Can anyone explain to me why first one is not working and second one is working?
First Statement
function test(n) {
switch (n) {
case (n == 0 || n == 1):
console.log("Number is either 0 or 1");
break;
case (n >= 2):
console.log("Number is greater than 1")
break;
default:
console.log("Default");
}
}
Second Statement
function test(n) {
switch (true) {
case (n == 0 || n == 1):
console.log("Number is either 0 or 1");
break;
case (n >= 2):
console.log("Number is greater than 1")
break;
default:
console.log("Default");
}
}
The parameter which is given to the switch will be compared using ===. In cases which you have, you have expressions which result to boolean type: n==0 || n==1 or n >= 2. When you pass a number , it tries to compare your number with a result given from the expression in cases. So for example with the given number 1 it tries to compare 1 === (1 == 0 || 1 == 1) -> 1 === true which returns false (strict comparison). So you get the Default text every time.
For the first case, you need to have numbers in the cases of your switch , not a boolean (n==0 || n==1 results to boolean).
With the second case, you have in the switch value true of type boolean.When you pass again 1 the comparing goes like true === (1 == 0 || 1 == 1) -> true === true and it returns true. So you get the desired result according to your value n. But the second case has no goals with using true as the value. You can replace it with a if else if statement.
If you want to get the same result for many cases you need to write 2 cases above each other. See this
case 0:
case 1:
result
Here the cases have type number, not boolean.
Code example.
function test(n){
switch (n) {
case 0:
case 1:
console.log("Number is either 0 or 1");
break;
case 2:
console.log("Number is 2")
break;
default:
console.log("Default");}
}
test(0);
test(1);
test(2)
switch is shorthand for a bunch of ifs.
switch(n) {
case x:
a();
break;
case y:
b();
break;
}
... is equivalent to:
if(n == x) {
a();
} else if(n == y) {
b();
}
So your first piece of code:
switch (n) {
case (n==0 || n==1):
console.log("Number is either 0 or 1");
break;
case (n>=2):
console.log("Number is greater than 1")
break;
default:
console.log("Default");}
}
... is equivalent to:
if(n == (n==0 || n==1)) {
console.log("Number is either 0 or 1");
} else if ( n == ( n >= 2)) {
console.log("Number is greater than 1");
} else {
console.log("Default");
}
I hope you can see that n == (n==0 || n==1) and n == ( n >= 2) are both nonsense. If n is 0, for example, the first will evaluate to 0 == true. In many languages this will cause a compiler error (comparing different types). I don't especially want to think about what it does in Javascript!
Your second example:
switch (true) {
case (n==0 || n==1):
console.log("Number is either 0 or 1");
break;
case (n>=2):
console.log("Number is greater than 1")
break;
default:
console.log("Default");
}
Is equivalent to:
if(true == (n==0 || n==1)) {
console.log("Number is either 0 or 1");
} else if(true == (n>=2)) {
console.log("Number is greater than 1");
} else {
console.log("Default");
}
... in which at least the condition statements true == (n==0 || n==1) and true == (n >=2) make sense.
But this is an unconventional way to use switch in most languages. The normal form is to use the value you're testing as the parameter to switch and for each case to be a possible value for it:
switch(n) {
case 0:
case 1:
console.log("n is 0 or 1");
break;
case 2:
console.log("n is 2);
break;
default:
console.log("n is some other value");
}
However switch doesn't provide a cleverer case than a full equality check. So there's no case >2 && <5.
Your can either use your trick using switch(true) (in Javascript -- there are many languages in which this won't work), or use if/else.
switch uses strict comparison.
You take a number in the switch statement and in cases, just comparsions which return a boolean value.
A switch statement first evaluates its expression. It then looks for the first case clause whose expression evaluates to the same value as the result of the input expression (using strict comparison, ===) and transfers control to that clause, executing the associated statements. (If multiple cases match the provided value, the first case that matches is selected, even if the cases are not equal to each other.) If no matching case clause is found, the program looks for the optional default clause, and if found, transfers control to that clause, executing the associated statements. If no default clause is found, the program continues execution at the statement following the end of switch. By convention, the default clause is the last clause, but it does not need to be so.
Related
This question already has answers here:
javascript switch(true)
(5 answers)
Closed 5 years ago.
Good afternoon!
Why does the first option work - switch (true), and the second option does not work - switch (a)?
First:
var a= prompt('Enter value', '');
switch(true)
{
case a>10:
alert('a>10');
break;
case a<10:
alert('a<10');
break;
default:
alert('a===10');
Second:
var a= prompt('Enter value', '');
switch(a)
{
case a>10:
alert('a>10');
break;
case a<10:
alert('a<10');
break;
default:
alert('a===10');
Why does the first option work - switch (true), and the second option
does not work - switch (a)?
