#define S(x)4 *(x)*x+1 main() {int k=5,j=2;printf("%d\n",S(k+j));}
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#defineS(x)4*(x)*x+1main(){intk=5,j=2;printf("%d\n",S(k+j));}#defineS(x)4*(x)*x+1main(){intk=5,j=2;p
#define S(x)4 *(x)*x+1 main() {int k=5,j=2;printf("%d\n",S(k+j));}
#define S(x)4 *(x)*x+1 main() {int k=5,j=2;printf("%d\n",S(k+j));}
#define S(x)4 *(x)*x+1 main() {int k=5,j=2;printf("%d\n",S(k+j));}
将k+j带到宏函数里面就可以了,4*(5+2)*5+2+1,结果143
#define s(x) 3
#define S(x) 3
#define S(x) 4*(x)*x+1 s(4)怎么计算
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]
若有宏定义# define s(x) x*x-x,设int k=3; 问cout
)define f(x)(x*x) 和 define f(x) x*x 之间的差别.
#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 S(x) 4*x*x+1 main() { int i=6,j=8; printf(%d
,S(i+j)); }
#define S(x) 4*x*x+1 main() { int i=6,j=8; printf(%d
,S(i+j)); }
#define S(x)4 *(x)*x+1 main() {int k=5,j=2;printf(%d
,S(k+j));}