# arithmetic exception

ronak mehta

Greenhorn

Posts: 23

posted 14 years ago

1) ArithmeticException will occur after execution of the following code

int i = 10;

float f = 0.0;

double d = i/f;

NO.

this is the answer and i can't understand it .i think that float f is 0 so divide by zero give arithmetic exception. may be this answer is becoz it give runtime error not exception am i right?

ronak please help.

int i = 10;

float f = 0.0;

double d = i/f;

NO.

this is the answer and i can't understand it .i think that float f is 0 so divide by zero give arithmetic exception. may be this answer is becoz it give runtime error not exception am i right?

ronak please help.

Marilyn de Queiroz

Sheriff

Posts: 9059

12

posted 14 years ago

integers give an ArithmeticException when division by zero occurs. A float divided by zero will give positive infinity, negative infinity or NaN.

This question does not belong in the JavaRanch forum. The JavaRanch forum is for discussion of the JavaRanch site, not questions about Java. I'm moving this to Java in General (beginner).

This question does not belong in the JavaRanch forum. The JavaRanch forum is for discussion of the JavaRanch site, not questions about Java. I'm moving this to Java in General (beginner).

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Cindy Glass

"The Hood"

Sheriff

Sheriff

Posts: 8521

posted 14 years ago

In theory 0.0f is a rounding of some number with a zillion decimal places but that is not necessarily zero. Therefore the division is legal, just results in a VERY SMALL or VERY LARGE numbers. That is why the answer comes back as POSITIVE_INFINITY.

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