main()
{
int x=-3;
unsigned int y=2;
if ( x + y < 0)
printf("compiler converted unsigned to signed");
else
printf("compiler converted signed to unsigned");
}
In this program , during statement x+ y < 0 , compiler converts signed int x to unsigned int ( -3 equals big unsigned value) and hence it prints
"compiler converted signed to unsigned".
My question is what is the intention of writing c compiler int shi fashion that it converts signed to unsigned. Is there any logic behind that. Inthe given program the intention of programmer may be to get -3 +2 = -1 and hence it will lead to wrong results.
Thanks,
av
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It makes no sense, and half of the time silently wrecks the whole comparison, but that's the way it works in C.
It's usually a sign that you havent thought through the variable declarations. An unsigned can take on twice the positive values of a signed of the same type. What does this imply about the expected range of the signed? Why are you even ccomparing two numbers of wildly different ranges? At the very least, you should add a few comments regarding the expected range of values and why they're being compared. Or fix the declarations to be more compatible.
When comparing signed and unsigned there isn't really a conversion mechanism that would reliably obtain the programmers intent. What if x were 3, and y were 65000? Then you'd want it to work the way it does.
I think there is sense in it working the way it does. And the sense is in line with C's overall philosophy.
First, the C language is biased towards "trusting" the programmer to know what he is doing. Right or wrong this is the language philosophy. C lets you do all kinds of crazy things, mostly on the understanding that its the programmers responsibility to understand what is going on.
Secondly C is intended to be pretty close to assembly language. So the compiler is not going to go to any extrordinary lengths to do something with this comparison (like promoting them both to long).
Now as language designers you are looking at a comparison between int and unsigned int. Given 2 (close to assembly) there are two possibilities for doing a comparison: either treat both as signed or both as unsigned. Which is more likely to be intended?
Given that signed is the default and unsigned is an explicit "extra" bit of declaration, and given 1 (trust the programmer) it is reasonable to conclude that the programmer *needs* the unsigned variable to be unsigned. That in turn means that one is constrained to assume that it is likely that the unsigned variable is going to contain a value that cannot be represented in a signed variable. And furthermore, it is expected (given 1) that a programmer knows that you cannot usefully compare an unsigned variable to a negative number.
So, given all that, you would write the third promotion rule as it stands. That is, when faced with a signed and unsigned variable of the same rank, convert the signed one to unsigned.
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by: Infinity08Posted on 2006-01-12 at 00:39:16ID: 15679420
That's because the compiler follows the C99 standard rules of integer promotion :
1) If both operands have the same type, then no further conversion is needed.
2) Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank.
3) Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then the operand with signed integer type is converted to the type of the operand with unsigned integer type.
4) Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, then the operand with unsigned integer type is converted to the type of the operand with signed integer type.
5) Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type.
In this case rule 1 doesn't apply since both operands are of different type (signed int and unsigned int). Rule 2 doesn't apply either since we have both a signed and an unsigned integer type. Rule 3 does apply, as the rank of signed int and unsigned int is the same.
So, the signed integer is converted to an unsigned integer.