A. rise
B. raise
C. rose
D. raised
第1题:
#include "stdio.h"void half(void *pval,char type);main(){ int i=20; long l=100000; float ff=12.456; double d=123.044444; printf("%d\n",i); printf("%ld\n",l); printf("%f\n",ff); printf("%lf\n",d); half(&i,'i'); half(&l,'l'); half(&ff,'ff'); half(&d,'d'); printf("\n%d",i); printf("\n%ld",l); printf("\n%f",ff); printf("\n%lf",d); return 0; }void half(void *pval,char type){ switch(type) { case 'i': { *((int *)pval)/=2; //我想问一下,这个语法怎么理解,太复杂了 break; } case 'l': { *((long *)pval)/=2; break; } case 'ff': { *((float *)pval)/=2; break; } case 'd': { *((double *)pval)/=2; break; } } }
第2题:
b)
main()
{
union{ /*定义一个联合*/
int i;
struct{ /*在联合中定义一个结构*/
char first;
char second;
}half;
}number;
number.i=0x4241; /*联合成员赋值*/
printf("%c%c\n", number.half.first,
mumber.half.second);
number.half.first='a'; /*联合中结构成员赋值
*/
number.half.second='b';
printf("%x\n", number.i);
getch();
}
AB (0x41 对应'A',是低位;Ox42 对应'B',
是高位)6261 (number.i 和number.half 共用一块地址空
间)
第3题:
A. Half-duplex Ethernet operates in a shared collision domain.
B. Full-duplex Ethernet has a lower effective throughput.
C. Half-duplex Ethernet operates in a private collision domain.
D. Full-duplex Ethernet allows two-way communication.
E. Half-duplex Ethernet operates in a private broadcast domain.
第4题:
I_______ to get up late, but now I get up early every morningl.
A、used to
B、is used to
C、use
D、use to
第5题:
A) still late
B) too late
C) so late
D) past
第6题: