输入7个字母单词，输出7位电话号码

Question

我正在为作业编写代码。我输入 7 个字母，它会在小键盘上输出相应的数字。到目前为止，它可以很好地转换字母，它忽略了我想要的第 7 个字母之后的所有内容，并在您键入 stop 时停止。但是，如果我包含空格（这是分配所必需的）它会引发错误，我写了一些应该让它忽略它们但它不起作用的东西。此外，如果我输入的任何内容少于 7​​ 个字母，我也会收到错误消息。 “在抛出 'std::out_of_range' 的实例后调用终止 what(): basic_string::at: _n (which is 1) >= this->size() (which is 1) Aborted (core dumped)

我很确定这是因为 while 循环，在调试器中，它总是停在那里，但我不知道如何解决这个问题。我期待收到任何人的来信。提前谢谢！

#include <iostream>
#include <string>

using namespace std;

int main(){
    //declares variables
    string str1, str2;
    int i;
    //first input
    cout << "Type 'stop' to stop or type your 7 letter message to be converted to numbers: " << endl;
    cin >> str1;
    
    //while loop so that you can input multiple numbers one after the other until stop is read
    while (str1 != "stop"){
        //resets i if str1 does not equal stop
        i=0;
        while(str2.length() != 7){
            //determines what number to assign to the charcter at the index of the i variable
            if (str1.at(i) == 'a' || str1.at(i) == 'A' || str1.at(i) == 'b' || str1.at(i) == 'B' || str1.at(i) == 'c' || str1.at(i) == 'C'){
                str2 = str2 + '2';
                i=i+1;
            }else if (str1.at(i) == 'd' || str1.at(i) == 'D' || str1.at(i) == 'e' || str1.at(i) == 'E' || str1.at(i) == 'f' || str1.at(i) == 'F'){
                str2 = str2 + '3';
                i=i+1;
            }else if (str1.at(i) == 'g' || str1.at(i) == 'G' || str1.at(i) == 'h' || str1.at(i) == 'H' || str1.at(i) == 'i' || str1.at(i) == 'I'){
                str2 = str2 + '4';
                i=i+1;
            }else if (str1.at(i) == 'j' || str1.at(i) == 'J' || str1.at(i) == 'k' || str1.at(i) == 'K' || str1.at(i) == 'l' || str1.at(i) == 'L'){
                str2 = str2 + '5';
                i=i+1;
            }else if (str1.at(i) == 'm' || str1.at(i) == 'M' || str1.at(i) == 'n' || str1.at(i) == 'N' || str1.at(i) == 'o' || str1.at(i) == 'O'){
                str2 = str2 + '6';
                i=i+1;
            }else if (str1.at(i) == 'p' || str1.at(i) == 'P' || str1.at(i) == 'q' || str1.at(i) == 'Q' || str1.at(i) == 'r' || str1.at(i) == 'R'|| str1.at(i) == 's' || str1.at(i) == 'S'){
                str2 = str2 + '7';
                i=i+1;
            }else if (str1.at(i) == 't' || str1.at(i) == 'T' || str1.at(i) == 'u' || str1.at(i) == 'U' || str1.at(i) == 'v' || str1.at(i) == 'V'){
                str2 = str2 + '8';
                i=i+1;
            }else if (str1.at(i) == 'w' || str1.at(i) == 'W' || str1.at(i) == 'x' || str1.at(i) == 'X' || str1.at(i) == 'y' || str1.at(i) == 'Y'|| str1.at(i) == 'z' || str1.at(i) == 'Z'){
                str2 = str2 + '9';
                i=i+1;
            }else if(str1.at(i) == ' '){
                //if there is a space, it is ignored
                i=i+1;
            }else{
                //in case they put somthing weird, sets str2 to length of 7 to end while
                cout << "incorrect input";
                str2 = "abcdefg";
            }
        }
        //outputs converted number
        cout << str2.substr(0,3) << "-" << str2.substr(3,7) << endl << endl;
        cout << "Type 'stop' to stop or type your 7 letter message to be converted to numbers: " << endl;
        cin >> str1;
        //resets str2
        str2 = "";
        
