#define MA(x) x*x-1 int a=1,b=2; cout
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#defineMA(x)x*x-1inta=1,b=2;cout#defineMA(x)x*x-1inta=1,b=2;cout#defineMA(x)x*x-1inta=1,b=2;cout#def
#define MA(x) x*x-1 int a=1,b=2; cout
#define MA(x) x*x-1 int a=1,b=2; cout
#define MA(x) x*x-1 int a=1,b=2; cout
#define MA(x) x*x-1 int a=1,b=2; cout
上面给的答案是错的,如果你拿到机器上运行一下(我用的是VS2010),结果应该是8!!!
MA(1+a+b) = 1+a+b*(1+a+b-1) = 1+1+2*(1+1+2-1) = 2+2*(3) = 2 + 6 = 8
宏替换先直接替换字符,然后再计算的
1+1+2*1+1+2-1=6
#define MA(x) x*x-1 int a=1,b=2; cout
#define SETBIT(x,y) (x|=(1
#includestdio.h #define MA(x) x*(x-1) void main() { int a=1,b=2; printf(%d
,MA(1+a+b)); }
#define s(x) 3
#define S(x) 3
define fun(x,
#define configASSERT( x )
# define MA(x,y) (x)*(y) int i=2; i=3/MA(i,i+2)+5; printf(“%d”,i); 输出是什么,为什么?
#define min(x,y) (x
#define MIN(x,y)(x)
#define __T(x) L ## x
#define get2byte(x) ((x)[0]
#define SWP_TYPE(x) (((x).val >> 1) & 0x3f)
#define set_bit(x,b) (x) |= (1U
#include #define MA(x) x*(x-1) void main() { int a=1,b=2; printf(“%d
”,MA(1+a+b)); }请问诸位大侠,上面式子的结果为多少?
结果是() A.10 B.11 C.12 D.13 #define MA(x) x*(x+3) main() {int a=1,b=2;printf(%d
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)define f(x)(x*x) 和 define f(x) x*x 之间的差别.
#define MEM_B( x ) ( *( (byte *) (x) ) 我看不懂