Good morning. Sorry for the indelicate question, but how to create an expression that is not greater than 2 and less than 1 in Objective C
My code dosn't work
if([(UIPinchGestureRecognizer*)sender scale]<=2.0 || [(UIPinchGestureRecognizer*)sender scale]>=1.0)
|| is the operator for logical OR, which isn't what you want. You need &&, the operator for logical AND.So now your code will look like this:
if([(UIPinchGestureRecognizer*)sender scale]<=2.0 && [(UIPinchGestureRecognizer*)sender scale]>=1.0)
Let's break that down...
an expression that is not greater than 2
if (! (someValue > 2)) {
// someValue is not greater than 2
}
However, "not greater than 2" is the same thing as "less-than-or-equal to 2" so...
if (someValue <= 2) {
// someValue is not greater than 2
}
Now, for the second part...
an expression that is less than 1
if (someValue < 1) {
// someValue is less than 1
}
And...
an expression that is not greater than 2 and less than 1.
if ((someValue <= 2) && (someValue < 1)) {
// someValue is not greater than 2 and less than 1
}
However, if you think about it, any number that is less than 1 will also be "not greater than 2" so you don't even need that part.
if (someValue < 1) {
// someValue is less than 1... and it is also not greater than 2
}
Related
sorry for asking this question but I couldn't understand it
-- but i don't understand this code
ballDX = math.random(2) == 1 and 100 or -100
--here ballDY will give value between -50 to 50
ballDY = math.random(-50, 50)
I don't understand the structure what is (2) and why it's == 1
Thank you a lot
math.random(x) will randomly return an integer between 1 and x.
So math.random(2) will randomly return 1 or 2.
If it returns 1 (== 1), ballDX will be set to 100.
If it returns 2 (~= 1), ballDX will be set to -100.
A simple way to make a 50-50 chance.
That is a very common way of assigning variables in Lua based on conditionals. It’s the same you’d do, for example, in Python with “foo = a if x else b”:
The first function, math.random(2), returns either 1 or 2. So, if it returns 1 the part math.random(2) == 1 is true and so you assign 100 to the variable ballDX. Otherwise, assign -100 to it.
In lua
result = condition and first or second
basically means the same as
if condition and first ~= nil and first ~= false then
result = first
else
result = second
end
So in your case
if math.random(2) == 1 then
ballDX = 100
else
ballDX = -100
end
in other words, there is a 50/50 chance for ballDX to become 100 or -100
For a better understanding, a look at lua documentation helps a lot :
https://www.lua.org/pil/3.3.html
You can read:
The operator or returns its first argument if it is not false; otherwise, it returns its second argument:
So if the random number is 1 it will return the first argument (100) of the "or" otherwise it will return the second argument (-100).
I've take a break since I posted this and have read through half of the C programming book I'm studying (Harvard cs50 book). I should be able to solve this by now, yet am unable.
The program runs in a continuous loop, no matter what integer is entered; prints "Good for you..." ad infinitum.
Example code:
//example 3 version2 from chapter 11, beginner programming in c
#include <cs50.h>
#include <stdio.h>
int main ()
{
int prefer;
printf("On a scale from 1 to 10, how happy are you?\n");
scanf(" %d", &prefer);
while(prefer >= 1 || prefer <= 10)
//goal is for program to run while entered int "prefer" is between 1 - 10
if (prefer > 10)
{
printf("Oh really, now? Can't follow simple directions, can you?\n");
printf("want to try that again? 1 through 10...?\n");
scanf(" %d", &prefer);
}
else if (prefer >= 8)
{
printf("Good for you!\n");
}
else if (prefer <= 5)
{
printf("Cheer up : )\n");
}
else if (prefer <= 3)
{
printf("Cheer up, Buttercup!\n");
}
else
{
printf("Get in the RIVER with that attitude!\n");
}
return 0;
}
Operator < and && are binary operators. When we use them, it compares the left and right side values. The above while would look like this.
while(prefer <= 10 && prefer > 0);
This is the loop in my app:
for var i = column - 1; i >= 0 && burgers[i, row]?.burgerType == burgerType;
i -= 1, horzLength += 1 {
}
What would be the best way to implement this loop in Swift 2.2.1?
Try this:
var i = column - 1
while i >= 0 && burgers[i, row]?.burgerType == burgerType {
i -= 1
horzLength += 1
}
This sort of abuse of the for loop syntax was the exact reason it was deprecated in Swift 2.2. Even if a for syntax was available, this would still be more clear than that abomination
The program works fine with var dig = 0 and it doesn't work with var dig:Int I get an error: Variable "dig" used before being initialized Could you explain me why?
func myFunc(a:Int, b:Int) {
var c = a / b
var o = a % b
var v = 0
var dig = 0
if o != 0 {println("\(a)/\(b) = \(c) и \(o)/\(b)")}
else {println("\(a)/\(b) = \(c)")}
if a > b {
v = b
}
else {
v = a
}
for var i = 1; i <= v; ++i {
if a % i == 0 && b % i == 0 {dig = i}
}
println("\(dig) - greatest common denominator of \(a) and \(b)")
}
myFunc(27,81)
The only place you set the value of dig is inside of an if statement that is inside of a for loop. The Swift compiler does not know if the body of the for loop will be executed, and it doesn't know if the if statement will ever be true, so it has to assume that there is a path in which dig is not initialized.
