更多"有以下程序:
struc STU {
char na"的相关试题:
[单项选择]
有以下程序
struc STU {
char name[10];
int num; };
void f1(struct STU c)
{ struct STU b={“LiSiGuo”,2042};
c=b; }
void f2(struct STU *c)
{ struct STU b={“SunDan”,2044};
*c=b; }
main( )
{ struct STU a={“YangSan”,2041},b={“WangYin”,2043 };
f1(a);f2(&b);
printf(“%d %d/n”,a.num,b.num); }
执行后的输出结果是()
A. 2041 2044
B. 2041 2043
C. 2042 2044
D. 2042 2043
[单项选择]有以下程序:
struc STU {
char name[10];
int num; };
void f1(struct STU c)
{ struct STU b={“LiSiGuo”,2042};
c=b; }
void f2(struct STU *c)
{ struct STU b={“SunDan”,2044};
*c=b; }
main( )
{ struct STU a={“YangSan”,2041},b={“WangYin”,2043 };
f1(a);f2(&b);
printf(“%d %d/n”,a.num,b.num); }
执行后的输出结果是( )。
A. 2041 2044
B. 2041 2043
C. 2042 2044
D. 2042 2043
[单项选择]有以下程序:
#include
struct STU
{char name[10];
int num;
};
void f(char *name, int num)
{ struct STU s[2]={{"SunDan",20044},{"Penghua",20045}};
num=s[0].num;
strcpy(name,s[0].name);
}
main( )
{ struct STU s[2]={{"YangSan",20041},{"LiSiGao",20042}},*p;
p=&s[1]; f(p->name,p->num);
printf("%s %d/n",p->name,p->num);
}
程序运行后的输出结果是______。
A. SunDan 20042
B. SunDan 20044
C. LiSiGuo 20042
D. YangSan 20041
[单项选择]有以下程序
struct STU
char name[10];
int num;
int Score;
main( )
struct Stu s[5]="YangSan",20041,703,"LiSiGuo",20042,580,
"WangYin",20043,680,"SunDan",20044,550,
"Penghua",20045;537,*p[5],*t;
int i,j;
for(i=0;i<5;i++)p[i]=&s[i];
for(i=0;i<4;i++)
for(j=i+1;j<5;j++)
if(p[i]->Score>p[j]->Score)
t=p[i];p[i]=p[j];p[i]=t;
printf("%d%d/n",s[1].Score,p[1]->Score);
执行后输出结果是
A. 550550
B. 680680
C. 580550
D. 580680
[单项选择]有以下程序:
struct STU
{ char name[10]; int num; float TotalScore; };
void f(struct STU *p)
{ struct STU s[2]={{"SunDan",20044,550},{"Penghua",20045,537}},*q=s;
++p; ++q; *p=*q;
}
main( )
{ struct STU s[3]={ {"YangSan",20041,703},{"LiSiGuo",20042,580}};
f(s);
printf("%s %d %3.0f/n",s[1].name, s[1].num,s[1].TotalScore);
}
程序运行后的输出结果是______。
A. SunDan 20044 550
B. Penghua 20045 537
C. LiSiGuo 20042 580
D. SunDan 20041 703
[单项选择]有以下程序:
struct STU
{ char name[10];v int num;
int Score; };
main( )
{ struct STU s[5]={ {“YangSan”,20041,703},{“LiSiGuo”,20042,580},
{“wangYin”,20043,680},{“SunDan”,20044,550},
{“Penghua”,20045,537}},*p[5],*t;
int i,j;
for(i=0;i<5;i++) p[i]=&s[i];
for(i=0;i<4;i++)
for(j=i+1;j<5;j++)
if(p[i]->Score>p[j]->Score)
{ t=p[i];p[i]=p[j];p[j]=t;}
printf(“5d %d/n”,s[1].Score,p[1]->Score); }
执行后输出结果是( )。
A. 550 550
B. 680 680
C. 580 550
D. 580 680