Question

I found a function to resize an array and am having trouble understanding how it works (or if it’s working properly). For testing I set the “temp” array to a new value and “startCounter” does get assigned to that value, however the memory location for startCounter doesn’t change. Here is my code:

int * startCounter;


void resizeArray(int *&arraySent,int origSize,int newSize) {
    output << "&arraySent " << &arraySent << endl;
    output << "arraySent[0] " << arraySent[0] << endl;
    int* temp = new int[newSize];
    output << "&temp " << &temp << endl;
    for (int i=0; i<origSize; i++) {
        temp[i] = arraySent[i];
    }
    temp[0]=744;
    delete [] arraySent;
    arraySent = temp;
    output << "&arraySent " << &arraySent << endl;
}

//....

startCounter = new int [3];
startCounter[0]=345;
output << &startCounter << endl;
resizeArray(startCounter,3,10);
output << "startCounter[0]" << startCounter[0] << endl;
output << "&startCounter" << &startCounter << endl;

Here is the output I get from this:

    &startCounter 0x60fab8
    &arraySent 0x60fab8
    arraySent[0] 345
    &temp 0x82cfe54
    &arraySent 0x60fab8
    startCounter[0] 744
    &startCounter 0x60fab8

My question is, why does the memory location of startCounter not change from 0x60fab8 after deleting it and assigning it to the new “temp” array? Shouldn’t it now become 0x82cfe54?

P.S. I understand about vectors and such but am mainly concerned with understanding how this particular function works.

Answer 1

void resizeArray(int *&arraySent,int origSize,int newSize) {
    output << "&arraySent " << &arraySent << endl;

outputs the address of the pointer variable, not the address that it holds.

simply omit the address operator to get the (probably) intended effect