I think I read somewhere that #define constants in the style
#define SOMETHING 6
are evaluated to long or unsigned long, depending on the situation and
not int.
7  2014 
Ioannis Vranos wrote:
I think I read somewhere that #define constants in the style
#define SOMETHING 6
are evaluated to long or unsigned long, depending on the situation and
not int.
.... Is that true?
Ioannis Vranos wrote:
Ioannis Vranos wrote:
>I think I read somewhere that #define constants in the style
#define SOMETHING 6
are evaluated to long or unsigned long, depending on the situation
and not int.
... Is that true?
I don't know whether it's true that you read that nonsense somewhere
or whether you dreamed it. I know that an integral literal has the
type 'int' if it fits in it, and if it doesn't, it has type long int,
unless overridden by a suffix (absent in your example). This all is
defined in [lex.icon]/2.
V
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Victor Bazarov wrote:
Ioannis Vranos wrote:
>Ioannis Vranos wrote:
>>I think I read somewhere that #define constants in the style
#define SOMETHING 6
are evaluated to long or unsigned long, depending on the situation
and not int.
... Is that true?
I don't know whether it's true that you read that nonsense somewhere
or whether you dreamed it. I know that an integral literal has the
type 'int' if it fits in it, and if it doesn't, it has type long int,
unless overridden by a suffix (absent in your example). This all is
defined in [lex.icon]/2.
My mistake. It is about the "identifier" of the macro expressions
defined identifier
and
defined ( identifier)
Errata for K&R2 page 232 (from http://www-db-out.research.bell-labs...ediffs.html):\
232(§A12.5): The result of the defined operator is not replaced
literally by 0L or 1L, nor are undefined names literally by 0L, but just
by plain 0 or 1. However, the constant expression is nevertheless
evaluated as if these and other constants appearing have long or
unsigned long type.
Ioannis Vranos wrote:
Victor Bazarov wrote:
>Ioannis Vranos wrote:
>>Ioannis Vranos wrote:
I think I read somewhere that #define constants in the style
#define SOMETHING 6
are evaluated to long or unsigned long, depending on the situation
and not int.
... Is that true?
I don't know whether it's true that you read that nonsense somewhere
or whether you dreamed it. I know that an integral literal has the
type 'int' if it fits in it, and if it doesn't, it has type long int,
unless overridden by a suffix (absent in your example). This all is
defined in [lex.icon]/2.
My mistake. It is about the "identifier" of the macro expressions
defined identifier
and
defined ( identifier)
Errata for K&R2 page 232 (from
http://www-db-out.research.bell-labs...ediffs.html):\
232(§A12.5): The result of the defined operator is not replaced
literally by 0L or 1L, nor are undefined names literally by 0L, but just
by plain 0 or 1. However, the constant expression is nevertheless
evaluated as if these and other constants appearing have long or
unsigned long type.
The fact that the "identifier" in a "defined" preprocessor statement is
evaluated as long or unsigned long, doesn't it imply that this
identifier which was defined with #define" preprocessor statement is
evaluated as long or unsigned long everywhere?
Ioannis Vranos wrote:
Ioannis Vranos wrote:
>Victor Bazarov wrote:
>>Ioannis Vranos wrote:
Ioannis Vranos wrote:
I think I read somewhere that #define constants in the style
>
>
#define SOMETHING 6
>
>
are evaluated to long or unsigned long, depending on the situation
and not int.
... Is that true?
I don't know whether it's true that you read that nonsense somewhere
or whether you dreamed it. I know that an integral literal has the
type 'int' if it fits in it, and if it doesn't, it has type long int,
unless overridden by a suffix (absent in your example). This all is
defined in [lex.icon]/2.
My mistake. It is about the "identifier" of the macro expressions
defined identifier
and
defined ( identifier)
Errata for K&R2 page 232 (from
http://www-db-out.research.bell-labs...ediffs.html):\
232(§A12.5): The result of the defined operator is not replaced
literally by 0L or 1L, nor are undefined names literally by 0L, but just
by plain 0 or 1. However, the constant expression is nevertheless
evaluated as if these and other constants appearing have long or
unsigned long type.
The fact that the "identifier" in a "defined" preprocessor statement is
evaluated as long or unsigned long, doesn't it imply that this
identifier which was defined with #define" preprocessor statement is
evaluated as long or unsigned long everywhere?
My mistake again, the _result_ of the defined operator is replaced by 0
or 1 which are evaluated as 0L and 1L. Strange stuff.
OK, lets get it from the start again:
Errata for K&R2 page 232 (from http://www-db-out.research.bell-labs...ediffs.html):\
232(§A12.5): The result of the defined operator is not replaced
literally by 0L or 1L, nor are undefined names literally by 0L, but just
by plain 0 or 1. However, the constant expression is nevertheless
evaluated as if these and *other constants* appearing have long or
unsigned long type.
When it mentions "these and other constants", what are the other
constants implied?
Ioannis Vranos wrote:
I think I read somewhere that #define constants in the style
#define SOMETHING 6
are evaluated to long or unsigned long, depending on the situation and
not int.
Nope. Not true. #define has little to do with (unless you are
throwing in some token pasting operators), an decimal integer literal
has type the smaller of int or long int that can hold it.
Other constraints on the language make 6 HAVE to be of type int.
There's no way a decimal literal ever has type unsigned. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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