有以下程序 struct STU { char name[10]; int num; float TotalScore; }; vold f(struct STU *p) { struct STU s[2]={{"SunDan",20044,550),{"Penghua".20045,537}},*q=s ++p; ++q; *p=*q; } main() { struct SrU s[3]={{"YangSan",20041,703),{"LiSiGuo",20042,580}}; f(s); printf("%s %d %3.of\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.USiGuo 20042 580
D.SunDan 20041 703
第1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(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
第2题:
有以下程序:#include <stdio, h>#include <string, h>struet STU{ int nam; float TotalSeore;};void f( strnct STU p){ struct STU s[2] = { {20044,550} ,{20045,537} }; p.num = s [1]. num; p.TotalScore = s [1]. TotalScore;main( ){ struct STU s[2] = {{20041,703} ,{20042,580}}; f(s[0] ); printf( "%d %3.Of\n" ,s[0].num,s[0].TotalSeore); }程序运行后的输出结果是( )。
A.20045 537
B.20044 550
C.20042 580
D.20041 703
第3题:
【单选题】有以下程序输出结果是()。 #include<stdio.h> struct stu {int num; char name[10]; int age; }; void fun(struct stu *p) {printf("%sn",(*p).name);} main() {struct stu students[3]= {{9801,"zhang",20},{9802,"Wang",19},{9803,"zhao",18}}; fun(students+2);}
A.Zhang
B.Zhao
C.Wang
D.18
第4题:
有以下程序#include <string.h>struct STU{ int num; float TotalScore; };void f(struct STU p){ struct STU s[2]={{1047,530},{1048,531}}; p.num = s[1].num; p.TotalScore = s[1].TotalScore;}main(){ struct STU s[2]={{2041,730},{2042,731}}; f(s[0]); printf("%d,%3.0f\n",s[0].num,s[0].TotalScore);}程序的运行结果是A.1047,530 B.1048,531C.2041,730 D.2042,731
第5题:
有以下程序:#include <stdio.h>#include <strine.h> struct STU { char name[10]; int hum;};void f(char * name,iht num){ struct STU s[2] = {{ "SunDan" ,20044} , {" Penghua" ,20045}}; num= s[0]. nnm; strepy(name,s[0], name);}main( ){ struct STU s[2] = {{"YangSan" ,20041 }, { "LiSiGao" ,20042}}, * P; p = &s[1]; f(p->name,p->hum); 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