RangePtr和Range *有什么区别? [英] What is the difference between RangePtr and Range *?

问题描述

我对excel自动化还比较陌生，而且我必须做自动化项目.
此代码表明RangePtr和Range *不同.

I am relatively new to excel automation, And I have to do automation projects.
This code shows that RangePtr and Range * differ.

Exell::_WorksheetPtr pSheet;
Excel::RangePtr pRangeA1 = pSheet->Cells->Item[1][1];
Excel::RangePtr pRangeB2 = pSheet->Cells->Item[2][2];
Excel::RangePtr pRange1 = pSheet->Range[(Range*)pRangeA1][(Range*)pRangeB2];//ok
Exell::RangePtr pRange2 = pSheet->Range[pRangeA1][pRangeA2];//causes error:
/*
error C2664: ''Excel::_Worksheet::GetRange'' : cannot convert parameter 1 from ''Excel::RangePtr'' to ''const _variant_t &''
*/

这意味着Range *会转换为_variant_t &，而RangePtr不会.
我没有找到RangePtr iGwtRangen的声明* .tlh，*.tli，VC ++包含目录.
有人可以帮我吗?
在此先感谢
abzadeh

It means that Range * converts to _variant_t & and RangePtr doesn''t.
I didn''t find Declaration for RangePtr iGwtRangen *.tlh, *.tli, VC++ include directories.
Can anybody help me?
Thanks in advance
abzadeh

推荐答案

您是否检查过RangePtr是否是指向Range实例的智能指针?从IUnknown继承的Range指针将适合_variant_t的构造函数，而智能指针在没有强制转换或取消引用的情况下将无法使用.

Did you examine if the RangePtr is a smart pointer to a Range instance? A Range pointer inheriting from IUnknown would fit a constructor of _variant_t whereas the smart pointer wouldn''t without a cast or dereference.

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