如何在MATLAB中正确使用screenRect函数？

编程问答 2023-03-20

问题描述

当我试图在屏幕上的随机位置显示数字时，我运行以下代码：

% Screen parameters
center = [screenRect(3)/2 screenRect(4)/2)];
[mainwin,screenRect] = Screen(0,'OpenWindow');

% potential locations to place sequences
nrow = 6; ncolumn = 6; cellsize = 100;
for ncells = 1:nrow*ncolumn
    xnum = (mod(ncells-1,ncolumn)+1)-3.5;
    ynum = ceil(ncells/nrow)-3.5;
    cellcenter(ncells,1) = center(1)+xnum.*cellsize;
    cellcenter(ncells,2) = center(2)+ynum.*cellsize;
end

% randomise position of numbers
cellindex = Shuffle(1:nrow*ncolumn); % randomize the position of the sequences within the grid specified earlier
     itemloc = [cellcenter(cellindex(1),1)-cellsize/2,cellcenter(cellindex(1),2)-cellsize/2,1)+cellsiz

我收到以下错误：

Unrecognized function or variable 'screenRect'.

Error in First_draft (line 4)
center = [screenRect(3)/2 screenRect(4)/2];

解决方法

如果您确定正确安装了toolbox，那么

    1. 使用screenRect方法返回Screen变量。
    • [mainwin,screenRect] = Screen(0,'OpenWindow');
    1. 然后使用screenRect变量
    • center = [screenRect(3)/2 screenRect(4)/2];

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