Swapping two pointers to char in c++ by real call by reference - c++

Question

I am trying to swap two points on const char with real call-by-reference. But I have problems.
void swap(const char *&str1, const char *&str2) { //swap char pointers
const char *one = str1;
str1 = str2;
str2 = one;
}
int main(void){
const char *str1 = "Apple";
const char *str2 = "Potato";
swap(*str1, *str2);
return 0.0;
}
I keep on getting this error:
invalid conversion from 'char' to 'const char'

Answer 1

You shouldn't be dereferences the pointers when calling swap. You need to call:
swap(str1, str2);
Or, better still, use std::swap.
Also, if you're trying to run the code you've written they you'll need to either prototype swap or swap the functions around:
void swap(const char *&str1, const char *&str2)
{
const char *one = str1;
str1 = str2;
str2 = one;
}
int main(void)
{
const char *str1 = "Apple";
const char *str2 = "Potato";
swap(str1, str2);
return 0;
}
Also, main returns an int, not a float

Answer 2

You should use std::swap (located in either the <algorithm> or <utility> header) instead of rolling your own:
std::swap(str1, str2);
Also, you should consider using std::string instead of const char* in general:
std::string str1 = "Apple";
std::string str2 = "Potato";
of course the std::swap algorithm will still work just fine.
And finally, neither void in the argument list of main nor return 0.0 are necessary in C++.
Here's the code revisited with the advices above:
#include <algorithm>
#include <string>
#include <iostream>
int main() {
std::string str1 = "Apple";
std::string str2 = "Potato";
std::swap(str1, str2);
}
and here's the live example.
In response to Mr. Cthulhu down here, I'll try to answer the question more explicitly.
Your error is caused by the fact that by dereferencing the pointers of type const char* you are actually getting an expression of type const char& which is obviously incompatible with the type expressed in your swap function. Here's the correct call to the function:
swap(str1, str2);
But the again, why replicating the code of std::swap? (this is a rhetorical question, in case you were wondering wether to flag this as "not an answer")

Answer 3

You're defining swap() after you're calling it. In C++, functions should be defined before they are called. Move the definition of swap() to the top of the file, above main(). You will then get this error:
test.cpp:11: warning: converting to ‘int’ from ‘double’
Your main() function should return 0 (an int), not 0.0 which is a double.
Fixing this, you'll finally get this error:
test.cpp: In function ‘int main()’:
test.cpp:10: error: invalid initialization of reference of type ‘const char*&’ from expression of type ‘const char’
test.cpp:1: error: in passing argument 1 of ‘void swap(const char*&, const char*&)’
This is because you're dereferencing your arguments to swap(). Remove the * and the program now works fine.
However, you should simply remove your swap() function altogether, #include <utility>, and use std::swap().