[英]How is the memory layed out in Eigen?
I have written this little parser that reads in some numbers from text file. 我写了这个小解析器,它从文本文件中读取一些数字。
data.resize(7,datapoints); //Eigen::Matrix<float,7,-1> & data
dst = data.data();
while( fgets(buf,255,fp) != 0 && i/7 < datapoints)
{
int n = sscanf(buf,"%f \t%f \t%f \t%f \t%f \t%f \t%f",dst+i++, dst+i++,dst+i++,dst+i++,dst+i++,dst+i++,dst+i++);
i = i - 7 * (n<=0);
}
fclose(fp);
return !(datapoints == i/7);
The thing is, when i do a std::cout on the data it flipped. 问题是,当我对翻转的数据执行std :: cout时。
Data in: 数据在:
0 4 0.35763609 0.64077979 0 0 1
0 4 0.36267641 0.68243247 1 0 2
0 4 0.37477320 0.72945964 2 1 3
data.col(3) is data.col(3)是
0.64077979
0.68243247
0.72945964
and data.col(4) is 并且data.col(4)是
0.35763609
0.36267641
0.37477320
I cant see the logic why it have flipped the data horizontal? 我看不到逻辑为什么将数据水平翻转?
To illustrate the problem: 为了说明问题:
#include <cstdio>
void f(int i, int j, int k)
{
printf("i = %d\tj = %d\tk = %d\n", i, j, k);
}
int main()
{
int i=0;
f(i++, i++, i++);
}
Executing this, returns here (g++ 4.3.4 on Cygwin): 执行此操作,返回此处(Cygwin上的g ++ 4.3.4):
i = 2 j = 1 k = 0
The execution order of the i++
calls inside the function call is entirely implementation defined (ie arbitrary). 函数调用内部的i++
调用的执行顺序完全由实现定义(即任意)。
Are you sure than? 你确定吗?
int i=0;
sscanf(buf,"%f \t%f \t%f \t%f \t%f \t%f \t%f",dst+i++, dst+i++,dst+i++,dst+i++,dst+i++,dst+i++,dst+i++);
is equal to: 等于:
sscanf(buf,"%f \t%f \t%f \t%f \t%f \t%f \t%f",dst+0,dst+1,dst+2,dst+3,dst+4,dst+5,dst+6 );
I think the variable list arg is geting evaluated back in this case, and as @Christian Rau comment in general is Undefined the order of evaluation. 我认为在这种情况下,变量列表arg正在重新评估,因为@Christian Rau的注释通常是未定义评估的顺序。 Oft it is not a good idea to realy on side -effect order. 要实现副作用顺序,通常不是一个好主意。
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