This is my code. I am not getting any Fizz buzz printed. I am only getting numbers. Could anyone explain why? Thanks
printOut = "";
for (var x=1; x < 101 ; x++) {
switch(x) {
case((x%3) == 0):
printOut+="\n"+ "Fizz" ;
break;
case((x%5) == 0):
printOut+="\nBuzz";
break;
default:
printOut+="\n" + x ;
break;
}
}
console.log(printOut);
check how you're using the switch statement: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/switch
in the switch line x is your expression and ((x%5) == 0) is your value. I think what you mean to do is a handful of if/else statements.
You're using the switch statement improperly. Each case (value): is basically supposed to be run whenever x equals value.
To solve this problem, simply remove the switch statement entirely, and substitute ifs for each case:
for (var x = 1; x < 101; x++) {
if ((x % 3) == 0)
printOut += "\n" + "Fizz";
else if ((x % 5) == 0)
printOut += "\nBuzz";
else
printOut += "\n" + x;
}
You are trying to match the value of x with expressions whose values are either true or false. You can pass true in the switch and the switch will "match" with the first case statement that evaluates as true.
While this sort-a works, I would recommend just doing if/else statements. This won't work for number 30 which is both True for X%3 and x%5. It will match with the x%3 first and stop there.
printOut = "";
for (var x=1; x < 101 ; x++) {
switch(true) {
case((x%3) == 0):
printOut+="\n"+ "Fizz" ;
break;
case((x%5) == 0):
printOut+="\nBuzz";
break;
default:
printOut+="\n" + x ;
break;
}
}
console.log(printOut);
Related
I'm new to programming and am curious so I thought I'd reach out.
I have tested myself with 2 outputs to the console, one being FizzBuzz and the other being a chessboard.
My question is why does my chessboard naturally (without "\n") print horizontally and FizzBuzz print out vertically even though they both use the same for loops?
Thanks, Luke.
FizzBuzz,
for(let i= 1; i<= 100; i++){
let message= "";
if(i %3 === 0 && i%5 === 0){
message= "FizzBuzz"
}else if(i % 3 === 0){
message = "Fizz";
}else if(i % 5 === 0){
message = "Buzz"
}
console.log(message || i);
};
Chessboard
let size= 8;
let board= '';
for (let x = 0; x < size; x++){
for (let y= 0; y < size; y++){
if((x+y) %2 == 0){
board += '#';
}else{
board += " ";
}
}
board += "\n"
}
console.log(board);
I think I have the answer.
(A) The code,
console.log("Fizz");
console.log("Buzz");
gives output
"Fizz"
"Buzz"
(B) But,
console.log("Fizz " + "Buzz");
gives output
"Fizz Buzz"
What you are doing with the FizzBuzz is similar to case (A) and in chessboard is similar to case (B). You are logging in every iteration of the for loop in case (A), but only once for every row in case (B).
Since you are new to programming, remember that each console.log("whatever"); logs "whatever" in a new line every time it is run.
This question already has answers here:
JavaScript switch with logical operators?
(5 answers)
Closed 2 years ago.
I'm running two seemingly identical statements to check whether a number is positive, negative or zero.
The if statement returns the expected result.
The switch statement always returns the default.
Why?
In this example I pass "2" (type: number) and I get + in the if statement and ? in the switch statement.
// Verify whether a number is positive, negative or Zero:
function myFuncIf(y) {
let result = null;
console.log("y is " + y);
console.log("y is " + typeof(y));
if (y > 0) {
return "+";
} else if (y < 0) {
return "-";
} else {
return "?";
}
}
function myFuncSwitch(y) {
let result = null;
console.log("y is " + y);
console.log("y is " + typeof(y));
switch (y) {
case (y > 0):
result = "+";
break;
case (y < 0):
result = "-";
break;
default:
result = "?";
}
return result;
}
console.log(myFuncIf(2));
console.log(myFuncSwitch(2));
switch matches the value.
The value of y is 2.
Your first case will match True.
