When you declare an array
what type does myarray get? "Array of hundred ints"?Code:int myarray[100];
Is that why you get sizeof(myarray) == 400 while you would get sizeof(myarrayhead) = 4 if you declare
?Code:int *myarrayhead = malloc(100 * sizeof(int));
So if you send myarray as an argument, the whole array will be copied and sent to the function, while if you send myarrayhead as an argument to another function, just the pointer will be copied and sent?
Yup.Originally Posted by TriKri
The array is an array, the pointer is a pointer. Taking the sizeof an array gives you the size of the array; taking the sizeof a pointer gives you the size of the pointer.
No, the whole array does not get copied. A pointer to the array is passed.
http://c-faq.com/aryptr/index.html
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.*
No. Arrays are kindof passed by reference. Only a pointer to the first element is passed to functions.
That means inside the function there is no difference if you pass myarray or myarrayhead.
Kurt
EDIT: too late again.
