The following C++ Code relate to pointer and reference

Question

I have seen this code

  void path(
    const char * p);
  » more...

void path(
    const std::string & p);
  » more...

Can I do the following modification, basically reversing the * and &.

void path(
    const char & p);
  » more...

void path(
    const std::string * p);
  » more...

the syntax is ok in both cases so yes you can but the meaning is completely different. check: http://stackoverflow.com/questions/57483/what-are-the-differences-between-pointer-variable-and-reference-variable-in-c — a.lasram, Aug 03 '13 at 19:34
Read a good C++ programming book. We don't have hours to explain the issues of C++ (which is a difficult programming language, so take many hours to learn it). — Basile Starynkevitch, Aug 03 '13 at 19:34
If you need to change the argument itself in the function, you need void path(char *& p) — huseyin tugrul buyukisik, Aug 03 '13 at 19:36
@BasileStarynkevitch Good comment but I'd say C++ is a rich language rather than difficult — a.lasram, Aug 03 '13 at 19:40

score 3 · Answer 1 · answered Aug 03 '13 at 19:40

The first is passing an array of characters as a pointer to the first one. This is how strings are conventionally handled in C, and you sometimes see it in C++ due to the common heritage of the two languages. While you could pass a reference and carefully document that it's actually supposed to be an array, it would be confusing to anyone used to that convention.

The second is passing a single object by reference. While you could emulate the reference with a pointer, this would be (mildly) confusing to anyone used to the reference semantics of C++.

score 1 · Answer 2 · answered Aug 03 '13 at 19:57

You should understand what the function is trying to do before changing.

Lets see what happens when you change void path(const char * p); to void path(const char &p);

Probably the reason a char * is passed could be that the function needs to access an array of characters. And if you change the parameter to char &p, then the function only accepts a char by reference.

score -1 · Answer 3 · answered Aug 03 '13 at 19:44

-1

yes you can. but you must be careful with the use of . and -> in these function

and keep in mind that the meaning compelitly changes

answered Aug 03 '13 at 19:44

Hossein Nasr

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The following C++ Code relate to pointer and reference

3 Answers3