If you put the following code in a .NET 4.5 application:
public const long MAXIMUM_RANGE_MAGNITUDE = int.MaxValue + 1;
A compiler error is generated stating "The operation overflows at compile time in checked mode". I know that I could put this in an "unchecked" block and be fine, but my question is why does the error appear in the first place? Clearly a long can hold a int's max value plus one.
Note that using Int32 and Int64 instead of long and int does not seem to help.
It is because the calculations on the right hand side of assignment is being done in integer type. And it is overflowing integer
You can fix that with:
public const long MAXIMUM_RANGE_MAGNITUDE = int.MaxValue + (long)1; // or 1L
By casting at least one of the operand to long
The reason you get the error is specified in C# specifications.
See C# Specification Section 4.1.5 (Integral types)
For the binary +, –, *, /, %, &, ^, |, ==, !=, >, <, >=, and <=
operators, the operands are converted to type T, where T is the first
of int, uint, long, and ulong that can fully represent all possible
values of both operands. The operation is then performed using the
precision of type T, and the type of the result is T (or bool for the
relational operators). It is not permitted for one operand to be of
type long and the other to be of type ulong with the binary operators.
In your case since both operands of addition can be represented in int therefore the calculation is done in integer type. Explicitly casting one of the operand to long would result in long result and thus no overflow error.
Your code in fact looks like this:
(long)(int.MaxValue + 1)
But because .Net framework has an inbuilt implicit conversion between int and long you do not have to explicitly put a cast to long in your code.
So firstly this part of the code is executed:
int.MaxValue + 1
and the result of this operation is an int value which causes and overflow exception.
So your code does not even have a chance to start the conversion from int to long.
I think this has to do with the value of int.MaxValue + 1 being calculated before the cast to long is made. Certainly a long can hold the value, but because you are doing integer addition, there is no way to store the integer value of int.MaxValue + 1 in an int until the cast is made.
try
public const long MAXIMUM_RANGE_MAGNITUDE = (long)int.MaxValue + 1L;
Cast the constant value to a long.
public const long MAXIMUM_RANGE_MAGNITUDE = (long) int.MaxValue + 1;
Related
Based on this interesting question: Addition of int and uint and toying around with constant folding as mentioned in Nicholas Carey's answer, I've stumbled upon a seemingly inconsistent behavior of the compiler:
Consider the following code snippet:
int i = 1;
uint j = 2;
var k = i - j;
Here the compiler correctly resolves k to long. This particular behavior is well defined in the specifications as explained in the answers to the previously referred question.
What was surprising to me, is that the behavior changes when dealing with literal constants or constants in general. Reading Nicholas Carey's answer I realized that the behavior could be inconsistent so I checked and sure enough:
const int i = 1;
const uint j = 2;
var k = i - j; //Compile time error: The operation overflows at compile time in checked mode.
k = 1 - 2u; //Compile time error: The operation overflows at compile time in checked mode.
k in this case is resolved to Uint32.
Is there a reason for the behavior being different when dealing with constants or is this a small but unfortunate "bug" (lack of a better term) in the compiler?
From the C# specification version 5, section 6.1.9, Constant Expressions only allow the following implicit conversions
6.1.9 Implicit constant expression conversions
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.
• A constant-expression of type long can be converted to type ulong, provided the value of the constant-expression is not negative.
Note that long is not on the list of int conversions.
The other half the problem is that only a small number of numeric promotions happen for binary operations:
(From Section 7.3.6.2 Binary numeric promotions):
If either operand is of type decimal, the other operand is converted to type decimal, or a binding-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 binding-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.
REMEMBER: The int to long conversion is forbidden for constants, meaning that both args are instead promoted to uints.
Check out this answer here
The problem is that you are using const.
At run time when there is a const the behavior is exactly as with literals, or as if you had simply hard coded those numbers in the code, so since the numbers are 1 and 2 it casts to a Uint32 since 1 is within the range of uint32. Then when you try to subtract 1 - 2 with uint32 it overflows, since 1u - 2u = +4,294,967,295 (0xFFFFFFFF).
The compiler is allowed to look at litterals, and interpret them different than it would other variables. Since const will never change, it can make guarantees that it otherwise couldn't make. in this instance it can guarentee that 1 is within the range of a uint, therfore it can cast it implicitly. In normal circumstances(without the const) it cannot make that guarantee,
a signed int ranges from -2,147,483,648 (0x80000000) to +2,147,483,647 (0x7FFFFFFF).
an unsigned int ranges from 0 (0x00000000) to +4,294,967,295 (0xFFFFFFFF).
Moral of the story, be careful when mixing const and var, you may get something you don't expect.
