C++:将结构传递给函数

Question

这是我在做的事情:

在single.h文件中定义了一个struct:

typedef struct
{
    double x,y,z;
}PhysicalSize;

此结构用于single.cpp文件中

PhysicalSize physicalSize;
physicalSize.x = xstageStepSize;
physicalSize.y = ystageStepSize;
physicalSize.z = zstageStepSize;

long SaveTIFFWithOME(wchar_t *filePathAndName, char * pMemoryBuffer, long width, long height, unsigned short * rlut, unsigned short * glut,unsigned short * blut,
                 double umPerPixel,int nc, int nt, int nz, double timeIncrement, int c, int t, int z,string * acquiredDateTime,double deltaT, 
                 string * omeTiffData, AcquireSingle::PhysicalSize physicalSize)    

我介绍了这个功能,看起来很好,所有人physicalSize.x, physicalSize.y and physicalSize.z都有正确的价值.但是,当我迈步时SaveTIFFWithOME(),所有physicalSize.x, physicalSize.y and physicalSize.z看起来都很奇怪,好像它们从未被初始化.

我想知道我做错了什么,以及将结构作为参数传递给函数的正确方法是什么？谢谢.

Answer 1

上面的代码没有任何问题.

你传递结构就好了!

要么SaveTIFFWithOME被破坏(可能超出那些缓冲区参数？),要么你误解了你的结果.