C-指向数组的指针：如何仅将一个指针的值（而非地址）复制到另一指针？

Question

If a pointer pr points to an array aa = {1, 2, 3, 4, 5, 0, 0, 0, 0, 0} , then we can access the array through the pointer pr[0] , pr[1] , pr[2] ,....
How can I shift the array (operated through the pointer) such that I can get {1, 1, 2, 3, 4, 5, 0, 0, 0, 0} ? obviously pr[i] = pr[i-1] won't work.

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Here is my code:

#include <stdio.h>

int main() {
    int aa[10] = {1, 2, 3, 4, 5, 0, 0, 0, 0, 0};
    int i, m, *pr = aa;
    printf("\n pr[0] = %d, pr[1] = %d, pr[2] = %d, pr[3] = %d", pr[0], pr[1], pr[2], pr[3]);

    for(i = 0; i < 9; i++) {
        m = *(pr + i);
        pr[i+1] = m;
    }

    printf("\n \n pr[0] = %d, pr[1] = %d, pr[2] = %d, pr[3] = %d \n", pr[0], pr[1], pr[2], pr[3]);
    printf("\n \n aa[0] = %d, aa[1] = %d, aa[2] = %d, aa[3] = %d \n", aa[0], aa[1], aa[2], aa[3]);
    return(1);
}

I am writing a C function for R using .Call , all the arrays in the C function have to be accessed through this type of the pointers. And I am very confused by the grammar of pointers in C.

Answer 1

You basically want to prepend a value, in your example 1 , and remove the last value from the array, in your example a 0 .

If you take a look at what happens you will see the following:

position:    0  1  2  3  4  5  6
starting: { a0 a1 a2 a3 a4 a5 a6 }
              \  \  \  \  \  \
final:    { b0 a0 a1 a2 a3 a4 a5 }

So you want to do as you proposed shifting each value, but you have to start at the end (or you will overwrite everything with the same value).

for(i = 9; i > 0; i--) {
    pr[i]=pr[i-1];
}
pr[0] = NEWVALUE;