Macro With Arguments
This simple macro calculates the arithmetic mean of two numbers:
main.c
12345678#include <stdio.h> #define MEAN(X,Y) ((X)+(Y))/2 int main() { printf("Mean: %d\n", MEAN(1053, 6037)); return 0; }
To make the macro able to handle non-integer numbers, we change the specifier and change the arguments to non-integer numbers:
main.c
123456789#include <stdio.h> #define MEAN(X,Y) ((X)+(Y))/2 int main() { double x = MEAN(1057.3, 6038); printf("Mean: %0.1f\n", x); return 0; }
This flexibility can be both an advantage and a problem.
main.c
12345678#include <stdio.h> #define SUM(a,b) (a + b) int main() { printf("sum: %d\n", SUM(4+5, 2+3)); return 0; }
In this case, the compiler perceives the arguments not as integers 9 and 6, but as the expression 4+5+2+3
.
An additional pair of parentheses for the arguments when creating a macro does not solve this problem.
For the next task, let's remember what a ternary operator is:
If the expression a > b
is true, the program will return a
, if the expression is false, the program will return b
.
Why not just use functions?
Speed
Macros work at the preprocessor level, meaning they substitute code directly before compilation. This eliminates the overhead associated with calling functions (saving/restoring registers, jumping by address, etc.).
Using macro:
Using function:
Universality of types
Functions require specifying the types of arguments and return values, while macros simply substitute code, so they can work with any type.
main.c
123456789#include <stdio.h> #define MAX(a, b) ((a) > (b) ? (a) : (b)) int main() { int maxInt = MAX(3, 7); // 7 float maxFloat = MAX(3.14, 2.71); // 3.14 return 0; }
Key takeaways:
Swipe to start coding
- Create a functional macro
MAX
with two argumentsa
andb
; - Specify a ternary operator for comparing two numbers as the body of the macro;
- Display the larger number.
Solution
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Macro With Arguments
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This simple macro calculates the arithmetic mean of two numbers:
main.c
12345678#include <stdio.h> #define MEAN(X,Y) ((X)+(Y))/2 int main() { printf("Mean: %d\n", MEAN(1053, 6037)); return 0; }
To make the macro able to handle non-integer numbers, we change the specifier and change the arguments to non-integer numbers:
main.c
123456789#include <stdio.h> #define MEAN(X,Y) ((X)+(Y))/2 int main() { double x = MEAN(1057.3, 6038); printf("Mean: %0.1f\n", x); return 0; }
This flexibility can be both an advantage and a problem.
main.c
12345678#include <stdio.h> #define SUM(a,b) (a + b) int main() { printf("sum: %d\n", SUM(4+5, 2+3)); return 0; }
In this case, the compiler perceives the arguments not as integers 9 and 6, but as the expression 4+5+2+3
.
An additional pair of parentheses for the arguments when creating a macro does not solve this problem.
For the next task, let's remember what a ternary operator is:
If the expression a > b
is true, the program will return a
, if the expression is false, the program will return b
.
Why not just use functions?
Speed
Macros work at the preprocessor level, meaning they substitute code directly before compilation. This eliminates the overhead associated with calling functions (saving/restoring registers, jumping by address, etc.).
Using macro:
Using function:
Universality of types
Functions require specifying the types of arguments and return values, while macros simply substitute code, so they can work with any type.
main.c
123456789#include <stdio.h> #define MAX(a, b) ((a) > (b) ? (a) : (b)) int main() { int maxInt = MAX(3, 7); // 7 float maxFloat = MAX(3.14, 2.71); // 3.14 return 0; }
Key takeaways:
Swipe to start coding
- Create a functional macro
MAX
with two argumentsa
andb
; - Specify a ternary operator for comparing two numbers as the body of the macro;
- Display the larger number.
Solution
Thanks for your feedback!
Awesome!
Completion rate improved to 5.56single