问题描述
是否可以通过 lambda 表达式中的 const 引用来捕获?
Is it possible to capture by const reference in a lambda expression?
我希望下面标记的作业失败，例如:
I want the assignment marked below to fail, for example:
#include <algorithm>
#include <string>
using namespace std;
int main()
{
string strings[] =
{
"hello",
"world"
};
static const size_t num_strings = sizeof(strings)/sizeof(strings[0]);
string best_string = "foo";
for_each( &strings[0], &strings[num_strings], [&best_string](const string& s)
{
best_string = s; // this should fail
}
);
return 0;
}
更新: 由于这是一个老问题，如果 C++14 中有工具可以帮助解决这个问题，最好更新它.C++14 中的扩展是否允许我们通过常量引用捕获非常量对象?(2015 年 8 月)
Update: As this is an old question, it might be good to update it if there are facilities in C++14 to help with this. Do the extensions in C++14 allow us to capture a non-const object by const reference? (August 2015)
推荐答案
const 不在 n3092 中捕获的语法中:
const isn't in the grammar for captures as of n3092:
capture:
identifier
& identifier
this
文本仅提及按复制捕获和按引用捕获，并未提及任何类型的常量.
The text only mention capture-by-copy and capture-by-reference and doesn't mention any sort of const-ness.
对我来说感觉像是疏忽，但我并没有非常密切地遵循标准化流程.
Feels like an oversight to me, but I haven't followed the standardization process very closely.
这篇关于Lambda 捕获作为常量引用?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跟版网！