在这种情况下如何正确释放双指针的内存?

我的程序是通过创建新的arr来增加结构数组的大小,然后将旧数组的双指针赋给新数组。在将旧数组分配给新数组之前或之后,我一直在努力释放它的内存。如果我不收费(),它会给出正确的结果,否则,它会自动拾取并打印一些垃圾值。请帮忙,谢谢。

*不允许重新分配。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct car {
    char name[30];
    int quantity;
};

void increaseSize(int *cap, int size, struct car **carArray) {
    printf("cap = %d size = %d\n", *cap, size);
    *cap *= 2;
    printf("cap x 2 = %d\n", *cap);
    struct car *newCarArray = (struct car * ) malloc(*cap * sizeof(struct car));
    memcpy(newCarArray, carArray, sizeof(struct car));
    free(*carArray); // If I free old carArray here, it will give incorrect results. How can I release/clear/delete/free the memory of the old array here?
    carArray = &newCarArray;
    printf("increased cap = %d size = %d\n", *cap, size);
}

void printArray(int cap, int size, struct car *carArray) {
    printf("Capacity = %d, Size = %d \n", cap, size);
    for(int i = 0; i < size; i++) {
        printf("Car %d, Name: %s, Quantity: %d\n", i, carArray[i].name, carArray[i].quantity);
    }
}

int main() {

    int cap = 2;
    int size = 0;
    struct car *carArray = (struct car *) malloc(cap * sizeof(struct car));

    struct car car_0 = {"Corolla", 5};
    carArray[0] = car_0;
    size++;

    struct car car_1 = {"Accord", 8};
    carArray[1] = car_1;
    size++;

    printArray(cap, size, carArray);

    increaseSize(&cap, size, &carArray);

    struct car car_2 = {"Forte", 3};
    carArray[2] = car_2;
    size++;

    printArray(cap, size, carArray);
    free(carArray);

    return 0;
}