只返回第一條if語句？（C++）

寫，提示用戶輸入贏得了一個療程的點，並確定等級的程序。該程序應顯示等級和等級的消息字符串。使用'if else'語句來確定打印消息字符串的等級和switch語句。

我編寫了代碼，它執行時沒有任何調試錯誤。問題是，無論我輸入什麼作爲我的PointsEarned，我只得到F的輸出！例如，如果我輸入「90」：

Try Harder Next Time. 
GRADE: F

什麼問題？我是否把我的switch語句放在錯誤的地方？代碼如下：

#include <iostream> 
using namespace std; 
int main() 
{ 
    int PointsEarned; 
    char Grade; 
    cout << "Please input the points earned in the course: "; 
    cin >> PointsEarned; 
    if (0 <= PointsEarned < 60) { 
     Grade = 'F'; 
    } 
    else if (60 <= PointsEarned < 70) { 
     Grade = 'D'; 
    } 
    else if (70 <= PointsEarned < 80) { 
     Grade = 'C'; 
    } 
    else if (80 <= PointsEarned < 90) { 
     Grade = 'B'; 
    } 
    else if (90 <= PointsEarned) { 
     Grade = 'A'; 
    } 
    else{ 
     cout << "That is not a valid input."; 
    } 
    switch (Grade) 
    { 
    case 'F': 
    case 'D': 
     cout << "Try Harder Next Time." << endl; 
     break; 
    case 'C': 
     cout << "Good." << endl; 
     break; 
    case 'B': 
     cout << "Very good." << endl; 
     break; 
    case 'A': 
     cout << "Excellent." << endl; 
     break; 
    default: 
     cout << "Please choose a valid input to receive your grade." << endl; 
    } 
    cout << "GRADE: " << Grade << endl; 
}

來源

2016-02-22 Naomi Jacobson

I''m敢肯定你需要的東西就像'if（0 <= PointsEarned && 60> PointsEarned）'一樣。 –

另請參閱http://stackoverflow.com/questions/8889522/is-4-y-1-a-valid-statement-in-c-how-do-you-evaluate-it-if-so – iksemyonov

你不能測試多個值，像你這樣的有：

if (0 <= PointsEarned < 60) {

這不是檢查，如果PointsEarned是0和60之間。只有那些<的ONE將在同一時間，這意味着你有效地測試的東西鐵道部elike

if (0 <= true/false) {

，而不是進行評估。

你需要的東西更像

if ((0 <= PointsEarned) && (PointsEarned < 60)) {

代替。

來源

2016-02-22 21:48:31

比較表格0 <= PointsEarned < 60在Python中工作，但不是C++。您應該使用0 <= PointsEarned && PointsEarned < 60。

來源

2016-02-22 21:48:02 BingsF

這if (0 <= PointsEarned < 60) { 更是打破了C++語言，更改爲2有效的比較 - if (0 <= PointsEarned && PointsEarned < 60) {

來源

2016-02-22 21:51:33

什麼都沒有*打破C++語言* – SergeyA

錯誤的是，它編譯和評估很好，但不是人們可能期望的結果。 – iksemyonov

我重寫代碼，做一些改進：在該行否則，如果(90 <= PointsEarned)我限制它100.Also其他評議指出你不能測試多個值，所以我重寫，因爲他們說。它現在工作正常！

#include <iostream> 
using namespace std; 
int main() 
{ 
    int PointsEarned; 
    char Grade; 
    cout << "Please input the points earned in the course: "; 
    cin >> PointsEarned; 
    if ((0 <= PointsEarned) && (PointsEarned < 60)) { 
     Grade = 'F'; 
    } 
    else if ((60 <= PointsEarned) && (PointsEarned < 70)){ 
     Grade = 'D'; 
    } 
    else if ((70 <= PointsEarned) && (PointsEarned < 80)) { 
     Grade = 'C'; 
    } 
    else if ((80 <= PointsEarned) && (PointsEarned < 90)){ 
     Grade = 'B'; 
    } 
    else if ((90 <= PointsEarned) && (PointsEarned <= 100)) { 
     Grade = 'A'; 
    } 
    else{ 
     cout << "That is not a valid input."; 
    } 
    switch (Grade) 
    { 
    case 'F': 
    case 'D': 
     cout << "Try Harder Next Time." << endl; 
     break; 
    case 'C': 
     cout << "Good." << endl; 
     break; 
    case 'B': 
     cout << "Very good." << endl; 
     break; 
    case 'A': 
     cout << "Excellent." << endl; 
     break; 
    default: 
     cout << "Please choose a valid input to receive your grade." << endl; 
    } 
    cout << "GRADE: " << Grade << endl; 
}

來源

2016-02-22 22:07:30 Zsombi

表達：if (0 <= PointsEarned < 60)意味着if ((0 <= PointsEarned) < 60)
這意味着if ((true/false) < 60)

上述表達式是不一樣的，你可能想表達：
if ((0 <= PointsEarned) && (PointsEarned < 60))

來源

2016-02-22 23:05:15

只返回第一條if語句？ （C++）

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只返回第一條if語句？（C++）