I recently created a function that checks if a pair exists in my swap smart contract.
The functoin looks like this:
function checkIfPairExists(address _token1, address _token2) internal returns(uint, bool) {
for (uint index = 0; index < tokenPairs.length; index++) {
if (tokenPairs[index].token1 == _token1 && tokenPairs[index].token2 == _token2) {
return (index, true);
}
else if (tokenPairs[index].token2 == _token1 && tokenPairs[index].token1 == _token2) {
return (index, true);
} else {
return (0, false);
}
}
}
This function works fine but then when I try to use the function in an if statement like this:
if (checkIfPairExists(_token1, _token2) == (uint256, true))
How do I write it so it is correct? I am trying to receive index of the pair for my array and bool to see if the pair exists. I then need to save that index to find which pair it should add to.
Hope it makes sense.
Let me know if I should rephrase the question so more people will understand it and it can help them.
Thanks
You need to assign the returned values to two separate variables. Then you can validate either of them.
(uint256 index, bool exists) = checkIfPairExists(_token1, _token2);
if (exists == true) {
// do something with `index`
}
As said in the above answer by #pert-hejda, you will need to assign the function return values then you can use those to check the condition. Why? Because multiple returns are represented as tuples and currently the feature you want is not supported in solidity. So, you will need to assign the return values and use those values in conditionals. Thank you.
I am getting the below error while trying to redact pdf document using itext7
I am calling pdfCleanupTool.cleanup() method for redaction and sometimes I am getting the below error from the cleanup method:
Index was out of range. Must be non-negative and less than the size of the collection.\r\nParameter name: index
Any help appreciated.
Thanks!
Updates:
Error Log:
There is a bug in the iText 7 PdfTextArray class which generates stack traces like yours. As you don't share your PDF, though, I cannot be sure whether that's the bug bothering you currently.
The Bug
The bug can be provoked quite easily, in Java like this
PdfTextArray textArray = new PdfTextArray();
textArray.add(1);
textArray.add(-1);
textArray.add(1);
(CancelingAdjustments test testCancelingAdjustments)
and similarly in C#.
This essentially may be what happens in the OP's case; redaction involves removal of text pieces from such text arrays and replacement by equivalent numeric adjustments, so such situations may be more probable during redaction than in general.
The Cause
When adding multiple numbers to a PdfTextArray, it attempts to combine them to a single number, and if that single number is zero, remove it altogether:
public boolean add(float number) {
// adding zero doesn't modify the TextArray at all
if (number != 0) {
if (!Float.isNaN(lastNumber)) {
lastNumber = number + lastNumber;
if (lastNumber != 0) {
set(size() - 1, new PdfNumber(lastNumber));
} else {
remove(size() - 1);
}
} else {
lastNumber = number;
super.add(new PdfNumber(lastNumber));
}
lastString = null;
return true;
}
return false;
}
(PdfTextArray method add)
But this code forgets to reset the lastNumber variable to "not a number" after removal due to cancelation. Thus, this bug can be fixed like this:
public boolean add(float number) {
// adding zero doesn't modify the TextArray at all
if (number != 0) {
if (!Float.isNaN(lastNumber)) {
lastNumber = number + lastNumber;
if (lastNumber != 0) {
set(size() - 1, new PdfNumber(lastNumber));
} else {
remove(size() - 1);
lastNumber = Float.NaN;
}
} else {
lastNumber = number;
super.add(new PdfNumber(lastNumber));
}
lastString = null;
return true;
}
return false;
}
(One could improve this some more by testing whether there is some string at the now last position of the array and initialize lastString accordingly.)
The iText/.Net code is very similar here.
A couple of days ago I had an interview for an internship and I've been asked to write a function that tests if another function (which was adding 2 numbers together) is working. My answer was something similar to this:
bool addTest(double a, double b){
if((a+b) == addFunction(a, b))
return true;
else
return false;
}
However, they did not seem impressed with my answer. How else could I do it?
