I have code like this in C:
if (count == 5 || count == 8 || count == 9 || count == 10)
{
// Do something
}
Is there a possibility to write this shorter? Something like:
if (count == 5, 8, 9, 10)
{
// Do something
}
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5 Answers
In some cases, it can make sense to use a bit map.
For example, your test:
if (count == 5 || count == 8 || count == 9 || count == 10)
is equivalent to:
if ((1<<count) & 0x720)
The values 5, 8, 9, and 10 are encoded in the bits of the value 0x720.
Typically this would make sense if you have meaningful symbolic constants for the values 5, 8, 9, and 10, with 0x720 constructed from a bitwise "or" (|) of those constants -- which would result in more verbose code than the simple if you have in your question.
In practice, real-world code should minimize the use of "magic numbers". It's difficult to tell what the values 5, 8, 9, and 10 mean.
Possibly you should consider using a different algorithm, but it's impossible to tell without more information.
2 Comments
Gugi
Yes, maybe there is a better solution to my problem, but it doesn't matter since this is just a small piece from my programming homework, and i dont want to post the whole task.
M.M
To avoid undefined behaviour, only do this if it is known that
count < 31 ! (or whatever your int size is)Depending on how many different values and different branches you have it can sometimes be neater/easier to use a switch:
switch (count)
{
case 5:
case 8:
case 9:
case 10:
do_something();
break;
default:
do_something_else();
break;
}
1 Comment
Gugi
I have just my 4 values. Just wanted to shorten the code. count is max. 16.
There is no way to insert a comma-separated list into an if statement itself as in your example, but you may write something like this to use the comma-separated list format.
int allowed_counts[] = {5, 8, 9, 10};
int i;
for (i = 0; i < 4; i++) {
if (count == allowed_counts[i]) {
...
break;
}
}
Although, a switch statement is more computationally efficient for all list sizes.
1 Comment
Keith Thompson
@Gugi: Depending on what you're doing, the straightforward
if statement in your question might be the best way to do it. It's more readable than most of the alternatives, and for something this small performance isn't likely to be a significant issue.I think you're looking for something like this:
So the condition is testing if count is 5, 8, 9, 10:
if (count == 5 || (count >= 8 && count <= 10 ))
printf("count is 5, 8, 9 or 10");
6 Comments
Clifford
Unless
count equals 6 or 7!
Bill Lynch
@Rizier123: The original code wanted only 5, 8, 9 and 10 to succeed at the conditional. Your code also succeeds at 6 and 7.
Keith Thompson
@Rizier123: No, it doesn't. Try it yourself, with
count == 6. The code in the question tests for count equal to 5, 8, 9, or 10, not for a range.Gugi
yes, this would work, but it's just a bit better, i wanted something very short and for every scenario. so it should also work when count is 4,6, 8, or 11.
Rizier123
@SimonGuggi i don't think you can get that spesific condition any smaller! (What you mean it also should work if
count is 4,6,8 or 11? You mean in the condition that the same should happen like if count is 5?) |
E.g EXIST(count, (5,8,9,10) expand by macro of boost
#include <boost/preprocessor/tuple/to_seq.hpp>
#include <boost/preprocessor/seq/for_each_i.hpp>
#include <boost/preprocessor/control/if.hpp>
#include <stdio.h>
#define F(r, data, i, elem) BOOST_PP_IF(i, ||, ) (data == elem)
#define EXIST(var, values) BOOST_PP_SEQ_FOR_EACH_I(F, var , BOOST_PP_TUPLE_TO_SEQ(values) )
int main(){
int count;
scanf("%d", &count);
if(EXIST(count, (5,8,9,10)))//if( (count == 5) || (count == 8) || (count == 9) || (count == 10) )
printf("yes\n");
else
printf("no\n");
return 0;
}
