Confusion about Pointers C++

Question

Task: I want to swap pointers without reference passing the pointers in the function.

Problem 1: The following code does the job but I think its swapping the "data" instead of swapping the "pointers".

void swappointers(char *ptrX, char *ptrY) { char dummy; dummy = *ptrX; *ptrX = *ptrY; //LINE *ptrY = dummy; }

Problem 2: How the commented line is working? *ptrX means that ptrX is being de-referenced(accessing the value the pointer is directing to). Similar is the case with *ptrY. But why the value being directed by ptrX is being changed here? I mean its actually looking like this:

ValueofX = ValueofY

Thank you :)

Edit:

Solved the first problem.

void swappointers(char **ptrX, char **ptrY) { char *dummy = nullptr; dummy = *ptrX; *ptrX = *ptrY; *ptrY = dummy; }

For problem #1 try std::swap then you don't have to worry about all of this. Pointer to pointer instead of reference to pointer just makes using swappointers clunky. — Captain Obvlious
– Captain Obvlious, Commented Aug 28, 2014 at 2:31

Ben Voigt · Accepted Answer · 2014-08-26 15:39:55Z

7

I think its swapping the "data" instead of swapping the "pointers".

You are correct.

I want to swap pointers without reference passing the pointers in the function.

That's impossible. Unless you pass by reference, changes made inside the function will not be seen by the caller.

answered Aug 26, 2014 at 15:39

Ben Voigt

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3 Comments

Asad Over a year ago

I don't think Problem 1 is impossible. How about by passing **ptrX and **ptrY in the function? I got that somewhat hint. Thank you :)

Ben Voigt Over a year ago

@Asad: That generally falls under the category of "passing by reference" (In C, it's the only way to pass by reference, in C++ you have the choice of using a pointer or a reference)

Asad Over a year ago

Thank you for telling that, Ben.

GingerPlusPlus · Accepted Answer · 2014-08-26 21:42:45Z

2

template<typename T> void swap(T &a, T &b){ T backup=a; a=b; b=backup; }

*ptrX = *ptrY; means variable, to which points ptrX, have to have same value as variable to which points ptrY.

edited Aug 26, 2014 at 21:42

answered Aug 26, 2014 at 15:44

GingerPlusPlus

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1 Comment

Asad Over a year ago

You cleared the 2nd problem. Thanks. I don't know about that "auto". Moreover, I can not use that.

Jacques de Hooge · Accepted Answer · 2014-08-26 15:57:32Z

About your commented line:

*ptrX = *ptrY; //LINE

In plain English it means:

Put the value of the memory location pointed to by ptrY into the memory location pointed to by ptrX.

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