Ok I have the following code that works. My question is this: does the () around the *b isolate the pointer part from the member part? Trying to figure out why its nessary. Thanks.
Code:#include <stdio.h> struct x { int y; }; void test2(struct x * b) { (*b).y = (*b).y + 1; } void test(struct x a) { printf("%d\n", a.y); } int main (void) { struct x z; z.y = 5; test(z); test2(&z); test(z); return 0; }
(*b) is an object; accessing a member of a struct or union object is done with the . operator. The parentheses are necessary because the . operator has higher precedence than unary *.
b is a pointer; accessing a member of a struct or union through a pointer is done with the -> operator. I believe using the dereferenced pointer is equivalent to this.Code:void test2(struct x * b) { b->y = b->y + 1; }
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.*
Ok so it seems you can do it the b->y or (*b).y way.
Hmmmm looks like its back to testing. Thanks for the answer Dave
Yeah, the only time I bother to dereference and use parenthesis around the pointer is when it's a pointer to a pointer to a struct.
like: (*some_struct)->member;
