This question already has answers here:
C macros and use of arguments in parentheses
(2 answers)
Closed 3 years ago.
I have a doubt in the output of a code. The code uses preprocessors of c language. The code is given below,
#include <stdio.h>
#define sqr(x) x*x
int main() {
int x = 16/sqr(4);
printf("%d", x);
}
The result of sqr(4) is equal to 16. So, the value in x must be 1 (16/16=1). But if I print the value of x, the output is 16. Please explain me why this is happening.
I am attaching the screenshot of output pane.
Output Window
In C, macros are filled into your code upon compilation. Let's consider what happens in your case:
Your macro is:
#define sqr(x) x*x
When it gets filled into this:
int x = 16/sqr(4);
You would get:
int x = 16/4*4;
Operator precedence being equal for / and *, this gets evaluated as (16/4)*4 = 16, that is, from left to right.
However, if your macro were this:
#define sqr(x) (x*x)
then you would get:
int x = 16/(4*4);
... and that reduces to 1.
However, when the argument for your macro is more complex than a simple number, e.g. 2+3, it would still go wrong. A better macro is:
#define sqr(x) ((x)*(x))
After macro replaces your expression, it becomes int x = 16/4*4 and is evaluated accordingly. i.e. (16/4)*4 which is 16.
sqr(x) doesn't calculate 4*4, it is not a function.
You are actually doing 16/4*4 which is in fact 16.
You can change your define to
#define sqr(x) (x*x)
Now you will be doing 16/(4*4).
It is expanded to
int x = 16/4*4;
Something like
#define sqr(x) ((x) * (x))
would be more safe, precedence and associativity-wise, still works poorly in expressions like sqr(a++). Inline functions are simpler to use in this context.
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.
This question already has answers here:
The need for parentheses in macros in C [duplicate]
(8 answers)
Closed 2 years ago.
#include<stdio.h>
#define sqr(x) x*x
int main()
{
int a = 16/sqr(4);
printf("%d", a);
}
if I am not wrong the output should be 1
as ,a = 16/sqr(4) gives 16/16 => 1
but the output is 16
this is happening only when I take division operator(/) ,I am getting correct output when using other operators
may I know the reason? .. and thank you
16 / 4 * 4 is (16 / 4) * 4 = 16
Moral: always take care with macros. In your case you should enclose in parenthesis:
#define sqr(x) ((x)*(x))
There are still problems with this macro as it evaluates the argument twice.
For instance:
int read_int();
int a = 16 / sqr(read_int());
Will unexpectedly evaluate read_int() twice as it expands to:
int a = 16 / ((read_int() * (read_int());
The advice would be to use functions for this. Make it inline to give the compiler better chances to optimize it.
a = 16/sqr(4);
Expanded the above expression gives:
a = 16 / 4 * 4
Which is 16 based on normal mathematical order of operations.
That's why macros should always be enclosed in parentheses:
#define sqr(x) ((x) * (x))
Are you trying to make a function definiton? If yes, then i don't think it should be on a #define. Instead, make a function sqr(x). Here's how:
#include <stdio.h>
int sqr(int x)
{
return x*x;
}
int main()
{
int a = 16/sqr(4);
printf("%d", a);
}
And you'll get the result 1.
Edit: but anyway, you've already get the solution with macros. This is just another way to do what you want.
I am newbie to C and trying to understand the MACRO expansion logic in C.
I wonder why the first approach is not working but second works as expected.
First approach
#include <stdio.h>
#define square(x) x*x
int main()
{
int x = 36/square(6); // Expended as 36/6*6
printf("%d", x);
return 0;
}
// Output: 36
Second approach
#include <stdio.h>
#define square(x) x*x
int main()
{
int x = square(6)/36; // Expended as 6*6/36
printf("%d", x);
return 0;
}
// Output: 1
Could someone explain me the difference ?
square(6)/36
expands to
6*6/36
which is equivalent to
(6*6)/36
and obviously equals 1.
Even though this is apparently for understanding macros and you may be aware of that, one suggestion:
macros involving operators should be surrounded by parantheses!
First expansion
36/6*6
Using the rules of precedence and left to right http://en.cppreference.com/w/c/language/operator_precedence works it out as
36/6 * 6 -> (36 / 6) * 6 -> 6 * 6 -> 36
Second expansion
6*6/36 -> (6 * 6)/36 -> 36 / 36 -> 1
Using the precedence/left to right rules above.
Sorry for the link - Did not want the cutter. Multiplication has higher precedence than division
Your macro should be defined as
#define square(x) ((x)*(x))
It is necessary to enclose the x's in parenthesis to prevent any surprises about operator precedence. The outer parenthesis are for the same reason.
Please note that this macro even as corrected above will not work if the parameter is a self-modifying expression. So you may want to consider putting it in all-uppercase or something to alert the user of the macro that it will not behave equivalently to a function call.
The reason your expansion of 36/square(6) does not work as you expect is because of its expansion.
36/square(6)
36/6*6
6*6 <-- 36/6 evaluated
36 <-- 6*6 evaluated
The corrected macro would be expanded thus
36/((6)*(6))
36/(36)
1
Which is the answer you would expect. Also note that 5+1 would also work as an argument because of the inner parenthesis but y++ would not behave as you would expect if reading the macro as a function, hence the reason I recommend naming it SQUARE to alert the user that this is a macro not a function.
Macros only behave as functions if each of their parameters appears exactly once and the syntax is otherwise like an expression (i.e. no {}'s). Also the user cannot pass a pointer to a macro as they can to a function.
Your question is a good illustration of the kind of problem that arise with macros:
36/square(6) expands to 36/6*6, which is parsed according to the C grammar as
(36 / 6) * 6
evaluating to 36.
If you had defined the macro as
#define square(x) ((x)*(x))
Both expressions would be equivalent and evaluate to 1.
Yet there is still a problem with this macro:
square(i++) expands as ((i++) * (i++)).
Evaluating i++ twice in the same expression is undefined behavior.
This question already has answers here:
some error in output in using macro in C
(3 answers)
Closed 9 years ago.
Why doesn't this math work with macros in C?
#include <stdio.h>
#define FOO 6
#define BAR 32
#define MULT FOO * BAR
main() {
int x = 28296;
int y = x / MULT;
printf("%d / %d = %d\n", x, MULT, y);
return 0;
}
The result of this is:
28296 / 192 = 150912
Why isn't it 147? If I set a variable " int mult" equal to MULT, and use the variable in the expression (int y = x / mult) it works as expected.
#define tells the preprocessor to replace the code before compilation, so your line actually says:
int y = x / 6 * 32;
since * and / operators have the same precedence, and are evaluated from left to right, you get (x/6) * 32. The compiler would probably do this calculation for you since x is known to it.
Instead, use parenthesis when defining macros like this
Put a bracket around the macro:
#define MULT (FOO * BAR)
Now, you'll get 147.
The reason getting 150912 is that after macro expansion the expression is equivalent to:
y = 28296 / 6 * 32;
and hence it's evaluated as 28296/6 and then multiplied by32.
As #kevin points out, it's better to put brackets around FOO and BAR as well in general case to avoid surprises like this.
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.