Range based for loop with pointer to vector in C++11

Question

Consider the following example:

vector<vector<char>*> *outer = new vector<vector<char>*>(); { vector<char> *inner = new vector<char>(); inner->push_back(0); inner->push_back(1); inner->push_back(2); outer->push_back(inner); inner->push_back(3); } auto x = outer->at(0); for (auto c : x) { cout << c << ","; }

I would like to iterate through the values of the vector<char>*; how can I accomplish that?

This is terrible, terrible code. At the very least, use unique_ptr rather than owning raw pointers. Better yet, get rid of the unnecessary news altogether. — T.C.
– T.C., Commented Jul 31, 2015 at 10:42
The last time I opened book on C++ which was roughly before C++98 such code would be normal... ))) Thank you, will go through unique_ptr... — Lu4
– Lu4, Commented Jul 31, 2015 at 11:06
Basically it's not even about the raw pointers. It's about using new in the first place. — Sambuca
– Sambuca, Commented Jul 31, 2015 at 12:52
It was used to illustrate the problem, however what's wrong with it (in general)? — Lu4
– Lu4, Commented Jul 31, 2015 at 13:21
Use vector<vector<char>> outer; instead of that vector<vector<char>*> *outer = new vector<vector<char>*>(); — Marian Spanik
– Marian Spanik, Commented Jul 31, 2015 at 15:42

Telokis · Accepted Answer · 2015-07-31 10:44:34Z

9

Why don't you simply dereference x?

for (auto c : *x) { // Here cout << c << ","; }

It will take *xby reference and, so, won't make a copy.

edited Jul 31, 2015 at 10:44

answered Jul 31, 2015 at 10:38

Telokis

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3 Comments

Lu4 Over a year ago

will it produce the copy of the original vector?

Telokis Over a year ago

No, it won't. It will take it by reference.

crafet Over a year ago

@Telokis " It will take it by reference." any reference for this point?thx

Niall · Accepted Answer · 2015-08-02 19:15:34Z

The type of x is vector<char>* so you need to iterate over what is pointed to by x; you need to dereference x.

for (auto&& c : *x) { // ^ dereference the container for a range // ^ note the use of &&

As a side note;

the characters for the integer values 0, 1 etc. are not printable, I'd test with 0x31 etc or just push back characters '0', '1' etc.
it is generally preferable to use auto&& in the range based for loops to avoid copies or subtle bugs when the container returns proxy object.

Vlad from Moscow · Accepted Answer · 2015-07-31 11:07:26Z

Here is a demonstrative program

#include <iostream> #include <vector> #include <cstring> int main() { const char *s = "Hello Lu4"; std::vector<std::vector<char> *> *v = new std::vector<std::vector<char> *>( 1, new std::vector<char>( s, s + std::strlen( s ) ) ); for ( char c : *( *v )[0] ) std::cout << c; std::cout << std::endl; for ( auto p : *v ) delete p; delete v; }

The program output is

Hello Lu4

Also you could use the following loop

for ( auto inner : *v ) { for ( char c : *inner ) std::cout << c; std::cout << std::endl; }

If to use this loop in the demonstrative program you will get the same result as above because the "outer" vector contains only one element with index 0.

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