c ++：如何从函数返回一个shared_ptr [英] c++: how to return a shared_ptr from function

问题描述

当试图从函数返回一个shared_ptr时，我得到：
引用局部变量'recipe'返回[-Werror = return-local-addr]

<我哪里错了？

  shared_ptr< Recipe>& Group :: addRecipe（const string& groupName，unsigned int autherId，const string& recipeName）{
 
 shared_ptr< Recipe>食谱（新食谱（recipeName，autherId））; 
 
 recipes.push_back（recipe）; 
返回食谱; 
} 
  

返回shared_ptr的正确方法是什么？

解决方案

函数的签名没有显示，但它听起来像可能返回 shared_ptr< Recipe>& 。将引用返回给临时对象是一个很大的禁忌，因为一旦函数退出，被引用的对象就会被销毁。只需返回值。

when trying to return a shared_ptr from a function I get: reference to local variable 'recipe' returned [-Werror=return-local-addr]

where did I go wrong ?

shared_ptr<Recipe>& Group::addRecipe(const string& groupName, unsigned int autherId, const string& recipeName){

    shared_ptr<Recipe> recipe(new Recipe(recipeName, autherId));

    recipes.push_back(recipe);
    return recipe;
}

what is the right way to return a shared_ptr ?

解决方案

The function's signature isn't shown, but it sounds like it's probably returning shared_ptr<Recipe>&. Returning a reference to a temporary is a big no-no since the referenced object will be destroyed as soon as the function exits. Just return by value instead.

这篇关于c ++：如何从函数返回一个shared_ptr的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！