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assignments: too easy?

 
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Howdy ranchers,

the following fragment is not from a mock exam but easily could be:

What will be the output?

Bu.
 
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0.0

1 / 2 -> (int / int = 0.)
0 * 0.5 = 0.
[ July 18, 2007: Message edited by: Thitipong Suparurkrat ]
 
Burkhard Hassel
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But there's a double in the expression. Shouldn't they all become double then?

Bu.
 
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I think it will execute like this. Correct me if I am wrong. I have not tried running it .

double d = 1 / 2 * 0.5 ;

First of all , 2*0.5 will be executed. That will result in 1.0 .

Then 1/1.0 will execute. 1 is int and 1.0 is double hence int will be broadened to double and answer will be 1.0

Prints : 1.0
 
Vivian Josh
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Now, I tried running in eclipse .

Its giving 0.0

Why's that ? Is it first executing 1/2 as int and then multiplying 0 and 0.5 ?
 
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output: 0.25 (predicted output)

So I agree with Thitipong 1/2 int/int got excuted first result in 0 then got multiplied by 0.5
 
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Hi,

evaluation order is always from left to right.
 
Burkhard Hassel
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Manfred posted

evaluation order is always from left to right.



System.err.println(1 / 2 * 0.5);
System.out.println(0.5 * 1 / 2);


prints
0.0
0.25

Bu.
 
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when an expression involves operators of equal precedence then the evaluation always starts from left to right unless overriden by using parenthesis.
Hence in 1/2*.5 '/' is evaluated and since both operands are int, hence integer division takes place.Hence O/P is 0.

In 0.5 * 1 / 2---> '*' gets evalutaed and since one of the operand after multiplication is floating point hence 2 also gets promoted to 2.0
Hence O/P is .5
 
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well!!!
The above question is confusing me
please anyone can explain how integer division result in 0


as it is explained above int/int :>result in 0.......how this is possible?

what i think is if execution start from left the answer should be

1/2*0.5

first step

1/2........results in :>0.5......as this value is getting assigned to a double.......it will be double ie 0.5

later in second step..........0.5*0.5 results in 1.0...........
[ July 19, 2007: Message edited by: dhwani mathur ]
 
Burkhard Hassel
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Howdy ranchers,

dhwani mathur posted
T

he above question is confusing me
please anyone can explain how integer division result in 0
(...)
first step

1/2........results in :>0.5




There it is. The expression 1 / 2 * 0.5 is not evaluated all in once, but pairwise from left to right.
First comes 1/2.

The compiler finds two integer literals, therefore the result is also an integer (and not a double!)
1/2 ---> 0.5 , but only the integer part of it will reside in memory, and that is 0 without a dot and without the halve (not 0.5!).

Next step (the one with the X)

Here the compiler finds an integer zero in memory and a double literal from the code.
Therefore this pair is converted to double:
0.0 * 0.5

Hence 0.0



BTW, when dividing integers, you always loose the part right of the decimal point:
System.out.println (3/2);
results in 1


Bu.
 
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