Your second will match False.
You probably want your switch to look like this:
switch (Math.sign(y)) {
case (1):
result = "+";
break;
case (-1):
result = "-";
break;
default:
result = "?";
}
See docs here:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/sign
As part of exam preparations, students are given a set of possible questions. One of them baffles me and I cannot acquire answer to it by myself.
The question goes like this (literary copy-paste):
Given a statement,
switch (x) {
case 0: x=1; break;
default: x=0;
}
How could you use if statement to get the same result?
To this, there are four possible answers you can choose.
if (x===0) x=1; else x=0;
if (x) x=1; x = 0;
if (x) x=1; else x=0;
if (x===0) x=1; if (x!=0) x=0;
Can please somebody explain to me the proper solution to this question?
The answer there would be 1.
The switch statement you supplied is basically as follows.
Based on the value of X do the following;
If X is equal to 0, make x equal 1.
Otherwise, make x equal 0.
There are different situations which make one more ideal than the other.
Edit:
switch (x) - This translates roughly to the 'if (x === )' part of the if statement.
case 1 - This is the second paramater in the if, in this case it would be 'if (x === 1)'
case 2 - This is again the second paramater in the if, in this case it would be 'if (x === 2)'
case "3" - This is again second paramater in the if, in this case it would be 'if (x === "3")'
default - This defines the 'default' operation, which evaluates to the 'else' part of an if statement.
Taking this into account, the above statement could be as follows;
switch (x):
case 1:
x = 1;
break;
case 2:
x = 2;
break;
case "3":
x = "3";
break;
default:
x = 0
break;
As an if statement, it would be as follows;
if (x === 1) {
x = 1;
} else if (x === 2) {
x = 2
} else if (x === "3") {
x = "3"
} else {
x = 0;
}
1st seems the answer. It will check for type and value. If it is 0 then it will assign 1 to x else 0.
x===0 matches value and types of both operands.Your first condition will set x to 1 if its 0 , else 0
It ultimately sets x to 0
if x has false,null,undefined,0 it will be set to 0 else 1
x will always be 0
The correct answer is if (x===0) x=1; else x=0;
So only if the x is 0 x should be 1 otherwise (default: statement) assign 0
I have an odd problem where I am receiving computer generated equations (as a string) where multiplications/divisions with zero or one and lone zeros are occasionally present. These equations are to be presented to a user in string form.
I know that I can remove these redundant parts of the equation by implementing a kind of parser, but I was curious as to whether a regular expression could be used to remove them.
I came up with the following before I finally gave up on my (quite limited) regex skills:
/([^\+\-]*(?:0|0\*|\*0){1,}[^\+\-]*)|(?:1\*|\*1)/g
It seems to work only if:
there are no non-zero numbers with a zero in them (ie. no 10's,20's,etc.)
there are no negations.
It also doesn't work well with parentheses. Unfortunately parentheses are quite common.
Note that removing the redundant portions stated above may result in redundant parentheses or "zero" parentheses (ie it could turn out like ()*x, which is equivalent to 0*x). The redundant parentheses are not as much of an issue, but I assume the "zero" parentheses could be removed by a second pass similar to the first looking for (). If either of these could be done in the same regex as the one that solves the problem I would be extremely impressed.
So I turn to you regex gurus of Stack Overflow. Can it be done?
Assumptions
The following can be assumed true about the stringified equations:
There are no divisions by zero, equations will not have any occurrence of [expr]/0or even expressions that evaluate to [expr]/0 such as [expr]/sin(0).
The only operators within the equations themselves are + - * and /.
Minus operator (-) includes both subtraction and negation, although negation is always surrounded by parentheses.
Any operation other than the above (sin,cos, pow, etc.) will appear as a function call. (no ^ %, etc.)