Test cases should be written with parameters and expected answers hard coded into them to eliminate any uncertainty.
bool addTest(){
//2+2=4
if (addfunction(2,2) !=4){
return false;
}
//1.66666+7.9=9.56666
if (addFunction(1.66666, 7.9) != 9.56666){
return false;
}
//0+0=0
if (addFunction(0,0) != 0){
return false;
}
//all tests passed
return true;
}
If a bug is found later on in the development, the conditions that caused the bug should be added to the test function. For example, one day you try to add two negative numbers, only to find they return a positive number. Adding a test case for this will ensure this bug is caught again if the function changes in the future.
//addFunction(-8, -1) //Oops! We return positive 9! Let's fix the bug and make a new test
//-8 + -1 = -9
if (addFunction(-8, -1) != -9){
return false;
}
It also wouldn't hurt to add test cases that verify two values don't produce a wrong answer, though this is harder to cover all test cases.
//-8 + -1 = -9
if (addFunction(-8, -1) == 9){
return false;
}
i was looking some examples of interactions with the keyboard and stumbled upon this code that i found interesting. But i'm having trouble understanding a certain part of it(it's marked down below).I don't get how all this whole ''boolean'' declaration, ''switch'' and ''CASE'' works, i tried to look in the reference but still. Could someone explain in a simple maner how these work?
float x = 300;
float y = 300;
float speed = 5;
boolean isLeft, isRight, isUp, isDown;
int i = 0;
void keyPressed() {
setMove(keyCode, true);
if (isLeft ){
x -= speed;
}
if(isRight){
x += speed;
}
}
void keyReleased() {
setMove(keyCode, false);
}
boolean setMove(int k, boolean b) {// <<<--- From this part down
switch (k) {
case UP:
return isUp = b;
case DOWN:
return isDown = b;
case LEFT:
return isLeft = b;
case RIGHT:
return isRight = b;
default:
return b; }
}
Questions like these are best answered by the reference:
Works like an if else structure, but switch() is more convenient when you need to select between three or more alternatives. Program controls jumps to the case with the same value as the expression. All remaining statements in the switch are executed unless redirected by a break. Only primitive datatypes which can convert to an integer (byte, char, and int) may be used as the expression parameter. The default is optional.
The rest of the code is setting the corresponding variable to whatever value you passed in as the b parameter, and then returning it.
You should get into the habit of debugging your code. Add print statements to figure out exactly what the code is doing.
I'm writing a function to find triangle numbers and the natural way to write it is recursively:
function triangle (x)
if x == 0 then return 0 end
return x+triangle(x-1)
end
But attempting to calculate the first 100,000 triangle numbers fails with a stack overflow after a while. This is an ideal function to memoize, but I want a solution that will memoize any function I pass to it.
Mathematica has a particularly slick way to do memoization, relying on the fact that hashes and function calls use the same syntax:
triangle[0] = 0;
triangle[x_] := triangle[x] = x + triangle[x-1]
That's it. It works because the rules for pattern-matching function calls are such that it always uses a more specific definition before a more general definition.
Of course, as has been pointed out, this example has a closed-form solution: triangle[x_] := x*(x+1)/2. Fibonacci numbers are the classic example of how adding memoization gives a drastic speedup:
fib[0] = 1;
fib[1] = 1;
fib[n_] := fib[n] = fib[n-1] + fib[n-2]
Although that too has a closed-form equivalent, albeit messier: http://mathworld.wolfram.com/FibonacciNumber.html
I disagree with the person who suggested this was inappropriate for memoization because you could "just use a loop". The point of memoization is that any repeat function calls are O(1) time. That's a lot better than O(n). In fact, you could even concoct a scenario where the memoized implementation has better performance than the closed-form implementation!
You're also asking the wrong question for your original problem ;)
This is a better way for that case:
triangle(n) = n * (n - 1) / 2
Furthermore, supposing the formula didn't have such a neat solution, memoisation would still be a poor approach here. You'd be better off just writing a simple loop in this case. See this answer for a fuller discussion.
