Markov chain. Implementation add method

Question

I am a bit confused with the next statment in implementation:

void add(char *prefix[NPREF], char *suffix) {
    State *sp;

    sp = lookup(prefix, 1);
    //addsuffix(sp, suffix);

    printf("size of prefix %d",&prefix);
    printf("size of prefix %s", prefix + 1);
    printf("size of prefix %d \n", &prefix+1);
    for (int i = 0; i < NPREF; i++)
        printf("%s \n" , prefix[i]);


    printf("memmove \n");

    memmove(prefix, prefix + 1, (NPREF - 1) * sizeof(prefix[0]));
    prefix[NPREF - 1] = suffix;

    for (int i = 0; i < NPREF; i++)
        printf("%s \n", prefix[i]);

}

mmemove(prefix, prefix + 1, (NPREF - 1) * sizeof(prefix[0]));

So prefix is a pointer of type char. In this terms prefix + 1 is the link to the next char in the array, isn't it?

How is it works in the right way? I've read about memmove and read about pointers but couldn't explore this add function behaviore for myself.

I am just trying get answer. Please dont take care about my printf functions. Question is not about it. — Роман Иванов, Jul 11 '16 at 09:09

4pie0 · Accepted Answer · 2016-07-15T00:19:58.697

So prefix is a pointer of type char. In this terms prefix + 1 is the link to the next char in the array, isn't it?

No and yes. prefix is not a pointer to char, it is an array of NPREF pointers to char. The size of this array is

assert(sizeof(prefix) == NPREF * sizeof(char*)

and yes prefix + 1 is address of next element after &prefix[0] because prefix (which is array) decays to pointer to array's first element. So

memmove(prefix, prefix + 1, (NPREF - 1) * sizeof(prefix[0]));

will mommove NPREF - 1 elements by one location. Sounds reasonable taking into account what Markov chain is, though we can't see your implementation. If it was instead written as

memmove(&prefix, &prefix + 1, sizeof(prefix)/sizeof(prefix[0]);

then it would memmove whole array.

Example:

[0][1][2][3][4][5]...[n] /* prefix was */
[A][B][C][D][E][F]...[N]
memmove(prefix, prefix + 1, (NPREF - 1) * sizeof(prefix[0]));
[0][1][2][3][4][5]...[n] /* prefix is */
[A][A][B][C][D][E]...[M]

score 0 · Answer 2 · answered Jul 11 '16 at 09:12

memmove will move n-1 pointers one step backwards.

Consider it for a second as memcpy that operates on 2 different arrays:

prefix
 V    
[ ][ ][ ][ ][ ][ ]
 ^  ^  ^  ^  ^
 \  \  \  \  \
  \  \  \  \  \
   \  \  \  \  \
    \  \  \  \  \
[ ][ ][ ][ ][ ][ ]
    ^
    prefix+1

Markov chain. Implementation add method

2 Answers2