#define f(x) x*x main() {int a=6,b=2,c; C=f(a)?f(b); printf("%d\n",c);}

来源：学生作业帮助网编辑：六六作业网时间：2024/11/21 21:26:09

#definef(x)x*xmain(){inta=6,b=2,c;C=f(a)?f(b);printf("%d\n",c);}#definef(x)x*xmain(){inta=6,b=2,c;C=

#define f(x) x*x main() {int a=6,b=2,c; C=f(a)?f(b); printf("%d\n",c);}
#define f(x) x*x main() {int a=6,b=2,c; C=f(a)?f(b); printf("%d\n",c);}

#define f(x) x*x main() {int a=6,b=2,c; C=f(a)?f(b); printf("%d\n",c);}
宏要加括号不然有风险
变量要初始化
c是区分大小写的

)define f(x)(x*x) 和 define f(x) x*x 之间的差别. #define F(x) x - 2 #define D(x) x*F(x) printf(%d,%d,D(3),D(D(3))) ; #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] #define f(x)x*x f(4+4)/(2+2) 为什么等于21 #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? #define SETBIT(x,y) (x|=(1 #define MEM_B( x ) ( *( (byte *) (x) ) 我看不懂 #define get_u8(X,O) (*(u8 *)(((u8 *)X) + #define f(x) x*x main( ) { int i; i=f(4+4)/f(2+2); printf(%d ,i); } 若有宏定义：#define f(x) x*x,则表达式f(4+4)/f(2+2)的值为( ).求详解 #define zero(x) (((x) > 0 (x) :(-x)) < eps) 是什么作用?