將 2 位整數分配給 char 變量 (Assigning a 2 digit integer to a char variable)

    char desired_score;
    desired_score = 95;

    cout << "I will work hard to get a grade of " << desired_score << " in this course!\n";

此代碼產生：

I will work hard to get a grade of _ in this course!

如果我將值分配給單個數字，它輸出就好了。所以我能夠編寫一個解決方法，我只需將變量值從 '9' 重新分配給 '5' 並使用兩個不同的 cout 語句.

有人知道為什麼會這樣嗎？這是家庭作業的一部分，我的教科書說這在 C++ 中應該是可能的。這是 Xcode 的特性嗎？

提前致謝。

參考解法

方法 1:

When you write a char to a stream, it's assumed that you wish to write an actual character instead of its integer value. To output an integer, cast it instead:

cout << "I will work hard to get a grade of "
     << static_cast<int>(desired_score)
     << " in this course!\n";

方法 2:

The reason this is happening is because the ascii code for _ (underscore) is 95. A simple fix would be to use either a string, an int, or a short instead of a char.

Edit: It works for single digits because you used a character literal, NOT because characters are also bytes in c++. That was my mistake.

(by MagicGAT、paddy、Aggs123)

參考文件

1. Assigning a 2 digit integer to a char variable (CC BY‑SA 2.5/3.0/4.0)