在遍历字符串时，即使我已经超出了长度，为什么还没有超出范围错误？

Question

My code is literally this:

int main(){
    string s = "Success!\n";
    for (int i=0; i<10; ++i) cout<<s[i];
    return 0;
}

string s is of length 8; and even though I've used an iterator 'i' that goes beyond the string's length, I do not get any error. Why? Output is:

Success!

WITH A NEWLINE, I have not even put a newline character

Answer 1

string s is of length 8

no it's not, it's 9.

and even though I've used an iterator 'i' that goes beyond the string's length, I do not get any error

You're not going beyond the string's length, you go equal to the string's length. This is defined as returning a null character. Even if you went beyond the string's length with operator[] , this is undefined behavior , not expected to throw an exception.

I have not even put a newline character

... yes you did.

Answer 2

As per the definition of subscripting (bracket) operator for string :

char& operator[] (size_t pos);

If pos is equal to the string length, the function returns a reference to the null character that follows the last character in the string (which should not be modified).

"Success!\\n" is a string of length 9. With S at index 0, ! at index 7, and \\n at index 8. So when you reference s[9], it's returning back the null \\0 char.

Answer 3

代替cout << s[i]使用cout << s.at(i) ，您将获得所需的结果。