有关常见的#define定义的函数的陷阱和修正（详解）

一、#define f(x) x*x

#include
#define f(x) x*x
int main()
{
	int a=6,b=2,c;
	c=f(a)/f(b);
	printf("f(a)=%d\n",f(a));//6*6
	printf("f(b)=%d\n",f(b));//2*2
	printf("f(b+1)=%d\n",f(b+1));//2+1*2+1; 
	printf("f(b+2)=%d\n",f(b+2));//2+2*2+2  
	printf("f(b+3)=%d\n",f(b+3));//2+3*2+3
	printf("f(a)/f(b)=%d\n",c);//36/(2+1)*(2+1)=36;
	return 0;
}

 

1.按照我们定义f(x)函数的初衷，我们希望f(b+1)=3*3=9，但现实是f(b+1)=5;

f(b+1)=5的运算逻辑则是    2+1*2+1=5

f(b+2)=8的运算逻辑则是    2+2*2+2=8

f(b+3)=11的运算逻辑则是   2+3*2+3=11

即 f(b+n)=b+n*b+b;

2.同样按照我们定义f(x)的函数的初衷，我们希望f(a)/f(b)=36/9=4,但现实是f(a)/f(b)=36;

f(a)/f(b)=36，背后的逻辑则是 36/(2+1)*(2+1)=36;相等于等于其本身

3.我们如果想修正这个函数为我们定义f(x)的初衷，就应该在整个函数外面加个括号，例如g(x)

#include
#define f(x) x*x
#define g(x) ((x)*(x))//修正 
int main()
{
	int a=6,b=2,c;
	c=f(a)/f(b);
	printf("f(a)=%d\n",f(a));//6*6
	printf("f(b)=%d\n",f(b));//2*2
	printf("f(b+1)=%d\n",f(b+1));//2+1*2+1; 
	printf("f(b+2)=%d\n",f(b+2));//2+2*2+2  
	printf("f(b+3)=%d\n",f(b+3));//2+3*2+3
	printf("f(a)/f(b)=%d\n",c);//36/(2+1)*(2+1)=36;
	printf("\n");
	printf("g(a)=%d\n",g(a));
	printf("g(b)=%d\n",g(b));
	printf("g(b+1)=%d\n",g(b+1));
	printf("g(a)/g(b)=%d\n",g(a)/g(b));
	return 0;
}