C++結(jié)構(gòu)體數(shù)組詳細(xì)解析

更新時(shí)間：2013年10月16日 10:09:19 作者：

定義結(jié)構(gòu)體數(shù)組和定義結(jié)構(gòu)體變量類似，定義結(jié)構(gòu)體數(shù)組時(shí)只需聲明其為數(shù)組即可

1.定義結(jié)構(gòu)體數(shù)組

和定義結(jié)構(gòu)體變量類似，定義結(jié)構(gòu)體數(shù)組時(shí)只需聲明其為數(shù)組即可。如：

struct Student{
     int num;
     char name[20];
     char sex[5];
     int age;
     float score;
     char addr[30];
};
Student stu[3]; //定義Student類型的數(shù)組stu

2.結(jié)構(gòu)體數(shù)組的應(yīng)用舉例

題目：對(duì)候選人的票的統(tǒng)計(jì)程序。

設(shè)有3個(gè)候選人，最終只能有一個(gè)當(dāng)選為領(lǐng)導(dǎo)。今有10個(gè)人參加投票，從鍵盤先后輸入這10個(gè)人所投的候選人的名字，要求最后能輸出這3個(gè)候選人的的票結(jié)果。

復(fù)制代碼代碼如下:

#include<iostream>
using namespace std;
struct Person{
&nbsp;&nbsp; &nbsp;char name[20];&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; //姓名 
&nbsp;&nbsp; &nbsp;int count;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; //票數(shù)計(jì)數(shù)器 
};
int main(){
&nbsp;&nbsp; &nbsp;Person leader[3]={"Tom",0,"Neo",0,"Marry",0};
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;//定義Person類型的數(shù)組，內(nèi)容為3個(gè)候選人的姓名和票數(shù)
&nbsp;&nbsp; &nbsp;int i,j,k=0;
&nbsp;&nbsp; &nbsp;bool tag;
&nbsp;&nbsp; &nbsp;cout<<"please input the name of the leader : Tom Neo Marry\n\n";
&nbsp;&nbsp; &nbsp;char leadername[20];&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; //該數(shù)組為每次輸入的候選人的名字
&nbsp;&nbsp; &nbsp;for(i=0;i<10;i++){&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; //循環(huán)輸入這10個(gè)人選的候選人的名字 
&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;cout<<"input name "<<i+1<<" :"; 
&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;cin>>leadername; 
&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;tag=1;
&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;for(j=0;j<3;j++){
&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;if(strcmp(leadername,leader[j].name)==0){
&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;leader[j].count++;
&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;tag=0;
&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;}
&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;} 
&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;if(tag==1)k++;
&nbsp;&nbsp; &nbsp;} 
&nbsp;&nbsp; &nbsp;cout<<endl;
&nbsp;&nbsp; &nbsp;for(i=0;i<3;i++){
&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; cout<<leader[i].name<<":"<<leader[i].count<<endl;&nbsp;&nbsp; &nbsp;
&nbsp;&nbsp; &nbsp;} &nbsp;
&nbsp;&nbsp; &nbsp;cout<<"Abandoned tickets:"<<k<<endl;
&nbsp;&nbsp; &nbsp;return 0;
}

當(dāng)然，如果不使用結(jié)構(gòu)體也可以解決這個(gè)問題：

復(fù)制代碼代碼如下:

#include<iostream>
#include<string>
using namespace std;
int main(){
 char *name[3]={"Tom","Neo","Marry"};
 int    count[3]={0,0,0};
 int    i,j,k=0;
 bool tag=1;
 cout<<"please input the name of the leader : Tom Neo Marry\n\n";
 char leadername[20];                
 for(i=0;i<10;i++){                 
  cout<<"input name "<<i+1<<" :"; 
  cin>>leadername; 
  for(j=0;j<3;j++){
   if(strcmp(leadername,name[j])==0){
    count[j]++;
    tag=0;
   }
  } 
  if(tag==1)k++;
  tag=1;
 } 
 cout<<endl;
 for(i=0;i<3;i++){
    cout<<name[i]<<":"<<count[i]<<endl; 
 }
 cout<<"Abandoned tickets:"<<k<<endl;
 return 0;
}

或者

復(fù)制代碼代碼如下:

#include<iostream>
#include<string>
using namespace std;
int main(){
 string name[3]={"Tom","Neo","Marry"};
 int    count[3]={0,0,0};
 int    i,j,k=0;
 bool tag=1;
 cout<<"please input the name of the leader : Tom Neo Marry\n\n";
 string leadername;                
 for(i=0;i<10;i++){                 
  cout<<"input name "<<i+1<<" :"; 
  cin>>leadername; 
  for(j=0;j<3;j++){
   if(leadername==name[j]){
    count[j]++;
    tag=0;
   }
  } 
  if(tag==1)k++;
  tag=1;
 } 
 cout<<endl;
 for(i=0;i<3;i++){
    cout<<name[i]<<":"<<count[i]<<endl; 
 }
 cout<<"Abandoned tickets:"<<k<<endl;
 return 0;
}