这个宏是什么意思#define div_ac_power（y，x）* y / x

Question

#define div_ac_power(y,x)           *y/x

Shall y be an address? As far as I understand it returns the result of division between a pointer (y) and whatever object is x.

Answer 1

As far as I understand it returns the result of division between a pointer (y) and whatever object is x.

That is probably the intention.

The macro needs more parenthesis to secure it from misuse.

Consider the following cases:

Case 1. Compound parameter:

result= div_ac_power(y,a+b);

which is expanded to:

result = *y/a+b; // This equals (*y/a)+b because of operator precedence.

Case 2. Fooling around:

result= a div_ac_power(y,b); // Note the lack of operator between a and macro

which is expanded to:

result= a*y/b; // This may well be a valid expression depending on the types of a, b, y.

To fix it, you could add parenthesis:

#define div_ac_power(y,x)           (*(y)/(x))

Answer 2

1. Dereference the value pointed-to by y pointer
2. Divide this value by x value

Same as:

int *y, x; // init them
int z = *y;
int res = z / x;