C++ 二维数组到一维数组
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C++ 2D array to 1D array
提问by Anton Antonov
I am attempting to convert a 2D array to 1D. I'm extremely new to C/C++ but I think it's very important to learn how to convert a 2D array to 1D. So here I am stumbling upon this problem.
我正在尝试将 2D 数组转换为 1D。我对 C/C++ 非常陌生，但我认为学习如何将 2D 数组转换为 1D 非常重要。所以在这里我偶然发现了这个问题。
My code so far is http://ideone.com/zvjKwP
到目前为止我的代码是 http://ideone.com/zvjKwP
#include<iostream>
using namespace std;
int main()
{
int n=0,m=0; // 2D array nRow, nCol
int a[n][m];
int i,j; // цикъл въвеждане 2D
int q,p,t; // for 2D=>1D
int b[100];
int r; // for cout
cout<<"Enter the array's number of rows and columns: ";
cin>>n>>m;
// entering values for the 2D array
for (i = 0;i<=n;i++)
{
for (j = 0;j<=m;j++)
{
cout<<"a["<<i<<"]["<<j<<"]="<<endl;
cin>>a[i][j];
cin.ignore();
}
}
// Most likely the failzone IMO
for (q = 0;q<=i;q++)
{
for (t = 0;t<=i*j+j;t++)
{
b[t] = a[i][j];
}
}
// attempting to print the 1D values
cout<<"The values in the array are"<<"\n";
for(r=0;r<=t;r++)
{
cout<<"b["<<r<<"] = "<<b[r]<<endl;
}
cin.get();
return 0;
}
I wrote a comment at where I think I fail.
我在我认为我失败的地方写了一条评论。
I must also limit the numbers that get into the 1D to numbers who's value^2 is greater than 50. But for sure I must solve the problem with the conversion 2D=>1D Can you help me out?
我还必须将进入 1D 的数字限制为 value^2 大于 50 的数字。但我肯定必须解决转换 2D=>1D 的问题你能帮我吗？
采纳答案by Boris Strandjev
You are right with your supposition:
你的假设是对的：
The cycle should be like:
循环应该是这样的：
for (q = 0; q < n; q++)
{
for (t = 0; t < m; t++)
{
b[q * m + t] = a[q][t];
}
}
It is always easier to consider such conversions from the view point of the higher dimension array. Furthermore with your code you did not actually modify ior jin the bassigning cycle, so you should not expect different values to be assigned to the different array members of b.
从更高维数组的角度考虑这种转换总是更容易。此外，对于您的代码，您实际上并未修改i或j在b分配周期中，因此您不应期望将不同的值分配给b.
回答by spellmansamnesty
http://www.cplusplus.com/doc/tutorial/arrays/
http://www.cplusplus.com/doc/tutorial/arrays/
In that link look at the section on pseudo-multidimensional arrays.
在该链接中查看伪多维数组部分。
I've seen many examples that that get the subscripting algorithm wrong. If in doubt, trace it out. The order of sub-scripting a 2D array should go sequentially from 0-(HEIGHT*WIDTH-1)
我已经看到很多例子表明下标算法是错误的。如有疑问，请追查。二维数组的下标顺序应该从 0-(HEIGHT*WIDTH-1) 开始
#define WIDTH 5
#define HEIGHT 3
int jimmy [HEIGHT * WIDTH];
int n,m;
int main ()
{
for (n=0; n<HEIGHT; n++)
for (m=0; m<WIDTH; m++)
{
jimmy[n*WIDTH+m]=(n+1)*(m+1);
}
}
回答by Vlad from Moscow
This code
这段代码
int n=0,m=0; // 2D array nRow, nCol
int a[n][m];
is invalid. First of all the dimensions shall be constant expressions and there is no sense to set them to 0.
是无效的。首先，维度应该是常量表达式，将它们设置为 0 是没有意义的。
And the more simple way to do your task is to use pointer. For example
完成任务的更简单方法是使用指针。例如
int *p = b;
for ( const auto &row : a )
{
for ( int x : row ) *p++ = x;
}
回答by Shubham
First of all, the size of the 1D array should be n*m.
首先，一维数组的大小应该是n*m.
The cycle can be as follows-
循环可以如下-
int lim = n*m;
for(q = 0; q<lim; ++q) {
b[q] = a[q/m][q%m];
}
