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  • JavaScript slice vs shift functions
    I was looking at a solution for the Array Addition problem, posted in another CoderByte Question: <a href=''><;/a> My question is regarding this piece of code: var n = array[0]; array = array.slice(1); This piece of code is within a recursive function. I tried to rewrite these steps into one line, ultimately trying these two methods to directly modify the array (instead of slicing it): var n = array.shift(); and var n = array.splice(0, 1)[0]; However, I do not get the same answer when using these methods, and I'm not sure why. Why would modifying the array directly (using shift or splice) be different than re-assigning it using array = slice(1)? Thanks!
    hokennethk posted this question on 2/15/15 | javascript, array
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  • Have you studied C? Or any language with pointers? If not, hopefully this will still make sense... Here's what's going on: var array = [1,2,3,4]; //<Made an array var n = array[0]; //<Saved first element in n. array = array.slice(1); //Created new array from slice, destroyed reference to original What you've done in line 3 is made a copy, then destroyed the *pointer* to the original. If the original has another pointer, it still exists unchanged. Like this: var A = B = [1,2,3,4]; var n = A[0]; A = A.slice(1); //B still exists, and it's still [1,2,3,4] //but... var C = D = [1,2,3,4]; var n = C.shift(); //C and D are now both [2,3,4]; In the particular example you gave: function recursion( array ){ //"array" here is a local variable, pointing to an array created elsewhere // ... stuff ... var n = array[0]; //new local variable //array[1] = 123; //Don't Do This! You'll Change The Original Array! array = array.slice(1); //This is OK. You've made a new array, // and pointed the local variable "array" at it. // The original array is still alive and well and unharmed // somewhere outside of this function! :-) // ... stuff ... } Does that make sense? Arrays and variables started to make sense to me after a little practice. If you've got it now, you can stop reading. If not, this code might help: var a = [1,2,3,4]; a[0] = 5; console.log(a); //logs: [5,2,3,4]. Changed the original array! a = [1,2,3,4]; //a is *pointing* to [1,2,3,4] (stored somewhere in memory) function change( array ) { //"array" is also *pointing* to [1,2,3,4] (stored somewhere in memory) array[0]=5; //array[0] tells JavaScript: change the 0th element of [1,2,3,4]. (Now [5,2,3,4]) array = "Hello!"; //This just tells JavaScript: //"array" isn't pointing at [5,2,3,4] anymore. //now, it's pointing at "hello" (stored somewhere else in memory). //[5,2,3,4] is not lost though! Var "a" outside this function still points to it. } change( a ); //runs the function above on variable "a" console.log(a); //logs: [5,2,3,4]. The function changed the original array! But, it did not destroy it. a = [1,2,3,4] function dontChange( arr ) { var b = array[0]; //b is copied from 0th of [1,2,3,4], which is 1. arr = arr.slice(1); //a is a new array, copied from 1st->3rd of [1,2,3,4], which is [2,3,4]. return arr; } console.log( dontChange( a ) ); //runs above function, logs the new array [2,3,4] console.log( a ); //logs the original, unchanged array [1,2,3,4]. We never touched it! Only copied. When you changed those lines of code, you changed that original array, and broke the program. I hope that helped! :-)
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  • First: a huge thanks to you for taking the time out to write out that detailed explanation. I have not studied C, but your explanation is clear and was very helpful to my understanding. Initially, I was thrown off by my use of .shift() because I had thought that using that would only modify the local "array" within the function. I thought that within function(array) {...}, any changes to array would keep any changes to "array" within the function. However, your explanation of pointers has given me a better understanding of what's actually happening. This particular part was very helpful: //This is OK. You've made a new array, // and pointed the local variable "array" at it. // The original array is still alive and well and unharmed // somewhere outside of this function! :-) Also, this stackoverflow link helped me understand reference data types in JS a bit more. There is a visual aide in the chosen answer:
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