This question already has answers here:
Integer summing blues, short += short problem
(5 answers)
Closed 6 years ago.
Why is
byte someVar;
someVar -= 3;
valid but
byte someVar;
someVar = someVar - 3;
isnt?
Surprisingly, when you perform operations on bytes the computations will be done using int values, with the bytes implicitly cast to (int) first. This is true for shorts as well, and similarly floats are up-converted to double when doing floating-point arithmetic.
The second snippet is equivalent to:
byte someVar;
someVar = (int) someVar - 3;
Because of this you must cast the result back to (byte) to get the compiler to accept the assignment.
someVar = (byte) (someVar - 3);
Here's a copy of a table in the CLI specification (Ecma 335) that specifies which operands are valid on the binary numeric operators of the type A op B, where A and B are the operands and "op" is the operator, like Opcodes.Sub that you are using in your snippet:
Some annotations are required with this:
"native int" is IntPtr in a C# program
F represents a floating point type, double or float in C#
& represents a pointer value, the boxes are shaded because they are unsafe operations
O represents an object reference
x is an operation that's not permitted.
Note the row and column for F, both operands must be floating point, you cannot directly add, say, an int to a double. The C# compiler deals with that limitation by automatically converting the int operand to double so that the operator is valid.
Relevant to your question: also note that the byte, sbyte, char, short and ushort types are not present. Same approach, the compiler converts the operands to the smallest type that can represent the value so that the operator can be used. Which will be int32. According to the table, the result of the operation will be int32.
Now here's the rub: the result is int32 but assigning that back to a byte value requires a narrowing conversion. From 32 bits to 8 bits. That's trouble because it loses significant bits. The C# compiler requires you to make that explicit. You essentially acknowledge that you know what you're doing and that you are aware of the potentially surprising result. Like this one:
byte v = 255;
v = (byte)(v + 1);
The -= operator is a problem because there is no effective way to apply that required cast. It isn't expressible in the language syntax. Using (byte)3 doesn't make sense, the literal gets converted to int32 anyway to make the operator work.
They punted the problem, the compiler automatically emits the cast without your help.
Related
First, this question has related posts:
Why Int32 maximum value is 0x7FFFFFFF?
However, I want to know why the hexadecimal value is always treated as an unsigned quantity.
See the following snippet:
byte a = 0xFF; //No error (byte is an unsigned type).
short b = 0xFFFF; //Error! (even though both types are 16 bits).
int c = 0xFFFFFFFF; //Error! (even though both types are 32 bits).
long d = 0xFFFFFFFFFFFFFFFF; //Error! (even though both types are 64 bits).
The reason for the error is because the hexadecimal values are always treated as unsigned values, regardless of what data-type they are stored as. Hence, the value is 'too large' for the data-type described.
For instance, I expected:
int c = 0xFFFFFFFF;
To store the value:
-1
And not the value:
4294967295
Simply because int is a signed type.
So, why is it that the hexadecimal values are always treated as unsigned even if the sign type can be inferred by the data-type used to store them?
How can I store these bits into these data-types without resorting to the use of ushort, uint, and ulong?
In particular, how can I achieve this for the long data-type considering I cannot use a larger signed data-type?
What's going on is that a literal is intrinsically typed. 0.1 is a double, which is why you can't say float f = 0.1. You can cast a double to a float (float f = (float)0.1), but you may lose precision. Similarly, the literal 0xFFFFFFFF is intrinsically a uint. You can cast it to an int, but that's after it has been interpreted by the compiler as a uint. The compiler doesn't use the variable to which you are assigning it to figure out its type; its type is defined by what sort of literal it is.
They are treated as unsigned numbers, as that is what the language specification says to do
This question already has answers here:
How to use C#'s ternary operator with two byte values?
(4 answers)
Closed 7 years ago.
The following code works.
byte b = 1;
But I noticed the following code doesn't work
byte b = BooleanProperty ? 2 : 3; // error
The compiler says
Cannot convert source type 'int' to target type 'byte'
I understand that int type cannot be converted to byte type implicitly.
But why the former code works, and the latter doesn't?