Sample Equation
"(0+(0/0.5+(0+1*cos(p)+0*0+0*sin(p))*cos(k)+(0+1*0+0*1+0*0)*(-sin(k))+(0+1*(-sin(p))/0.5+0*0+0*cos(p))*0)*x+(0+(0+1*cos(p)+0*0+0*sin(p))*sin(k)+(0+1*0+0*1+0*0)*cos(k)+(0+1*(-sin(p))/0.5+0*0+0*cos(p))*0)*y+(0+(0+1*cos(p)+0*0+0*sin(p))*0+(0+1*0+0*1+0*0)*0+(0+1*(-sin(p))/0.5+0*0+0*cos(p))*1)*z)"
Quite cluttered isn't it?
After leaving the comment I couldn't resist having a crack at it :)
You're biggest problem is nested parentheses. Regexes themselves are really bad at handling nested structures. This is a prime example of my mantra "Regular expressions are a tool, not a solution".
With regexes as your tool, you can apply kind of a "leaf-first" (or bottom-up) approach for this tree-structure, that's what I do in the first part while (sEq.match(...)) {...}. After that I can walk through the created array and do some simple text edits.
I've also added that 1*, *1 and /1 are deleted as they similarly don't affect the equation. You could probably expand this to make it smart enough to replace sin(0)/cos(0) with 0 and 1 respectively, then the solution would be even smaller in some cases.
(As mentioned in the comments of the code, this breaks if the equation contains stuff like 5.0*4 because JavaScript regexes don't have lookbehind so I'm trusting the \b word boundary to do that work for me. Simply adding logic that deletes unnecessary decimals would solve this though. Something like sEq = sEq.replace(/\.0+\b/g, ''); but I don't know if that's necessary for your use-case.) Edit: now fixed, 5.0*4 should remain in tact
This is not thoroughly tested though, feedback welcome.
var sEq = "(0+(0/0.5+(0+1*cos(p)+0*0+0*sin(p))*cos(k)+(0+1*0+0*1+0*0)*(-sin(k))+(0+1*(-sin(p))/0.5+0*0+0*cos(p))*0)*x+(0+(0+1*cos(p)+0*0+0*sin(p))*sin(k)+(0+1*0+0*1+0*0)*cos(k)+(0+1*(-sin(p))/0.5+0*0+0*cos(p))*0)*y+(0+(0+1*cos(p)+0*0+0*sin(p))*0+(0+1*0+0*1+0*0)*0+(0+1*(-sin(p))/0.5+0*0+0*cos(p))*1)*z)";
var aParts = [];
document.getElementById('output').value = sEq + '\n';
while (sEq.match(/\([^()]*\)/)) {
// while there are still "leafs", save them to aParts and replace with
// a reference to their index in aParts, making their parent a new
// "leaf" because it now doesn't contain the round brackets anymore
sEq = sEq.replace(/([a-z]*)\(([^()]*)\)/gi, function(m, f, a) {
var n = aParts.length;
aParts[n] = {
'found':m,
'funct':f,
'arith':a
};
return '[' + n + ']';
});
}
for (var i = 0; i < aParts.length; i++) {
// isolate divisions/multiplications
var dms = aParts[i]['arith'].split(/(?=[+-])/);
for (var j = 0; j < dms.length; j++) {
// if the isolated part is multiplied by or divided into 0, replace with 0
if (dms[j].match(/([^.]|^)\b0[*\/]|\*0(?!\.?0*[1-9])/)) {
dms[j] = dms[j].replace(/([+-]?).*/, '$1'+'0');
}
// remove /1, *1 and 1*
dms[j] = dms[j].replace(/[\/*]1\b(?!\.0*[1-9])(?:\.0*)?/g, '').replace(/([^.]|^)\b1\*/g, '$1');
}
// join back together, remove 0+, +0 and -0; 0- results in negation
aParts[i]['arith'] = dms.join('').replace(/[+-]0(?!\.?0*[1-9])(?:\.?0*)?/g, '').replace(/([^.]|^)\b0(?:\+|(-))/g, '$1$2');
// if after this the part contains just 0, perpetuate down to further eliminate
if (aParts[i]['funct']=='' && aParts[i]['arith']=='0') {
