based on this code, will the null termination byte be assigned to the destination? If yes why?Code:for (i = 0; source[i] != '\0'; i++) destination[i] = source[i];
Secondly, can i just simply use this
instead of the above code?Code:destination = source
and what about?Code:*destination++ = *source++
and what about this?Code:*destination = *source
all work expected the following:
*destination++ = *source++
Not all of my questions are answered.. Please
>> based on this code, will the null termination byte be assigned to the destination? If yes why?
No.
Some of them might work depending on the rest of the code. As it's written, none of them work.
Sounds like it might be homework.
Why not try them and find out?
The first one is the correct way to copy a string (although you'll have to manually add the '\0' to the end, I'll let you figure out why).
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For heaven's sake, this is not my homework of any kind. I am having doubt on it.Sounds like it might be homework.
Why not try them and find out?
The first one is the correct way to copy a string (although you'll have to manually add the '\0' to the end, I'll let you figure out why).
And please.. Please quote because i do not know which first one you are talking about.
That's the first one where you have to manually copy the '\0' into destination[i] after the loop.Originally Posted by valthyx
Yes, that works as well, but only if destination is declared as "char *". If you declare destination as an array of char, you can't assign a new starting address to this array.
No, that doesn't work, because the postfix-expression isn't a lvalue expression, so that you can't assign a value to it. For reference, please read the ANSI C standard ;-)
yes, that should work as well, but you need then to manually increment the pointers, e.g. destination++;
- Andi -
With the var names 'source' and 'destination' I'm assuming you want a copy of the string, so no you can't use the second example. The first code example copies the string, the second just copies the pointer to the string.
What? No, this should work.
It's a common way of copying strings.
This is a typical implementation of strcpy:
Or even:Code:while (*src) *src++ = *dst++;
(Incomplete, however.)Code:while (*src++ = *dst++) ;
Last edited by Elysia; 01-26-2010 at 11:15 AM.
To explain Elysia's example: it is certainly true that the result of destination++ is not an lvalue. However, the result of *destination is a modifiable lvalue. Since the result of *destination++ is the result of *destination, the result of *destination++ is a modifiable lvalue, hence contrary to your assertion, the statement is valid C.
Unfortunately, Elysia mixed up the src and dst, and I would argue that the fun canonical implementation of strcpy is:
(Of course, it is also incomplete since it is missing stuff like the return statement, but it is more complete than Elysia's example: even if src and dest were correctly swapped, one would need to append a null character after the loop.)Code:while (*destination++ = *source++);
EDIT:
Oh wait, Elysia did list this as the second example.
Last edited by laserlight; 01-26-2010 at 11:24 AM.
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Didn't say it was right
Yes, you're indeed right. I originally thought this should work (especially when working with pointers), but I got a compiler error that an lvalue is required. But I must admit, that I was to fast evaluating the error message. The error message was just saying, that an lvalue is required for the increment operand which was a char array instead of a char pointer.
Though, sorry for the confusion :-)
- Andi -
