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ex12_17_18.cpp
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ex12_17_18.cpp
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/***************************************************************************
* @file The code is for the exercises in C++ Primmer 5th Edition
* @author Alan.W
* @date 24 DEC 2013
* @remark
***************************************************************************/
//
// Exercise 12.17:
// Which of the following unique_ptr declarations are illegal or likely to
// result in subsequent program error? Explain what the problem is with each
// one.
//
// Exercise 12.18:
// Why doesn’t shared_ptr have a release member?
// Because other shared_ptr that points the same object can still delete this
// object.Thus, it's meaningless to provide this member
// more detail can be found a thread on Stack Overflow:
// http://stackoverflow.com/questions/1525764/how-to-release-pointer-from-boostshared-ptr
#include <iostream>
#include <vector>
#include <string>
#include <memory>
int main()
{
int ix = 1024, *pi = &ix, *pi2 = new int(2048);
typedef std::unique_ptr<int> IntP;
/**
* @brief error: invalid conversion from 'int' to 'std::unique_ptr<int>::pointer { aka int* }' [-fpermissive]
*/
//IntP p0(ix);
/**
* @brief The code below can compile, but will cause error at run time.
* The reason is that when the unique_ptr p1 is out of scope, delete will be called
* to free th object. But the object is not allocate using new.Thus, an error
* would be thrown by operating system.
* @badcode
*/
//IntP p1(pi);
/**
* @brief This code can compile, but cause a dangling pointer at run time.
* The reason is that the unique_ptr will free the object the raw pointer
* is pointing to.
* @badcode
*/
//{ IntP p2(pi2); }
/**
* @brief When the unique_ptr goes out of scope, it will call delete to free an
* obeject not allocated using new.
* @badcode
*/
//IntP p3(&ix);
/**
* @brief Recommended.
*/
//IntP p4(new int(2048));
/**
* @brief error: double free or corruption at run time
* two unique_ptr are pointing to the same object. Thus, when both are out of
* scope, Operating system will throw double free or corruption.
* @badcode
*/
//IntP p2(new int(555));
//IntP p5(p2.get());
return 0;
}