Some detail description for my question:

1. Create an iterator of vector or string in c 11.
2. Do iterator arithmetic, iterator minus n

Question: Will c 11 keep the iterator minus n bigger than begin()? (if n is big enough, will the compiler ensure that the iter - n do not exceed the legal range of iterator?)

CodePudding user response：

According to cppreference, iter - n is effectively the same as:

vector<T>::iterator temp = iter;
while(n--) --temp;
return temp;

Assuming iter was a iterator from a vector named container. If n is larger than distance(container.begin(), iter), then at some point from the last while loop, --temp would be equivalent of:

--container.begin();

And according to cppreference, that line would be undefined behavior.

Since an iterator cannot know any information of the originated container, it does not have a way to detect if it is currently container.begin(), thus it cannot ensure it to be remained in the legal range without manually checking against the range.

CodePudding user response：

It will not, it will simply perform the arithmetic operations and print the values in negative number.

Performed the code using VSCode.

#include <conio.h>
#include <iostream>
#include <vector> 
using namespace std;

int main() {
    vector<int> v = {1,2,3,4,5,6};
    vector<int>::iterator it;

    for(it = v.begin(); it != v.end(); it  ) {
        cout << *it-6 << endl;
    }

    return 0;
}

and the results were :

 -5
 -4
 -3
 -2
 -1
  0

CodePudding user response：

It will not! I wrote following code, then the VS2019 reported "degug assert failed"!

std::vector<int> v1 = { 1,2,3,4,5,6,7 };
std::vector<int> v_assign;
v_assign.assign(v1.begin() - 2, v1.begin()   3);

CodePudding user response：

No, iterator arithmetic will not do any bounds checking on the result. You can easily end up with an invalid iterator.

A random access iterator such as given by std::vector will allow you to do subtraction, so you can impose your own bounds checking.

it2 = it - min(n, it - container.begin());