what does "->", "&=" and "|=" means?

[c beginner1986]

if (a) 
{
b->c |= d;
}
else
{
b->c &= ~d;
}		

Above code means, if a is true, things inside bracket will execute right?
Then what does "->", "&=" and "|=" means?

Thank you.

[maeriden]

-> is an operator used to access members of classes and structs. It is used when the variable is a pointer to a data structure. So in that example, b is a pointer to something and you are accessing member c of b
The arrow operator is a less verbose way of writing (*b).c which does the exact same thing

c &= ~d

is the same thing as writing

c = c & ~d

And

c |= d

is the same thing as writing

c = c | d

[c beginner1986]

Hi maeriden,

Thanks indeed for your explanation. It is indeed very helpful and clear. :)

as for "c &= ~d" my understanding is as below.

~d meaning invert bits of d
c &= ~d will do bit wise AND operation between c and inverted bits of d, then result will store in c.

Please correct me if i am wrong.

Thanks in advance. :)

[maeriden]

Glad to help

~d meaning invert bits of d
c &= ~d will do bit wise AND operation between c and inverted bits of d, then result will store in c.

That's correct. The same applies to operator |=

[Stewbond]

regarding ->, it means that b is a pointer to some structure or class that has c as a member. You're just using c from that class.

like so:

struct SomeType
{
  int a;
  int b;
  int c;
};

SomeType o; // Creates object o;
cout << o.c; // outputs the value of c

SomeType* p = &o; // Creates pointer p, it points to object o
cout << p->c; // outputs the value of c 

Edit: Fixed &o as maeriden noticed.

[c beginner1986]

Stewbond,

Thanks a lot. It is a very good example :)

[maeriden]

There's a typo, it should be
SomeType* p = &o