char.name[20];int num;
scanf(“name=%s num=%d”,name,&num);
当执行上述程序段,并从键盘输入:name=Lili num=1001<回车>后,name的值为
A.Lill
B.name=Lili
C.Lili.num=
D.name=Lili num—1001
[单选题]有以下程序段char.name[20];int num;scanf(“name=%s num=%d”,name,&num);当执行上述程序段,并从键盘输入:name=Lili num=1001<回车>后,name的值为A.LillB.name=LiliC.Lili.num=D.name=Lili num—1001
[单选题]有以下程序段:char name[20);int num;scanf("name=%S num=%d",name,&num);当执行上述程序段,并从键盘输入:name=Lili mum=1001<回车>后,name的值为( )。A.LiliB.name=LiliC.Lili num=D.name=Lili num=1001
[单选题](16)有以下程序段char name[20]; int num;scanf("name=%s num=%d",name;&num);当执行上述程序段,并从键盘输入:name=Lili num=1001<回车>后,name的值为A.)LiliB.)name=LiliC.)Lili num=D.)name=Lili num=1001
[单选题]有以下程序段:char name[20];int num;scanf(″name=%s num=%d″,name,&num):当执行上述程序段,并从键
[单选题]有以下程序段 int j;float y; char name[50]; scanf("%2d%f%s",&j,&y,name); 当执行上述程序段,从键盘上输入55566 7777abc后,y的值为( )。A.55566.0B.566.0C.7777.0D.566777.0
[单选题]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]=&am
[单选题]有以下程序: struct 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(&
[单选题]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/
[单选题]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(
[单选题]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]=&a