I was wondering if I can create macro that creating value with custom "shift" (or other way to create it in compile time). I mean, concatenate two numbers or something...
Something like that (of course it's not working):
#define CUSTOM_DOUBLE(shift) 1.05E shift
I know I can do:
#define CUSTOM_DOUBLE(shift) 1.05 * pow(10.0, shift)
But I'm aware that it isn't calculated in compile time.
As long as shift argument is passed as integer constant (of decimal form) this can be accomplished by ## operator, that concatenates preprocessing tokens. For instance, it may be implemented as:
#include <stdio.h>
#define CUSTOM_DOUBLE(shift) (1.05E ## shift)
int main(void)
{
double d = CUSTOM_DOUBLE(3);
printf("%E\n", d);
return 0;
}
You want this:
#define CUSTOM_DOUBLE(shift) 1.05E##shift
## is the concatenation operator.
Related
This question already has answers here:
C macros and use of arguments in parentheses
(2 answers)
Closed 5 years ago.
I tried to play with the definition of the macro SQR in the following code:
#define SQR(x) (x*x)
int main()
{
int a, b=3;
a = SQR(b+5); // Ideally should be replaced with (3+5*5+3), though not sure.
printf("%d\n",a);
return 0;
}
It prints 23. If I change the macro definition to SQR(x) ((x)*(x)) then the output is as expected, 64. I know that a call to a macro in C replaces the call with the definition of the macro, but I still can’t understand, how it calculated 23.
Pre-processor macros perform text-replacement before the code is compiled so
SQR(b+5) translates to
(b+5*b+5) = (6b+5) = 6*3+5 = 23
Regular function calls would calculate the value of the parameter (b+3) before passing it to the function, but since a macro is pre-compiled replacement, the algebraic order of operations becomes very important.
Consider the macro replacement using this macro:
#define SQR(x) (x*x)
Using b+5 as the argument. Do the replacement yourself. In your code, SQR(b+5) will become: (b+5*b+5), or (3+5*3+5). Now remember your operator precedence rules: * before +. So this is evaluated as: (3+15+5), or 23.
The second version of the macro:
#define SQR(x) ((x) * (x))
Is correct, because you're using the parens to sheild your macro arguments from the effects of operator precedence.
This page explaining operator preference for C has a nice chart. Here's the relevant section of the C11 reference document.
The thing to remember here is that you should get in the habit of always shielding any arguments in your macros, using parens.
Because (3+5*3+5 == 23).
Whereas ((3+5)*(3+5)) == 64.
The best way to do this is not to use a macro:
inline int SQR(int x) { return x*x; }
Or simply write x*x.
The macro expands to
a = b+5*b+5;
i.e.
a = b + (5*b) + 5;
So 23.
After preprocessing, SQR(b+5) will be expanded to (b+5*b+5). This is obviously not correct.
There are two common errors in the definition of SQR:
do not enclose arguments of macro in parentheses in the macro body, so if those arguments are expressions, operators with different precedences in those expressions may cause problem. Here is a version that fixed this problem
#define SQR(x) ((x)*(x))
evaluate arguments of macro more than once, so if those arguments are expressions that have side effect, those side effect could be taken more than once. For example, consider the result of SQR(++x).
By using GCC typeof extension, this problem can be fixed like this
#define SQR(x) ({ typeof (x) _x = (x); _x * _x; })
Both of these problems could be fixed by replacing that macro with an inline function
inline int SQR(x) { return x * x; }
This requires GCC inline extension or C99, See 6.40 An Inline Function is As Fast As a Macro.
A macro is just a straight text substitution. After preprocessing, your code looks like:
int main()
{
int a, b=3;
a = b+5*b+5;
printf("%d\n",a);
return 0;
}
Multiplication has a higher operator precedence than addition, so it's done before the two additions when calculating the value for a. Adding parentheses to your macro definition fixes the problem by making it:
int main()
{
int a, b=3;
a = (b+5)*(b+5);
printf("%d\n",a);
return 0;
}
The parenthesized operations are evaluated before the multiplication, so the additions happen first now, and you get the a = 64 result that you expect.
