How can this code be explained ? byte b1 = 5; works fine . When Integral literals are considered by default to be 32 bit literals then 5 is represented in 32 bits . Is it not an error to put 32 bits explicitly into a 8 bit variable . Somebody please help me in understanding this . (With reference to float f = 1.3 giving an error because all floating point literals are 64 bit values )
Cheers !
nupur dhawan
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this works fine as byte b1 = 5, assigns b1 a value of 5 which is within the range of a byte , that is , -127 to 128 .
Nupur. <br />SCJP2.
nupur dhawan
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4.2.1 Integral Types and Values The values of the integral types are integers in the following ranges: For byte, from -128 to 127, inclusive For short, from -32768 to 32767, inclusive For int, from -2147483648 to 2147483647, inclusive For long, from -9223372036854775808 to 9223372036854775807, inclusive For char, from '\u0000' to '\uffff' inclusive, that is, from 0 to 65535
Howard Roarke
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I understand the limits for Integer data types and all . My question is this In Java float f = 1.0 is not allowed . Why ? Because 1.0 is considered by default to be a 64 bit double. It has also been told that integer literals are by default 32 bit literals . What I am asking is why does the same rule not apply here ?
Dirk Schreckmann
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Marilyn de Queiroz
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Originally posted by Smart Word: I understand the limits for Integer data types and all .
My question is this
In Java float f = 1.0 is not allowed . Why ? Because 1.0 is considered by default to be a 64 bit double. It has also been told that integer literals are by default 32 bit literals . What I am asking is why does the same rule not apply here ?
Because they are stored differently. An int is stored like 0000 0000 0000 0000 0000 0000 0000 0005 so losing a few leading zeros to store it (the constant) in a byte is ok.
However, floats and doubles are stored in something similar to scientific notation so they can't be "chopped off" to fit into a smaller slot.
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