    }
    return 0;
}

Answer 1

• 您应该使用std::getline() 来读取整行，即使它包含空白字符。
• 您应该检查输入字符串的长度并在字符串末尾停止以避免出现std::out_of_range 错误。

#include <iostream>
#include <string>

using namespace std;

int main(){
    //declares variables
    string str1, str2;
    int i;
    //first input
    cout << "Type 'stop' to stop or type your 7 letter message to be converted to numbers: " << endl;
    getline(cin, str1); // *** use std::getline()
    
    //while loop so that you can input multiple numbers one after the other until stop is read
    while (str1 != "stop"){
        //resets i if str1 does not equal stop
        i=0;
        while(str2.length() != 7){
            //determines what number to assign to the charcter at the index of the i variable
            if (i >= str1.length()) { // *** length check
                cout << "input too short";
                str2 = "abcdefg";
            }else if (str1.at(i) == 'a' || str1.at(i) == 'A' || str1.at(i) == 'b' || str1.at(i) == 'B' || str1.at(i) == 'c' || str1.at(i) == 'C'){
                str2 = str2 + '2';
                i=i+1;
            // *** omit ***
            }else{
                //in case they put somthing weird, sets str2 to length of 7 to end while
                cout << "incorrect input";
                str2 = "abcdefg";
            }
        }
        //outputs converted number
        cout << str2.substr(0,3) << "-" << str2.substr(3,7) << endl << endl;
        cout << "Type 'stop' to stop or type your 7 letter message to be converted to numbers: " << endl;
        getline(cin, str1); // *** use std::getline()
        //resets str2
        str2 = "";
        
    }
    return 0;
}

Answer 2

1. 问题是由cin引起的： cin 将空格（空格、制表符等）视为终止字符，这意味着它只能显示一个单词（即使您输入了很多单词）。[https://www.w3schools.com/cpp/ cpp_strings_input.asp]
2. 您还必须将 while 条件从 while(str2.length() != 7) 更改为 while(i != 7)，因为 else if(str1.at(i) == ' ') 只会增加 i 而不是 str2。

Answer 3

您的代码中的主要问题是输入，因为当遇到空格字符时 cin 会处理字符串的输入。另一个问题是关于不总是满足的最小长度的假设。

这不是您的问题的原因，但命名变量可以帮助您的代码更清晰（例如input, output 而不是str1 and str2）。最后，还有比长 if 更简单的方法来获得正确的编码：

int main(){
    string input, output, mapin ="abcdefghijklmnopqrstuvwxyz", 
                          mapout="22233344455566677778889999";
    cout << "Type 'stop' to stop or type your 7 letter message to be converted to numbers: " << endl;
    
    bool input_ok=true; 
    while (cin >> input) {
        for (auto &c: input)
            c = tolower(c);    // lowercase,  just to simplify
        if (input=="stop" || output.size()==7) 
            break;              // reasons to stop reading 
        
        for (auto &c:input) {
            auto pos=mapin.find(c); 
            if (pos==string::npos) {
                input_ok=false; 
                output += '?';
            }
            else output += mapout[pos];
            if (output.size()==7)
                break;
        }
    }
    if (input_ok) {
        cout << output.substr(0,3)  ;
        if (output.size()>3) 
            cout << "-" << output.substr(3,7) ;
        cout << endl << endl;
    }
    else {
        cout << "incorrect input" << endl <<output<<endl;
    }
    return 0;
}

Online demo

为了便于处理，我首先将字符串转换为小写，然后 STOP、StOp 或 stop 都可以。然后我运行几个输入，以捕获可能由空格分隔的几组字母（空白，制表符，换行符，...）。然后对于每个字母，我在输入表中查找它的偏移量，并在输出表中使用相同偏移量的数字。

另一种变体是计算输出字符：

  for (auto &c:input) {
        if (isalpha(c)) 
            output += (c>='w'? '9' : (c>='p'? "78"[(c-'p')/4]:'2'+(c-'a')/3));
        else {
            input_ok=false; 
            output += '?';
        }
        if (output.size()==7)
            break;
    }

Online demo

但您也可以坚持原来的版本，但选择 switch 语句而不是长长的 if 级联。