum (the converted version of doubNum) is the smallest/largest/same value entered so far
    if (testNum < smallestNum) {
        smallestNum = doubNum;
        cout << smallestNum << " " << unitOfDistance << " is the smallest distance entered so far.n";
    }
    else if (testNum > largestNum) {
        largestNum = doubNum;
        cout << largestNum << " " << unitOfDistance << " is the largest distance entered so far.n";
    }
    else {
        cout << smallestNum << " " << unitOfDistance << " is the smallest distance entered so far.n";
        cout << largestNum << " " << unitOfDistance << " is the largest distance entered so far.n";
    }
    cout << "Enter another distance with unit: n";
}}

试试这个

#include <iostream>
#include <limits>
using namespace std;
int main() {
    double num, result, smallest, largest;
    smallest = numeric_limits<double>::max();
    largest = numeric_limits<double>::min();
    string unit;
    cout << "Enter a distance with a unit of measure (ft, in, cm, m): ";
    while (cin >> num >> unit) {
        if (unit == "in")       // in to cm
            result = num * 2.54;
        else if (unit == "ft")  // ft to cm
            result = (num * 12) * 2.54;
        else if (unit == "cm")  // cm
            result = num;
        else if (unit == "m")   // m to cm
            result = num * 100;
        else {
            cout << "I don't know that unit.n";
            break;
        }
        smallest = min(smallest, result);
        largest = max(largest, result);
        cout << smallest << " cm is the smallest distance entered so far.n";
        cout << largest << " cm is the largest distance entered so far.n";
    }
    return 0;
}

输入

2 m
3 ft
6 in

输出

Enter a distance with a unit of measure (ft, in, cm, m):
200 cm is the smallest distance entered so far.
200 cm is the largest distance entered so far.
91.44 cm is the smallest distance entered so far.
200 cm is the largest distance entered so far.
15.24 cm is the smallest distance entered so far.
200 cm is the largest distance entered so far.

你的代码几乎没有问题：
1.double smallestNum = ' '应替换为双smallestNum = DBL_MAX（或非常大的值）。
2.正如您使用largestNum和smallestNum来跟踪最大值和最小值一样，您还需要使用unitOfLargestDistance和unitOfSmallestDistance来跟踪其相应的单位。
3. 

if (testNum < smallestNum) {
    smallestNum = doubNum;
    cout << smallestNum << " " << unitOfDistance << " is the smallest distance entered so far.n";
}

在这里，您需要更新smallestNum testNum 不doubNum .所以它应该是这样的：

if (testNum < smallestNum) {
    smallestNum = testNum;
    cout << doubNum << " " << unitOfDistance << " is the smallest distance entered so far.n";
}

其他 2 个条件也是如此。

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