在函数内声明一个指针，c ++答案

【问题标题】：Declaring a pointer inside a function, C++在函数内声明一个指针，c ++
【发布时间】：2021-05-16 10:02:36
【问题描述】：

当我在函数中注册一个新指针时会发生什么？我承认，应该销毁指针的名称。但是分配的内存会发生什么？会不会是内存泄漏？在我的示例中，数据对象在函数终止后是否保持有效，或者它只是偶然工作，在更复杂的实例中容易出现未定义的行为？在这个例子中应该在哪里正确放置删除？还是应该避免这种写法？

int * one (void);

int main (void){
    int *two = new int;
    *two = *one ();
    cout << "Address: " << one() << "|" << *one() << endl; // Address: 0xe51a78|5555
    cout << "Address: " << two << "|" << *two << endl;  // Address: 0xe519d8|5555
}

int * one (void){
    int * pFun = new int; // declaration of a pointer inside a function
    *pFun = 5555;
    return pFun;
}

【问题讨论】：

这里的讨论可能是相关的cplusplus.com/forum/beginner/60724/#msg329931

标签： c++11 pointers

【解决方案1】：

您编写的代码会泄漏。

==365== Memcheck, a memory error detector
==365== Copyright (C) 2002-2017, and GNU GPL'd, by Julian Seward et al.
==365== Using Valgrind-3.17.0.GIT and LibVEX; rerun with -h for copyright info
==365== Command: ./test
==365== 
==365== 
==365== HEAP SUMMARY:
==365==     in use at exit: 8 bytes in 2 blocks
==365==   total heap usage: 3 allocs, 1 frees, 72,712 bytes allocated
==365== 
==365== 4 bytes in 1 blocks are definitely lost in loss record 1 of 2
==365==    at 0x484D103: operator new(unsigned long) (vg_replace_malloc.c:342)
==365==    by 0x10919E: main (in /petsc/test)
==365== 
==365== 4 bytes in 1 blocks are definitely lost in loss record 2 of 2
==365==    at 0x484D103: operator new(unsigned long) (vg_replace_malloc.c:342)
==365==    by 0x1091CC: one() (in /petsc/test)
==365==    by 0x1091A7: main (in /petsc/test)
==365== 
==365== LEAK SUMMARY:
==365==    definitely lost: 8 bytes in 2 blocks
==365==    indirectly lost: 0 bytes in 0 blocks
==365==      possibly lost: 0 bytes in 0 blocks
==365==    still reachable: 0 bytes in 0 blocks
==365==         suppressed: 0 bytes in 0 blocks
==365== 
==365== For lists of detected and suppressed errors, rerun with: -s
==365== ERROR SUMMARY: 2 errors from 2 contexts (suppressed: 0 from 0)

上面的注意事项我删除了两个打印调用，因为第一个调用 one() 两次以上，导致 2 个额外的泄漏

您永远不会直接存储one() 返回的指针，而是立即取消引用它。这具有丢弃该地址的效果。更好的方法是执行以下操作：

// two is an uninitialized pointer
int *two;
// Assign the returned pointer to two
two = one();
// Assign the value pointed to by two (and in effect one()) to three and use it
int three = *two;
functionThatExpectsAnInt(three);
// Alternatively use two directly, it points to valid memory whose value has not changed
functionThatExpectsAnInt(*two);
// Clean up resources from one()
delete two;

对于返回分配内存的函数，调用者是函数返回的资源的有效“所有者”。所以调用者必须要么释放资源，要么将它们传递给另一个函数。

【讨论】：

【解决方案2】：

动态分配的对象在被释放或“释放”之前一直存在。这正是动态分配的目的：创建其生命周期不依赖于范围的对象（即，不会在下一个右花括号结束，如自动变量）但也不是无限的，如全局或静态变量。相反，动态分配对象的生命周期完全由在运行时“动态”决定。

返回指向动态分配对象的指针是一种常见模式，例如factory functions。

Jacob 是正确的，您的程序没有释放分配的对象。在程序结束之前的某个时间释放所有资源（这包括打开的文件和套接字以及系统资源的其他句柄）是一种很好的风格。但这会有什么不同吗？不，因为all operating systems clean up the memory after program termination（同时链接的答案使得清理无论如何 :-) 的理由）。您的程序可以运行任意多次，而不会耗尽内存。

【讨论】：