C指针算术和数组访问

Question

I have a function that takes a pointer to an array of structs

typedef struct {
    bool isUsed;
    int count;
} MyStructure;

void Process(MyStructure *timeStamps, int arrayLength){
    for (int i = 0; i < arrayLength; i++){
        MyStructure *myStructure = &(*(timeStamps + i));
        if (myStructure->isUsed == true){
          /*do something*/
        }
    }

}

The way that I am accessing the array seems to be a little off.

&(*(timeStamps + i))

Is there a better way to do this?

Answer 1

&(*(timeStamps + i))

is equivalent with

&timeStamps[i]

which is, in turn, simply

timeStamps + i

That's all :)

Answer 2

The argument timeStamps is of type MyStructure* , which means that this line:

MyStructure *myStructure = &(*(timeStamps + i));

is equivalent to:

MyStructure *myStructure = timeStamps + i;

which is also equivalent to:

MyStructure *myStructure = &timeStamps[i];

Note that in this expression: &(*(timeStamps + i)) , the timeStamps + i is a pointer to the element at index i (ie the address of this element), which is then dereferenced by using dereference operator ( * ) that returns an l-value of type MyStructure and then you retrieve the address of this element by using the address-of operator ( & ) which is equal to the address that timeStamps + i held at the beginning.

Answer 3

MyStructure *myStructure = &timeStamps[i];

Answer 4

Been a while since I worked in C, but as far as I remember the square brackets indexing operator ([]) is the same as adding to the pointer. So:

timeStamps[i] should be equivalent to *(timeStamps + i)

Therefore you should be able to use

myStructure = &timeStamps[i]

Answer 5

I would suggest:

void Process(MyStructure *timeStamps, int arrayLength){
    for (MyStructure *i = timeStamps; i < timeStamps + arrayLength; i++) {
        if (i->isUsed == true) {