Side Effects of macros

Question 3: Write a "standard" macro min, this macro enters two parameters and returns the smaller one. Also, what happens when you write the following code.

least = MIN (*p++, b);

Answer:

#define MIN (A,b) ((A) <= (B)? (A): (B))

MIN (*p++, b) can create side effects of macros

Analysis:

This question mainly examines the interviewer's use of the macro definition, a macro definition can implement functions similar to a function, but it is not a function at all, and the "arguments" in parentheses in a macro definition are not real arguments, and a one-to-one substitution is made for "parameters" when the macro is expanded.

Programmers should be very careful about using macro definitions, with particular attention to two issues:

(1) Carefully enclose the "parameters" and the entire macro in the macro definition in parentheses. So, strictly speaking, the following answers:

#define MIN (A,b) (A) <= (B)? (A): (B)

#define MIN (a,b) (A <= B?) A:B)

Should be sentenced to 0 points;

(2) To prevent the macro side effects.

Macro definition #define MIN (a,b) ((A) <= (B)? (A): (b) The result of the effect on min (*p++, b) is:

((*p++) <= (b)? (*p++): (*p++))

This expression will have side effects, and the pointer P will do three + + self-increment operations.

In addition, another answer that should be 0 points is:

#define MIN (A,b) ((A) <= (B)? (A): (B));

This solution adds ";" to the macro definition, showing that the creator's concept of the macro is ambiguous and can only be ruthlessly sentenced to 0 points and eliminated by the interviewer.

Eliminate side effects of macros

#include <stdio.h> #define Min_i (x, y) ((x) < (y)? (x): (y))//(1) #define MIN_T (type, x, y) ({type __x = x; \//(2) Type __y = y; \ __x < __y? __x: __y; \ }) #define MIN (x, y) ({const typeof (x) _x = (x); \//(3) Const typeof (Y) _y = (y); \ (void) (&_x = = &_y); \//(4) _x < _y? _x: _y; \ }) int main () { int a = 10; int B = 20; printf ("Min_i (a++, b++) =%d\n", Min_i (a++, b++)); 11 printf ("A =%d\n", a); 12 printf ("B =%d\n", b); 21st A = 10; b = 20; printf ("min_t (int, a++, b++) =%d\n", min_t (int, a++, b++)); 10 printf ("A =%d\n", a); 11 printf ("B =%d\n", b); 21st A = 10; b = 20; printf ("min (a++, b++) =%d\n", min (a++, b++)); 10 printf ("A =%d\n", a); 11 printf ("B =%d\n", b); 21st } (1). This definition calculates x and y respectively two times (the small in x and Y are counted two times), and when the parameter has side effects, it produces an incorrect result (2). Use statement expressions to calculate only one argument at a time, avoiding possible errors; Statement expressions are typically used for macro definitions (3). typeof (X) represents the value type of x (4). Check that the types of parameters x and Y are the same (if the types of x and Y are different compilers will emit warning and do not affect the execution of subsequent statements)