简易可变范围问题...... [英] Easy Variable Scope question...
问题描述
我应该是一个简单的范围问题。你可以帮我解决这个问题吗？
我想这样结束：
originalArray = [1,2,7]和newArray = [6,7,12]。
相反，我这样结束：
originalArray = [6,7,12]和newArray = [6， 7,12]。
这是我的代码：
var originalArray = [1,2,7]; / *我从表单中提取这个并希望
保持原样，但是函数修改它* /
var newArray = createNewArray（originalArray）; / *我想结束
with newArray = [6,7,12] * /
function createNewArray（x）{
for（var i = 0; i< x.length; i ++）{x [i] + = 5}
返回x;
}
我也试过这个：
函数createNewArray（x）{
var y = x;
for（var i = 0; i< y.length; i ++）{y [i] + = 5}
返回y;
}
和这个：
var originalArray = [1,2,3];
var tmpArray = originalArray;
var newArray = createNewArray（tmpArray）;
函数createNewArray（x）{
for（var i = 0; i< x.length; i ++）{x [i] + = 5}返回x;
}
所有具有相同的灾难性结果。
谢谢！
Mike
I have, what should be, a simple scope problem. Can you help me fix this?
I''m trying to end up like this:
originalArray = [1,2,7] and newArray = [6,7,12].
Instead I wind up like this:
originalArray = [6,7,12] and newArray = [6,7,12].
Here''s my code:
var originalArray = [1,2,7]; /* I''m pulling this from a form and want
to keep it as is, but the function modifies it */
var newArray = createNewArray( originalArray ); /* I want to wind up
with newArray = [6,7,12] */
function createNewArray( x ) {
for ( var i = 0; i < x.length; i++ ) { x[i] += 5 }
return x;
}
I''ve also tried this:
function createNewArray( x ) {
var y = x;
for ( var i = 0; i < y.length; i++ ) { y[i] += 5 }
return y;
}
and this:
var originalArray = [1,2,3];
var tmpArray = originalArray;
var newArray = createNewArray( tmpArray );
function createNewArray( x ) {
for ( var i = 0; i < x.length; i++ ) { x[i] += 5 } return x;
}
all with the same disasterous results.
Thanks!
Mike
推荐答案
Mike P写道：
Mike P wrote:
我有，应该是什么，一个简单的范围问题。你能帮我解决这个问题吗？
我想这样结束：
originalArray = [1,2,7]和newArray = [6,7,12]。
相反，我这样结束：
originalArray = [6,7,12]和newArray = [6,7,12]。
这里''我的代码：
var originalArray = [1,2,7]; / *我是从表单中提取这个并希望
保持原样，但函数修改它* /
var newArray = createNewArray（originalArray）; / *我想结束使用newArray = [6,7,12] * /
函数createNewArray（x）{
for（var i = 0; i< ; x.length; i ++）{x [i] + = 5}
返回x;
}
我也试过了：
function createNewArray（x）{
var y = x;
for（var i = 0; i< y.length; i ++）{y [i] + = 5}
返回y;
}
这个：
var originalArray = [1,2,3];
var tmpArray = originalArray;
var newArray = createNewArray（tmpArray）;
函数createNewArray（x）{
for（var i = 0; i< x.length; i ++）{x [i] + = 5}返回x;
}
所有相同的灾难结果。
谢谢！
Mike
I have, what should be, a simple scope problem. Can you help me fix this?
I''m trying to end up like this:
originalArray = [1,2,7] and newArray = [6,7,12].
Instead I wind up like this:
originalArray = [6,7,12] and newArray = [6,7,12].
Here''s my code:
var originalArray = [1,2,7]; /* I''m pulling this from a form and want
to keep it as is, but the function modifies it */
var newArray = createNewArray( originalArray ); /* I want to wind up
with newArray = [6,7,12] */
function createNewArray( x ) {
for ( var i = 0; i < x.length; i++ ) { x[i] += 5 }
return x;
}
I''ve also tried this:
function createNewArray( x ) {
var y = x;
for ( var i = 0; i < y.length; i++ ) { y[i] += 5 }
return y;
}
and this:
var originalArray = [1,2,3];
var tmpArray = originalArray;
var newArray = createNewArray( tmpArray );
function createNewArray( x ) {
for ( var i = 0; i < x.length; i++ ) { x[i] += 5 } return x;
}
all with the same disasterous results.
Thanks!
Mike
嗨Mike，
您非常接近解决方案，从上面的
获取您的解决方案之一并修改它如下工作：
函数createNewArray（x）
{
var y = new Array（）;
for（ var i = 0;我< x.length; i ++）
{
y [i] = x [i] + 5
}
返回y;
}
var originalArray = [1,2,7];
var newArray = createNewArray（originalArray） ;
你错过了以下内容：
var y = new Array（）;
这是因为你引用了导致
问题的相同数组。如果您明确创建了一个新数组，那将解决您的问题。
Hi Mike,
You were very close to the solution, taking one of your solutions from
above and modified it as follows will work:
function createNewArray(x)
{
var y = new Array();
for (var i = 0; i < x.length; i++)
{
y[i] = x[i] + 5
}
return y;
}
var originalArray = [1, 2, 7];
var newArray = createNewArray(originalArray);
You were missing the following:
var y = new Array();
It was because you were referencing the same Array that was causing the
problem. If you explicitly created a new array, that will solve your
problem.
Mike P说：
我有，应该是什么，一个简单的范围问题。你能帮我解决这个问题吗？
我想这样结束：
originalArray = [1,2,7]和newArray = [6,7,12]。
相反，我这样结束：
originalArray = [6,7,12]和newArray = [6,7,12]。
这里''我的代码：
var originalArray = [1,2,7]; / *我是从表单中提取这个并希望
保持原样，但函数修改它* /
var newArray = createNewArray（originalArray）; / *我想结束使用newArray = [6,7,12] * /
函数createNewArray（x）{
for（var i = 0; i< ; x.length; i ++）{x [i] + = 5}
返回x;
}
I have, what should be, a simple scope problem. Can you help me fix this?
I''m trying to end up like this:
originalArray = [1,2,7] and newArray = [6,7,12].
Instead I wind up like this:
originalArray = [6,7,12] and newArray = [6,7,12].
Here''s my code:
var originalArray = [1,2,7]; /* I''m pulling this from a form and want
to keep it as is, but the function modifies it */
var newArray = createNewArray( originalArray ); /* I want to wind up
with newArray = [6,7,12] */
function createNewArray( x ) {
for ( var i = 0; i < x.length; i++ ) { x[i] += 5 }
return x;
}
无论你怎么分配一个数组变量到另一个变量，
你只是分配对原始数组的引用。
你需要创建一个全新的数组：
函数createNewArray（x）{
var y = new Array（x.length）;
for（var i = 0; i< x.length; i ++）{
y [i] = x [i] +5;
}
返回y;
}
No matter how you assign an array variable to another variable,
you are only assigning a reference to the original array.
You need to create an entirely new array:
function createNewArray(x) {
var y=new Array(x.length);
for(var i=0;i<x.length;i++) {
y[i]=x[i]+5;
}
return y;
}
谢谢......得到了它。
Mike
Thanks... got it :)
Mike
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