Python 3 for f, b in zip(foo, bar): print(f, b) zip stops when the shorter of foo or bar stops. In Python 3, zip returns an iterator of tuples, like itertools.izip in Python2. To get a list of tuples, use list(zip(foo, bar)). And to zip until both iterators are exhausted, you would use itertools.zip_longest. Python … Read more
Not only the commented line, but another will also fail if there is space in the begin of string. First line fail (to compile) because string::find_first_not_of return size_t. And construct string from two size just make no sense. Second line may fail because string::substr accept length (not end) as it’s second parameter.
One possible solution: #include <iostream> using namespace std; int main() { float a[] = { 1.5, 55, 5, 4.4,7.1,9,8,2,4.2, -3, -3.2, 5.2, -9 }; float sumofnums = 0; int size = sizeof(a) / sizeof(a[0]); // size is 13 here, this allows you // to determine the size automatically int first = 0; int last = … Read more
This might be a wrong answer to your question. I’ll remove it in that case, but maybe you can use it for what you want to achieve: int[][] it = {{1,2}, {3,4,5}}; OfInt iterator = Arrays.stream(it).flatMapToInt(x -> IntStream.of(x)).iterator(); iterator.forEachRemaining((IntConsumer) System.out::print); Stream the jagged array, flatmap it into one single IntStream and then do what you … Read more