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I am making a program that shifts values left or right depending on the value of the 2nd argument. If it is positive then it shifts left, else it shifts right. N is the number of times it shifts left or right. I am having trouble in implementation of my Macros.

#include <stdio.h>

#define SHIFT(value, n) 

#if (n) > 0
    (unsigned int value) <<= (int n);
#else 
    ( (unsigned int value)) >>= (int -n);



int main()
{
    printf("%d\n", SHIFT(1, 4));
}

Currently I am getting a conditional directive error.

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1
  • Try ` #define SHIFT(value, n) \ `. A backslash at the end indicates next line is part of current line. Commented Jan 6, 2014 at 9:58

3 Answers 3

Reset to default
6

The C preprocessor doesn't really work the way you intend to use it. In particular, you cannot use other CPP directives (like #if, ...) in the expansion of a macro. Besides, since macro expansion is a static compile-time feature, this wouldn't help anyway, when the actual shift-values are only known at run-time:

int value_to_shift = read_some_integer_from_user();
int amount_to_shift_by = read_another_integer_from_user();
int shifted_value = SHIFT(value_to_shift, amount_to_shift_by);

If you don't mind the potential double-evaluation of macro arguments, go with the ternary operator:

#define SHIFT(value, n)   ( (n) < 0? ((value) >> (-n)) : ((value) << (n)) )

Note, that using <<= (and >>=) as you do in the code is most likely not what you want, given, that you pass literal numbers as value arguments to your the SHIFT macro.

I would probably go for a tiny helper function here instead of a macro:

static int 
shift(int value, int nbits)
{
    return nbits < 0? (value >> -nbits) : (value << nbits);
}
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4 Comments

Okay. Is there any way to do this with using data types in arguments?
@XiJiaopin: And why do you want that?
It would be appropriate in this specific program which is derived from shifting with unsigned integers
@XiJiaopin -- In C++, you could use a template. In C, there is currently no way I could think of. Depending on your compiler, you might be able to cook something up using the mechanism used to implement the (type generic math)[en.cppreference.com/w/c/numeric/tgmath] stuff of C99. I haven't used that myself at all, and cannot tell, whether this would be a viable way at all.
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Your #define isn't doing what you think it does. You need to continue the lines with \. And since n is known at runtime (assuming your case is just a simplification) you can use a regular function, and inline it if needed:

inline int shift(int value, int n) {
    if (n < 0) {
        return (unsigned int) value << n;
    }
    else {
         return (unsigned int) value << -n;
    }
}
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1 Comment

Your cast syntax looks odd. Did you mean (unsigned int)value?
0

You need to combine the lines with a \. you also missed thed #endif

Inplace macro:

#define SHIFT(value, n)        \
     value = ((n) > 0)  ?      \
             value << (n) :    \
             value >> -(n)

Return shifted value:

#define SHIFT(value, n)        \
     ( ((n) > 0)  ?            \
             (value) << (n) :  \
             (value) >> -(n) )
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1 Comment

I know that's why I wrote it wouldn't work. Better to remove it then.

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