I bet something like this should work with variable argument lists in Lua:
local function varg_tostring(...)
local s = select(1, ...)
for n = 2, select('#', ...) do
s = s..","..select(n,...)
end
return s
end
local function memoize(f)
local cache = {}
return function (...)
local al = varg_tostring(...)
if cache[al] then
return cache[al]
else
local y = f(...)
cache[al] = y
return y
end
end
end
You could probably also do something clever with a metatables with __tostring so that the argument list could just be converted with a tostring(). Oh the possibilities.
In C# 3.0 - for recursive functions, you can do something like:
public static class Helpers
{
public static Func<A, R> Memoize<A, R>(this Func<A, Func<A,R>, R> f)
{
var map = new Dictionary<A, R>();
Func<A, R> self = null;
self = (a) =>
{
R value;
if (map.TryGetValue(a, out value))
return value;
value = f(a, self);
map.Add(a, value);
return value;
};
return self;
}
}
Then you can create a memoized Fibonacci function like this:
var memoized_fib = Helpers.Memoize<int, int>((n,fib) => n > 1 ? fib(n - 1) + fib(n - 2) : n);
Console.WriteLine(memoized_fib(40));
In Scala (untested):
def memoize[A, B](f: (A)=>B) = {
var cache = Map[A, B]()
{ x: A =>
if (cache contains x) cache(x) else {
val back = f(x)
cache += (x -> back)
back
}
}
}
Note that this only works for functions of arity 1, but with currying you could make it work. The more subtle problem is that memoize(f) != memoize(f) for any function f. One very sneaky way to fix this would be something like the following:
val correctMem = memoize(memoize _)
I don't think that this will compile, but it does illustrate the idea.
Update: Commenters have pointed out that memoization is a good way to optimize recursion. Admittedly, I hadn't considered this before, since I generally work in a language (C#) where generalized memoization isn't so trivial to build. Take the post below with that grain of salt in mind.
I think Luke likely has the most appropriate solution to this problem, but memoization is not generally the solution to any issue of stack overflow.
Stack overflow usually is caused by recursion going deeper than the platform can handle. Languages sometimes support "tail recursion", which re-uses the context of the current call, rather than creating a new context for the recursive call. But a lot of mainstream languages/platforms don't support this. C# has no inherent support for tail-recursion, for example. The 64-bit version of the .NET JITter can apply it as an optimization at the IL level, which is all but useless if you need to support 32-bit platforms.
If your language doesn't support tail recursion, your best option for avoiding stack overflows is either to convert to an explicit loop (much less elegant, but sometimes necessary), or find a non-iterative algorithm such as Luke provided for this problem.
function memoize (f)
local cache = {}
return function (x)
if cache[x] then
return cache[x]
else
local y = f(x)
cache[x] = y
return y
end
end
end
triangle = memoize(triangle);
Note that to avoid a stack overflow, triangle would still need to be seeded.
Here's something that works without converting the arguments to strings.
The only caveat is that it can't handle a nil argument. But the accepted solution can't distinguish the value nil from the string "nil", so that's probably OK.
local function m(f)
local t = { }
local function mf(x, ...) -- memoized f
assert(x ~= nil, 'nil passed to memoized function')
if select('#', ...) > 0 then
t[x] = t[x] or m(function(...) return f(x, ...) end)
return t[x](...)
else
t[x] = t[x] or f(x)
assert(t[x] ~= nil, 'memoized function returns nil')
return t[x]
end
end
return mf
end
I've been inspired by this question to implement (yet another) flexible memoize function in Lua.
https://github.com/kikito/memoize.lua
Main advantages:
Accepts a variable number of arguments
Doesn't use tostring; instead, it organizes the cache in a tree structure, using the parameters to traverse it.
Works just fine with functions that return multiple values.