There's an implicit conversion from int constants (not just literals, but any compile-time constant expression of type int) to byte, so long as the value is in range. This is from section 6.1.9 of the C# 5 specification:
An implicit constant expression conversion permits the following conversions:
A constant-expression (§7.19) of type int can be converted to type sbyte, byte, short, ushort, uint, or ulong, provided the value of the constant-expression is within the range of the destination type.
However, there's no implicit conversion from a "general" expression of type int to byte - and that's what you've got in your second case. It's a bit like this:
int tmp = BooleanProperty ? 2 : 3;
byte b = tmp; // Not allowed
Note that the use of the conditional expression doesn't play any part in inferring its type - and as both the second and third operands are of type int, the overall expression is of type int as well.
So if you understand why the snippet above where I've separated the code into two statements doesn't compile, that explains why your single-line version with the conditional doesn't either.
There are two ways of fixing it:
Change the second and third operands to expressions of type byte so that the conditional expression has an overall type of byte:
byte b = BooleanProperty ? (byte) 2 : (byte) 3;
Cast the result of the conditional expression:
byte b = (byte) (BooleanProperty ? 2 : 3);
When it comes to literal integers in C#.
If the literal has no suffix, it has the first of these types in which
its value can be represented: int, uint, long, ulong.
Compiler is smart enough to deduce that in case of byte b = 1; the literal fits the byte type. But it's not smart enough to figure it out in case of the conditional (ternary) operator ?:.
Could someone point me, why here:
Byte b = 100;
b = (Byte)(b+200);
I have to use explicit type conversion. But here
Byte b = 100;
b += 200;
I don't need to do this?
Does compiler generate different IL code for this two cases? And which case is better?
Because the standard permits it (see the second case below):
14.14.2 Compound assignment
An operation of the form x op= y is processed by applying binary operator overload resolution (§14.2.4) as if the operation was written x op y. Then,
If the return type of the selected operator is implicitly convertible to the type of x, the operation is evaluated as x = x op y, except that x is evaluated only once.
Otherwise, if the selected operator is a predefined operator, if the return type of the selected operator is explicitly convertible to the type of x, and if y is implicitly convertible to the type of x or the operator is a shift operator, then the operation is evaluated as x = (T)(x op y), where T is the type of x, except that x is evaluated only once.
Otherwise, the compound assignment is invalid, and a compile-time error occurs.
The IL code should be essentially identical in this case. Of course, if evaluating b has side effects, it will be evaluated twice in the b = (byte)b + 200 case and only once when using compound assignment.
This is a FAQ in the C# tag, hard to find the duplicate. The need for the cast is relevant first. The underlying reason is that the CLI only specifies a limited number of valid types for the Opcodes.Add IL instruction. Only operands of type Int32, Int64, Single, Double and IntPtr are supported. IntPtr is special as well, the C# language forbids using that one.
So the C# compiler has to use an implicit conversion to uplift the byte to a type that the operator supports. It will pick Int32 as the closest compatible type. The result of the addition is Int32. Which does not fit back into a byte without truncating the result, throwing away the extra bits. An obvious example is 255 + 1, the result is 256 in Int32 but doesn't fit a Byte and yields 0 when stored.
That's a problem, the language designers didn't like that truncation to happen without you explicitly acknowledging that you are aware of the consequences. A cast is required to convince the compiler that you're aware. Bit of a cop-out of course, you tend to produce the cast mechanically without thinking much about the consequences. But that makes it your problem, not Microsoft's :)
The rub was the += operator, a very nice operator to write condense code. Resembles the brevity of the var keyword. Rock and a hard place however, where do you put the cast? It just doesn't work so they punted the problem and allowed truncation without a cast.
Notable is the way VB.NET works, it doesn't require a cast. But it gives a guarantee that C# doesn't provide by default, it will generate an OverflowException when the result doesn't fit. Pretty nice, but that check doesn't come for free.
Designing clean languages is a very hard problem. The C# team did an excellent job, warts not withstanding. Otherwise the kind of warts brought on by processor design. IL has these type restrictions because that's what real 32-bit processors have too, particularly the RISC designs that were popular in the 90s. Their internal registers can only handle 32-bit integers and IEEE-754 floating point. And only permit smaller types in loads and stores. The Intel x86 core is very popular and actually permits basic operations on smaller types. But that's mostly a historical accident due to Intel keeping the design compatible through the 8-bit 8080 and 16-bit 8086 generations. It doesn't come for free, 16-bit operations costs an extra cpu cycle. To be avoided.
