Question

 /* Write a program that would mix-and-merge two given strings (s1 and s2) into string s3 as follows:
 first character of s1, first character of s2, second character of s1, second character of s2, etc. */

我在互联网上找到了这个示例练习，我想我会尝试一下。一切似乎都没问题，代码编译没有错误，但我的＆＃34; s3＆＃34;字符串变量不会输出任何内容，它只会保持空白。

有趣的是，但问题似乎是在返回0之前的最后一行;

cout << s3;

如果我做这样的事情：

cout << s3[0];

或者我想要的任何索引，它将在运行代码时显示正确的字符，该代码是从其他字符串连接的。那么问题是什么？代码在此供参考：

#include <iostream>
#include <string>

using namespace std;

int main ()
{
    string s1, s2, s3;
    int i;
    int j = 0;

    cout << "Type in the first string: ";
    getline(cin, s1);
    cout << "Type in the second string: ";
    getline(cin, s2);

    for(i = 0; j < s1.size(); i += 2) // Merge first string by starting at index 0 and moving in 2s hence i += 2.
    {
        s3[i] = s1[j];
        ++j;
    }

    j = 0;

    for(i = 1; j < s1.size(); i +=2) // Merge second string by starting at index 1 hence i + 1 and again moving in 2s as to not overwrite an index that s1 put in.
    {
        s3[i] = s2[j];
        ++j;
    }


    cout << s3; // Problem is here?

    return 0;
}

Answer 1

使用大于当前长度的运算符s3访问[i]字符串是未定义的行为。

您可以尝试在循环之前使s3足够大（例如填充空格）。请注意，您的当前实现仅在两个字符串长度相同时才有效。

或者，尝试以不同的方式思考。例如如果你在你面前的小纸叠（每件上有一个字母）上有字符串s1和s2并希望将它们合并到一个堆栈中，那么你会这样做。< / p>

Answer 2

在您的代码中只添加了一行并使其正常工作：

#include <iostream>
#include <string>

using namespace std;

int main ()
{
    string s1, s2, s3;
    size_t i;
    size_t j = 0;

    cout << "Type in the first string: ";
    getline(cin, s1);
    cout << "Type in the second string: ";
    getline(cin, s2);

    s3.resize(s1.size() + s2.size());
    for(i = 0; j < s1.size(); i += 2) // Merge first string by starting at index 0 and moving in 2s hence i += 2.
    {
        s3[i] = s1[j];
        ++j;
    }

    j = 0;

    for(i = 1; j < s1.size(); i +=2) // Merge second string by starting at index 1 hence i + 1 and again moving in 2s as to not overwrite an index that s1 put in.
    {
        s3[i] = s2[j];
        ++j;
    }


    cout << s3; // Problem is here?

    return 0;
}

虽然这仅适用于大小相同的s1和s2。

当将2个字符串连接到第三个字母时，为什么最终字符串不会显示正确的输出？

2 个答案: