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Created January 29, 2012 14:34
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assignment of object properties

140byt.es

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Object.prototype.pro = function r(o, t, k) {
t = t || this;
for (k in o) o[k] == {} + '' ? r(o[k], t[k]) : t[k] = o[k]
}
Object.prototype.pro=function r(o,t,k){t=t||this;for(k in o)o[k]=={}+''?r(o[k],t[k]):t[k]=o[k]}
DO WHAT THE FUCK YOU WANT TO PUBLIC LICENSE
Version 2, December 2004
Copyright (C) 2011 Evghenusi
Everyone is permitted to copy and distribute verbatim or modified
copies of this license document, and changing it is allowed as long
as the name is changed.
DO WHAT THE FUCK YOU WANT TO PUBLIC LICENSE
TERMS AND CONDITIONS FOR COPYING, DISTRIBUTION AND MODIFICATION
0. You just DO WHAT THE FUCK YOU WANT TO.
{
"name": "assignment of object properties",
"description": "assignment of object properties",
"keywords": [
"object",
"css",
"properties",
"recursion",
]
}
<!DOCTYPE html>
<title>Foo</title>
<p id="p">color</p>
<script>
Object.prototype.pro=function r(o,t,k){t=t||this;for(k in o)o[k]=={}+''?r(o[k],t[k]):t[k]=o[k]}
var p = document.getElementById('p');
p.pro({
innerHTML:'new text/html',
style:{backgroundColor:'Silver',color:'red',border:'1px solid',fontStyle:'italic'}
});
</script>
@Evghenusi
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or function is not worth the attention?

@tsaniel
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tsaniel commented Jan 29, 2012

Try this:
Object.prototype.prop=function f(a,b,c){b=b||this;for(c in a)a[c]=={}+''?f(a[c],b[c]):b[c]=a[c]}

By the way, i prefer prop over pro.

@Evghenusi
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Thank tsaniel.
I missed the attention that it is possible register the function Object.prototype.prop=function f()
{}+"" tried, but with typeof..... my inattention ((

@Evghenusi
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,c can be removed

@atk
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atk commented Jan 30, 2012

I suggest using some spare bytes to check for hasOwnProperty; since you already added one prototypic method to Object...

@tsaniel
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tsaniel commented Jan 30, 2012

Also, k leaks to global scope.

@atk
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atk commented Jan 30, 2012

@tsaniel: nope, because it's defined as function argument.

@tsaniel
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tsaniel commented Jan 30, 2012

@atk: true. Because @Evghenusi has just changed it.

@Evghenusi
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guilty)))

@atk
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atk commented Jan 31, 2012

that's ok. Not convicted, in that case ;-)

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