指针帮助：将指向结构的指针取消引用以访问其中的数据

Question

I'm having some trouble wrapping my head around how a certain line of code works. For some reason it's just not clicking. This line of code is used generally with abstracting file handles in unix domain sockets.

Context:

typedef struct myStruct {
    char charArray[10];
} myStruct; 

myStruct myStructure;
myStruct *ptrToStruct = &myStructure;

/* This should change myStructure.charAarray[0] to equal a */
*(ptrToStruct.charArray) = 'a'; 

I understand that an array is essentially a pointer that is pointing to the first index in the array but the pointer has no data ( charArray ).

The reason this is so hard for me to understand is because the ptrToStruct is trying to access the pointer's data member charArray but the pointer has no data member charArray and then it's dereferencing it.

Is this sort of like (*ptrToStruct).(*charArray) = 'a' ? But the dereferencing operator is being factored out? I apologize for being at all unclear.

---

UPDATE : The question has been answered. I was misreading code, the code was actually *(myStructure.charArray) and that's how it was altering the first index of the array. I should have also figured this out because as Sid explained pointers do not have the . operator.

Answer 1

ptrToStruct isn't a struct, so

ptrToStruct.charArray

should be

(*ptrToStruct).charArray

or

ptrToStruct->charArray

Then, yes, you can set the character using

*( ptrToStruct->charArray ) = 'a';

or

( ptrToStruct->charArray )[0] = 'a';

This is no different than

char charArray[10];
*charArray = 'a';

and

char charArray[10];
charArray[0] = 'a';