A fuzzy logic is the multivariate logic form, in which the truth values of the variables can be any real number between 0 (false) and 1 (true). Already Boolean logic the variables and functions can only have values 0 and 1, in this logic we also have the logical ports: and: You have to have all the values for the sentence to be true. or: It is necessary only one of the values for the sentence to be true. xor: returns true only when the input values are different, otherwise false.

I realized a possible problem in the function check RoundInstead of using forEach, use find function verificarRodada(rodada, rodadas, partida) { mandante = partida.mandante; visitante = partida.visitante; rodadas.find(item => { return (item[0] == rodada) && (item[1] == mandante || item[1] == visitante || item[2] == mandante || item[2]== visitante) }); } ForEach is not stopped with return false. Use find or every.

All examples here are in https://ideone.com/LdTMfV .It has several ways. Here are some:Switching with subtraction$alterna = 1; while ( ... ) { $alterna = 1 - $alterna; // Ou vai ser 1-0, ou 1-1 sempre if ($alterna) { // var1 } else { // var2 } Applied to your case:$alterna = 1; while ( ... ) { $alterna = 1 - $alterna; // se for true vira false, se for false vira true $veiculo1 = $request->input($alterna?'veiculo1':'veiculo2'); } Switching with logical denial$alterna = true; while ( ... ) { $alterna = !$alterna; $veiculo1 = $request->input($alterna?'veiculo1':'veiculo2'); } This line here:$alterna = !$alterna; is equivalent to other constructions (a crazy medium) that you find on the web:$alterna = ($alterna == false); $alterna = $alterna ? false : true; // essas são de doer, mas já vi piores Switching with bit reversal$alterna = 0; while ( ... ) { $alterna = ~$alterna; $veiculo1 = $request->input($alterna?'veiculo1':'veiculo2'); } Or even inverting the first bit:$alterna = 0; while ( ... ) { $alterna ^= 1; $veiculo1 = $request->input($alterna?'veiculo1':'veiculo2'); } In the above examples we are creating a situation where we only have two possible states. It has many other ways.Switching with rest operatorIf it were in a loop with incremental counter 1 in 1 (like most of the loopsif ($contador % 2) { // var1 } else { // var2 } in your case:$veiculo1 = $request->input($contador%2?'veiculo1':'veiculo2'); This works because the rest operator % will always be zero or a sequential counter (working with integers).

Three things:You overwrite the value of soma to each iteration of the loop;You define soma as whole;You must first multiply the fat, then add;To fix (1), you will need to change soma = For soma +=. To correct (2) just set soma as float. The item (3) needs to be done because otherwise you will have 2 times the denominator equal to 1. fat starts in 1, generating the term (x+1)/1!, then it is multiplied by i which is also 1, generating in the next iteration (x+2)/1!. Next time (x+3)/2!, etc. That is, the result would be different than expected.#include <stdio.h> int main(){ float x = 0.0; int n = 0; float soma = 0; int fat = 1; printf("Digite um valor:"); scanf("%f", &amp;x); printf("Digite a quantidade de operações:"); scanf("%d", &amp;n); for(int i = 1; i &lt;= n; i++){ fat = fat * i; soma += (x + i)/fat; } printf("A soma n pra S = x+n/n!\nÉ: %f", soma); return 0; }

You can create a function for this:function get(obj, path) { const attributes = path.split('.'); for (let attribute of attributes) { if (!obj.hasOwnProperty(attribute)) { return null; } obj = obj[attribute]; } return obj; } It basically goes through the path of attributes you want and if any of them does not exist returns null. Weak points of this solution is that if the desired value is null you will not be able to distinguish whether returned the value correctly or if returned null because it did not find the attribute; another is that as the point character is used as a separator, none of the attributes can possess it in the name (this can be contoured by changing the tab to another character).function get(obj, path) { const attributes = path.split('.'); for (let attribute of attributes) { if (!obj.hasOwnProperty(attribute)) { return null; } obj = obj[attribute]; } return obj; } const obj = { _embedded: { "wp:featuredmedia": [ { media_details: { sizes: { full: { source_url: 'Conteudo' } } } } ] } }; const existe = get(obj, '_embedded.wp:featuredmedia.0.media_details.sizes.full.source_url'); const naoExiste = get(obj, '_embedded.wp:featuredmedia.1.media_details.sizes.full.source_url'); console.log(existe, naoExiste);

