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byte b = (byte) 27; // Explicitly cast the int literal to a byte
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It looks as though the compiler gives you a break, and let s you take a shortcut with assignments to integer variables smaller than an int. (Everything we re saying about byte applies equally to char and short, both of which are smaller than an int.) We re not actually at the weird part yet, by the way. We know that a literal integer is always an int, but more importantly the result of an expression involving anything int-sized or smaller is always an int. In other words, add two bytes together and you ll get an int even if those two bytes are tiny. Multiply an int and a short and you ll get an int. Divide a short by a byte and you ll get an int. OK, now we re at the weird part. Check this out:
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byte b = 3; // No problem, 3 byte c = 8; // No problem, 8 byte d = b + c; // Should be // fits in a fits in a byte fits in a byte no problem, sum of the two bytes byte
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The last line won t compile! You ll get the following error:
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TestBytes.java:5: possible loss of precision found : int required: byte byte c = a + b; ^
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We tried to assign the sum of two bytes to a byte variable, the result of which (11) was definitely small enough to fit into a byte, but the compiler didn t care. It knew the rule about int-or-smaller expressions always resulting in an int. It would have compiled if we d done the explicit cast:
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byte c = (byte) (a + b);
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Assigning Floating-Point Numbers Floating-point numbers have slightly different assignment behavior than integer types. We ve already discussed this in 1, but we ll do another quick review here while we re on the subject. First, you must know that every floating-point literal is implicitly a double (64 bits), not a float. So the literal 2.3, for example, is considered a double. If you try to assign a double to a float, the compiler knows you don t have enough room in a 32-bit float container to hold the precision of a 64-bit double, and it lets you know. The following code looks good, but won t compile:
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float f = 32.3;
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You can see that 32.3 should fit just fine into a float-sized variable, but the compiler won t allow it. In order to assign a floating-point literal to a float variable, you must either cast the value or append an f to the end of the literal. The following assignments will compile:
float f = (float) 32.3; float g = 32.3f; float h = 32.3F;
Assigning a Literal That Is Too Large for the Variable We ll also get a compiler error if we try to assign a literal value that the compiler knows is too big to fit into the variable.
byte a = 128; // byte can only hold up to 127
3: Operators and Assignments
The preceding code gives us this error:
TestBytes.java:5: possible loss of precision found : int required: byte byte a = 128;
We can fix it with a cast:
byte a = (byte) 128;
But then what s the result When you narrow a primitive, Java simply truncates the higher-order bits that won t fit. In other words, it loses all the bits to the left of the bits you re narrowing to. Let s take a look at what happens in the preceding code. There, 128 is the bit pattern 10000000. It takes a full 8 bits to represent 128. But because the literal 128 is an int, we actually get 32 bits, with the 128 living in the right-most (lower-order) 8 bits. So a literal 128 is actually
Take our word for it; there are 32 bits there. To narrow the 32 bits representing 128, Java simply lops off the leftmost (higher-order) 24 bits. We re left with just the 10000000. But remember that a byte is signed, with the leftmost bit representing the sign (and not part of the value of the variable). So we end up with a negative number (the 1 that used to represent 128 now represents the negative sign bit). Remember, to find out the value of a negative number using two s complement notation, you flip all of the bits and then add 1. Flipping the 8 zeroes give us: 01111111, and adding 1 to that gives us 10000000, or back to 128! And when we apply the sign bit, we end up with -128. You must use an explicit cast to assign 128 to a byte, and the assignment leaves you with the value -128. A cast is nothing more than your way of saying to the compiler, Trust me. I m a professional. I take full responsibility for anything weird that happens when those top bits are chopped off. That brings us to the compound assignment operators. The following will compile,
byte b = 3; b += 7; // No problem - adds 7 to b (result is 10)
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