F
Francis Moreau
Hello,
I'm a bit puzzled by the ISO C99 and its section 6.3.1.8.
In particular when I reach the following part:
,----
| Otherwise, the integer promotions are performed on both
| operands. Then the following rules are applied to the promoted
| operands: ...
`----
Then the sub sections are talking about the poreands and their
type. For example the first sub section tells:
,----
| If both operands have the same type, then no further conversion is
| needed.
`----
This is where I don't understand. At this point, in my understanding,
both operands have always the same type and rank since a) an integer
promotion have been performed on both of them, b) the rank of any
unsigned integer type shall equal the rank of the corresponding signed
integer type.
Could anybody enlight me ?
thanks
I'm a bit puzzled by the ISO C99 and its section 6.3.1.8.
In particular when I reach the following part:
,----
| Otherwise, the integer promotions are performed on both
| operands. Then the following rules are applied to the promoted
| operands: ...
`----
Then the sub sections are talking about the poreands and their
type. For example the first sub section tells:
,----
| If both operands have the same type, then no further conversion is
| needed.
`----
This is where I don't understand. At this point, in my understanding,
both operands have always the same type and rank since a) an integer
promotion have been performed on both of them, b) the rank of any
unsigned integer type shall equal the rank of the corresponding signed
integer type.
Could anybody enlight me ?
thanks