Difference between char *str1 & const char *str2

Gudday all
On a previous thread http://cboard.cprogramming.com/c-pro...rs-string.html I was given some help to extract and correctly terminate strings.
My next step is to concatenate two strings, comprising a path and a file name, and then move the file to the new location using rename() and so move the file to its new location.
I thought that I would make a new variable to hold the concatenated strings (path and file) and then put that new variable into rename().

Code:

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#include <cstring.h>
const size_t size = 64;
...
char move_tiff_file[size+1] = {0};  //final destination of TIFF file  /* this var. added 05/08 */
...
/* these next 2 lines added 05/08 */
move_tiff_file = strncat(dest, tiff_file, size);
rename(tiff_file, move_tiff_file);

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When I build the code I get the error message in the strncat line " error: incompatible types in assignment of `char*' to `char[65]' ".
Looking at a reference site for C and looking up strncat I see the following

Quote:

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#include <cstring>
char *strncat( char *str1, const char *str2, size_t count );

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dest and tiff_file are both char*

Q: What is the difference between char *str1 & const char *str2?
Q: What is wrong with either my definition of move_tiff_file or my handling of strncat?

Const has nothing to do with your issue. You are trying to assign a pointer to array address. You can't do this, but you can do vice versa:

Code:

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char ff[10];
char *pp

pp = ff; //ff is similar to &ff[0]

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Next,

Quote:

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strncat(dest, tiff_file, size);

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already means that concntenated string ends up in dest, and dest value is actually what gets returned. Either use dest, or strcpy(move_tiff_file, dest); to achieve your goal.

An array is NOT a pointer. Remember those words.
Furthermore, you CANNOT assign anything to an array. Remember that too.
So the problem is that you're trying to assign a pointer to an array, which is a no-no.
Another interesting thing is that if you put the array name on the right-hand-side of a assignment, you will get a pointer.

And lastly, the const part: const simply means it cannot be changed.
So char* is a pointer to char, which you can change.
Const char* is a pointer to const char - in other words, one or more chars which you cannot change.

The difference is that the array actually has no pointer pointing to it. The compiler saves the location of the beginning of the array and can convert it to a pointer, but it can't modify it.

"arr" is exactly equivalent to "&arr[0]" and results in a rvalue pointer. I think it's like taking the address of any other local variable. You can't do "&i = malloc(sizeof(int))", can you?

&arr[0]=&(*arr)=arr
&* cancel each other