1: The size is printed as 10, which is the specified size of the array.Code:1: char parray[10]; strcpy(parray, "abcdefghij"); printf("%s, %lu\n", parray, sizeof(parray)/sizeof(char)); 2: char *marray = malloc(10*sizeof(char)); strcpy (marray, "abcdefghij"); printf("%s, %lu\n", marray, sizeof(marray)/sizeof(char));
2: The size is printed as 8, no matter what number I initialize it to.
Why?
You cant size dynamic memory with 'sizeof'.
http://c-faq.com/malloc/querysize.html
http://c-faq.com/malloc/sizeof.html
1: The size is printed as 10, which is the specified size of the array.
When declare an array, certain bytes are allocated for the array by the compiler and the while block is refered by base array name. When you try to find or sizeof the array, your're bound to get the size of the array. As this is predefined and not at the runtime allocation.
2: The size is printed as 8, no matter what number I initialize it to.
Where as in pointers, char *marray is pointer, a pointer something which can point to some part in the memory location. So when you find the sizeof a pointer it always give you the size of pointer variable rather than where its pointing. After all the a pointer variable hold just an address.
Hence array's and pointer are not equivalent.
ssharish
Life is like riding a bicycle. To keep your balance you must keep moving - Einstein
So how can I get the size of an array created by using malloc()?
You know the size because you passed it to malloc to begin with!
You know how much memory you allocated when you call malloc. Sounds like you may need to keep track of that value yourself through the lifetime of the program.Originally Posted by cnewbie1
"Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
-Christopher Hitchens
Logical, of course.
