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A more compact way to do this?

I have a couple of functions that move around the surrounding cells of a cell. The grid is contained within the array.

In my code, I have checks to make sure it is not one of the extreme cells, as checking a cell for undefined throws an error.

Thus, I have code like this:

if(x > 0) {
    var firstX = x - 1;
} else {
    var firstX = x;
}
if(x < 199) {
    var lastX = x + 1;
} else {
    var lastX = x;
}

if(y > 0) {
    var firstY = y - 1;
} else {
    var firstY = y;
}
if(y < 199) {
    var lastY = y + 1;
} else {
    var lastY = y;
}

Many lines of code do very little. Is there a more elegant way to do this?

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4 answers


You can use the conditional operator:

var firstX = x > 0 ? x - 1 : x;
var lastX = x < 199 ? x + 1 : x;
var firstY = y > 0 ? y - 1 : y;
var lastY = y < 199 ? y + 1 : y;


You can remove the redundancy by writing a function to calculate the "first" given the value as well as the "last" - but I think that would be overkill in this case.

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Or more clearly:



var firstX = Math.max(x - 1, 0);
var lastX = Math.min(x + 1, 199);
var firstY = Math.max(y - 1, 0);
var lastY = Math.min(y + 1, 199);
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You can use the conditional operator:

var firstX = x - (x > 0 ? 1:0);
var lastX = x + (x < 199 ? 1:0);
var firstY = y - (y > 0 ? 1:0);
var lastY = y + (y < 199 ? 1:0);

Edit:
An alternative way of using it is suggested as John has already posted "my" code.;)

Edit 2:
As Raphael pointed out, the condition can be implicitly converted to a number, so the conditional statement is not needed:

var firstX = x - (x > 0);
var lastX = x + (x < 199);
var firstY = y - (y > 0);
var lastY = y + (y < 199);

However, it is less obvious what this code actually does. From my tests it seems that Javascript consistently uses the value 1 for true, but in all programming languages ​​the value -1 is also widely used.

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Use the variables you are checking (x and y) instead of First / lastX and First / lastY

if(x > 0 && x < 199) x-=1;
else if(x > 0) x+=1;

if(y > 0 && y < 199) y-=1;
else if(y > 0) y +=1;

Just check x and y after that. :)

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