#define S(x) 4*x*x+1 main() { int i=6,j=8; printf("%d\n",S(i+j)); }
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/13 03:32:30
#define S(x) 4*x*x+1 main() { int i=6,j=8; printf("%d\n",S(i+j)); }
#define S(x) 4*x*x+1 main() { int i=6,j=8; printf("%d\n",S(i+j)); }
#define S(x) 4*x*x+1 main() { int i=6,j=8; printf("%d\n",S(i+j)); }
#define宏定义是直接替换
所以S(i+j)是4*i+j*i+j+1(没有括号),所以结果是4*6+8*6+8+1=81
如果要加括号,应改为
#define S(x) (4*(x)*(x)+1)
#define s(x) 3
#define S(x) 3
#define S(x) 4*(x)*x+1 s(4)怎么计算
#define SETBIT(x,y) (x|=(1
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 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));}
若有宏定义# define s(x) x*x-x,设int k=3; 问cout
求此程序的解释,就是#define s(x) 4*(x)*x+1这语句是什么意思啊,怎么运算#include#define s(x) 4*(x)*x+1main(){int k=5, j=2;printf(%d
,s(k+j));}
#define SWP_TYPE(x) (((x).val >> 1) & 0x3f)
#define set_bit(x,b) (x) |= (1U
)define f(x)(x*x) 和 define f(x) x*x 之间的差别.