'C++11 range based loop: How does it really work
I know how this loop works, and how I can use it in practical problems. But I want to know what is happening under the hood.
I thought that this loop was similar to a regular for loop in which for example
for(int i = 0 ; i < 5 ; i ++){
// instructions
}
Variable i is initialized only once, so I thought that this was the same for range based loops. But if I for example write this code:
for(const int x : vec) {
cout << x << endl;
}
The compiler lets me to do this, but I don't understand how this is possible. If variable x is const, how come in every iteration the x value is different?
Solution 1:[1]
Every iteration of the loop creates a local variable x and initializes it to the next element of vec. When the loop iteration ends, x goes out of scope. A single x is never modified.
See this link for the precise semantics.
Solution 2:[2]
The range-based for-loop is indeed somewhat different than the classical for-loop in this regard. The declaration you provide (const int x) is declared for every iteration separately, in contrast to classical for-loops.
To be more precise:
for (const int x : vec) {
cout << x << endl;
}
is just a shorthand for (and simply replaced with) the following "classical iterator loop":
for (auto it = vec.begin(), e = vec.end(); it != e; ++it) {
const int x = *it;
cout << x << endl;
}
(except that it and e are not available in the body; also vec is actually "saved" in a separate variable; but let's not focus on unimportant details here; the exact definition of range-based for loop can be looked up here)
Note that const int x is declared and initialized to *it inside the loop body! So it is initialized in every iteration, rather than changed.
Solution 3:[3]
For unterstanding purpose you can think of it as if the compiler is replacing for (auto x: y) {...} with for (auto i = begin(y), end = end(y); i != end; ++i) { auto x = *i; {...} }.
For std::vector begin(y)/end(y) will resolve (via the adl) to std::begin(y)/std::end(y) versions that would call y.begin()/y.end() respectively.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | John Smith |
| Solution 2 | |
| Solution 3 |
