[c++] Use the auto keyword in C++ STL

I have seen code which use vector,

vector<int>s;
s.push_back(11);
s.push_back(22);
s.push_back(33);
s.push_back(55);
for (vector<int>::iterator it = s.begin(); it!=s.end(); it++) {
    cout << *it << endl;
}

It is same as

for (auto it = s.begin(); it != s.end(); it++) {
    cout << *it << endl;
}

How safe is in this case the use of the auto keyword? And what about if type of vector is float? string?

This question is related to c++ stl c++11

The answer is


This is new item in the language which I think we are going to be struggling with for years to come. The 'auto' of the start presents not only readability problem , from now on when you encounter it you will have to spend considerable time trying to figure out wtf it is(just like the time that intern named all variables xyz :)), but you also will spend considerable time cleaning after easily excitable programmers , like the once who replied before me. Example from above , I can bet $1000 , will be written "for (auto it : s)", not "for (auto& it : s)", as a result invoking move semantics where you list expecting it, modifying your collection underneath .

Another example of the problem is your question itself. You clearly don't know much about stl iterators and you trying to overcome that gap through usage of the magic of 'auto', as a result you create the code that might be problematic later on


It's additional information, and isn't an answer.

In C++11 you can write:

for (auto& it : s) {
    cout << it << endl;
}

instead of

for (auto it = s.begin(); it != s.end(); it++) {
    cout << *it << endl;
}

It has the same meaning.

Update: See the @Alnitak's comment also.


If you want a code that is readable by all programmers (c++, java, and others) use the original old form instead of cryptographic new features

atp::ta::DataDrawArrayInfo* ddai;
for(size_t i = 0; i < m_dataDraw->m_dataDrawArrayInfoList.size(); i++) {
    ddai = m_dataDraw->m_dataDrawArrayInfoList[i];
    //...
}

The auto keyword gets the type from the expression on the right of =. Therefore it will work with any type, the only requirement is to initialize the auto variable when declaring it so that the compiler can deduce the type.

Examples:

auto a = 0.0f;  // a is float
auto b = std::vector<int>();  // b is std::vector<int>()

MyType foo()  { return MyType(); }

auto c = foo();  // c is MyType

auto keyword is intended to use in such situation, it is absolutely safe. But unfortunately it available only in C++0x so you will have portability issues with it.


Examples related to c++

Method Call Chaining; returning a pointer vs a reference? How can I tell if an algorithm is efficient? Difference between opening a file in binary vs text How can compare-and-swap be used for a wait-free mutual exclusion for any shared data structure? Install Qt on Ubuntu #include errors detected in vscode Cannot open include file: 'stdio.h' - Visual Studio Community 2017 - C++ Error How to fix the error "Windows SDK version 8.1" was not found? Visual Studio 2017 errors on standard headers How do I check if a Key is pressed on C++

Examples related to stl

Why is it OK to return a 'vector' from a function? How to remove all the occurrences of a char in c++ string How to use the priority queue STL for objects? use std::fill to populate vector with increasing numbers What does iterator->second mean? How to set initial size of std::vector? Sorting a vector in descending order How do I reverse a C++ vector? Recommended way to insert elements into map Replace an element into a specific position of a vector

Examples related to c++11

Remove from the beginning of std::vector Converting std::__cxx11::string to std::string What exactly is std::atomic? C++ How do I convert a std::chrono::time_point to long and back Passing capturing lambda as function pointer undefined reference to 'std::cout' Is it possible to use std::string in a constexpr? How does #include <bits/stdc++.h> work in C++? error::make_unique is not a member of ‘std’ no match for ‘operator<<’ in ‘std::operator