I came across this funny behavior of the compiler:
If I have
public int GetInt()
{
Random rnd = new Random();
double d = rnd.NextDouble();
int i = d % 1000;
return i;
}
I get an error of:
Cannot implicitly convert type 'double' to 'int'. An explicit conversion exists (are you missing a cast?)
which actually makes sense as 1000 can be a double, and the result of the modulo operator might be a double as well.
But after changing the code to:
public int GetInt()
{
Random rnd = new Random();
double d = rnd.NextDouble();
int i = d % (int)1000;
return i;
}
The error persists.
As far as I can tell, the compiler has all of the information in order to determine that the output of the modulo operator will be an int, so why doesn't it compile?
if d is == to 1500.72546 then the result of the calculation int d % (int)1000 would be 500.72546 so then implicitly casting to an int would result in a loss of data.
This is by design. See C# language specification:
7.3.6.2 Binary numeric promotions
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:
...
Otherwise, if either operand is of type double, the other operand is converted to type double.
The compiler can tell the result is an integer here, but is being your friend.
Sure, it can automatically convert it for you and not tell you about it and in your trivial example this would be just fine. However, given that converting a double to an int means a LOSS of data, it might well have not been your intention.
If it had not been your intention and the compiler had just gone ahead and done the conversion for you, you could have ended up in a marathon debugging session, devoid of sanity - trying to figure out why a rather esoteric bug as been reported.
This way, the compiler is forcing you to say, as a programmer, "I know I will lose data, it's fine".
Compiler will assume you don't want to loose precision of the operation and implicitly use double.
http://msdn.microsoft.com/en-us/library/0w4e0fzs.aspx
From the documentation:
http://msdn.microsoft.com/en-us/library/0w4e0fzs.aspx
The result of a modulo of a double will be a double. If you need an integer result, then you must:
int i = (int)(d % 1000);
But bear in mind that you are liable to lose data here.
As I slowly engage my brain here - your code doesn't make any sense. NextDouble() will return a value between 0 and 1. There are some logical issues with what you are doing, the result will always be zero, e.g.:
0.15 % 1000 = 0.15
Cast 0.15 to int (always rounds towards zero) -> 0
double d = rnd.NextDouble();
int i = d % (int)1000;
This code doesn't makes sense. (int)1000 says 1000 is int, but d % (int)1000 says that oh d is double, so compiler has to convert both into a common type Binary numeric promotions mentioned in another answer to make it work.
One thing to understand is that you can't apply any operations with different types, so compiler will convert implicitly for you if there is no loss of data. So (int)1000 will be still converted to double by the compiler before applying operation. so the result will be of type double not int.
The following two statements will cause a compiler overflow error ( reason being that constant expressions are by default checked for overflow):
int i=(int)(int.MaxValue+100); // error
long l=(long)(int.MaxValue+100); // error
But if compiler is able to figure out that adding the two values causes an overflow, why doesn't it then promote both int.MaxValue and 100 to long and only then try to add them together? As far as I can tell, that shouldn't be a problem since according to the following quote, the integer literal can also be of type long:
When an integer literal has no suffix,
its type is the first of these types
in which its value can be represented:
int, uint, long, ulong.
thanx
The literal 100 can be represented as an int, which is the first of those four types in that order, so it's made an int.
int.MaxValue is not a literal. It is a public constant field of type int.
So, the addition operation is int + int, which results in an int, which then overflows for this case.
To turn the literal 100 into a long so you perform long integer addition, suffix it with L:
long l = int.MaxValue + 100L;
The rules are:
int + int is int, not long
the default for arithmetic on constants is "checked"; the default for arithmetic on non-constants is "unchecked"
100 and int.MaxValue are constants
Therefore the correct behaviour according to the specification is to do overflow checking at compile time and give an error.
If instead you said:
int x = 100;
int y = int.MaxValue;
long z = x + y;
then the right behaviour is to do an unchecked addition of two integers, wrap around on overflow, and then convert the resulting integer to long.
If what you want is long arithmetic then you have to say so. Convert one of the operands to long.
The reason is all in the sequence. If you read your code, it does: Take an int variable with value MAX, and add 100 to it. This is true in both cases, and this is the code that will be executed before anything else will happen.
If you want to make it work. do
long l = ((long)int.MaxValue)+100;
The short answer, I would imagine, is because it hasn't been designed to. Whenever MS adds features to the C# compiler (or when anyone adds features to anything), there has to be a cost-benefit analysis. People have to want the feature, and the cost of implementing the feature (in terms of time coding and testing and the opportunity cost of some other feature that could be implemented) must be outweighed by the potential benefit that the feature provides to the developer.
