c++ - 如何在字符串索引为空时停止检查

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. */ 

但是我的代码的问题是:如果我输入 s1 "John" 然后输入 s2 "Stevens" 结果将是 = JSothenv e n s.

如何修复其中一个字符串结束后留下的空格？

我想我会修复它的方法是我会检查我在下面的 for 循环中的 ifs 以查看索引是否为 null 或 '\0' 但这不起作用，因为字符串在之后保存随机值字符串结束。

#include <iostream>
#include <string>

using namespace std;

int main()
{
    string s1, s2; 
    string s3;
    int i; // For index
    int j = 0; // For second index and loop checking

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

    s3.resize(s1.size() + s2.size() + 100); // The + 100 is used so we have space for all the characters. The + 100 is not needed if i fix my problem.

    for(i = 0; j <= s1.size(); i += 2)
    {
        if(s1[j] == null) // With what do i check it?
        {
            break;
        }
        else
        {
            s3[i] = s1[j];
            ++j;
        }
    }

    j = 0;

    for(i = 1; j <= s2.size(); i += 2)
    {
        if(s2[j] == null)
        {
            break;
        }
        else
        {
        s3[i] = s2[j];
        }
        ++j;
    }

    for( i = 0; i <= s3.size(); ++i)
    {
        cout << s3[i];  
    }

    return 0;   

Answer 1

试试这个:

#include <iostream>
#include <string>

using namespace std;

int main(int argc, char const *argv[])
{
    string s1, s2, s3;

    int i,j,k;
    cout << "Type first string: ";
    getline(cin, s1);
    cout << "Type second string: ";
    getline(cin, s2);
    s3.resize(s1.size()+s2.size());

    for(i = 0, j = 0, k = 0; j < s1.size() && k < s2.size(); i++) {
        if(i & 1) {
            s3[i] = s2[k++];
        } else {
            s3[i] = s1[j++];
        }
    }
    if(j == s1.size()) {
        while(k < s2.size()) {
            s3[i++] = s2[k++];
        }
    } else {
        while(j < s1.size()) {
            s3[i++] = s1[j++];
        }
    }

    cout << s3 << endl;

    return 0;
}