c - 读取图像

Question

我正在尝试使用 OpenCV 2.1(今天才开始)读取图像，C++ 使用下面的代码。我看不出下面有什么问题。如果您能提供一些帮助，谢谢。

int main( int argc, char** argv )
{
int height,width,step,channels;

//Load the image and make sure that it loads correctly
IplImage* im = cvLoadImage("kermit.jpg", 1); 
if(!im )
{
    //Drop out if the image isn't found
    cout << "Failed to load: "<<"kermit.jpg"<<"\n";
return 0;
}
else
{
    cout<<"Image was loaded with success "<<"kermit.jpg"<<"\n";
return (0);
}

height=im->height;
width=im->width;
step=im->widthStep;
channels=im->nChannels;

cout<<"(height, width)"<<height<<width<<"\n";


cvNamedWindow("kermit.jpj",CV_WINDOW_AUTOSIZE);

cvShowImage( "kermit.jpg", im );
cvWaitKey(0);
cvDestroyWindow ("kermit.jpg");

return 0;
}

Answer 1

既然你没有告诉我们真正的问题是什么，我想指出以下代码:

if(!im )
{
    //Drop out if the image isn't found
    cout << "Failed to load: "<<"kermit.jpg"<<"\n";
    return 0;
}
else
{
    cout<<"Image was loaded with success "<<"kermit.jpg"<<"\n";
    return (0);
}

这意味着如果加载失败你退出应用程序，如果加载成功你也退出。听起来不对吧？你不应该在成功时返回。

您不需要else block ，所以请删除它并重试。其余代码似乎没问题。