Hello,
We came across a mystery/bug yesterday. This was tested with C51 v9.53 (Simplicity Studio) and 9.56 (uVision).
unsigned char data c; unsigned int data d; // The values 0x7B and 0xFF7B are wrong c = (200 - 50) * (255 - 0) / (255 - 50); 0000738d: MOV 65H, #7BH d = (200 - 50) * (255 - 0) / (255 - 50); 00007390: MOV 66H, #0FFH 00007393: MOV 67H, #7BH //These are correct c = (200 - 50) * (255u - 0) / (255 - 50); 0000738d: MOV 65H, #0BAH d = (200 - 50) * (255u - 0) / (255 - 50); 00007390: MOV 66H, #00H 00007393: MOV 67H, #0BAH
The uVision docs say that numeric constants default to 16 bits. Is this not true? Or is this an issue with the "Constant Folding" optimizing step?
Any insights appreciated, Darren
So look up the rules in your C textbook (or the Standard itself).
I pulled out my 1988 K&R.
page 198, section A2.5.1 ------------------------------------------ "An integer constant may be suffixed by the letter u or U, to specify that it is unsigned. It may also be suffixed by the letter l or L to specify that it is long.
The type of an integer constant depends on its form, value, and suffix. If it is unsuffixed and decimal, it has the first of these types which its value can be represented: int, long int, unsigned long int.
If it is unsuffixed octal or hexadecimal, it has the first possible of these types: int, unsigned int, long int, unsigned long int.
If it is suffixed by u or U, then unsigned int, unsigned long int.
If it is suffixed by l or L, then long int, unsigned long int." ------------------------------------------
Does that contradict anybody else's observations/understanding?
Make that page 193, and there's a typo in this sentence, it should read:
"If it is unsuffixed and decimal, it has the first of these types in which its value can be represented: int, long int, unsigned long int."
That's a little more readable.
I suppose I was expecting that, at compile time, it would catch the signed size issue and upgrade to a 32bit evaluation.
for the '51 an int is 16 bits. 32 bit int would be ridiculous for the '51
Yes, it does. K&R is not really used as a spec by compilers anymore.
Hexadecimal constants are always unsigned (without suffix). So it is unsigned int then unsigned long int.
Decimal constants are always sign integer (without suffix). They will not be upgraded to unsigned constants.
Ridiculous or not, a long int / unsigned long int needs to be at least 32-bits on an 8051 and decimal constants will be upgraded to long int if need be.
Hans-Bernhard Broker correctly pointed this out earlier, referencing a more recent spec than K&R.
(200 - 50) * (255 - 0) / (255 - 50)
These are all int constants and all calculations will to be performed as int, hence the original unexpected, but correct result from the compiler.
38250 / (255 - 50)
This would produce the OP's expected value. The 38250 is a long int constant, 255 and 50 are int constants.
Notice here that the constant 38250 and the value of the compile time constants (200 - 50) * (255 - 0) are not the same. 38250 is an long int (32-bits) and the result fits. (200 - 50) * (255 - 0) is an int (16-bits) and the result does not fit. The compiler can easily know this, but as Hans-Bernhard Broker pointed out, the compiler has 4 int values and is not allowed to do anything but int calculations with these numbers. So (38250 != (25 - 50) * (255 - 50))
ADC0CF = (((SYSCLK/2)/3000000)-1)<<3;
As the OP stated, this calculation is proper, and it should be because it will be done at the 32-bit level - it may be signed or unsigned calculation based on type of SYSCLK, but it is likely a decimal constant which will be signed long int just as 3000000 is signed long int.
(200 - 50) * (255u - 0) / (255 - 50)
Here we have 5 int constants and 1 unsigned int constants. Even though the 0 is an int constant, when the calculation is done between an int and an unsigned int, the int is converted to an unsigned int for the calculation and the calculation is done at that level, producing an unsigned int result. 39250 is an long int constant so any calculations using this will be done at least the long int level.
K&R is not really used as a spec by compilers anymore. While correct, that is irrelevant for the case at hand, because this aspect hasn't changed since K&R 2nd edition, 1988, which still accurately reflects the current international standard in these aspects.
Hexadecimal constants are always unsigned (without suffix). No, they're not. It used to be like that back when K&R 1st edition was the only de-facto standard, but it has been wrong for 27 years now. The type of 0xff is signed int. Yes, that may be surprising.
Decimal constants are always sign integer (without suffix). They will not be upgraded to unsigned constants. Depending on what kind of "being upgraded" you refer to here, that's wrong in one of two ways.
Decimal, unsuffixed constants have signed type, correct. But there is no "upgrading" of constants involved in getting that type. They get a type big enough for their value, and that's that.
But constants, like any other integer operand, will be upgraded (the correct term is: "implicitly converted") to unsigned if they appear in expression contexts like the ones described in this threads.
Hans-Bernhard Broker correctly pointed this out earlier, referencing a more recent spec than K&R. That statement is misleading, because this discussion hasn't touched any difference between K&R1 and ANSI/ISO standard C except your mention of the type of hex constants.