这是结合std :: generate_n和std :: back_inserter的正确方法吗？ [英] Is this correct way to combine std::generate_n and std::back_inserter?

问题描述

在我的追求:)中，我想使用尽可能多的STL，我想知道是否可以将std :: generate和std :: back_inserter结合使用，以便我可以执行以下代码： / p>

In my quest :) to use as much of STL as I can I came to wonder is it possible to use std::generate and std::back_inserter combined so that I can do the same thing as the following code :

static const size_t nitems=1024*1024;
std::string mrbig;
for (size_t pos=0; pos<nitems; ++pos)
    mrbig.push_back('a'+ (rand()%26));

我尝试过

 std::generate_n(std::back_inserter(mrbig),nitems,[](){return 'a'+(rand()%26);});

似乎正常，但是我想确定我不会弄乱什么。

and it seems to work OK, but I would like to be sure Im not messing up something.

推荐答案

generate_n 要求其第一个参数满足 OutputIterator ，其中 back_insert_iterator 起作用（其 iterator_category 是 output_iterator_tag ）。

generate_n requires that its first argument satisfy OutputIterator, which back_insert_iterator does (its iterator_category is output_iterator_tag).

您的代码可能存在的问题：

Potential issues with your code:

std::generate_n(std::back_inserter(mrbig),nitems,[](){return 'a'+(rand()%26);});

• 调用 mrbig.reserve（nitems）会更有效率


• 您应使用 c ++ 03 std :: rand 或 c ++ 11 uniform_int_distribution<> 从< random> 而不是 c rand 。


• 括号之间的括号（）对于不接受任何参数的lambda，lambda捕获和lambda主体是不必要的，尽管有些人喜欢它们。


• 不能保证在实现中字符集字母 az 连续块例如，EBCDIC中的， i 和 j 不相邻。我相信唯一可移植的形式是在程序中存储字符串 abcd ... xyz 并对其进行索引：如何编写一个


• Calling mrbig.reserve(nitems) would be more efficient
• You should use c++03 std::rand or c++11 uniform_int_distribution<> from <random> instead of c rand.
• The parentheses () between the lambda capture and lambda body are unnecessary for a lambda taking no arguments, although some people prefer them
• It is not guaranteed that in implementation character set the letters a-z form a contiguous block; for example, in EBCDIC, i and j are not adjacent. I believe the only portable form is to store a string "abcd...xyz" in your program and index into it: How can I write a single for loop running from a to z and A to Z in C?

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