Javascript some() returns false on zero
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I need to check if numbers exist in my array. Using the some() function I find that zero comes back false. It's a problem because I am working with a ton of different numbers and zero being one of them.
var array = [0, 1, 3, 4, 5];
var test = function(element) {
return 0;
};
console.log(array.some(test));
// expected output: true on 0 <-- getting false
// expected output: true on 1
// expected output: false on 20
In short how can I get 0 to return true?
javascript
asked Nov 17 '18 at 7:16
limitlimit
352419
add a comment |
I need to check if numbers exist in my array. Using the some() function I find that zero comes back false. It's a problem because I am working with a ton of different numbers and zero being one of them.
var array = [0, 1, 3, 4, 5];
var test = function(element) {
return 0;
};
console.log(array.some(test));
// expected output: true on 0 <-- getting false
// expected output: true on 1
// expected output: false on 20
In short how can I get 0 to return true?
javascript
asked Nov 17 '18 at 7:16
limitlimit
352419
3
Your function always only returns0, a falsey value. The function is supposed to do whatever comparison you want, and returntrueorfalse!
– deceze♦
Nov 17 '18 at 7:18
add a comment |
I need to check if numbers exist in my array. Using the some() function I find that zero comes back false. It's a problem because I am working with a ton of different numbers and zero being one of them.
var array = [0, 1, 3, 4, 5];
var test = function(element) {
return 0;
};
console.log(array.some(test));
// expected output: true on 0 <-- getting false
// expected output: true on 1
// expected output: false on 20
In short how can I get 0 to return true?
javascript
asked Nov 17 '18 at 7:16
limitlimit
352419
I need to check if numbers exist in my array. Using the some() function I find that zero comes back false. It's a problem because I am working with a ton of different numbers and zero being one of them.
var array = [0, 1, 3, 4, 5];
var test = function(element) {
return 0;
};
console.log(array.some(test));
// expected output: true on 0 <-- getting false
// expected output: true on 1
// expected output: false on 20
In short how can I get 0 to return true?
javascript
javascript
asked Nov 17 '18 at 7:16
limitlimit
352419
asked Nov 17 '18 at 7:16
limitlimit
352419
asked Nov 17 '18 at 7:16
limitlimit
352419
asked Nov 17 '18 at 7:16
limitlimit
352419
asked Nov 17 '18 at 7:16
limitlimit
352419
352419
3
Your function always only returns0, a falsey value. The function is supposed to do whatever comparison you want, and returntrueorfalse!
– deceze♦
Nov 17 '18 at 7:18
add a comment |
3
Your function always only returns0, a falsey value. The function is supposed to do whatever comparison you want, and returntrueorfalse!
– deceze♦
Nov 17 '18 at 7:18
3
3
Your function always only returns
0, a falsey value. The function is supposed to do whatever comparison you want, and return true or false!– deceze♦
Nov 17 '18 at 7:18
Your function always only returns
0, a falsey value. The function is supposed to do whatever comparison you want, and return true or false!– deceze♦
Nov 17 '18 at 7:18
add a comment |
2 Answers
2
active
oldest
votes
the test function always returns zero.
var test = function(element) {
return element == 0
};
This way the function should work properly.
The test function should return true/false.
In your case you always return 0 which is evaluated to boolean false.
answered Nov 17 '18 at 7:17
user2693928user2693928
1,9211716
use strict comparison operator, nlelement === 0
– Icepickle
Nov 17 '18 at 7:19
Yes, strict comparison is better thx
– user2693928
Nov 17 '18 at 7:20
add a comment |
For what you're trying to implement, it might make more sense to use the .includes function, which tests whether an array includes a value:
var array = [0, 1, 3, 4, 5];
console.log(array.includes(0));
console.log(array.includes(1));
console.log(array.includes(2));Though, .includes (along with all array iteration methods) is an O(N) process - if you need to carry out a bunch of tests for the same array, you might convert the array to a Set first, so that you can then use Set.has (which is generally O(1)):
var array = [0, 1, 3, 4, 5];
const set = new Set(array);
console.log(set.has(0));
console.log(set.has(1));
console.log(set.has(2));answered Nov 17 '18 at 7:19
CertainPerformanceCertainPerformance
101k166393
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
the test function always returns zero.
var test = function(element) {
return element == 0
};
This way the function should work properly.
The test function should return true/false.
In your case you always return 0 which is evaluated to boolean false.
answered Nov 17 '18 at 7:17
user2693928user2693928
1,9211716
use strict comparison operator, nlelement === 0
– Icepickle
Nov 17 '18 at 7:19
Yes, strict comparison is better thx
– user2693928
Nov 17 '18 at 7:20
add a comment |
the test function always returns zero.
var test = function(element) {
return element == 0
};
This way the function should work properly.
The test function should return true/false.
In your case you always return 0 which is evaluated to boolean false.
answered Nov 17 '18 at 7:17
user2693928user2693928
1,9211716
use strict comparison operator, nlelement === 0
– Icepickle
Nov 17 '18 at 7:19
Yes, strict comparison is better thx
– user2693928
Nov 17 '18 at 7:20
add a comment |
the test function always returns zero.
var test = function(element) {
return element == 0
};
This way the function should work properly.
