I'm trying to understand the correct use of the -> operator in C. I think I have the understanding, but need to have that validated. so...

if i have an 32 bit data structure (eg. MyStruct) to store some status data (1s or 0s) which is pointed to by pnt. there are 3 members of that data structure ('first', 'second' both 8 bit, and 'third' 16-bit)

If I have

pnt -> second ^= (1<<3) 

this can be written as

pnt -> second = pnt -> second ^ (1<<2) 

then this is saying that; get the value of 'second' from the structure MyStruct that is being pointed to by pnt. XOR that value with a bit that has been shifted left (i.e. 00000100). Whatever the result, put that back into MyStruct at the member 'second'

yes?

Thanks.

CodePudding user response：

Yes, the compound assignment

A ^= B;

has the same effect as

A = A ^ B;

Note that pnt -> second ^= (1<<3) can not be written pnt -> second = pnt -> second ^ (1<<2) though. 1<<3 is not the same as 1<<2.

CodePudding user response：

pnt->second is a shorthand for (*pnt).second which means "take the struct pointed to by pnt and from that struct, get the element second".

CodePudding user response：

As for the assignment part of your question, the statements A=A XOR B is identical to A XOR= B, as with many other operators.

just make sure to change the (1<<2)(1<<3) difference between the lines.

As for the header of your question regarding the arrow(->) symbol: Given a struct A, you can reference a field (second) within the struct by two ways -

1. A.second
2. (&A)->second - that is, if you have a pointer to A, call it B (*B=A) - so referencing will be done thus: B->second.

hope that clarifies your question.