shared_ptr用于原始指针参数 [英] shared_ptr for a raw pointer argument

问题描述

当函数需要char *时，是否可以传入shared_ptr？

When the function requires a char*, can you pass in a shared_ptr?

我正在读取整个文本文件将字符存储到char []数组中。我使用的天真的方式是：

I'm reading in a whole text file (length = 100), and want to store the char's into a char[] array. The naive way I used was this:

ifstream dictFile(fileName);
size_t fileLength = 100;
char* readInBuffer(new char[fileLength]);
dictFile.read(readInBuffer, fileLength);
//processing readInBuffuer..............
delete[] readInBuffer;
dictFile.close();

当然，如果在delete []语句之前抛出异常，我想知道如果我可以使用
shared_ptr readInBuffer（new char [fileLength]）;
但函数原型

Of course there is memory leak if an exception is thrown before the delete[] statement. I'm wondering if I can use shared_ptr readInBuffer(new char[fileLength]); But the function prototype

read（char * s，streamsize n）

read ( char* s, streamsize n )

不会接受智能指针作为输入？任何技巧？

won't accept a smart pointer as input? Any tricks?

编辑：我想写如下：

shared_ptr<char[]> readInBuffer(new char[fileLength]);
dictFile.read(readInBuffer.get(), fileLength);

但不会编译。

推荐答案

BIG FAT WARNING ：创建指向数组的 shared_ptr 会引发未定义的行为，指针将 delete 指针，而不是 delete [] 。使用向量代替！

BIG FAT WARNING: creating a shared_ptr that points to an array provokes undefined behaviour, because the smart pointer will delete the pointer, not delete[] it. Use a vector instead!

在这里留下这里，因为它可能作为一个有用的警告。原始答案如下...

Leaving this here because it might serve as a useful warning. Original answer follows...

get（）函数返回底层原始指针。

The get() function returns the underlying raw pointer. You already wrote this in your code!

shared_ptr<char> readInBuffer(new char[fileLength]);
dictFile.read(readInBuffer.get(), fileLength);

使用& * readInBuffer

当然，你必须确定 dictFile.read（） code> delete 指针，或者恶魔可能飞出你的鼻子。

Of course, you have to be certain that dictFile.read() doesn't delete the pointer, or demons might fly out of your nose.

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