较长的 getline cin 输入的问题

Question

我正在编写一个 C++ 程序，它要求用户输入一个单词或句子，遍历单词/句子，用 'aoa' 或 'AoA' 替换所有 'a' 或 'A' 实例，然后输出结果。但是，如果我尝试输入更长的句子，我会遇到问题。例如，如果我输入“程序为什么不运行”，程序会输出奇怪的字母而不是预期的结果。 这是我的代码：

#include <iostream>
#include <string>

using namespace std;

int main(int argc, const char * argv[])
{
string mening, temp; //The mening string is the word/sentence the user will input.
int play = 1, add;

while (play == 1) {
cout<<"Type in the sentence: ";

getline(cin, mening); //The input is saved in the string variable mening.

unsigned long y = mening.size(); //Grabs the amount of characters in input; this number is saved in the unsigned long variable y.
add = 0; //Makes sure the int variable add is reset to 0 if the loop restarts.

for (int k = 0, n = 1;n<=y;k++, n++) {
    if (mening[k] == 'a' || mening[k] == 'A') {
        k++;


        for (int i = k, m = 1;m<=y - n;i++, m++) {
            temp[i] = mening[i];
        } //The characters after the one that has been checked are stored in temp array indexes, if the character that has been checked is an a or A.

        for (int i = k, m = 1, j = k + 2;m<=y - n;i++, m++, j++) {
            mening[j] = temp[i];
        } //The characters after the one that has been checked move two steps to the right, to allow the two extra letters.

        mening[k] = 'o';
        mening[k + 1] = mening[k - 1];

        k++;

        add = add + 2; //The int variable add is increased by 2 during each aoa/AoA to avoid strange characters being outputted at the very end.

    }
    else { }

}

for (int k = 0;k<=y + add - 1;k++) {
    cout<<mening[k];
}

cout<<endl<<"Do you want to do it again? (yes/no): ";

getline(cin, mening);

    cin.clear();
    cout << flush;
    cout.flush();
    cout.clear();

if (mening == "Yes" || mening == "yes" || mening == "YES") {

}
else {
    play = 2;
}
}


cout<<endl<<"The program will now close.";

return 0;
}

什么可能导致问题？

Answer 1

一个直接的问题是您从[1,n] 索引， 而 C++（std::string，std::vector，还有 C 风格 数组）使用[0,n)。这意味着您将访问 字符串的结尾。你将字符存储到temp 使用temp[x]，尽管temp 的大小始终为0。两者 其中一些是未定义的行为，可能会产生任何影响 （包括使程序崩溃）。

你应该使用标准库的调试模式 开发代码。在 Visual Studios 中，我认为这是 默认;使用 g++，您需要将 -D_GLIBCXX_CONCEPT_CHECKS -D_GLIBCXX_DEBUG -D_GLIBCXX_DEBUG_PEDANTIC 添加到您的命令中 线。

处理这个最简单的方法是复制到一个新的字符串中， 随时进行更改：

std::string results;
for ( auto current = mening.cbegin(); current != mening.cend(); ++ current ) {
    switch ( *current ) {
    case 'a':
        results += "aoa";
        break;

    case 'A':
        results += "AoA";
        break;

    default:
        results += *current;
        break;
    }
}

如果您确实想就地更换，那就很棘手了。 当您插入比开始时更多的文本时，迭代器是 无效。所以你需要这样的东西：

static std::string const Ao( "Ao" );
static std::string const ao( "ao" );
for ( auto current = mening.begin(); current != mening.end(); ++ current ) {
    switch ( *current ) {
    case 'a':
        current = mening.insert( current, ao.begin(), ao.end() ) + 2;
        break;

    case 'A':
        current = mening.insert( current, Ao.begin(), Ao.end() ) + 2;
        break;
    }
}

就个人而言，我更喜欢复制到一个新的字符串中。

Answer 2

我建议您应该使用标准函数std::string::insert 插入字符'oA' 或'oa'。它将使您的代码更易于处理和调试。

或者你可以简单地这样做：

#include <iostream>
#include <string>

using namespace std;

int main(int argc, const char * argv[])
{
string mening, temp; //The mening string is the word/sentence the user will input.
int play = 1, add;

while (play == 1) {
cout<<"Type in the sentence: ";

getline(cin, mening); //The input is saved in the string variable mening.

unsigned long y = mening.size(); //Grabs the amount of characters in input; this number is saved in the unsigned long variable y.
add = 0; //Makes sure the int variable add is reset to 0 if the loop restarts.

for (int n = 0; n<y; n++ ) {
    cout << mening[n];
    if (mening[n] == 'a' || mening[n] == 'A') 
        cout << "o" << mening[n];
}

cout<<endl<<"Do you want to do it again? (yes/no): ";

getline(cin, mening);

    cin.clear();
    cout << flush;
    cout.flush();
    cout.clear();

if (mening == "Yes" || mening == "yes" || mening == "YES") {

}
else {
    play = 2;
}
}


cout<<endl<<"The program will now close.";

return 0;
}

Answer 3

#include <iostream>
#include <string>


std::string ReplaceA(std::string s) {
    std::string temp = "";
    for (unsigned int k = 0; k < s.size(); k++) {
        if (s[k] == 'a' || s[k] == 'A') {
            temp = temp + s[k];
            temp = temp + "o";
            temp = temp + s[k];
            }
            else { 
                temp = temp + s[k];
            }

        }
    return temp;
}

int main(int argc, const char * argv[])
{
    std::string mening; //The mening string is the word/sentence the user will input.
    int play = 1;

    while (play == 1) {
        std::cout<<"Type in the sentence: ";

        getline(std::cin, mening); //The input is saved in the string variable mening.

        std::cout << ReplaceA(mening) << std::endl;

        std::cout << "Do you want to do it again? (yes/no): ";

        std::cin >> mening;
        if( std::cin.fail() || ( mening != "yes" && mening != "no" ) ) {
            std::cout << "Bad Input\nDo you want to do it again? (yes/no): ";
            std::cin.clear();
            std::cin.ignore('256','\n');
            std::cin >> mening;
        }else{
            if(mening == "no") break;
        }
        std::cin.clear();
        std::cin.ignore('256','\n');

    }


    std::cout << "The program will now close.";

    return 0;
}

更有条理，带有处理返回字符串的转换的函数。

你也可以试试这个功能，两者都可以，一个看起来更漂亮一点。

std::string ReplaceA(std::string s) {
    std::string temp = "";
    for(std::string::iterator k = s.begin(); k != s.end(); k++) {
        switch(*k) {
        case 'a': temp += "aoa"; break;
        case 'A': temp += "AoA"; break;
        default: temp += *k; break;
        }
    }
    return temp;
}

Answer 4

实际上最后一个cin还剩下一个尾随字符'\n'>>，所以getline取这个'\n'字符并被终止；在getline之前忽略buffer的内容...

cin.ignore();
getling(cin, string_name)