4 ms·

I am quite aware of that keep mind I said generally. However, still the compiler is not wrong returning 1 in the example. Let me show the parts of standard as t

by anfilt 7y ago

I am quite aware of that keep mind I said generally. However, still the compiler is not wrong returning 1 in the example. Let me show the parts of standard as to why. So first the operation is a relation expression. Both signed Char and Unsigned Char of the arithmetic type. So we will follow usual arithmetic conversions. https://i.ibb.co/92bHJMj/1.png https://i.ibb.co/92bHJMj/1.png So the when converting integer types the following applies. Particularly, the section I have highlighted. Both signed and unsigned char are also the same rank. So the signed char is converted to unsigned char. However, we have a problem -1 can't be represented in unsigned so we need to look a bit further. https://i.ibb.co/jRhqd9H/2.png https://i.ibb.co/jRhqd9H/2.png If we look here we can see the 3 applies. As you can see implementation defined. https://i.ibb.co/8XrXt34/3.png https://i.ibb.co/8XrXt34/3.png However, if we promote to a signed int as you suggest happens; we still run into the same thing. https://i.ibb.co/RppGK5D/4.png https://i.ibb.co/RppGK5D/4.png

jjnoakes 7y ago
You are close, but a little off. Your first citation is correct - we have to perform the usual arithmetic conversions. In your second citation, though, you missed the very first paragraph in your picture: > The integer promotions are performed on both operands, then the following rules apply: (emphasis is mine). Before you apply those rules you have to perform the "integer promotions"; they are defined in section 6.3.1.1 subsection 2 and they say, in part, that operands of rank smaller than int (which we have in this example) are converted to int first (which is what my original reply was pointing out). Once both operands are converted to int, no more "usual arithmetic conversions" are needed, so the resulting operation is done on two ints, and the result has to be 0. This is not implementation-defined. (By the way your last citation is for extended integer types, which we are not dealing with here - we are dealing with the basic integer types here).

by anfilt 7y ago

jjnoakes 7y ago

You are close, but a little off. Your first citation is correct - we have to perform the usual arithmetic conversions. In your second citation, though, you missed the very first paragraph in your picture: > The integer promotions are performed on both operands, then the following rules apply: (emphasis is mine). Before you apply those rules you have to perform the "integer promotions"; they are defined in section 6.3.1.1 subsection 2 and they say, in part, that operands of rank smaller than int (which we have in this example) are converted to int first (which is what my original reply was pointing out). Once both operands are converted to int, no more "usual arithmetic conversions" are needed, so the resulting operation is done on two ints, and the result has to be 0. This is not implementation-defined. (By the way your last citation is for extended integer types, which we are not dealing with here - we are dealing with the basic integer types here).