The page I am working on has a javascript function executed to print parts of the page. For some reason, printing in Safari, causes the window to somehow update. I say somehow, because it does not really refresh as in reload the page, but rather it starts the "rendering" of the page from start, i.e. scroll to top, flash animations start from 0, and so forth. The effect is reproduced by this fiddle: http://jsfiddle.net/fYmnB/ Clicking the print button and finishing or cancelling a print in Safari causes the screen to "go white" for a sec, which in my real website manifests itself as something "like" a reload. While running print button with, let's say, Firefox, just opens and closes the print dialogue without affecting the fiddle page in any way. Is there something with my way of calling the browsers print method that causes this, or how can it be explained - and preferably, avoided? P.S.: On my real site the same occurs with Chrome. In the ex
As always with these questions, the JLS holds the answer. In this case §15.26.2 Compound Assignment Operators. An extract:
ReplyDeleteA compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.
And an example:
For example, the following code is correct:
short x = 3;
x += 4.6;
and results in x having the value 7 because it is equivalent to:
short x = 3;
x = (short)(x + 4.6);
In other words, your assumption is correct.
A good example of this casting is using *= or /=
ReplyDeletebyte b = 10;
b *= 5.7;
System.out.println(b); // prints 57
or
byte b = 100;
b /= 2.5;
System.out.println(b); // prints 40
or
char ch = '0';
ch *= 1.1;
System.out.println(ch); // prints '4'
or
char ch = 'A';
ch *= 1.5;
System.out.println(ch); // prints 'a'
Very good question. The Java Language specification confirms your suggestion.
ReplyDeleteFor example, the following code is correct:
short x = 3;
x += 4.6;
and results in x having the value 7 because it is equivalent to:
short x = 3;
x = (short)(x + 4.6);
Yes,
ReplyDeletebasically when we are writing
i+=l;
compiler is converting this to
i = (int)(i + l);
I just checked the .class file code.
really a good thing to know
you need to cast from long to int explicitly in case of i = i + l then it will compile and give correct output. like
ReplyDeletei = i + (int)l;
or
i = (int)((long)i + l); // this is what happens in case of += , dont need (long) casting since upper casting is done implicitly.
but in case of += it just works fine because the operator implicitly does the type casting from type of right variable to type of left variable so need not cast explicitly.
The problem here is of type casting.
ReplyDeleteWhen you add int and long,
The int object is casted to long & both are added and you get long object.
but long object cannot be implicitly casted to int. So, you have to that explicitly.
But += is coded in such a way that it does type casting. i=(int)(i+m)