The test function should return true/false.
In your case you always return 0 which is evaluated to boolean false.
answered Nov 17 '18 at 7:17
user2693928user2693928
1,9211716
the test function always returns zero.
var test = function(element) {
return element == 0
};
This way the function should work properly.
The test function should return true/false.
In your case you always return 0 which is evaluated to boolean false.
answered Nov 17 '18 at 7:17
user2693928user2693928
1,9211716
answered Nov 17 '18 at 7:17
user2693928user2693928
1,9211716
answered Nov 17 '18 at 7:17
user2693928user2693928
1,9211716
answered Nov 17 '18 at 7:17
user2693928user2693928
1,9211716
1,9211716
use strict comparison operator, nlelement === 0
– Icepickle
Nov 17 '18 at 7:19
Yes, strict comparison is better thx
– user2693928
Nov 17 '18 at 7:20
add a comment |
use strict comparison operator, nlelement === 0
– Icepickle
Nov 17 '18 at 7:19
Yes, strict comparison is better thx
– user2693928
Nov 17 '18 at 7:20
use strict comparison operator, nl
element === 0– Icepickle
Nov 17 '18 at 7:19
use strict comparison operator, nl
element === 0– Icepickle
Nov 17 '18 at 7:19
Yes, strict comparison is better thx
– user2693928
Nov 17 '18 at 7:20
Yes, strict comparison is better thx
– user2693928
Nov 17 '18 at 7:20
add a comment |
For what you're trying to implement, it might make more sense to use the .includes function, which tests whether an array includes a value:
var array = [0, 1, 3, 4, 5];
console.log(array.includes(0));
console.log(array.includes(1));
console.log(array.includes(2));Though, .includes (along with all array iteration methods) is an O(N) process - if you need to carry out a bunch of tests for the same array, you might convert the array to a Set first, so that you can then use Set.has (which is generally O(1)):
var array = [0, 1, 3, 4, 5];
const set = new Set(array);
console.log(set.has(0));
console.log(set.has(1));
console.log(set.has(2));answered Nov 17 '18 at 7:19
CertainPerformanceCertainPerformance
101k166393
add a comment |
For what you're trying to implement, it might make more sense to use the .includes function, which tests whether an array includes a value:
var array = [0, 1, 3, 4, 5];
console.log(array.includes(0));
console.log(array.includes(1));
console.log(array.includes(2));Though, .includes (along with all array iteration methods) is an O(N) process - if you need to carry out a bunch of tests for the same array, you might convert the array to a Set first, so that you can then use Set.has (which is generally O(1)):
var array = [0, 1, 3, 4, 5];
const set = new Set(array);
console.log(set.has(0));
console.log(set.has(1));
console.log(set.has(2));answered Nov 17 '18 at 7:19
CertainPerformanceCertainPerformance
101k166393
add a comment |
For what you're trying to implement, it might make more sense to use the .includes function, which tests whether an array includes a value:
var array = [0, 1, 3, 4, 5];
console.log(array.includes(0));
console.log(array.includes(1));
console.log(array.includes(2));Though, .includes (along with all array iteration methods) is an O(N) process - if you need to carry out a bunch of tests for the same array, you might convert the array to a Set first, so that you can then use Set.has (which is generally O(1)):
var array = [0, 1, 3, 4, 5];
const set = new Set(array);
console.log(set.has(0));
console.log(set.has(1));
console.log(set.has(2));answered Nov 17 '18 at 7:19
CertainPerformanceCertainPerformance
101k166393
For what you're trying to implement, it might make more sense to use the .includes function, which tests whether an array includes a value:
var array = [0, 1, 3, 4, 5];
console.log(array.includes(0));
console.log(array.includes(1));
console.log(array.includes(2));Though, .includes (along with all array iteration methods) is an O(N) process - if you need to carry out a bunch of tests for the same array, you might convert the array to a Set first, so that you can then use Set.has (which is generally O(1)):
var array = [0, 1, 3, 4, 5];
const set = new Set(array);
console.log(set.has(0));
console.log(set.has(1));
console.log(set.has(2));var array = [0, 1, 3, 4, 5];
console.log(array.includes(0));
console.log(array.includes(1));
console.log(array.includes(2));var array = [0, 1, 3, 4, 5];
console.log(array.includes(0));
console.log(array.includes(1));
console.log(array.includes(2));var array = [0, 1, 3, 4, 5];
const set = new Set(array);
console.log(set.has(0));
console.log(set.has(1));
console.log(set.has(2));var array = [0, 1, 3, 4, 5];
const set = new Set(array);
console.log(set.has(0));
console.log(set.has(1));
console.log(set.has(2));answered Nov 17 '18 at 7:19
CertainPerformanceCertainPerformance
101k166393
answered Nov 17 '18 at 7:19
CertainPerformanceCertainPerformance
101k166393
answered Nov 17 '18 at 7:19
CertainPerformanceCertainPerformance
101k166393
answered Nov 17 '18 at 7:19
CertainPerformanceCertainPerformance
101k166393
101k166393
add a comment |
add a comment |
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3
Your function always only returns
0, a falsey value. The function is supposed to do whatever comparison you want, and returntrueorfalse!– deceze♦
Nov 17 '18 at 7:18