lol streams

[RetroX]

That's const char*

char* would mean something like this:

Code: [Select]

char x[]={'h','e','l','l','o',',',' ','\0'};
char y[]={'d','a','m','n','i','t','\0'};
char *z=x+y;or

Code: [Select]

char *z=const_cast<char*>("hello, ")+const_cast<char*>("damnit");

[Josh @ Dreamland]

The only difference between const char* and char* is that const char* points to read-only memory. And even that's only on systems that support such (ie, not Windows). '\0' == 0, and I'd be willing to bet that if you were to compare {'n','o',0} and "no" you'd get that they were equivalent. And I do mean as-is.

"Test" == "test" compares correctly for string literals. "Test" == "test" will return false, "test" == "test" will return true. This is because they point to the same location in memory when GCC is done with them.

Also, you don't need to use const_cast to get it represented as a const char.
Const char* can be set to a char* without cast. Vice-versa requires cast, but is dangerous on Linux and the like.

[RetroX]

The only difference between const char* and char* is that const char* points to read-only memory. And even that's only on systems that support such (ie, not Windows). '\0' == 0, and I'd be willing to bet that if you were to compare {'n','o',0} and "no" you'd get that they were equivalent. And I do mean as-is.

"Test" == "test" compares correctly for string literals. "Test" == "test" will return false, "test" == "test" will return true. This is because they point to the same location in memory when GCC is done with them.

Also, you don't need to use const_cast to get it represented as a const char.
Const char* can be set to a char* without cast. Vice-versa requires cast, but is dangerous on Linux and the like.

Whenever I have tried to do it without const_cast, G++ bugs me about a depreciated conversion (that works).  I always use -Wall and purge all warnings and errors, and that is one that I happen to get into the habit of doing.

[score_under]

That's const char*

char* would mean something like this:

Code: [Select]

char x[]={'h','e','l','l','o',',',' ','\0'};
char y[]={'d','a','m','n','i','t','\0'};
char *z=x+y;or

Code: [Select]

char *z=const_cast<char*>("hello, ")+const_cast<char*>("damnit");

Wait, what?

[RetroX]

That's const char*

char* would mean something like this:

Code: [Select]

char x[]={'h','e','l','l','o',',',' ','\0'};
char y[]={'d','a','m','n','i','t','\0'};
char *z=x+y;or

Code: [Select]

char *z=const_cast<char*>("hello, ")+const_cast<char*>("damnit");

Wait, what?

Exactly.  It's pointless.

[score_under]

That's const char*

char* would mean something like this:

Code: [Select]

char x[]={'h','e','l','l','o',',',' ','\0'};
char y[]={'d','a','m','n','i','t','\0'};
char *z=x+y;or

Code: [Select]

char *z=const_cast<char*>("hello, ")+const_cast<char*>("damnit");

Wait, what?

Exactly.  It's pointless.

Tell me...

Code: [Select]

char x[]={'h','e','l','l','o',',',' ','\0'};
char y[]={'d','a','m','n','i','t','\0'};
char *z=x+y;
printf("%c",z[2]);...What happens when you run that? If you say "prints l" I'll shoot your face off. If you say "access violation because you added 2 random pointers", then I'll forgive you... for now.

[Micah]

Of course it's adding two random pointers. He was explaining what I meant when I asked if you could overload operator+ for two char *s.

[score_under]

I see, though overloading the + for a char* is really pointless IMO.

[Josh @ Dreamland]

Yeah, compiler presently concatenates "" + "" automatically. For "" + anything else, I want it to cast first "" to variant.

[Rusky]

It concatenates "" "" automatically. someCharPtr + "" is what might be useful to overload, to avoid std::string constructor noise. However, I agree that it's not very useful. Allowing operator overloading on arbitrary types would make the language completely unpredictable; it'd be worse than monkeypatching in Ruby.