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Use std::swap between vectors or vector::swap?


Use std::swap between vectors or vector::swap?

By : Lamb42
Date : November 15 2020, 06:54 AM
I wish this help you You should not use std::swap() directly in any case! Instead, you should use something like this:
code :
using std::swap;
swap(x, y);


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Does C++11 change the behavior of explicitly calling std::swap to ensure ADL-located swap's are found, like boost::swap?

Does C++11 change the behavior of explicitly calling std::swap to ensure ADL-located swap's are found, like boost::swap?


By : Andrea Bahe
Date : March 29 2020, 07:55 AM
I wish this helpful for you I would have had to vote against your proof-of-concept implementation had it been proposed. I fear it would break the following code, which I'm pretty sure I've seen in the wild at least once or twice over the past dozen years.
code :
namespace oops
{

    struct foo
    {
        foo() : i(0) {}
        int i;

        void swap(foo& x) {std::swap(*this, x);}
    };

    void swap(foo& lhs, foo& rhs)
    {
        lhs.swap(rhs);
    }

}
template <typename T>
void do_swap(T& lhs, T& rhs)
{
    std::iter_swap(&lhs, &rhs); // internally does what do_swap did above
}
Does vector.swap() and vector.erase() free all the allocated memory if the vector is a vector of vectors?

Does vector.swap() and vector.erase() free all the allocated memory if the vector is a vector of vectors?


By : Nenad Pavicevic
Date : March 29 2020, 07:55 AM
help you fix your problem 1) yes, the object (vector here) is destroyed. But it is worth noting that the outer vector (the one you called erase() on) will not change its capacity.
2) yes, it would be emptied. You can also call a.clear() but it won't change your outer vector's capacity.
I want to swap two numbers by using another class with a swap method and not the general swap function which is commonly

I want to swap two numbers by using another class with a swap method and not the general swap function which is commonly


By : Malluf
Date : March 29 2020, 07:55 AM
hope this fix your issue You are essentially swapping the two numbers twice, and therefore not swapping them at all:
code :
swapper s = new swapper(a[j],a[j+1]); // this assigns a[j] to s.x and a[j+1] to s.y
s.swap(); // this swaps s.x and s.y
a[j] = s.y; // this assigns the original value of s.x (a[j]) to a[j]
a[j+1] = s.x; // this assigns the original value of s.y (a[j+1]) to a[j+1]
swapper s = new swapper(a[j],a[j+1]); 
s.swap();
a[j] = s.x;
a[j+1] = s.y;
Is it safe to swap two different vectors in C++, using the std::vector::swap method?

Is it safe to swap two different vectors in C++, using the std::vector::swap method?


By : concernedthirdparty
Date : March 29 2020, 07:55 AM
will be helpful for those in need It is safe because nothing is created during the swap operation. Only data members of the class std::vector are swapped.
Consider the following demonstrative program that makes it clear how objects of the class std::vector are swapped.
code :
#include <iostream>
#include <utility>
#include <iterator>
#include <algorithm>
#include <numeric>

class A
{
public:
    explicit A( size_t n ) : ptr( new int[n]() ), n( n )
    {
        std::iota( ptr, ptr + n, 0 );   
    }

    ~A() 
    { 
        delete []ptr; 
    }

    void swap( A & a ) noexcept
    {
        std::swap( ptr, a.ptr );
        std::swap( n, a.n );
    }

    friend std::ostream & operator <<( std::ostream &os, const A &a )
    {
        std::copy( a.ptr, a.ptr + a.n, std::ostream_iterator<int>( os, " " ) );
        return os;
    }

private:    
    int *ptr;
    size_t n;
};

int main() 
{
    A a1( 10 );
    A a2( 5 );

    std::cout << a1 << '\n';
    std::cout << a2 << '\n';

    std::cout << '\n';

    a1.swap( a2 );

    std::cout << a1 << '\n';
    std::cout << a2 << '\n';

    std::cout << '\n';

    return 0;
}
0 1 2 3 4 5 6 7 8 9 
0 1 2 3 4 

0 1 2 3 4 
0 1 2 3 4 5 6 7 8 9 
std::vector<Widget> WidgetVector;

std::vector<Widget2> Widget2Vector;
How to swap two row vectors inside a row vector?

How to swap two row vectors inside a row vector?


By : pedro ribeiro
Date : March 29 2020, 07:55 AM
around this issue The proper and safe way to what you're trying to do is by assigning your variables to a cell, permuting the elements of the cell, and finally concatenating the result.
Imagine a specific permutation, say, [c, b, a, d]. This permutation can be coded as [3, 2, 1, 4] in terms of a mapping. The corresponding code to generate your array is then:
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