How to use c++ swap function? - c++

Question

When I write like this swap(a,b); it is ok.
When I write like this swap(&c[0],&d[0]); there is a error.
Somebody can tell me why?
#include<iostream>
#include<algorithm>
using namespace std;
int main(void){
int *a;
int *b;
int c[]={1,2};
int d[]={3,4};
a=&c[0];
b=&d[0];
swap(a,b);// it is ok
//swap(&c[0],&d[0]);// it is error why
cout<<a[0]<<" "<<b[0]<<endl;
cin.get();
}

Answer 1

user3365922 is correct on the syntax if you wish to swap the contents of c[0] and d[0].
Just to add on because you code sample is a bit weird.
std::swap(a,b) in your code isn't actually swapping the contents c[0] and d[0]. It's swapping the pointers a & b. I mention this because it looks like you replaced the std::swap(&c[0], &d[0]) with std::swap(a,b)--which isn't actually equivalent (this is an assumption, my bad if they weren't meant to be equivalent).
I'm not totally sure what your goal is, but std::swap(c[0], d[0]) will achieve swapping the first entry of c with the first entry of d (leaving the second entry as-is). If you actually wanted to swap the contents of arrays entirely in the example above, you could also do std::swap(c, d).

Answer 2

Replace swap(&c[0],&d[0]) with swap(c[0], d[0]).
&c[0] and &d[0] are rvalues (temporary objects of type int*), but you can swap lvalues only.
You're able to swap(a, b) because a and b are lvalues, a after swap will be pointing to d[0] and b to c[0], swap(a, b) wouldn't swap values in array, this is a difference in behaviour (thanks to M.M's comment with a notice).

Answer 3

swap function
template <class T> void swap ( T& a, T& b )
{
T c(a); a=b; b=c;
}
And a=&c[0]; b=&d[0]; after swap(a,b);
a will pointing to d[0]
b will pointing to c[0]
and here swap(&c[0],&d[0]); the prototype dose not match the current swap function.The required prototype is
swap(int*,int*)
so there is no matching function to call so the compiler throws error.

Answer 4

The arguments to std::swap() are two integer pointers, not two addresses of an array with an index.
For swapping normal singular int values you will have to use a pointer (call by reference), but for an array you could directly pass it (as base address of array is taken) for which you could have used c[0],d[0]. Using & to reference the array (i.e. using &c[0],&d[0]) with the function accepting a int * type would cause argument type mismatch, with your passed values being rvalues of type int *, which cannot be swapped. (whereas lvalues like c,d and c[index],d[index] can be swapped)
The error you recieved should be:
main.cpp:12:13: error: no matching function for call to ‘swap(int*, int*)’
swap(&c[0],&d[0]);
^
This pertains to invalid initialization of a non-const reference of type int*& from an rvalue of type int*, which is included in the error message.
If what you want to consider as the parameters are indeed addresses of the arrays, this would be your function:
void swap_array_elements(int a, int b, int arr[], int arr2[])
{
int temp = arr[a];
arr[a] = arr2[b];
arr2[b] = temp;
}
Note that the first two arguments used here are indices of the two arrays passed respectively as the third and fourth argument. This will swap in between any two elements you require among the two arrays.
Example:
#include <iostream>
void swap_array_elements(int a, int b, int arr[], int arr2[])
{
int temp = arr[a];
arr[a] = arr2[b];
arr2[b] = temp;
}
int main()
{ int array1[]={42,56}, array2[]={63,89};
swap_array_elements(0,1,array1,array2); // swapping element with index 0 in array1 (42) with element with index 1 in array2 (89)
std::cout<<array1[0]<<" "<<array2[1]; // 89 42
return 0;
}