It has several issues in the code, but the main thing is that it is not converting a text to number. Whenever you use a function you should read all of her documentation in detail and learn how to use it. A prompt() says it will return one string, so you can't handle it. Must use parseFloat() in this case to convert to a number that can be calculated. Note that this conversion can give error and break the application if you type something that cannot be converted, and this needs to be handled, this becomes exercise for you.I used more significant variable names, simplified the calculation and eliminated unnecessary variables, and reduced their scope. I left the function because it may have a purpose, but in the current code it is simpler to make the account that create a function. I didn't use one array because everything that is not necessary only creates potential for problems. It would be useful in another context, in this is only complication, the result is the same without it. The statement does not speak in array or something that implies your need.function calcMedia(a, b) { return (a + b) / 2; } var numAlunos = parseFloat(prompt("Quanto alunos você deseja verficar? ")); for (var i = 0; i < numAlunos; i++) { var aluno = prompt("Qual o seu nome? "); var nota1 = parseFloat(prompt("Digite sua 1ª nota: ")); var nota2 = parseFloat(prompt("Digite sua 2ª nota: ")); console.log(aluno); console.log(nota1); console.log(nota2); console.log(calcMedia(nota1, nota2)); console.log(""); }

Instead of using 2 .find, just do so: $(xml).find('entry state')... that already searches all tags <state> inside <entry>.You can go doing the append by checking if one state has been inserted or not, resulting in unique options with the names of states:$.ajax({ type: "GET", url: "googleshoppingcustom.xml", dataType: 'xml', success: function(xml){ $(xml).find('entry state').each(function(i){ var state = $(this).text(); // verifica se já existe um option com o nome do Edtado if( !$('#region option[value="'+ state +'"]').length ){ $('#region').append(`&lt;option value='${state}'&gt;${state}&lt;/option&gt;`); } }); } });

And iterative you understand?#include <stdio.h> int func(int n) { return n == 0 ? 1 : func(n - 1) - n; } int main() { int a; printf("Digite um valor inteiro: "); scanf("%d", &a); printf("%d\n", func(a)); //agora iterativo int n = a; int temp = 1; while (1) { temp -= n; //faz a acumulação na mão n--; //faz o n - 1 guardando seu próprio estado if (n == 0) break; //condição de saída, poderia estar no próprio while com condição invertida } printf("%d\n", temp); } See https://ideone.com/4LDMOB . And https://repl.it/join/hqyeknvm-maniero . Also https://github.com/maniero/SOpt/blob/master/C/Function/RecursiveExplain.c .I wrote the iterative code in a way that is not ideal only to view easier. And I wrote how to write the functional code in just one line.The codes are doing the same thing. Only in reverse. In a more functional style, which is the recursive you eliminate a variable because the state is going to be passed every call of the function. In the most imperative style you need the result control variable.What is doing in your code and calling the same function once behind the other passing the value of the variable always subtracted 1, and considering the result always subtracting the value of the state of the moment, in the case represented by n. In each flame n it will be worth less until it arrives at 0, when it considers only to be 1, it is the exit condition.So call it:func(5) flame func(4) - 5 that calls func(3) - 4 that calls func(2) - 3 that calls func(1) - 2 that calls func(0) - 1 and there like func() receives 0 he falls into the other condition and returns 1 without calling anything else. Note that it will form a stack of function calls (the term officially used is this same).There on the return of the call func(1) - 2 is -1, since func(0) We know it's 1. And the return of this is -2, and the return of the following is -3, and then returns -4, and finally returns -5. If you accumulate all this will have -14.My suggestion to understand better is to see running, put this code to run on an IDE, turn on an breakpoint, stop at the first line and go running step by step, and see how everything goes being called, how the values change. It has no better way to understand the functioning of the computer in the algorithm that is trying to identify what it does.

It is even possible to use this but, as you can see, the readability of the code gets horrible."Identifying" the condition gives to see where the error is (in the control of the parentheses):echo ($estado == 'SP' ? 'São Paulo' : ($estado == 'RJ' ? 'Rio de Janeiro') : ($estado == 'PR' ? 'Paraná') : 'teste'); The correct assembly would look like this:echo ($estado == 'SP' ? 'São Paulo' : ($estado == 'RJ' ? 'Rio de Janeiro' : ($estado == 'PR' ? 'Paraná' : 'teste') ) ); //em uma linha só: echo ($estado == 'SP' ? 'São Paulo' : ($estado == 'RJ' ? 'Rio de Janeiro' : ($estado == 'PR' ? 'Paraná' : 'teste'))); remembering: in cases with more than one validation, it is more interesting to use if you don't know or switch.

