Is The for Loop Condition Evaluated Each Loop in Swift

Swift: For Loop End Condition Evaluated Only Once?

0 ..< self.numberOfRows is a Range and in particular a Sequence. Iterating over a sequence is done by creating an iterator, and then calling its next() method until the iterator is exhausted, compare IteratorProtocol:

Whenever you use a for-in loop with an array, set, or any other collection or sequence, you’re using that type’s iterator. Swift uses a sequence’s or collection’s iterator internally to enable the for-in loop language construct.

for i in 0 ..< self.numberOfRows {
    ...
}

is equivalent to

let range = 0 ..< self.numberOfRows
var it = range.makeIterator()
while let i = it.next() {
    ...
}

Modifying numberOfRows during the iteration does not mutate the range (which is a value type) or the iterator, and therefore does not affect the number of iterations.

does Swift array.count get evaluated each time in a loop

while loops (and do while loops) have their predicates evaluated on each iteration.

for loops evaluate the sequences once.

Here's is a demonstration:

var array: [Int]

print("Test Case 1 - while i < array.count")
array = [1, 2, 3, 4, 5, 6]
var i = 0
while i < array.count {
    print(array[i])
    if i < 3 { array.append(123) }
    i += 1
}
print("\r\nTest Case 2 - for i in array.indices")
array = [1, 2, 3, 4, 5, 6]
for i in array.indices {
    print(array[i])
    if i < 3 { array.append(123) }
}

print("\r\nTest Case 3 - for i in 0 ..< array.count")
array = [1, 2, 3, 4, 5, 6]
for i in 0 ..< array.count {
    print(array[i])
    if i < 3 { array.append(123) }
}

Test Case 1 - while i < array.count
```
1
2
3
4
5
6
123
123
123
```
Test Case 2 - for i in array.indices
```
  1
  2
  3
  4
  5
  6
```
Test Case 3 - for i in 0 ..< array.count
```
1
2
3
4
5
6
```

In a loop, do any operations in the end-condition get evaluated in every iteration?

It obviously depends on the language. For JavaScript, the spec (ECMAScript §12.6.3) requires it always be evaluated each time. As an optimization, a specific JavaScript runtime could skip one or more of the length calls, if it could prove that the result would not change.

For Loop condition

The all_mem macro is returning a size_t value; integer promotion rules mean the comparison of i <= (all_mem - 2) is promoting i to a size_t, which means the value is huge, rather than -1. Try casting to ensure signed comparison:

for(i = -1; i <=(ssize_t)(all_mem- 2); ++i)

Does a C++11 range-based for loop condition get evaluated every cycle?

It is only evaluated once. The standard defines a range-based for statement as equivalent to:

{
    auto && __range = range-init;
    for ( auto __begin = begin-expr, __end = end-expr; __begin != __end; ++__begin ) {
        for-range-declaration = *__begin;
        statement
    }
}

where range-init is the expression (surrounded by parentheses) or braced-init-list after the :

how for loop is getting evaluated when printf is used inside the loop in C?

Before going to the actual problem you should know how the for loop works in steps:
For loop has 4 steps that is :

For example:

for(i=0;i<10;i++){

statement;

}

Step 1: first it will initialize i as 0;
Step 2: It will check the condition is satisfying or not;
Step 3: The it will go to the statement block and then it will perform the statement
step 4: It will increment the value .

But Here in your Question:

 #include<stdio.h>  
    int main(){  
    int x=1,y=1;  
    for(;y;printf("%d %d\n",x,y))
    y=x++<=5;
    }

Step 1 : It enter into the loop(i.e. 1 because it is blank)

Step 2 : Check the condition block which is one(i.e the y value is one)

Step 3 : Here first the post increment Logic is implemented.

First the x value is initialized that is one and the incremented.
The the condition x++<=5 is checked.

If x++<=5 it will initialize y to the value of 1 because the condition satisfy
else it will initialize y to the value of 0 because the condition not satisfy.

****In First step of loop
it will check 1 <= 5 true
so y = 1 and so on till the condition satisfy otherwise y = 0****

Step 4:It will print the value of x and y

Next : Again the pointer will go to condition block and then goes to second because y = 1
when y = 0 the loop will stop;
**It will not go to the initialization step from the second loop....