From what I've read, the following two definitions do the same thing:
But what about when you want to define a string of a specific length, like this:Code:char str[] = "Example string"; char* str = "Example string";
Is it possible do this with pointer syntax? Or would I have do to something like this:Code:char str[20];
Code:char string[20]; char* str = string;
Those are not the same:
Furthermore, in the second case it's a char* to an immutable string array. It should be declared as const char*, and the compiler lets you get away without the const only for obscure legacy reasons.Code:char str[] = "Example string"; //the type of str is char[15] char* str = "Example string"; //the type of str is char*
Yes, here str is a pointer to the first character in the array of 20 characters called string.Code:char string[20]; char* str = string;
Last edited by anon; 12-05-2011 at 04:05 PM.
I might be wrong.
Quoted more than 1000 times (I hope).Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
Right, I realize that one is an array and one is a pointer. What I meant is that they both define strings set to the same thing, even if one is the string itself and one is a pointer to it.Originally Posted by anon
If that's literally what you want, then yes. To declare a pointer to the beginning of some fixed-size buffer, you would do the above.
Code://try //{ if (a) do { f( b); } while(1); else do { f(!b); } while(1); //}
It is not possible to use some special pointer syntax to indicate size because a pointer is just that - it points to something. It has no notion of size.
The above is only way to do what you want if you need a fixed-size buffer and a pointer to it (but why would you want a pointer to it?).
In the example I was reading, it used a pointer to a string so it could increment and loop through the characters, like this:
but I guess you could just do this instead:Code:char string[20]; char* str = string; cin.get(str, 20); for (int i = 0; i < 20; i++) { cout<< *(str++) << endl; }
Code:char string[20]; cin.get(string, 20); for (int i = 0; i < 20; i++) { cout<< *(string+i) << endl; }
You could, but then you might as well write string[i] instead of *(string+i). Incidentally, be careful about using the name string in case you conflict with std::string.
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