Because Macros are just string replacement and it is happens before the completion process. The compiler will not have the chance to see the Macro variable and its value. For example: If a macro is defined as
#define BAD_SQUARE(x) x * x
and called like this
BAD_SQUARE(2+1)
the compiler will see this
2 + 1 * 2 + 1
which will result in, maybe, unexpected result of
5
To correct this behavior, you should always surround the macro-variables with parenthesis, such as
#define GOOD_SQUARE(x) (x) * (x)
when this macro is called, for example ,like this
GOOD_SQUARE(2+1)
the compiler will see this
(2 + 1) * (2 + 1)
which will result in
9
Additionally, Here is a full example to further illustrate the point
#include <stdio.h>
#define BAD_SQUARE(x) x * x
// In macros alsways srround the variables with parenthesis
#define GOOD_SQUARE(x) (x) * (x)
int main(int argc, char const *argv[])
{
printf("BAD_SQUARE(2) = : %d \n", BAD_SQUARE(2) );
printf("GOOD_SQUARE(2) = : %d \n", GOOD_SQUARE(2) );
printf("BAD_SQUARE(2+1) = : %d ; because the macro will be \
subsituted as 2 + 1 * 2 + 1 \n", BAD_SQUARE(2+1) );
printf("GOOD_SQUARE(2+1) = : %d ; because the macro will be \
subsituted as (2 + 1) * (2 + 1) \n", GOOD_SQUARE(2+1) );
return 0;
}
Just enclose each and every argument in the macro expansion into parentheses.
#define SQR(x) ((x)*(x))
This will work for whatever argument or value you pass.
I need to write a macro that would count bits in odd positions (in ANSI C). For example, in 1010 the count of such bits is 2 while in 0101 the count is 0.
This is the macro I came up with:
#include <stdio.h>
#define BIT_IN_BYTE 8
#define size(x,t) { \
int i, sum = 0; \
for(i = 0; x; x >>= 1) { \
if( (x & 1) && (i % 2) != 0) { \
sum++; \
} \
i++; \
} \
t = sum; \
} \
int main() {
int b = 44444, result;
size(b, result);
printf("count = %d\n", result);
return 0;
}
I have 2 questions really: whether I could've written the macro in a way that it wouldn't need two arguments in its signature and more importantly if there's a relatively simple way to extend the macro onto float and double types.
As in the comments to your question, using a macro here really is not good practice. You should really use a function instead - inline or otherwise.
Since you're apparently forced to use a macro in your case for reasons outside of your control, here are some relevant points to your example:
You can use sizeof to get the number of bytes in whatever type you pass in your x parameter. This will potentially allow you to generalize the macros to different types (you mention float and double) that have different sizes.
Since the macro does not do type checking, if you have something that works for int in a generic way - probably using bit-wise operations - you'll have a fighting chance of making it work for other types. I again note, however, that the lack of type safety is a reason to avoid macros in many cases.
As noted in a comment on your question, your current approach destroys the value in whatever you pass as parameter x. That's likely undesired and unnecessary.
You would only be able to avoid passing the second parameter if you can rewrite this such that is evaluates to a number. I'm not motivated to figure out exactly how to do that for this contrived example or if it is actually possible.
I wonder if it is possible to build a gnu C macro which expands to a list of tokens (integer numbers) which are the arguments of the macro without duplicates. The number of arguments could be assumed fixed (for the moment). I.e. I want something like:
#define MAC(a,b,c) ???
which expands e.g.
MAC(1,2,1)
to 1,2.
The arguments are not sorted and the result does not have to be.
Based on the proposal below I built an example which does essentially what I want using the p99 includes:
#include <p99/p99_if.h>
#include <p99/p99_paste.h>
#define MAC2(a,b) double P99_PASTE2(myvar_, a) P99_IF_EQ(a,b)()(; double P99_PASTE2(myvar_, b))
#define MAC3(a,b,c) double P99_PASTE2(myvar_, a) P99_IF_EQ(a,b)()(; double P99_PASTE2(myvar_, b)) P99_IF_EQ(a,c)()(P99_IF_EQ(b,c)()(; double P99_PASTE2(myvar_, c)) )
MAC2(1,2);
MAC2(3,3);
MAC3(1,2,3);
MAC3(10,10,1);
If your arguments are always small decimal numbers as in your example, you could get away with what I provide in P99. It has macros like P99_IF_EQ that you could use as
#define MAC(A,B) unsigned P99_PASTE2(myvar_, A) P99_IF_EQ(A,B)()(; unsigned P99(unsigned P99_PASTE2(myvar_, B))
MAC(1,2); // -> myvar_1 and myvar_2
MAC(3,3); // -> myvar_3
to only expand the declaration for B if it is not equal to A. Obviously for three different arguments this already becomes a bit tedious, but would be doable.
