数组在通过引用传递给函数之前和之后显示不同的地址

Question

I was curious to know why the addresses of the same array ie array b in my case is changing before and after it has been passed to the function manipulation(), as shown in the output picture. Please share your thoughts and help me figure this out. Thank you!

#include <stdio.h>
void manipulation(int *pa,int *pb){   
    int i;
    for(i=0;i<10;i++){
        *(pb+i)=*(pa+i);
        printf("%d\t  %04x\n",*pb+i,&pb+i);
    }
}

int main(){
    int a[10],b[10];
    int i;

    int *point;
    point = &b[0];
    printf("Enter the array elements\n");
    for(i=0;i<10;i++){
        scanf("%d\n",&a[i]);    
    }

    for(i=0;i<10;i++){
        printf("%04x\n",&point+i);
    }

    manipulation(&a[0],&b[0]);
    return 0;
}

Answer 1

because you print ( address of your pointer) + i:

printf("%04x\n",&point+i);
printf("%d\t  %04x\n",*pb+i,&pb+i);

you want to print the pointer + i:

printf("%04x\n",point+i);
printf("%d\t  %04x\n",*pb+i,pb+i);

Also you have a bug in there:

printf("%d\t  %04x\n",*pb+i,pb+i);

*pb+i is interpreted as (*pb) + i that is the first value of the array plus i, or simply pb[0]+i .

you seem to get the right answer because your array is 1 2 3 ...

you probably want this:

printf("%d\t  %04x\n",*(pb+i),pb+i);

or simply:

printf("%d\t  %04x\n",pb[i],pb+i);

Answer 2

Yeah, the real problems is comparing the address of point to the address of pb .

Try using point + i (or &point[i] ) and pb + i (or &pb[i] ). I'm sure you will get the expected answer.

---

In case you are curious, &point + i is the location in memory of the local variable point plus the value of i . &pb + i is the location in memory of the parameter pb plus the value of i . They are different variables, so they have different locations in memory.

Answer 3

point stores the address of array a. &point prints the address of point and adding 4 through out the loop. same is the case with &pb. 28fee8 is the place where point is stored and 28fed4 is the place where pb is stored.