Hi gang,
My text is telling this can be done, but I do not see any mention of it in this particular chapter. Is there indeed a way to modify the actual address stored in a pointer? In this case, it is a pointer to an array. What is VERY strange is that if I pass a char array to a function, then inside that function,
pointerName++;
is valid, and seems to increment. However, this doesn't work in int main(). Any ideas why?
Nevermind, it didn't increment it, it just output differently while in the function (had a different address while in the function than in main()- WEIRD!!!). Any ideas on why
pointerName++
gives me an error in main(), but not inside a function? So, can an address in a pointer be incremented?
For help with real code, post real code.
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
I wouldn't say I need help with code, I just can't think of how to increment the address stored in a pointer variable. Isn't it fixed once it is set?
I would say it depends on the code.Originally Posted by patricio2626
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
> gives me an error in main(), but not inside a function? So, can an address in a pointer be incremented?
Now that sounds like you tried to increment an array name, not a pointer (to the array) inside another function.
Dave's right, post actual code and actual error messages.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Do you want to increment the memory address or the value that the pointer points to?
Kind of odd you want to change the memory address.
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[edit]
Last edited by patricio2626; 04-04-2007 at 06:55 PM.
I think what's not working for you is this:
The difference is that when passed to a function, the function recieves a a copy of the array pointer. This copy is mutable, but the orignal is not.Code:void func(char array[]){ ++array;//will compile fine } int main(){ char array[5]={0} array++;//Will not compile func(array);//will compile }
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
OK, that explains it, King Mir! I thought that the function received the actual array itself. Thanks for the help, that explains it!
-Patrick
