为什么这个 c 程序 output 会这样？

Question

#include<stdio.h>
main
{
    int x[]={1,2,3,4,5};
    int i,*j;
    j=x;
    for(i=0;i<=4;i++)
    {
        printf("%u",j);
        j++;
    }
}

output:

65512
65514
65516
65518
65520

But when I change the printf to"

printf("%u",&j[i]);

Output is:

65512
65516
65520
65524
65528

Why the address differ by 2 in first case and 4 in second casee?

What is wrong with just printing j and printing &j[i] ?

Answer 1

You get jumps of 4 in the second example because you are incrementing j and offsetting by i . Both of these contribute a difference of 2.

Note also that printf is not type-safe; it is up to you to ensure that the arguments match the format-specifiers. You have specified %u , but you've given it an int * , you should use %p for pointers.

Answer 2

First, just to make it clear, you are printing the pointer j , and not the pointed value, *j

Now, regarding the printed address. In your second example:

for(i=0;i<=4;i++)
{
  printf("%u",&j[i]); 
  j++;

&j[i] equals to (j+i) . i is incremented in each iteration, which contributes 2 to the pointer's value, and j is incremented too, which contributes another 2.