This question already has answers here:
C macros and use of arguments in parentheses
(2 answers)
Closed 5 years ago.
I tried to play with the definition of the macro SQR in the following code:
#define SQR(x) (x*x)
int main()
{
int a, b=3;
a = SQR(b+5); // Ideally should be replaced with (3+5*5+3), though not sure.
printf("%d\n",a);
return 0;
}
It prints 23. If I change the macro definition to SQR(x) ((x)*(x)) then the output is as expected, 64. I know that a call to a macro in C replaces the call with the definition of the macro, but I still can’t understand, how it calculated 23.
Pre-processor macros perform text-replacement before the code is compiled so
SQR(b+5) translates to
(b+5*b+5) = (6b+5) = 6*3+5 = 23
Regular function calls would calculate the value of the parameter (b+3) before passing it to the function, but since a macro is pre-compiled replacement, the algebraic order of operations becomes very important.
Consider the macro replacement using this macro:
#define SQR(x) (x*x)
Using b+5 as the argument. Do the replacement yourself. In your code, SQR(b+5) will become: (b+5*b+5), or (3+5*3+5). Now remember your operator precedence rules: * before +. So this is evaluated as: (3+15+5), or 23.
The second version of the macro:
#define SQR(x) ((x) * (x))
Is correct, because you're using the parens to sheild your macro arguments from the effects of operator precedence.
This page explaining operator preference for C has a nice chart. Here's the relevant section of the C11 reference document.
The thing to remember here is that you should get in the habit of always shielding any arguments in your macros, using parens.
Because (3+5*3+5 == 23).
Whereas ((3+5)*(3+5)) == 64.
The best way to do this is not to use a macro:
inline int SQR(int x) { return x*x; }
Or simply write x*x.
The macro expands to
a = b+5*b+5;
i.e.
a = b + (5*b) + 5;
So 23.
After preprocessing, SQR(b+5) will be expanded to (b+5*b+5). This is obviously not correct.
There are two common errors in the definition of SQR:
do not enclose arguments of macro in parentheses in the macro body, so if those arguments are expressions, operators with different precedences in those expressions may cause problem. Here is a version that fixed this problem
#define SQR(x) ((x)*(x))
evaluate arguments of macro more than once, so if those arguments are expressions that have side effect, those side effect could be taken more than once. For example, consider the result of SQR(++x).
By using GCC typeof extension, this problem can be fixed like this
#define SQR(x) ({ typeof (x) _x = (x); _x * _x; })
Both of these problems could be fixed by replacing that macro with an inline function
inline int SQR(x) { return x * x; }
This requires GCC inline extension or C99, See 6.40 An Inline Function is As Fast As a Macro.
A macro is just a straight text substitution. After preprocessing, your code looks like:
int main()
{
int a, b=3;
a = b+5*b+5;
printf("%d\n",a);
return 0;
}
Multiplication has a higher operator precedence than addition, so it's done before the two additions when calculating the value for a. Adding parentheses to your macro definition fixes the problem by making it:
int main()
{
int a, b=3;
a = (b+5)*(b+5);
printf("%d\n",a);
return 0;
}
The parenthesized operations are evaluated before the multiplication, so the additions happen first now, and you get the a = 64 result that you expect.
Because Macros are just string replacement and it is happens before the completion process. The compiler will not have the chance to see the Macro variable and its value. For example: If a macro is defined as
#define BAD_SQUARE(x) x * x
and called like this
BAD_SQUARE(2+1)
the compiler will see this
2 + 1 * 2 + 1
which will result in, maybe, unexpected result of
5
To correct this behavior, you should always surround the macro-variables with parenthesis, such as
#define GOOD_SQUARE(x) (x) * (x)
when this macro is called, for example ,like this
GOOD_SQUARE(2+1)
the compiler will see this
(2 + 1) * (2 + 1)
which will result in
9
Additionally, Here is a full example to further illustrate the point
#include <stdio.h>
#define BAD_SQUARE(x) x * x
// In macros alsways srround the variables with parenthesis
#define GOOD_SQUARE(x) (x) * (x)
int main(int argc, char const *argv[])
{
printf("BAD_SQUARE(2) = : %d \n", BAD_SQUARE(2) );
printf("GOOD_SQUARE(2) = : %d \n", GOOD_SQUARE(2) );
printf("BAD_SQUARE(2+1) = : %d ; because the macro will be \
subsituted as 2 + 1 * 2 + 1 \n", BAD_SQUARE(2+1) );
printf("GOOD_SQUARE(2+1) = : %d ; because the macro will be \
subsituted as (2 + 1) * (2 + 1) \n", GOOD_SQUARE(2+1) );
return 0;
}
Just enclose each and every argument in the macro expansion into parentheses.
