在C ++中是否有“循环遍历”数组的最佳实践？

Question

In Python, one can compare each element of an array with the "next" element (including the last element with the first element) with the following code:

a = [0, 1, 2, 3]
for i in range(-1, 3):
    if a[i] + 1 >= a[i+1]:    
        print a[i], 'works'

The output:

3 works
0 works
1 works
2 works

If I want to compare the first element of an array with the second, the second with the third, etc., and finally the last with the first , I can do so with just a loop in Python.

Can I do this in C++? Can I loop though elements in this manner whilst staying entirely in one loop? To further illustrate what I mean, here is some C++ code that has the same functionality as the above Python code.

int a[] = {0, 1, 2, 3};
std::cout a[3] << std::endl;
for(int i = 0; i < 2; i++)
    std::cout << a[i] << std::endl;

It has the same output. My point is, for certain algorithms, I'd have to manually duplicate the contents of my for loops for certain steps. For example, if I want to see if the first element equals the second, the second the third, etc., and finally if the last equals the first, I'd have to manually add this step after or before my for loop. --- if (a[0] == a[3]) foo();

Is this what I am supposed to do in C++? I am quite new to it and I don't want to get rooted in bad practices.

Answer 1

for (int i=3; i<(3 + size_of_array); ++i)
    std::cout << a[i % size_of_array] << '\n';

Answer 2

Use std::begin() and std::end() algorithm.

int a[3]={1,2,3};
for( auto x=end(a)-1; x>=begin(a); --x){
    cout<<*x<<endl;
}

The above code outputs 3 2 1 , the revrese order of the array a . Have a look at [begin()][1] .