Pasting the code here as reference:
local globalCache = {}
local function getFromCache(cache, args)
local node = cache
for i=1, #args do
if not node.children then return {} end
node = node.children[args[i]]
if not node then return {} end
end
return node.results
end
local function insertInCache(cache, args, results)
local arg
local node = cache
for i=1, #args do
arg = args[i]
node.children = node.children or {}
node.children[arg] = node.children[arg] or {}
node = node.children[arg]
end
node.results = results
end
-- public function
local function memoize(f)
globalCache[f] = { results = {} }
return function (...)
local results = getFromCache( globalCache[f], {...} )
if #results == 0 then
results = { f(...) }
insertInCache(globalCache[f], {...}, results)
end
return unpack(results)
end
end
return memoize
Here is a generic C# 3.0 implementation, if it could help :
public static class Memoization
{
public static Func<T, TResult> Memoize<T, TResult>(this Func<T, TResult> function)
{
var cache = new Dictionary<T, TResult>();
var nullCache = default(TResult);
var isNullCacheSet = false;
return parameter =>
{
TResult value;
if (parameter == null && isNullCacheSet)
{
return nullCache;
}
if (parameter == null)
{
nullCache = function(parameter);
isNullCacheSet = true;
return nullCache;
}
if (cache.TryGetValue(parameter, out value))
{
return value;
}
value = function(parameter);
cache.Add(parameter, value);
return value;
};
}
}
(Quoted from a french blog article)
In the vein of posting memoization in different languages, i'd like to respond to #onebyone.livejournal.com with a non-language-changing C++ example.
First, a memoizer for single arg functions:
template <class Result, class Arg, class ResultStore = std::map<Arg, Result> >
class memoizer1{
public:
template <class F>
const Result& operator()(F f, const Arg& a){
typename ResultStore::const_iterator it = memo_.find(a);
if(it == memo_.end()) {
it = memo_.insert(make_pair(a, f(a))).first;
}
return it->second;
}
private:
ResultStore memo_;
};
Just create an instance of the memoizer, feed it your function and argument. Just make sure not to share the same memo between two different functions (but you can share it between different implementations of the same function).
Next, a driver functon, and an implementation. only the driver function need be public
int fib(int); // driver
int fib_(int); // implementation
Implemented:
int fib_(int n){
++total_ops;
if(n == 0 || n == 1)
return 1;
else
return fib(n-1) + fib(n-2);
}
And the driver, to memoize
int fib(int n) {
static memoizer1<int,int> memo;
return memo(fib_, n);
}
Permalink showing output on codepad.org. Number of calls is measured to verify correctness. (insert unit test here...)
This only memoizes one input functions. Generalizing for multiple args or varying arguments left as an exercise for the reader.
In Perl generic memoization is easy to get. The Memoize module is part of the perl core and is highly reliable, flexible, and easy-to-use.
The example from it's manpage:
# This is the documentation for Memoize 1.01
use Memoize;
memoize('slow_function');
slow_function(arguments); # Is faster than it was before
You can add, remove, and customize memoization of functions at run time! You can provide callbacks for custom memento computation.
Memoize.pm even has facilities for making the memento cache persistent, so it does not need to be re-filled on each invocation of your program!
Here's the documentation: http://perldoc.perl.org/5.8.8/Memoize.html
Extending the idea, it's also possible to memoize functions with two input parameters:
function memoize2 (f)
local cache = {}
return function (x, y)
if cache[x..','..y] then
return cache[x..','..y]
else
local z = f(x,y)
cache[x..','..y] = z
return z
end
end
end
Notice that parameter order matters in the caching algorithm, so if parameter order doesn't matter in the functions to be memoized the odds of getting a cache hit would be increased by sorting the parameters before checking the cache.
But it's important to note that some functions can't be profitably memoized. I wrote memoize2 to see if the recursive Euclidean algorithm for finding the greatest common divisor could be sped up.
function gcd (a, b)
if b == 0 then return a end
return gcd(b, a%b)
end
As it turns out, gcd doesn't respond well to memoization. The calculation it does is far less expensive than the caching algorithm. Ever for large numbers, it terminates fairly quickly. After a while, the cache grows very large. This algorithm is probably as fast as it can be.
Recursion isn't necessary. The nth triangle number is n(n-1)/2, so...
public int triangle(final int n){
return n * (n - 1) / 2;
}
Please don't recurse this. Either use the x*(x+1)/2 formula or simply iterate the values and memoize as you go.
int[] memo = new int[n+1];
int sum = 0;
for(int i = 0; i <= n; ++i)
{
sum+=i;
memo[i] = sum;
}
return memo[n];