In your case, the way to get the compiler to do what you want is simple, obvious, and clear. If they add:
Infer the type of a numeric expression consisting only of constants and literals to be the minimal type that can contain the resulting value
That means that they now have more code paths to check and more unit tests to write. Changing expected behavior also means that there is probably someone who relies on this documented fact whose code will now be invalid because the inferences could be different.
It's not the compiler's place to promote types based on the result of run-time expressions.
Both int.MaxValue and 100 are integers. I would find the potential for problems if the compiler change the type based on the result of an expression.
Well, what do you expect? You are saying int.MaxValue+100 which goes over the maximum value allowed for an integer! To make it work for the long, do this:
((long)(int.MaxValue)) + 100;
Don't assume the compiler will promote the value to long automatically. That would be even stranger.
1) If one operand is of type ulong, while the other operand is of type sbyte/short/int/long, then compile-time error occurs. I fail to see the logic in this. Thus, why would it be bad idea for both operands to instead be promoted to type double or float?
long L = 100;
ulong UL = 1000;
double d = L + UL; // error saying + operator can't be applied
to operands of type ulong and long
b) Compiler implicitly converts int literal into byte type and assigns resulting value to b:
byte b = 1;
But if we try to assign a literal of type ulong to type long(or to types int, byte etc), then compiler reports an error:
long L = 1000UL;
I would think compiler would be able to figure out whether result of constant expression could fit into variable of type long?!
thank you
To answer the question marked (1) -- adding signed and unsigned longs is probably a mistake. If the intention of the developer is to overflow into inexact arithmetic in this scenario then that's something they should do explicitly, by casting both arguments to double. Doing so implicitly is hiding mistakes more often than it is doing the right thing.
To answer the question marked (b) -- of course the compiler could figure that out. Obviously it can because it does so for integer literals. But again, this is almost certainly an error. If your intention was to make that a signed long then why did you mark it as unsigned? This looks like a mistake. C# has been carefully designed so that it looks for weird patterns like this and calls your attention to them, rather than making a guess that you meant to say this weird thing and blazing on ahead as if everything were normal. The compiler is trying to encourage you to write sensible code; sensible code does not mix signed and unsigned types.
Why should it?
Generally, the 2 types are incompatible because long is signed. You are only describing a special case.
For byte b = 1; 1 has no implicit type as such and can be coerced into byte
For long L = 1000UL; "1000UL" does have an explicit type and is incompatible and see my general case above.
Example from "ulong" on MSDN:
When an integer literal has no suffix,
its type is the first of these types
in which its value can be represented:
int, uint, long, ulong.
and then
There is no implicit conversion from
ulong to any integral type
On "long" in MSDN (my bold)
When an integer literal has no suffix,
its type is the first of these types
in which its value can be represented:
int, uint, long, ulong.
It's quite common and logical and utterly predictable
long l = 100;
ulong ul = 1000;
double d = l + ul; // error
Why would it be bad idea for both operands to instead be promoted to type double or float?
Which one? Floats? Or doubles? Or maybe decimals? Or longs? There's no way for the compiler to know what you are thinking. Also type information generally flows out of expressions not into them, so it can't use the target of the assignment to choose either.
The fix is to simply specify which type you want by casting one or both of the arguments to that type.
The compiler doesn't consider what you do with the result when it determines the result type of an expression. The rules for how types are promoted in an expression only consider the values in the expression itself, not what you do with the value later on.
In the case where you assign the result to a variable, it could be possible to use that information, but consider a statement like this:
Console.Write(L + UL);
The Write method has overloads that take several different data types, which would make it rather complicated to decide how to use that information.
For example, there is an overload that takes a string, so one possible way to promote the types (and a good candidate as it doesn't lose any precision) would be to first convert both values to strings and then concatenate them, which is probably not the result that you were after.
Simple answer is that's just the way the language spec is written:
http://msdn.microsoft.com/en-us/library/y5b434w4(v=VS.80).aspx
You can argue over whether the rules of implicit conversions are logical in each case, but at the end of the day these are just the rules the design committee decided on.
Any implicit conversion has a downside in that it's doing something the programmer may not expect. The general principal with c# seems to be to error in these cases rather then try to guess what the programmer meant.
Suppose one variable was equal to 9223372036854775807 and the other was equal to -9223372036854775806? What should the result of the addition be? Converting the two values to double would round them to 9223372036854775808 and -9223372036854775808, respectively; performing the subtraction would then yield 0.0 (exactly). By contrast, if both values were signed, the result would be 1.0 (also exact). It would be possible to convert both operands to type Decimal and do the math exactly. Conversion to Double after the fact would require an explicit cast, however.
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.