for (var j = i + 1; j < aParts.length; j++) {
if (aParts[j]['arith'].indexOf('[' + i + ']') != -1) {
aParts[j]['arith'] = aParts[j]['arith'].replace('[' + i + ']', '0');
break;
}
}
}
// add back parts previously simplified by replacing [n] with the content of aParts[n]
aParts[i]['arith'] = aParts[i]['arith'].replace(/\[(\d+)\]/g, function (m, n) {
return aParts[parseInt(n)]['funct'] + '(' + aParts[parseInt(n)]['arith'] + ')';
});
// This is just to show the progress of the algorithm
document.getElementById('parts').value += i + '\t' + aParts[i]['found'] + '\n';
var tmp = [];
for (var a = 0; a < aParts.length; a++) {
tmp[a] = {
'funct':aParts[a]['funct'],
'arith':aParts[a]['arith'].replace(/\[(\d+)\]/g, function (m, n) {
return tmp[parseInt(n)]['funct'] + '(' + tmp[parseInt(n)]['arith'] + ')';
})
};
}
// some steps didn't change after analysing, only append when significant
if (document.getElementById('output').value.indexOf('\n' + tmp[tmp.length-1]['arith'] + '\n') ==-1)
document.getElementById('output').value += tmp[tmp.length-1]['arith'] + '\n';
}
document.getElementById('solution').innerHTML =
aParts[aParts.length-1]['funct'] +
'(' + aParts[aParts.length-1]['arith'] + ')';
<h3>Parts isolated:</h3>
<textarea id="parts" rows="10" style="width:100%" wrap="off"></textarea>
<h3>Steps that simplified the equation:</h3>
<textarea id="output" rows="10" style="width:100%" wrap="off"></textarea>
<h3>Solution:</h3>
<pre id="solution"></pre>
I ended up implementing a completely non-regex, recursive approach to the problem. The cleanupEqn() function essentially splits each string by operators (top level parentheses are grouped as a single operand), recursively operates on each sub part, then does another run through on the way back up the function call chain.
Comparing this with funkwurm's regex solution in jsperf shows it is significantly faster (at least in my personal chrome and firefox browsers).
It hasn't been thoroughly tested yet, and I'm sure there could be improvements made so I welcome any feedback.
Stealing funkwurm's snippet display to show my solution:
var sEq = "(0+(0/0.5+(0+1*cos(p)+0*0+0*sin(p))*cos(k)+(0+1*0+0*1+0*0)*(-sin(k))+(0+1*(-sin(p))/0.5+0*0+0*cos(p))*0)*x+(0+(0+1*cos(p)+0*0+0*sin(p))*sin(k)+(0+1*0+0*1+0*0)*cos(k)+(0+1*(-sin(p))/0.5+0*0+0*cos(p))*0)*y+(0+(0+1*cos(p)+0*0+0*sin(p))*0+(0+1*0+0*1+0*0)*0+(0+1*(-sin(p))/0.5+0*0+0*cos(p))*1)*z)";
var operators = ['+','-','*','/'];
var level = 0;
var result = cleanupEqn(sEq);
document.getElementById('solution').innerHTML = result;
function cleanupEqn(eqn){
var parts = removeRedundant(splitByParen(eqn));
level++;
document.getElementById('output').value += 'Level ' + level + ': Processing ' + eqn + '\n';
for(var i=0; i < parts.length; i++){
document.getElementById('parts').value += parts[i] + '\n';
if(parts[i].charAt(0) === '('){
// Clean up the expression inside the parentheses
var tmp = cleanupEqn(parts[i].substring(1,parts[i].length-1));
// If it was reduced to a zero, don't add the parentheses back
if(tmp === '0'){
parts[i] = '0';
}
else {
parts[i] = '(' + tmp + ')';
}
}
}
// Finally, remove redundancies again, since some might have bubbled up.
