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I'm super new with Swift (and programming). I need to compare if the result of a Remainder Operator (modulo operator) equals an integer.
This is my code:
if Int.random(in: 1...1000) % 4 >= "This number" {
print ("OK")
} else {
print ("WRONG")
}
If "This number" is replaced by an actual number e.g. 100 it works fine.
But if replaced by Int it crashes displaying: Type 'Int.Type' cannot conform to 'BinaryInteger'; only struct/enum/class types can conform to protocols
You have to compare Int with Int, not String or Int type (Int.Type). If you want to declare a value that you will use for comparison you can declare it as variable. To compare if it is equal use == operator.
let thisNumber: Int = 3 // or any other value
if Int.random(in: 1...1000) % 4 == thisNumber {
print ("OK")
} else {
print ("WRONG")
}
now both Int.random(in: 1...1000) % 4 and thisNumber is Int so you can compare them without getting an error.
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I don't know how to word the question to find help in a search engine. Basically I'm trying to figure out the meaning of this operation
self.isFromCurrentUser = fromId == Auth.auth().currentUser?.uid
The variables are NOT the question because I already know about Firebase and whatnot. Rather the question is why is there an assignment operator = and an == in this line of code?
variable1 = variable2 == variable3
What does this mean? What is variable1 getting assigned to?
The statement
variable1 = variable2 == variable
assigns variable1 the value of the boolean (true or false) of the result of variable2 == variable3 (if they are equal).
It's equivalent to
if variable2 == variable3 {
variable1 = true
} else {
variable1 = false
}
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In my while statement, I cannot understand why my output is printed twice ?
I would like to print i only one time, where is my error ?
func fetch2(){
var i: Int = 0
while i <= (self.returned-1) {
let itemLookUp = "https://shopping.yahooapis.jp/ShoppingWebService/V1/json/itemLookup?appid=\(self.appId)&itemcode=\(self.arrayCodeProduct[i])&responsegroup=large"
print(i)
i = i+1
}
}
Here is the output that I obtain :
0
1
2
3
0
1
2
3
Thank you in advance.
It looks like fetch2() is called twice.
Add a print(#function) before you var i and check that fetch2() is not called several times.
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I was studying about passing by reference. It made me wonder what would happen in the following example (Written in pseudo-C which supports "by reference"):
int foo(int a) {
a = 5;
return a * 2;
}
int main() {
int a = 1;
a = foo(a);
printf("%d",a);
return 0;
}
What should be printed? if we only did foo(a); without assigning into a then we would get 5. but what would be printed when assigning? should it be 5 or 10?
Since you have a = foo(a); in your main() function, a will contain the result returned by foo(a). foo(a) will always return 10 no matter what a is.
C does not support pass by reference. Changing a = foo(a); to just foo(a); would mean a would retain the value it had before it was passed to foo(), so it would be 1.
One variation of C that supports pass by reference is C++. In C++, you could write foo() as:
int foo(int &a) {
a = 5;
return a * 2;
}
The int &a syntax is used to denote that the parameter will be passed by reference. Now, foo will assign 5 to the referenced variable, but still always return 10.
In this case a = foo(a); will result in a having the value 10, and foo(a); alone will result in a having the value 5.
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Let's say you have an Int.
let int = 12345
I want it to only display some of the digits.
For example: print(firstTwoDigits) --> 12
How do I do this and thank you in advance.
It depends on your specific requirements.
This prints the first two digits of an integer number
let intVal = 12345
print(String(intVal).prefix(2))
Output: 12
Another way which only prints certain ones in the number:
let intVal = 12345
let acceptableValues = ["1", "2"]
let result = String(intVal).filter {
acceptableValues.contains(String($0))
}
print(result)
Output: 12
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The code below would print abc 5 times and then print 1024. As far as I understand, in any for, the "iterator " is automatically declared (the equivalent of a C(++)/Java for(int i=1; i<=5; i++) ). Is it possible to actually not automatically create that variable and use the one declared before the for so that it would print abc 5 times and then print 5, thus modifying it?
var i = 1024
for i in 1...5 {
print("abc")
}
print(i)
#DrummerB's answer works, but if you want a for...in loop, this will also work. It's the same principle - declare your variable outside the loop and increment it inside it:
var i:Int = 0
for _ in 0...5 {
print("abc")
i += 1
}
print(i)
Since you aren't referencing a loop variable, Swift syntax recommends an underscore.
You could just rewrite the for loop as a while loop like this:
var i = 1024
i = 1
while i <= 5 {
print("abc")
i = i+1
}
print(i)
If you really want to change the value of the i using the loop in that way, you can do:
var i=1024
for j in 1...5 {
print("abc")
i = j
}
print(i)
The j would be used strictly in the loop, so once finished, it's value is dumped. But a variable declared before (i in your case), could take it's value and maintain it.