The spec calls this out as a specific case of the compound assignment operator:
http://msdn.microsoft.com/en-us/library/aa691316%28v=vs.71%29.aspx
Specifically, bullet 2:
Otherwise, if the selected operator is a predefined operator, if the
return type of the selected operator is explicitly convertible to the
type of x, and if y is implicitly convertible to the type of x, then
the operation is evaluated as x = (T)(x op y), where T is the type of
x, except that x is evaluated only once.
The second rule above permits x op= y to be evaluated as x = (T)(x op
y) in certain contexts. The rule exists such that the predefined
operators can be used as compound operators when the left operand is
of type sbyte, byte, short, ushort, or char. Even when both arguments
are of one of those types, the predefined operators produce a result
of type int, as described in Section 7.2.6.2. Thus, without a cast it
would not be possible to assign the result to the left operand.
This because of the implicit conversion rules.
When you have a binary + operator the result is converted to the larger of the two types.
The literal 200 is of type int and therefore the type of the expression b+200 is int.
The assignment operator = doesn't do implicit conversion, but rather throws error. As in
int x = 10;
Byte b = x; //Error
In the second case the += operator expects byte, so 200 (which is of type int, but fits into byte) is implicitly converted to byte, because the compiler knows it can. The following won't compile, because the compile doesn't know if x will fit in a byte or not.
Byte b = 100;
int x = 200;
b += x; //Error
If you make x a const then it compiles:
Byte b = 100;
const int x = 200;
b += x; //OK
This question already has answers here:
byte + byte = int... why?
(16 answers)
Closed 5 years ago.
Why does the following raise a compile time error: 'Cannot implicitly convert type 'int' to 'byte':
byte a = 25;
byte b = 60;
byte c = a ^ b;
This would make sense if I were using an arithmentic operator because the result of a + b could be larger than can be stored in a single byte.
However applying this to the XOR operator is pointless. XOR here it a bitwise operation that can never overflow a byte.
using a cast around both operands works:
byte c = (byte)(a ^ b);
I can't give you the rationale, but I can tell why the compiler has that behavior from the stand point of the rules the compiler has to follow (which might not really be what you're interesting in knowing).
From an old copy of the C# spec (I should probably download a newer version), emphasis added:
14.2.6.2 Binary numeric promotions This clause is informative.
Binary numeric promotion occurs for
the operands of the predefined +, ?,
*, /, %, &, |, ^, ==, !=, >, <, >=, and <= binary operators. Binary
numeric promotion implicitly converts
both operands to a common type which,
in case of the non-relational
operators, also becomes the result
type of the operation. Binary numeric
promotion consists of applying the
following rules, in the order they
appear here:
If either operand is of type decimal, the other operand is
converted to type decimal, or a
compile-time error occurs if the other
operand is of type float or double.
Otherwise, if either operand is of type double, the other operand is
converted to type double.
Otherwise, if either operand is of type float, the other operand is
converted to type float.
Otherwise, if either operand is of type ulong, the other operand is
converted to type ulong, or a
compile-time error occurs if the other
operand is of type sbyte, short, int,
or long.
Otherwise, if either operand is of type long, the other operand is
converted to type long.
Otherwise, if either operand is of type uint and the other operand is of
type sbyte, short, or int, both
operands are converted to type long.
Otherwise, if either operand is of type uint, the other operand is
converted to type uint.
Otherwise, both operands are converted to type int.
So, basically operands smaller than an int will be converted to int for these operators (and the result will be an int for the non-relational ops).
I said that I couldn't give you a rationale; however, I will make a guess at one - I think that the designers of C# wanted to make sure that operations that might lose information if narrowed would need to have that narrowing operation made explicit by the programmer in the form of a cast. For example:
byte a = 200;
byte b = 100;
byte c = a + b; // value would be truncated
While this kind of truncation wouldn't happen when performing an xor operation between two byte operands, I think that the language designers probably didn't want to have a more complex set of rules where some operations would need explicit casts and other not.
Just a small note: the above quote is 'informational' not 'normative', but it covers all the cases in an easy to read form. Strictly speaking (in a normative sense), the reason the ^ operator behaves this way is because the closest overload for that operator when dealing with byte operands is (from 14.10.1 "Integer logical operators"):
int operator ^(int x, int y);
Therefore, as the informative text explains, the operands are promoted to int and an int result is produced.