removeRedundant(parts);
document.getElementById('output').value += 'Level ' + level + ': Completed ' + eqn + '\n' + JSON.stringify(parts, null, '\t') + '\n';
level--;
// Join it all into a string and return
return parts.join('');
}
function splitByParen(str){
var out = [];
var exprStart = 0;
var count = 0;
var i;
for (i = 0; i < str.length; i++) {
var t = str.charAt(i);
if(str.charAt(i) === '('){
if(i > exprStart && count === 0){
out.push(str.substring(exprStart, i));
exprStart = i;
}
count++;
}
else if(str.charAt(i) === ')'){
count--;
if(count === 0){
out.push(str.substring(exprStart, i+1));
exprStart = i+1;
}
}
else if(count === 0 && operators.indexOf(str.charAt(i)) > -1){
if(i > exprStart){
out.push(str.substring(exprStart, i));
}
out.push(str.charAt(i));
exprStart = i+1;
}
}
// Add the last part
if(i > exprStart){
out.push(str.substring(exprStart, i));
}
return out;
}
function removeRedundant(parts) {
for(var i=0; i < parts.length; i++){
if(parts[i] === '0'){
if(i === 0){
switch(parts[i+1]){
case '*':
case '/':
if(parts[i+1] === '*' || parts[i+1] === '/'){
parts.splice(i, 3, '0');
}
else {
parts.splice(i, 2);
}
i--;
break;
case '+':
parts.splice(i, 2);
i--;
break;
case '-':
parts.splice(i, 1);
i--;
}
}
else {
switch(parts[i-1]){
case '*':
if(parts[i+1] === '*' || parts[i+1] === '/'){
// Check if the prior portion is part of a function call
if(i > 2 && operators.indexOf(parts[i-3]) < 0){
// Check if the next portion is part of a function call (or undefined)
if(i+3 < parts.length && operators.indexOf(parts[i+3]) < 0){
parts.splice(i-3, 6, '0');
i -= 4;
}
else {
parts.splice(i-3, 5, '0');
i -= 4;
}
}
else {
parts.splice(i-2, 4, '0');
i -= 3;
}
}
else {
parts.splice(i-2, 3, '0');
i -= 3;
}
break;
case '+':
case '-':
if(parts[i+1] === '*' || parts[i+1] === '/'){
// Check if the next portion is part of a function call (or undefined)
if(i+3 < parts.length && operators.indexOf(parts[i+3]) < 0){
parts.splice(i, 4, '0');
}
else {
parts.splice(i, 3, '0');
}
i--;
}
else if(parts[i+1] === '+'){
parts.splice(i-1, 2);
i -= 2;
}
else {
parts.splice(i-1, 2);
i -= 2;
}
}
}
}
else if(parts[i] === '1'){
if(i === 0){
switch(parts[i+1]){
case '*':
parts.splice(i, 2);
i--;
break;
case '+':
case '-':
if(parts[i+1] === '*'){
parts.splice(i, 2);
i--;
}
}
}
switch(parts[i-1]){
case '*':
case '/':
if(parts[i+1] !== '/'){
parts.splice(i-1, 2);
i -= 2;
}
break;
case '+':
case '-':
if(parts[i+1] === '*'){
parts.splice(i, 2);
i--;
}
}
}
}
return parts;
}
<h3>Parts isolated:</h3>
<textarea id="parts" rows="10" style="width:100%" wrap="off"></textarea>
<h3>Steps that simplified the equation:</h3>
<textarea id="output" rows="10" style="width:100%" wrap="off"></textarea>
<h3>Solution:</h3>
<pre id="solution"></pre>
<script src="new.js"></script>
I'm currently taking the code academy course on Javascript and I'm stuck on the FizzBuzz task. I need to count from 1-20 and if the number is divisible by 3 print fizz, by 5 print buzz, by both print fizzbuzz, else just print the number. I was able to do it with if/ else if statements, but I wanted to try it with switch statements, and cannot get it. My console just logs the default and prints 1-20. Any suggestions?
