阵列和指数 [英] arrays and index
问题描述
我怎样才能输入数字到重复的条目这样一个数组不会被插入?例如,如果我试图把 6
入阵,那么 3
进入第二的第一个条目,然后, 6
进入第三个, 6
应该被拒绝(因为它会在数组中已存在)。请张贴在C ++中完成code。我的继承人code你能告诉我什么是失踪?
的#include<&iostream的GT;使用命名空间std;
诠释的main()
{
INT X,Y;
INT编号;
INT ARR [5]; 为(X = 0; X小于5)
{
COUT<<进入nmber<< ENDL;
CIN>>数;
布尔取代= TRUE;
对于(Y = 0; Y< X; Y ++)
{
如果(号码!= ARR [Y])
{
COUT<< 尝试下一次<< ENDL;
取代=假;
打破;
}
} 如果(替换)
{
ARR [X] =号;
X ++;
}
}
返回0;}
你有太多的 X ++'和
你没有preSET 改编
(比错误也许更多的风格)
你怎么知道它不工作?
(把一些调试code 内,如果(数字==常用3 [Y])
和如果(更换
)
How can I enter numbers into an array in such a way that duplicate entries will not be inserted? For example, if I try to put 6
into the first entry of the array, then 3
into the second, then 6
into the third, 6
should be rejected (since it will already exist in the array). Please post the complete code in C++. heres my code can you tell me whats missing?
#include <iostream>
using namespace std;
int main()
{
int x,y;
int number;
int arr[5];
for (x=0;x<5;)
{
cout<<"enter nmber"<<endl;
cin>>number;
bool replace=True;
for (y=0;y<x;y++)
{
if (number != arr[y])
{
cout << "try next time"<<endl;
replace=False;
break;
}
}
if (replace)
{
arr[x]=number;
x++;
}
}
return 0;
}
you have too many x++
's and you don't preset arr
(maybe more style than error)
how do you know it's not working?
(put some debug code inside of if (number == arr[y])
and if (replace
)
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