在char上使用toupper返回char的ascii号，而不是字符? [英] Using toupper on char returns the ascii number of the char, not the character?

问题描述

int main()
{

char hmm[1000];

cin.getline(hmm, 1000);
cout << hmm << endl;    //this was to test if I could assign my input to the array properly
for (int sayac = 0; hmm[sayac] != '@'; sayac++) {
    if (!isdigit(hmm[sayac])) {
        if (islower(hmm[sayac])) 
            cout << toupper(hmm[sayac]);
        else if (isupper(hmm[sayac]))
            cout << tolower(hmm[sayac]);
        else
            cout << hmm[sayac];
    }
}

编写一个程序，该程序将键盘输入读取到@符号并回显输入除数字外，将每个大写字符转换为小写，反之亦然.(不要忘记cctype家族.)"

"Write a program that reads keyboard input to the @ symbol and that echoes the input except for digits, converting each uppercase character to lowercase, and vice versa. (Don’t forget the cctype family.) "

我正在从入门书中进行此练习.但是，当我运行它时，它返回char的ascii顺序，而不是字符的大写/小写版本.无法解决问题.有人可以告诉我为什么吗?

I'm doing this exercise from the primer book. But when I run it, it returns the ascii order of the char, not the uppercase/lowercase version of the character. Couldn't figure out the problem. Can someone tell my why please?

(我可能对该练习有其他问题，如果有，请不要更正.我想自己修复它(我解释的问题除外)，但是我无法检查其他问题有这个问题.

(I may have other problems about the exercise, please don't correct them if I have. I want to fix it on my own (except the problem I explained), but I can't check the other ones as I have this problem.

推荐答案

编写时

std::cout << toupper('a');

发生以下情况:

1. int toupper(int ch) 会被调用，并返回一个整数，其值为'A'( 0x41 ).
2. std :: basic_ostream :: operator<<(std::cout，0x41) 被调用，这是 int (2)重载，因为提供了 int .

1. int toupper(int ch) is called, and returns an integer whose value is 'A' (0x41).
2. std::basic_ostream::operator<<(std::cout, 0x41) is called, that is the int (2) overload since an int was provided.

总体上，它显示"65".

Overall, it prints "65".

作为解决方案，您可以将大写字母回退为 char :

As a solution, you can cast back your upper case to a char:

std::cout << static_cast<char>(toupper('a'));

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