#define p(x) x*x int x=5,y=3,z; z=p(x+y); A) 64 B) 23 C) 46 D) 32

#define P 3 void F(int x){return(P*x*x);} main() {printf(%d ,F(3+5));} #define P3 void F(int x){return(P*x*x);} main() {printf(%d ,F(3+5));} #define p(x) x*x int x=5,y=3,z; z=p(x+y); A) 64 B) 23 C) 46 D) 32 若有宏定义# define s(x) x*x-x,设int k=3; 问cout #define X 3 #define Y X*2 #undef X #define X 2 int z=Y; z 的值为多少? #define MA（x） x*x-1 int a=1,b=2; cout #define s(x) 3 #define S(x) 3 define fun(x, #define configASSERT( x ) #define min(x,y) (x #define MIN(x,y)(x) #define __T(x) L ## x #define get2byte(x) ((x)[0] 若有定义∶# define P(x) x * x *x+1和说明语句：int a=2;则表达式P（2）的值是（ ）具体说说程序正义运行, #define N 10#define s(x) x*x#define f(x) (x*x)main(){int i1,i2;i1=1000/s(N);i2=1000/f(N);printf(%d %d ,i1,i2);}运行结果是? 求此题解题过程：#include #define N 8 void fun(int *x,int i) {*x=*(x+i);} main() {int a[N] #define f(x) x*x main( ) { int i; i=f(4+4)/f(2+2); printf(%d ,i); }