As per documentation
The switch statement evaluates an expression, matching the
expression's value to a case clause, and executes statements
associated with that case.
So, in your first option true will match to either a < 10 or a > 10, however in second option, a being a string may not match to either of them.
Edit: I just realize OP ask for the difference instead of why it won't work, sorry for misunderstanding the question
It should work nicely
var a = prompt('Enter value', '');
switch (true) {
case (a > 10):
alert("a > 10");
break;
case (a < 10):
alert("a < 10");
break;
default:
alert('a == 10');
}
It's because a > 10 is true, like the switch(true), while switch(a) was passed a, which is not true. Of course, you should cast a. a = +a or use parseInt() or parseFloat().
Here's what you probably meant to do:
var a = prompt('Enter value');
if(+a > 10){
alert('a > 10');
}
else if(a !== '' && +a < 10){
alert('a < 10');
}
else if(+a === 10){
alert('a === 10');
}
else{
alert('Really, you should avoid using prompt and alert!');
}
// notice this is less code than that pointless switch
You need to convert the user input from a string to an integer, like so
a = parseInt(a)
I have 2 variables : var1 and var2.
I found this answer https://stackoverflow.com/a/9235320/3917754 I try to implement it in my case :
switch (var1 | var2) {
case ('Contact' | true):
$('#btnCopyCompanyAddress').removeClass('hidden');
$('#btnCopyPersonalAddress').addClass('hidden');
break;
case ('Company' | true):
$('#btnCopyCompanyAddress').addClass('hidden');
$('#btnCopyPersonalAddress').removeClass('hidden');
break;
default:
$('#btnCopyCompanyAddress, #btnCopyPersonalAddress').addClass('hidden');
}
but always first case is executed even if var1 = Company and var2 = true.
It will always match the first case since 'any string string' | true === 1 results always 1.
Bitwise operators treated both value as string so parsing string result NaNor 0 and true result 1. The result would be calculates as NaN | 1, which results 1.(If the value is NaN or Infinity, it's converted to 0)
Refer : Bitwise operations on non numbers
So that would not work in this case, instead try concatenation.
switch (var1 + var2) {
case ('Contacttrue'):
$('#btnCopyCompanyAddress').removeClass('hidden');
$('#btnCopyPersonalAddress').addClass('hidden');
break;
case ('Companytrue'):
$('#btnCopyCompanyAddress').addClass('hidden');
$('#btnCopyPersonalAddress').removeClass('hidden');
break;
default:
$('#btnCopyCompanyAddress, #btnCopyPersonalAddress').addClass('hidden');
}
You can use "+" instead of "|" :
function Check(var1, var2) {
switch (var1 + var2) {
case ('Contact' + var2):
alert('1');
break;
case ('Company' + var2):
alert('2');
break;
default:
alert('0');
}
}
I'm trying this simple code and seems like the user's input is not going through all the comparisons and jumps to the default one right away. I'm guessing that JS is taking the user's input as a string instead. I did try to parseInt() but didn't work. Here is my code;
var number = prompt('What\'s your favority number?');
switch(number){
case (number < 10):
console.log('Your number is to small.');
break;
case (number < 100):
console.log('At least you\'re in the double digits.');
break;
case (number < 1000):
console.log('Looks like you\'re in three digits.');
break;
default:
console.log('Looks like you\'re in the fouth digits.');
}
Use true as an expression for switch.
The switch statement evaluates an expression, matching the expression's value to a case clause, and executes statements associated with that case.[Ref]
A switch statement first evaluates its expression. It then looks for the first case clause whose expression evaluates to the same value as the result of the input expression (using strict comparison, ===) and transfers control to that clause, executing the associated statements. (If multiple cases match the provided value, the first case that matches is selected, even if the cases are not equal to each other.) . If no matching case clause is found, the program looks for the optional default clause, and if found, transfers control to that clause, executing the associated statements.
var number = prompt('What\'s your favority number?');
number = Number(number); //Use `Number` to cast it as a number
switch (true) {
//----^^^^
case (number < 10):
console.log('Your number is to small.');
break;
case (number < 100):
console.log('At least you\'re in the double digits.');
break;
case (number < 1000):
console.log('Looks like you\'re in three digits.');
break;
default:
console.log('Looks like you\'re in the fouth digits.');
}
<script src="http://gh-canon.github.io/stack-snippet-console/console.min.js"></script>
Edit: As suggested by #bergi in the comments, an if/else cascade is the best approach for this problem.
var number = prompt('What\'s your favority number?');
number = Number(number); //Use `Number` to cast it as a number
if (number < 10)
console.log('Your number is to small.');
else if (number < 100)
console.log('At least you\'re in the double digits.');
else if (number < 1000)
console.log('Looks like you\'re in three digits.');
else
console.log('Looks like you\'re in the fouth digits.');
You're not understanding how the switch statement works. It is not a shortcut for checking dynamic values. It's a shortcut for checking known values.