FWIW
byte a = 25;
byte b = 60;
a = a ^ b;
does not work. However
byte a = 25;
byte b = 60;
a ^= b;
does work.
The demigod programmer from Microsoft has an answer: Link
And maybe it's more about compiler design. They make the compiler simpler by generalizing the compiling process, it doesn't have to look at operator of operands, so it lumped bitwise operations in the same category as arithmetic operators. Thereby, subjected to type widening
Link dead, archive here:
https://web.archive.org/web/20140118171646/http://blogs.msdn.com/b/oldnewthing/archive/2004/03/10/87247.aspx
I guess its because the operator XOR is defined for booleans and integers.
And a cast of the result from the integer result to a byte is an information-losing conversion ; hence needs an explicit cast (nod from the programmer).
It seems to be because in C# language specifications, it is defined for integer and long
http://msdn.microsoft.com/en-us/library/aa691307%28v=VS.71%29.aspx
So, what actually happens is that compiler casts byte operands to int implicitly because there is no loss of data that way. But the result (which is int) can not be down-cast-ed without loss of data (implicitly). So, you need to tell the compiler explicitly that you know what you are doing!
As to why the two bytes have to be converted to ints to do the XOR?
If you want to dig into it, 12.1.2 of the CLI Spec (Partition I) describes the fact that, on the evaluation stack, only int or long can exist. All shorter integral types have to be expanded during evaluation.
Unfortunately, I can't find a suitable link directly to the CLI Spec - I've got a local copy as PDF, but can't remember where I got it from.
This has more to do with the rules surrounding implicit and explicit casting in the CLI specification. An integer (int = System.Int32 = 4 bytes) is wider than a byte (1 byte, obviously!). Therefore any cast from int to byte is potentially a narrowing cast. Therefore, the compiler wants you to make this explicit.
This question already has answers here:
byte + byte = int... why?
(16 answers)
Closed 9 years ago.
I have to following code in VS2008 .net 3.5 using WinForms:
byte percent = 70;
byte zero = 0;
Bitmap copy = (Bitmap)image1.Clone();
...
Color oColor = copy.GetPixel(x, y);
byte oR = (byte)(oColor.R - percent < zero ? zero : oColor.R - percent);
When I leave the "(byte)" off the last line of code, I get a compiler error saying it "Cannot implicitly convert type 'int' to 'byte'." If everything is of type byte and byte is an integer type... then why do I need to have the cast?
Because subtraction is coercing up to an integer. As I recall, byte is an unsigned type in C#, so subtraction can take you out of the domain of bytes.
That's because the result of a byte subtraction doesn't fit in a byte:
byte - byte = (0..255) - (0..255) = -255..255
Arithmetic on bytes results in an int value by default.
Because arithmetic on bytes returns integers by default so of the two possible assignments the narrower type of zero (byte) is promoted to int (that of oColor.r - percent). Thus the type of the operation is an int. The compiler will not, without a cast, allow you to assign a wider type to a narrower type, because it's a lossy operation. Hence you get the error, unless you explicitly say "I know I'm losing some data, it's fine" with the cast.
This is because byte subtraction returns an int. Actually any binary arithmetic operations on bytes will return an int, so the cast is required.
Because arithmetic operations on sbyte, byte, ushort, and short are automatically converted to int. The most plausible reason for this is that such operations are likely to overflow or underflow.
Thus, in your ternary operation, the final oColor.R - percent, actually results in an int, not a byte. So the return type of the operation is int.
Because the arithmetic expression on the right-hand side of the assignment operator evaluates to int by default. In your example percent defaults to int. You can read more about it on the MSDN page.
Try to run this C# code:
object o = (byte)(1);
o = (int)o;
What do you expect? Now try :)
I think this is right:
Eric Lippert says, "I don't think of bytes as "numbers"; I think of them as patterns of bits that could be interpreted as numbers, or characters, or colors or whatever. If you're going to be doing math on them and treating them as numbers, then it makes sense to move the result into a data type that is more commonly interpreted as a number."
Perhaps byte is closer to char than to int.
FWIW, java also promotes bytes to int.