Thread: swapping within a char pointer array

While reading about char pointer arrays, I was told that these offer, among other things, the advantage of swapping 2 positions without the need to store one of them in a temporary location just like we have to do with normal arrays.

It gives no example as to how exactly one does it and I couldn't code it. Please help me out with the following code:

Code:

char* [] pstr = { "string1", "string2", "string3", "string4" };

How do I go about swapping say, pstr[2] contents with pstr[1]?

Hmm... that's exactly how I swap array elements in normal arrays (except of course that here I have a pointer as a temp variable).

It's obvious I misinterpret the text... at some point it reads:

From Wrox Press C++ Tutorials - Ivor Horton
If we had stored these simply as strings, as we did in Ex3_04.cpp, a great deal of copying would be necessary - we would need to copy the whole string 'Robert Redford' to a temporary location while we copied 'Oliver Hardy' in its place, and then we would need to copy 'Robert Redford' to the end position. This would require significantly more computer time to execute.

The Ex3_04.cpp he refers to is an array of strings.

(gosh! damn, pointers! )

From Wrox Press C++ Tutorials - Ivor Horton
If we had stored these simply as strings, as we did in Ex3_04.cpp, a great deal of copying would be necessary - we would need to copy the whole string 'Robert Redford' to a temporary location while we copied 'Oliver Hardy' in its place, and then we would need to copy 'Robert Redford' to the end position. This would require significantly more computer time to execute.

Not quite. std::string defines the member function swap, which has constant-time complexity. So:

Code:

std::string array[] = { "string1", "string2", "string3", "string4" };
array[1].swap(array[2]); // constant-time
std::swap(array[1], array[2]); // also constant-time, if not - complain to library vendor