#define SQR(x) ((x)*(x))
This will work for whatever argument or value you pass.
As the questions says, is the C preprocessor able to do it?
E.g.:
#define PI 3.1416
#define OP PI/100
#define OP2 PI%100
Is there any way OP and/or OP2 get calculated in the preprocessing phase?
Integer arithmetic? Run the following program to find out:
#include "stdio.h"
int main() {
#if 1 + 1 == 2
printf("1+1==2\n");
#endif
#if 1 + 1 == 3
printf("1+1==3\n");
#endif
}
Answer is "yes", there is a way to make the preprocessor perform integer arithmetic, which is to use it in a preprocessor condition.
Note however that your examples are not integer arithmetic. I just checked, and gcc's preprocessor fails if you try to make it do float comparisons. I haven't checked whether the standard ever allows floating point arithmetic in the preprocessor.
Regular macro expansion does not evaluate integer expressions, it leaves it to the compiler, as can be seen by preprocessing (-E in gcc) the following:
#define ONEPLUSONE (1 + 1)
#if ONEPLUSONE == 2
int i = ONEPLUSONE;
#endif
Result is int i = (1 + 1); (plus probably some stuff to indicate source file names and line numbers and such).
The code you wrote doesn't actually make the preprocessor do any calculation. A #define does simple text replacement, so with this defined:
#define PI 3.1416
#define OP PI/100
This code:
if (OP == x) { ... }
becomes
if (3.1416/100 == x) { ... }
and then it gets compiled. The compiler in turn may choose to take such an expression and calculate it at compile time and produce a code equivalent to this:
if (0.031416 == x) { ... }
But this is the compiler, not the preprocessor.
To answer your question, yes, the preprocessor CAN do some arithmetic. This can be seen when you write something like this:
#if (3.141/100 == 20)
printf("yo");
#elif (3+3 == 6)
printf("hey");
#endif
YES, I mean: it can do arithmetic :)
As demonstrated in 99 bottles of beer.
Yes, it can be done with the Boost Preprocessor. And it is compatible with pure C so you can use it in C programs with C only compilations. Your code involves floating point numbers though, so I think that needs to be done indirectly.
#include <boost/preprocessor/arithmetic/div.hpp>
BOOST_PP_DIV(11, 5) // expands to 2
#define KB 1024
#define HKB BOOST_PP_DIV(A,2)
#define REM(A,B) BOOST_PP_SUB(A, BOOST_PP_MUL(B, BOOST_PP_DIV(A,B)))
#define RKB REM(KB,2)
int div = HKB;
int rem = RKB;
This preprocesses to (check with gcc -S)
int div = 512;
int rem = 0;
Thanks to this thread.
Yes.
I can't believe that no one has yet linked to a certain obfuscated C contest winner. The guy implemented an ALU in the preprocessor via recursive includes. Here is the implementation, and here is something of an explanation.
Now, that said, you don't want to do what that guy did. It's fun and all, but look at the compile times in his hint file (not to mention the fact that the resulting code is unmaintainable). More commonly, people use the pre-processor strictly for text replacement, and evaluation of constant integer arithmetic happens either at compile time or run time.
As others noted however, you can do some arithmetic in #if statements.
Be carefull when doing arithmetic: add parenthesis.
#define SIZE4 4
#define SIZE8 8
#define TOTALSIZE SIZE4 + SIZE8
If you ever use something like:
unsigned int i = TOTALSIZE/4;
and expect i to be 3, you would get 4 + 2 = 6 instead.
Add parenthesis:
#define TOTALSIZE (SIZE4 + SIZE8)