for (var x = 0; x<=20; x++){
switch(x){
case x%3==0:
console.log("Fizz");
break;
case x%5===0:
console.log("Buzz");
break;
case x%5===0 && x%3==0:
console.log("FizzBuzz");
break;
default:
console.log(x);
break;
};
};
Switch matches the x in switch(x){ to the result of evaluating the case expressions. since all your cases will result in true /false there is no match and hence default is executed always.
now using switch for your problem is not recommended because in case of too many expressions there may be multiple true outputs thus giving us unexpected results. But if you are hell bent on it :
for (var x = 0; x <= 20; x++) {
switch (true) {
case (x % 5 === 0 && x % 3 === 0):
console.log("FizzBuzz");
break;
case x % 3 === 0:
console.log("Fizz");
break;
case x % 5 === 0:
console.log("Buzz");
break;
default:
console.log(x);
break;
}
}
I thought switch too,but no need.
for (var n = 1; n <= 100; n++) {
var output = "";
if (n % 3 == 0)
output = "Fizz";
if (n % 5 == 0)
output += "Buzz";
console.log(output || n);
}
Switch statement checks if the situation given in the cases matches the switch expression. What your code does is to compare whether x divided by 3 or 5 is equal to x which is always false and therefore the default is always executed. If you really want to use a switch statement here is one way you can do.
for (var i=1; i<=30; i++){
switch(0){
case (i % 15):
console.log("fizzbuzz");
break;
case (i % 3):
console.log("fizz");
break;
case (i % 5):
console.log("buzz");
break;
default:
console.log(i);
}
}
Not to too my own horn but this is much cleaner:
var numbers = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
for (var i = 1; i <= numbers.length; i++) {
if (i % 15 === 0) {
console.log("FizzBuzz");
} else if (i % 5 === 0) {
console.log("Buzz");
} else if (i % 3 === 0) {
console.log("Fizz");
} else {
console.log(i);
}
};
The switch(true) part of this statement helped me. I was trying to do a switch statement for fizzbuzz. My solution incorporates the coding style of Rosettacodes - general solution. Most significantly the use of force typing to shorten the primary conditionals. I thought, it was valuable enough to post:
var fizzBuzzSwitch = function() {
for (var i =0; i < 101; i++){
switch(true) {
case ( !(i % 3) && !(i % 5) ):
console.log('FizzBuzz');
break;
case ( !(i % 3) ):
console.log('Fizz');
break;
case ( !(i % 5) ):
console.log('Buzz');
break;
default:
console.log(i);
}
}
}
Here's what made it clear for me, might help :
It's a misinterpretation of what switch (x){} means.
It doesn't mean : "whenever whatever I put inbetween those brackets is true, when the value of x changes."
It means : "whenever x EQUALS what I put between those brackets"
So, in our case, x NEVER equals x%3===0 or any of the other cases, that doesn't even mean anything. x just equals x all the time. That's why the machine just ignores the instruction. You are not redefining x with the switch function. And what you put inbetween the brackets describes x and x only, not anything related to x.
In short :
With if/else you can describe any condition.
With switch you can only describe the different values taken by the variable x.
Here's a solution incorporating #CarLuvr88's answer and a switch on 0:
let fizzBuzz = function(min, max){
for(let i = min; i <= max; i++){
switch(0){
case i % 15 : console.log('FizzBuzz'); break;
case i % 3 : console.log('Fizz'); break;
case i % 5 : console.log('Buzz'); break;
default : console.log(i); break;
}
}
}
fizzBuzz(1,20)
We can use a function to find a multiple of any number and declare two variables to identify these multiples so that if you want to change the multiples you only need to change at max 2 lines of code
function isMultiple(num, mod) {
return num % mod === 0;
}
let a = 3;
let b = 5;
for(i=0;i<=100;i++){
switch(true){
case isMultiple(i,a) && isMultiple(i,b):
console.log("FizzBuzz")
case isMultiple(i,a):
console.log("Fizz");
case isMultiple(i,b):
console.log("Buzz");
default:
console.log(i);
}
}