For loops are used by every developer regularly. There is for-in as well as a for-each loop in Swift which has a bit different syntaxes. Both of them are used to iterate over a range, array, set or dictionary but have a bit different syntaxes.
While comparing with other languages, I realized in Swift, there are few concepts which are completely different from java or any other another language for loops.
Wait! They are not this tough. In fact, they are very easy, interesting and helpful.
Let’s check one by one.
1. Simple for loop in java that iterates from some number to some number incrementing one on each loop pass.
/*Java code*/
for (int i = 0; i <= 10; i++){
System.out.print(i);
}
its equivalent Swift code
/*Swift code*/
for i in 1...10 {
print(i)
}
Things to notice
- No need to declare the data type of variable.
- If iterating over a range, we can use
...(closed range) operator. - The lower and upper (including) limit can be defined on both the sides of
...operator.
2. Now let’s say if I don’t don’t want to include the upper limit in the loop and break the loop if it hits the upper limit.
/*Java code*/
for (int j = 0; j < 10; j++) {
System.out.print(j); //this will print only up to 9
}
We can use the half-open range operator (..<).
/*Swift code*/
for i in 1..<10 {
print(i)
}
3. I want to increment it by 2 or some other number.
/*Java code*/
for (int i = 0; i <= 10; i += 2) {
System.out.print(i);
}
We can use stride function here
/*Swift code*/
for i in stride(from: 0, through: 10, by: 2) {
print(i)
}
//this will run till last value, in our case till 10
for i in stride(from: 0, to: 10, by: 2) {
print(i)
}
//this will run less then last value, in our case till 8
4. Wait, what if I want to run the loop in reverse order. Can I use 10...1 ?
/*Java code*/
for (int i = 10; i > 0; i--) {
System.out.print(i);
}
No, you can not use 10...1 directly as ... operator never works on the reverse ranges.
It will give you below run time error.
//Error
error: Execution was interrupted, reason: EXC_BAD_INSTRUCTION (code=EXC_I386_INVOP, subcode=0x0).
The process has been left at the point where it was interrupted, use "thread return -x" to return to the state before expression evaluation.
You have to use reversed() function.
/*Swift code*/
for i in (0...10).reversed() {
print(i)
}
You can also change the step size with stride function.
/*Java code*/
for (int i = 10; i > 0; i -= 2) {
System.out.print(i);
}
/*Swift code*/
for i in stride(from: 0, to: 10, by: 2).reversed() {
print(i)
}
for i in stride(from: 0, through: 10, by: 2).reversed() {
print(i)
}
But please note that stride function overload to and through works on upper range, in both nomral as well as in reversed case.
5. What if I have a complex calculation instead of addition or subtraction in each step. Let’s say multiplication or division.
/*Java code*/
for (int k = 2; k <= 256; k *= 2) {
System.out.print(k);
}
Move to while loop, no other way
/*Swift code*/
var k = 2
while (k <= 256) {
print(k)
k *= 2
}
6. I want to iterate over an array now.
/*Java code*/
int[] arr = new int[5];
for (int item: arr) {
System.out.print(item);
}
Simple, use the in in Swift
/*Swift code*/
let arr = [0,1,2,3,4]
for element in arr {
print(element)
}
7. I want to use the position (index) also. How can I do that?
/*Java code*/
int[] arr = new int[5];
for (int i = 0; i < arr.length; i++) {
System.out.print(arr[i]);
}
Just use enumerated function.
/*Swift code*/
let arr = [0,1,2,3,4]
for (index, element) in arr.enumerated() {
print("The element at \(index) is \(element)")
}
8. I heard about some for foreach also. Can I use the same in Swift?
/*Java code*/
let arr = [0,1,2,3,4]
for element in arr {
print(element)
}
Just use foreach function.
/*Swift code*/
let arr = [0,1,2,3,4]
arr.forEach{ element in
print(element)
}
9. And what about Dicionary?
/*Java code*/
HashMap<String,String> map = new HashMap();
map.put("Suneet","Engineering");
map.put("Ballu","Sales");
map.put("John","Marketing");
for (Map.Entry<String, String> entry : map.entrySet()) {
System.out.println(entry.getKey() + " is working in " + entry.getValue());
}
Simple. Use in based iteration.
/*Swift code*/
let employee = ["Suneet": "Engineering", "Ballu": "Sales", "John": "Marketing"]
for (name, department) in employee {
print("\(name) is working in \(department) department")
}
No, I am a fan of foreach loop.
Ok, no problem, there you go.
/*Swift code*/
let employee = ["Suneet": "Engineering", "Ballu": "Sales", "John": "Marketing"]
employee.forEach { name, department in
print("\(name) is working in \(department) department")
}
And we are done. See I told you this will be very easy and interesting.