为什么cout不打印此字符串? [英] why cout is not printing this string?

问题描述

所以，我是C ++的新手，我不知道为什么会这样.我有一个包含所有字母的字符串，我将其中的10个字符复制到了新的字符串 s2 中，并使用如下所示的for循环逐个字符地复制了字符串，并在执行时将 cout 函数正在打印空白行.

So, I'm new to C++ and I can't figure why this is happening. I have a string with all the alphabets and I copied 10 characters from it into a new string s2, character by character using a for loop as follows and when I execute it the cout function is printing a blank line.

#include <iostream>

using namespace std;

int main(){
    string s = "abcdefghijklmnopqrstuvwxyz";
    string s2;
    for(int i=0; i<10; i++){
        s2[i] = s[i];
    }
    cout << s2 << endl;
    return 0;
}

但是当我逐字符打印此字符串 s2 时，我得到了正确的输出

But when I print this string s2 character by character I got the correct output

#include <iostream>

using namespace std;

int main(){
    string s = "abcdefghijklmnopqrstuvwxyz";
    string s2;
    for(int i=0; i<10; i++){
        s2[i] = s[i];
    }
    
    for(int i=0; i<10; i++){
        cout << s2[i];
    }
    return 0;
}

任何帮助将不胜感激！

推荐答案

两个代码都具有未定义的行为. string s2; 只是将 s2 初始化为一个空的 std :: string ，其中不包含任何元素.尝试以 s2 [i] 的形式访问元素会导致UB，这意味着一切皆有可能.(即使在您的第二个代码段中似​​乎也给出了正确的输出.)

Both your code have undefined behavior. string s2; just initializes s2 as an empty std::string which contains no elements. Trying to access element as s2[i] leads to UB, means anything is possible. (Even it appears to give the correct output in your 2nd code snippet.)

您可以使用 push_back 将元素附加为

You can use push_back to append element as

for(int i=0; i<10; i++){
    s2.push_back(s[i]);
}

或

for(int i=0; i<10; i++){
    s2 += s[i];
}

或者您可以使 s2 包含10个元素.

Or you can make s2 containing 10 elements in advance.

string s2(10, '\0'); // or... string s2; s2.resize(10);
for(int i=0; i<10; i++){
    s2[i] = s[i];
}

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