Question

我目前正试图将角色'i'大写，但仅限于它本身

这是我的代码

int main()
{
    string textMessage = "Yesterday’s Doctor Who broadcast made me 
    laugh and cry in the same episode! i can only wonder what the 
     Doctor will get into next. My family and i are huge fans.";


replace(textMessage.begin(), textMessage.end(), 'i', 'I');

 cout << textMessage;

}

我的输出是

Yesterday’s Doctor Who broadcast made me laugh and cry In the same epIsode! I can only wonder what the Doctor wIll get Into next. My famIly and I are huge fans.

这是我想要的输出

Yesterday’s Doctor Who broadcast made me laugh and cry in the same episode! I can only wonder what the Doctor will get into next. My family and I are huge fans.

Answer 1

一封信本身就是一个被空白包围的字母，而不是其他字母。一个字母本身也很容易解释，因为它总是被空格包围。

replace(textMessage.begin(), textMessage.end(), ' i ', ' I ');

如果您将程序中的代码更改为此（它替换了具有大写字母I＆＃39; s的空格），它应该可以正常工作。

Answer 2

您不必使用替换函数，而是必须遍历字符串并仅检查空格前面和后面的“i”，并将其替换为“I”。我想以下代码段可以实现您想要实现的目标： -

for (int iTraverse = 0; iTraverse< textMessage.length(); iTraverse++)
{
    if (textMessage[iTraverse] == 'i' && textMessage[iTraverse-1] == ' ' && textMessage[iTraverse+1]     == ' ')
    {
        textMessage[iTraverse] = 'I';
    }
}

C ++替换字符串中的字符

2 个答案: