so I am studying about Integer Promotion and I know that type promotions only apply to the values operated upon when an expression is evaluated.
I have this example where I take an sbyte that is equal to 127 and increment it by 1 and then I output it on the Console and I get a -128. Does this mean that the sbyte turns into an int during the process of incrementing and then it somehow transforms into a -128.
I would like to know how exactly this happens.
sbyte efhv = 127;
efhv++;
Console.WriteLine(efhv);
As #Sweeper pointed out in the comments, integer promotion does not apply to the ++ operator: ++ modifies the value in-place, which means that the modified value must be within the range of the
specified data type. Integer promotion would not make any sense here.
As #beautifulcoder explained in their answer, the effect you see is simply a two's complement number overflowing, since C# executes arithmetic operations in an unchecked context by default.
According to the C# language specification integer promotion does occur for the unary + and the binary + operators, as can be seen in the following code example:
sbyte efhv = 127;
sbyte one = 1;
Console.WriteLine((efhv).GetType().Name); // prints SByte
Console.WriteLine((efhv++).GetType().Name); // prints SByte
Console.WriteLine((+efhv).GetType().Name); // prints Int32
Console.WriteLine((efhv + one).GetType().Name); // prints Int32
(fiddle)
It is because this is a signed 8-bit integer. When the value exceeds the max it overflows into the negative numbers. It is not promotion but overflow behavior you are seeing.
To verify:
sbyte efhv = 127;
efhv++;
Console.WriteLine(efhv);
Console.WriteLine(sbyte.MaxValue);
Console.WriteLine(sbyte.MinValue);
Related
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;
As I remember we had learned that signed integer types (sbyte, short, int , long)
the first bit is for the sign and the latter 7 bit is for the value.
I saw that sbyte range is -128 to 127 while I thought it must be -127 to 127.
I tried some codes to understand how this is possible and I faced two strange things:
1- I tried the following code:
sbyte a = -128;
Console.Write(Convert.ToString(a, 2));
and the resutl was
1111111100000000
As if its a two byte variable.
2-Tried converting all numbers in the range to binary:
for(sbyte i=-128;i<=127;i++)
{
Console.WriteLine(Convert.ToString(i, 2));
if(i==127) break;
}
If I omit the if(i==127) break; the loop goes on. and with the break, the code in the loop does not execute, some how as if -128 is greater than 127.
My Conclusion:
As I thought that -128 must not fit in a unsigned byte variable and the first and second tries approves that (111111110000000 > 01111111)
but If it does not fit, then why range is -128 to 127?
I saw that sbyte range is -128 to 127 while I thought it must be -127 to 127.
The range is [-128, +127] indeed. The range [-127, +127] would mean that sbyte can represent only 255 different values, while 8 bits make 256 combinations.
And BTW if -128 would not be a legal value, compiler would complain about
sbyte a = -128;
There is no overload Convert.ToString(sbyte value, int toBase), so in your case Convert.ToString(int value, int toBase) is called, and sbyte is promoted to int.
To quickly check the value, you can print sbyte as a hexadecimal number:
sbyte s = -1;
Console.WriteLine("{0:X}", s); // FF, i.e. 11111111
If I omit the if(i==127) break; the loop goes on.
Sure, sbyte.MaxValue is 127, so i<=127 is always true. When i is 127 and gets incremented, an overflow occurs and the next value is -128.
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
I have a 32 bit int and I want to address only the lower half of this variable. I know I can convert to bit array and to int16, but is there any more straight forward way to do that?
It you want only the lower half, you can just cast it: (Int16)my32BitInt
In general, if you're extending/truncating bit patterns like this, then you do need to be careful about signed types - unsigned types may cause fewer surprises.
As mentioned in the comments - if you've enclosed your code in a 'checked' context, or changed your compiler options so that the default is 'checked', then you can't truncate a number like this without an exception being thrown if there are any non-zero bits being discarded - in that situation you'd need to do:
(UInt16)(my32BitInt & 0xffff)
(The option of using signed types is gone in this case, because you'd have to use & 0x7fff which then preserves only 15 bits)
just use this function
Convert.ToInt16()
or just
(Int16)valueasint
You can use implicit conversation to Int16 like;
(Int16)2;
but be careful when you do that. Because Int16 can't hold all possible Int32 values.
For example this won't work;
(Int16)2147483683;
because Int16 can hold 32787 as maximum value. You can use unchecked (C# Reference) keyword such this cases.
If you force an unchecked operation, a cast should work:
int r = 0xF000001;
short trimmed = unchecked((short) r);
This will truncate the value of r to fit in a short.
If the value of r should always fit in a short, you can just do a normal cast and let an exception be thrown.
If you need a 16 bit value and you happen to know something specific like that the number will never be less than zero, you could use a UINT16 value. That conversion looks like:
int x = 0;
UInt16 value = (UInt16)x;
This has the full (positive) range of an integer.
Well, first, make sure you actually want to have the value signed. uint and ushort are there for a reason. Then:
ushort ret = (ushort)(val & ((1 << 16) - 1));
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.