Consider this simpler example:
func myFunc(a:Int, b:Int) {
var dig: Int
if a >= b {
dig = 3
}
if a < b {
dig = 4
}
println("\(dig) - greatest common denominator of \(a) and \(b)")
}
This example also gives the same error, because Swift considers each if separately. It is obvious to us that a is either greater than or equal to b or it is less than b, but Swift doesn't go that far in evaluating the situation. It just considers that each if may not be true, and dig is only set inside of ifs, so it is possible (as far as Swift is concerned) that dig may not be set.
func myFunc(a:Int, b:Int) {
var dig: Int
if a >= b {
dig = 3
} else {
dig = 4
}
println("\(dig) - greatest common denominator of \(a) and \(b)")
}
If you change the second condition to an else, Swift is then happy because it can reason that the if must be true or false and dig is set in each path, so it will certainly have a value before the println statement.
The compiler does not know mathematics good enough to
recognize that the statement
if a % i == 0 && b % i == 0 {dig = i}
is actually executed at least once (for i == 1). Therefore
the compiler assumes that dig might be undefined at
println("\(dig) - greatest common denominator of \(a) and \(b)")
Assigning an initial value in
var dig = 0
is the correct solution.
Btw., the Euclidean Algorithm is a much more effective method to
compute the greatest common divisor, see for example
http://rosettacode.org/wiki/Greatest_common_divisor#Swift.
This question already has answers here:
What is a simple example of floating point/rounding error?
(9 answers)
Why is Lua arithmetic is not equal to itself? [duplicate]
(1 answer)
Closed 9 years ago.
Today morning I found a bug on my Lua Script, wich seem very weird. How can this evaluation fail this way? Examples can be tested in here
First example:
if( math.abs(29.7 - 30) <= 0.3 ) then
result = 1
else
result = 0
end
print("result = "..result )
-->> result = 0
Second example:
if( 0.3 <= 0.3 ) then
result = 1
else
result = 0
end
print("result = "..result )
-->> result = 1
Third example
if( math.abs(29.7-30) == 0.3 )then
print("Lua says: "..math.abs(29.7-30).." == 0.3")
else
print("Lua says: "..math.abs(29.7-30).." ~= 0.3")
end
-->> Lua says: 0.3 ~= 0.3 WHAT?
I am really confuse, and I would like to understand this to avoid similiar bugs in the future. Thanks
You are being hit by the fact that Lua uses (IEEE 754) 64-bit double-precision floating point numbers.
Look at the following examples
> print(0.3 == 0.3)
true
> print(0.3 <= 0.3)
true
> print(0.3 >= 0.3)
true
The actual value of 0.3 in memory is:
> print(string.format("%1.64f",math.abs(-0.3)))
0.2999999999999999888977697537484345957636833190917968750000000000
Now look at you example:
> print(math.abs(29.7-30) == 0.3)
false
> print(math.abs(29.7-30) >= 0.3)
true
> print(math.abs(29.7-30) <= 0.3)
false
The actual value of 29.7-30 is:
> print(string.format("%1.64f",29.7-30))
-0.3000000000000007105427357601001858711242675781250000000000000000
The actual value of math.abs(29.7-30) is:
> print(string.format("%1.64f", math.abs(29.7-30))
0.3000000000000007105427357601001858711242675781250000000000000000
And just for fun the value of math.abs(-0.3) is:
> print(string.format("%1.64f", math.abs(-0.3)))
0.2999999999999999888977697537484345957636833190917968750000000000
There are two solutions to you problem, the first is read What Every Computer Scientist Should Know About Floating-Point Arithmetic, and understand it :-). The second solution is to configure Lua to use another type for numbers, see Values and Types for hints.
Edit
I just thought of another way of "solving" the problem, but it is a bit of a hack, and not guarantied to always work. You can use fixed point numbers in lua by first converting the float to a string with a fixed precision.
In your case that would look something like:
a = string.format("%1.1f", math.abs(29.7 - 30))
print(a == "0.3")
or a bit more robust:
a = string.format("%1.1f", math.abs(29.7 - 30))
print(a == string.format("%1.1f", 0.3))
However you must make sure that you use a precision that is both adequate and the same for all you comparisons.
As we know, float point has a precision problem
Refer: http://lua-users.org/wiki/FloatingPoint
a = 1
if a < 1 then print("<1") end
Will never print "<1". Not unless you actually change a