Generally these systems have an encryption key that is used in a single way, just to not have how to undo and the goal is to actually make it different every cycle.I took a look and it changes the keys to each generation, should not be difficult to play this in php if you want to generate your own routine.

Today most projects with serious development in javascript end up doing so with nodejs and webpack as part of the process of build.With PHP, researching fast found this library, the https://github.com/tholu/php-packer .Basically with it you pass the javascript code and it returns the obfuscated code. $js = file_get_contents('test.js'); $packer = new Tholu\Packer\Packer($js, 'Normal', true, false, true); $packed_js = $packer->pack(); echo $packed_js;

She is not a constant, to be, she could not be changed at all. Although, as you put in the example, it is not changed during the code, yet it is not an accountant, because at any time it can be changed. It would be similar to saying that any member of the class is declared public, but by never being used out of class, be considered private.

In several languages (Javascript and C are also so), it is common that 0 numeric be treated as equivalent to false in flow deviations (IF). That's for design.In the specific case of PHP, an exresion conversion occurs within the parentheses of IF for a Boolean value. The following expressions saw false when converted:the boolean himself FALSE;the whole 0 (zero);the floating point 0.0 (zero);empty string ("");the string "0";an array without elements;an object without member elements ( PHP 4 only);the special type NULL (including undefined variables);the SimpleXML object created from empty tags;I stole that information from http://php.net/manual/pt_BR/language.types.boolean.php#language.types.boolean.casting .

To run a String as a command in PHP, you can use the "eval()" function.<?php eval('$var = (("TR1"=="TR" OR "TR2"=="TR" OR "TR"=="TR") AND ("10,0"=="10,1" OR "9,0"=="9,0"));'); echo $var; ?> http://php.net/manual/pt_BR/function.eval.php

I have made the basic idea of the program and will explain the logica, so this program fails in some cases that I will say so, it will have after having some validations for this code to work for the various names, it is useful to think about it and not to immediately make a copy paste in the code.There may be some more efficient method, or even a function that can do this, however logica must be the same.Logic1. You must first locate the last vowel 2. should pull forward qtd sometimes to get the vowel.Ex: qtd=2--> celula_ _r, to then in that space we put the letter we want3. Put the letter in the right placechar nome[100]="Celular"; char letra; int qtd=0; int i, j; printf("Nome definido: %s\n\n", nome); printf("\nCARACTER que deseja INCLUIR no nome "); scanf(" %c",&letra); printf("\nQuantidade de vezes que deseja colocar no nome? "); scanf(" %d",&qtd); for( i=strlen(nome) ; i>-1; i--){ /** 1 **/ if(nome[i]=='a' || nome[i]=='e' || nome[i]=='i' || nome[i]=='o' || nome[i]=='u' ) { i++; break; } } for(j=strlen(nome); j>=i; j--) /** 2 **/ nome[j+qtd]=nome[j]; for(j=i; j<qtd+i; j++) /** 3 **/ nome[j]=letra; printf("resultado = %s\n ", nome); https://ideone.com/Vd1iSt The name may not have any vowels, which may cause problems in this program (because of the 1st for, if you find nothing i will have value -1 and will add in the same)This program assumes it will have tiny vowels This program does not guarantee that it does not reach overflow in the name, because it can pull forward in such a way that it can reach more than 100. Ex: qtd=100

return 1; //Sempre retornará 1, aqui entendi Not always, it will return only if index is less than 7.return 0; //Não achei o porque do return 0 neste ponto He will perform whenever index is not less than 7. The function needs to have a return in any circumstance, can not return something only in some cases and in another does not.the same maybeNew(x) in any way affects the attribution of the count variable itself from the beginning of the line?It itself not, but the whole line is affecting count, after all this line is saying to keep in count, and this alone already affects something, a value that in the case is itself count added from an expressionSystem.out.println(count + " " + m4a[1].counter); //Porque m4a[1]? We have no way of knowing what is in the question, but for some reason he needs to take the second element of this array.The exit of this exercise would be 14 1, but I was lost and I can't see how it came to 14, the 1st itself return indicates the value.He was following the algorithm. I could explain line by line, but the best is yourself do the https://pt.stackoverflow.com/q/220474/101 and identify. However, in a confusing code so the best is neither trying. Learn from something more meaningful. Understanding this will be harmful to your learning.Rarely main() must be next to a class that needs to be instantiated. This confuses everything. There is no reason for every part of this class to exist and if you start working with meaningless things will get used to it. Note that in each step there is accumulation of 2. The first accumulation is very explicit and then the other happens because maybeNew() always returns 1, this could have been written on a line.It's not your fault that you have trouble understanding, it's the fault of the code.The code is very poorly written and undocumented. It teaches to do strange, confused and probably wrong things, so I would question the book. This series of books was well-recognized, so I don't know if it got worse with time or that was specifically miswritten.

