问题动态数组 [英] problem dynamic array

问题描述

嗨 我无法从动态数组中获取一个数字，然后打印出来.
我该怎么办?
输入:2
3
3
0
0
5
1
2
3

代码:

Hi I get a number from dynamic array and then print, I can not.
What do I do?
input:2
3
3
0
0
5
1
2
3

code:

#include<iostream>
using namespace std;
void s(int f[],int d)
{
	cout<<"matris  element i,j   :\ni=";
        cin>>f[0,0];cout<<"j=";cin>>f[0,1];
        f[0,2]=d;
        cout<<"add number not 0 ";
	for(int k=1;k<d+1;k++)
	{   
		cout<<"\n array i="; cin>>f[k,0];
		cout<<"array j=";cin>>f[k,1];
		cout<<"value array ij=";cin>>f[k,2];
		cout<<f[k,2];
	}
	for(int i=1;i<d+1;i++)
	 {
		for(int j=0;j<3;j++)
		cout<<f[i,j];
	    cout<<"\n";
	 }
}
void main()
{
    int w;
	cout<<"number not 0 :";
	cin>>w;
    int *b=new int [w+1,3];
    s(b,w);
}

推荐答案

逗号(，)是列表运算符.数组操作(operator [](int index);)只有一个参数.列表运算符将以以下方式工作:将执行所有操作(用逗号分隔)，但最后一个仍然是操作数.
例如，您可以编写int* px = new int [n=calcsize(),n+=8,push(n),42];，最终将像这样:n = calcsize(); n+=8; push(n); int* px = new int [42];.因此，在您的代码中，所有数组操作都使用最后一个逗号后面的参数:
int *b=new int [w+1,3];等于int *b=new int [3];
cin>>f[k,0];等于cin>>f[0];
等等.
也许您可以解释表达方式所能达到的目标.
最好的问候.

The comma (,) is a list operator. The array operation (operator [] (int index);) has only one argument. The list operator will work in following way: all operations (separated by comma) will been executed but still the last is the operand.
For example you can write int* px = new int [n=calcsize(),n+=8,push(n),42]; will finally be the same like: n = calcsize(); n+=8; push(n); int* px = new int [42];. So at your code all your array operations are using the parameter behind the last comma:
int *b=new int [w+1,3]; is equal to int *b=new int [3];
cin>>f[k,0]; is equal to cin>>f[0];
and so on.
Perhaps you can explan what to achive with your expressions.
Best regards.

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