英文:
Issue when writing a simple code using for loops
问题
这是错误：
在 Project1.exe 中的 0x00007FFD16B7FE7C 处发生未处理的异常: 内存位置 0x0000000B2FAFF760 处的 Microsoft C++ 异常: std::out_of_range。
这是代码：
#include <iostream>
using namespace std;
int main() {
string word;
cin >> word;
cout << word.size() << endl;
for (int i = 0; i <= word.size(); i++)
cout << word.at(i) << endl;
return 0;
}
请问有其他需要吗？
英文:
this is the error :
Unhandled exception at 0x00007FFD16B7FE7C in Project1.exe: Microsoft C++ exception: std::out_of_range at memory location 0x0000000B2FAFF760.
here is the code :
#include <iostream>
using namespace std;
int main() {
string word;
cin >> word;
cout << word.size() << endl;
for (int i = 0; i <= word.size(); i++)
cout << word.at(i) << endl;
return 0;
}
im also still not sure where the code breaks it, it just takes me to a weird window with A LOT OF LINES that i have no idea what they mean.
i tried debugging, rebuilding, trying some alternative ways to write the code that i know of and nothing worked, like the code is so simple HOW AM I OUT OF MEMORY XD.
答案1
得分: 3
在这个for循环中，
for (int i = 0; i <= word.size(); i++)
cout << word.at(i) << endl;
变量i的范围是[0, word.size()]。然而，成员函数at在其参数等于或大于成员函数size()返回的值时会抛出out_of_range异常。
因此，修改for循环如下：
for (int i = 0; i < word.size(); i++)
cout << word.at(i) << endl;
请注意，与成员函数at相反，下标运算符的索引可以等于size()的值。
也就是说，你可以编写：
for (int i = 0; i <= word.size(); i++)
cout << word[i] << endl;
尽管使用索引等于size()的情况没有太大意义。
为了避免这样的错误，最好使用基于范围的for循环，如下：
for (char c : word)
cout << c << endl;
或者
for (const auto &c : word)
cout << c << endl;
英文:
In this for loop
for (int i = 0; i <= word.size(); i++)
cout << word.at(i) << endl;
the variable i is being changed in the range [0. word.size()]. However the member function at throws exception out_of_range when its argument is equal to or greater than the value returned by the member function size().
So change the for loop like
for (int i = 0; i < word.size(); i++)
cout << word.at(i) << endl;
Pay attention to that opposite to the member function at the index of the subscript operator may be equal to the value of size().
That is you may write
for (int i = 0; i <= word.size(); i++)
cout << word[i] << endl;
though there is no great sense to use the index equal to the value of size().
To avoid such a mistake it is better to use range-based for loop like
for ( char c : word )
cout << c << endl;
or
for ( const auto &c : word )
cout << c << endl;
答案2
得分: 1
尝试使用：
for (int i = 0; i < word.size(); i++)
如果word中有size个元素，循环将运行size次，而不是size + 1次。
英文:
Try with:
for (int i = 0; i < word.size(); i++)
If you have size elements in word, the loop will run size times, not size + 1 times.
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