Question

在下面的程序中，当我在字符串中再添加一个字符时，其大小仍然保持不变（从str1.size（）函数可以明显看出）。为什么会这样？

#include <iostream>
#include <cstring>

using std::cout;
using std::endl;

int main() {

        std::string str1 = "hello";
        cout << "std::string str1 = \"hello\""<< endl;

        cout << "string is " << str1 << " with length " << str1.size() << endl;

        str1[5] = 'a';


        cout << "string is " << str1 << " with length " << str1.size() << endl;

   for (int i = 0 ; i < 7; i++) {
                cout << "str["<<i<<"] = " << str1[i] << " (int)(str[i])" << (int)str1[i] << endl;
        }
}

输出

std::string str1 = "hello"
string is hello with length 5
string is hello with length 5 //expected 6
str[0] = h (int)(str[i])104
str[1] = e (int)(str[i])101
str[2] = l (int)(str[i])108
str[3] = l (int)(str[i])108
str[4] = o (int)(str[i])111
str[5] = a (int)(str[i])97
str[6] =  (int)(str[i])0

Answer 1

操作符str1[5] = 'a';不会将某些内容“添加”到字符串中。它将值设置在特定位置，并且该位置必须在0..(length()-1)范围内；否则，行为是不确定的。

要在字符串后附加内容，请使用

str1 += "a";

或

str1.push_back('a');

请注意，与普通的“ C”样式字符串相反，std::string将length保留在单独的属性中（并且不仅仅依靠字符串终止符{{1 }}。

如果在字符串中添加其他字符，为什么字符串大小不变？

1 个答案: