From the sublime level of continuation passing style we go back to the mundane level of twiddling individual bits.int i = SomeBagOfBits();
ulong u = SomeOtherBagOfBits();
ulong result = u | i; // combine them together Whoops, that's an error. "Operator | cannot be applied to operands of type int and ulong." There are bitwise-or operators defined on int, uint, long and ulong, but none between int and ulong. You cannot use the int version because the ulong might not fit, and you cannot use the ulong version because the int might be negative. I demand that the compiler do my bidding regardless!ulong result = u | (ulong) i;There, now the compiler has to choose the ulong operator, and the explicit conversion from int to ulong we know never fails. Argh, now we've got a warning! "CS0675: Bitwise-or operator used on a sign-extended operand; consider casting to a smaller unsigned type first."I am often asked what the meaning of this warning is. The crux of the matter is that the conversion from int to ulong does sign extension. Let's make a more concrete example: int i = -17973521; // In hex that is FEEDBEEF
ulong u = 0x0123456700000000;
ulong result = u | (ulong)i;
Console.WriteLine(result.ToString("x"));What is the expected result? Most people expect that the result is 1234567FEEDBEEF. It is not. It is FFFFFFFFFEEDBEEF. Why? Because when converting an int to a ulong, first the int is converted to a long so that the sign information is not lost. The long -17973521 is in hex FFFFFFFFFEEDBEEF. That long is then converted to that ulong, which is then or'd in the natural way to produce the unexpected result. Read more: Fabulous Adventures In Coding
ulong u = SomeOtherBagOfBits();
ulong result = u | i; // combine them together Whoops, that's an error. "Operator | cannot be applied to operands of type int and ulong." There are bitwise-or operators defined on int, uint, long and ulong, but none between int and ulong. You cannot use the int version because the ulong might not fit, and you cannot use the ulong version because the int might be negative. I demand that the compiler do my bidding regardless!ulong result = u | (ulong) i;There, now the compiler has to choose the ulong operator, and the explicit conversion from int to ulong we know never fails. Argh, now we've got a warning! "CS0675: Bitwise-or operator used on a sign-extended operand; consider casting to a smaller unsigned type first."I am often asked what the meaning of this warning is. The crux of the matter is that the conversion from int to ulong does sign extension. Let's make a more concrete example: int i = -17973521; // In hex that is FEEDBEEF
ulong u = 0x0123456700000000;
ulong result = u | (ulong)i;
Console.WriteLine(result.ToString("x"));What is the expected result? Most people expect that the result is 1234567FEEDBEEF. It is not. It is FFFFFFFFFEEDBEEF. Why? Because when converting an int to a ulong, first the int is converted to a long so that the sign information is not lost. The long -17973521 is in hex FFFFFFFFFEEDBEEF. That long is then converted to that ulong, which is then or'd in the natural way to produce the unexpected result. Read more: Fabulous Adventures In Coding
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