Each case is a statement that gets evaluated to a value. If you look at the docs, you'll see that they have case: value, rather than what you are attempting, which is case: (expression). So, it's going to turn all your expressions into values.
So, for example, your first case is:
case (number < 10):
But what that really becomes is:
case false:
And of course, no number will evaluate to false (technically 0 is a falsey value, but the switch uses === comparison rather than == so 0 === false // false). Thus, all your cases are really case false, and so the switch is falling through all of them and landing on the default case.
So, for your situation, the switch statement is inappropriate. You should use if statements.
if(number < 10) {
} else if(number < 100) {
} else if(number < 1000) {
} else {
}
The switch statement is only appropriate when you know the values:
switch(number) {
case 10:
break;
case 100:
break;
case 1000:
break;
default:
}
(And yes, use parseInt to ensure you have integers.)
"original post" : This function should compare the value of 'a' with several other values, but always defaults. My test shows that the value of 'a' or 'b' is never changed. Do I have the case a > statement incorrect or elsewhere?
Now I understand that I can not use comparison in the case statement:
Should I use a bunch of if statements and a while (a <> = 0) to do the multiple checking and decrementing?
The snippit below shows 'a' with a particular value. In the full function, actually 'a' gets a value from a random number in another function. It must be checked against 16 possible values and decremented, then rechecked until it finally reaches 0. The comparison values are actually powers of 2 (1 through 16).
function solution() {
var a = 18000;
var b = 0;
switch (a) {
case a > 30000:
a = a - 30000;
b = b++;
break;
case a > 16000:
b = b++; a = a - 16000;
break;
case a > 8000:
b = b++; a = a - 8000;
break;
default:
c = "defaulted!, Why?";
break;
}
window.alert (a + " " + b + " " + c);
}
Don't use switch for range checks like this. It's possible with
switch (true) {
case (a > 30000):
a = a - 30000;
b = b++;
but just don't do that.
Use if/else instead. While switch is really just an abstract if/else construct, use it for things like this:
switch(a){
case 1: ...
}
In a nutshell, you can't use boolean expressions in switch case labels. You'll need to rewrite the code as a series of if statements.
Switch cases are usually like
Monday:
Tuesday:
Wednesday:
etc.
I would like to use ranges.
from 1-12:
from 13-19:
from 20-21:
from 22-30:
Is it possible? I'm using javascript/jquery by the way.
you could try abusing the switch fall through behaviour
var x = 5;
switch (x) {
case 1: case 2: case 3: case 4: ...
break;
case 13: case 14: case 15: ...
break;
...
}
which is very verbose
or you could try this
function checkRange(x, n, m) {
if (x >= n && x <= m) { return x; }
else { return !x; }
}
var x = 5;
switch (x) {
case checkRange(x, 1, 12):
//do something
break;
case checkRange(x, 13, 19):
...
}
this gets you the behaviour you would like. The reason i return !x in the else of checkRange is to prevent the problem of when you pass undefined into the switch statement. if your function returns undefined (as jdk's example does) and you pass undefined into the switch, then the first case will be executed. !x is guaranteed to not equal x under any test of equality, which is how the switch statement chooses which case to execute.
Late to the party, but upon searching for an answer to the same question, I came across this thread. Currently I actually use a switch, but a different way. For example:
switch(true) {
case (x >= 1 && x <= 12):
//do some stuff
break;
case (x >= 13 && x <= 19):
//do some other stuff
break;
default:
//do default stuff
break;
}
I find this a lot easier to read than a bunch of IF statements.
You can make interesting kludges. For example, to test a number against a range using a JavaScript switch, a custom function can be written. Basically have the function test a give n value and return it if it's in range. Otherwise returned undefined or some other dummy value.
<script>
// Custom Checking Function..
function inRangeInclusive(start, end, value) {
if (value <= end && value >= start)
return value; // return given value
return undefined;
}
// CODE TO TEST FUNCTION
var num = 3;
switch(num) {
case undefined:
//do something with this 'special' value returned by the inRangeInclusive(..) fn
break;
case inRangeInclusive(1, 10, num):
alert('in range');
break;
default:
alert('not in range');
break;
}
</script>
This works in Google Chrome. I didn't test other browsers.
Nope, you need to use an if/else if series to do this. JavaScript isn't this fancy. (Not many languages are.)