It has some ways to do that, it has the most confusing and the smartest, that I preferred:#include <stdio.h> #include <stdlib.h> #include <math.h> void parImpar (int num) { int digitos = floor(log10(abs(num))) + 1; for (int i = digitos - 1; i >= 0; i--) { int div = num / (int)pow(10, i); num = num % (int)pow(10, i); printf(div % 2 == 0 ? "\n%d eh Par" : "\n%d eh Impar", div); } } int main() { int num; printf("Digite um numero inteiro: "); scanf("%d", &num); parImpar(num); } See https://ideone.com/AcsNWy . And https://repl.it/join/rxhtvges-maniero . Also https://github.com/maniero/SOpt/blob/master/C/Algorithm/OddEven.c .First he discovers the number of digits through a known formula. You have other ways to get this, but I don't like any.Then we make a bond already knowing where it begins and where it ends. And in it we make the account to get the value of the digit by dividing by the power of 10 by raising to the position you are analyzing. And then you take the rest of that division to keep doing the same with the other digits.Exemplifying: the first power will result in 10000 (10 ^4) then split 12345 by 10000 gives 1 and this can be analyzed whether it is par or impar. Then you take the rest of this which gives 2345 and which will be divided by 1000, then gives 2, and then about 345 divided by 100 that gives 3, and left 45 divided by 10 gives 4 and finally 5 divided by 1 gives 5.

Having two conditions cannot, only one is possible that is the middle, as you already know. But you can use || or other operators, several times. It will continue to have a condition, but the expression can be much more complex, what matters is that in the end it results in a boolean. Then it can have several comparative expressions linked by relational operators.But in your case it doesn't make sense, it's impossible index being smaller than tam, after all she was declared worthwhile 0 soon before and tam is a size that cannot be smaller than 0. It could make simple like this:for (let index = 0; index < entrada.length; index++) https://github.com/maniero/SOpt/blob/master/JavaScript/Syntax/ForArray.js .It doesn't make sense, and it's usually worse, catch the size off the loop. He's got some other weird stuff and probably wrong with that code.

DEC2BIN converts a decimal number to a binary numberfunction dec2bin(dec){ return (dec >>> 0).toString(2); } console.log(dec2bin(256)); console.log(dec2bin(1024)); Resultado Formula 1010 = DEC2BIN(10) 01010 = DEC2BIN(10, 5) 1111100000 = DEC2BIN(-32) https://pt.stackoverflow.com/questions/163246/transformar-number-em-texto-bin%C3%A1rio/163252#163252 https://stackoverflow.com/questions/9939760/how-do-i-convert-an-integer-to-binary-in-javascript

The operator used is wrong, needs to be a <(lower than). If you start from 0 and go to 10, and will increment you cannot verify if the number is greater than a certain value to determine if it will continue. Maybe the confusion is because it didn't quite understand how the condition works for, there has to be true to continue performing, so when it is false it closes the loop of repetition. The correct thing is to check if the number is still smaller than the desired target value that is 10.Then you may wonder why it should be just "less than" and not "less or equal to". Because you started from 0, and it's right to start from this number, not that it makes so much difference, the important thing is that it performs 10 times, because that is the statement of the problem, then it can start from the number you want and go to the number when to this, but it was convinced to start from 0, then the condition is true until the number 9, since the 0 count too. Therefore the signal should be only smaller, when it is equal, or it is reached in 10, it must be false, because starting from 0 the number 10 is the eleventh, one more than you want. So use it <= is wrong and the answer accepted and voted is wrong.#include <iostream> #include <string> using namespace std; int main() { for (int i = 0; i < 10; i++) { string nome; cin >> nome; cout << nome << endl; } } See https://ideone.com/mJVGuQ . And https://repl.it/join/pjgvppno-maniero . Also https://github.com/maniero/SOpt/blob/master/CPP/Algorithm/Loop.cpp .How curious I could do this:for (int i = 1; i <= 10; i++) There <= would be correct because it started from 1, but this is not how programmers make, leaving the pattern creates difficulties of understanding. We write codes for everyone to understand and so it is better to follow the standards already adopted.