my doubt is a general C doubt.. we know if we are using enum the variables which we declare inside automatically increments by one than the previous variable.. but is there any method by which we could make the variables to decrement by one instead of incrementing...
Example
enum my_enum { var1=90, var2,var3 };
for this code var2 and var3 will be 91 & 92 respectively, is there any method (possible) to get them 89 & 88... It was asked in an interview.. any one knows the answer..?
Impressively horrible
Yes, something the preprocessor is good at :)
The reason for the %u was that I did test the code with values way above the range of signed integers.
I wanted to compare the two macros:
#define _(x) \ x##tmp1, \ x = 2*base-x##tmp1, \ x##tmp2 = x##tmp1
and
#define _(x) \ x##tmp1, \ x = base-(x##tmp1-base),\ x##tmp2 = x##tmp1
with values large enough that the construct "2*base" would overflow. There should be no difference between the two on a modern processor using two's complement numbers.
But one note here about the formatting character for printf() is that the standard says (in 2.7.2.2): "Each enumerated type shall be compatible with char, a signed integer type, or an unsigned integer type. The choice of type is implementation-defined [...]"
Well, it doesn't answer the specific question about having the compiler automatically assign decreasing values - although it does achieve the required end of the given example.
If I were the interviewer, I would certainly give credit for that!
Generally, interview questions aren't about showing what you know, but showing that you can think...
Next step up is to play with the preprocessor.
Impressively horrible. Note that the printf() format specifiers require the 'b' modifier for C51 (and ought to be 'd' rather than 'u').
I'm surprised nobody has suggested:
enum my_enum{var3=88, var2, var1};
or does that not fit with the spirit of the thing?
#include <stdio.h> #define _(x) \ x##tmp1, \ x = base-(x##tmp1-base),\ x##tmp2 = x##tmp1 enum { base = 90, _(alt1), _(alt2), _(alt3) }; int main() { printf("%u %u %u %u\n",base,alt1,alt2,alt3); return 0; }
Hello ninja, Did you bother to READ the previous posts? We stated clearly that it is possible if you assign the values yourself! But that was not what the original post asked!
... enum my_num{var1=90,var2=89,var3=88}; printf("%d,%d,%d",var1,var2,var3); ...
"...An enumerated type is a user-defined type consisting of a set of named constants called enumerators. By default, the first enumerator has a value of 0, and each successive enumerator is one larger than the value of the previous one, unless you explicitly specify a value for a particular enumerator."
Greetings, Tamir Michael
No, there isn't!
"my doubt is a general C doubt"
So you should look for your answer in a general 'C' textbook - or even the 'C' standard itself:
"Each subsequent enumerator with no '=' defines its enumeration constant as the value of the constant expression obtained by adding 1 to the value of the previous enumeration constant"
View all questions in Keil forum