C pointer arithmetic for 2D arrays

问题:

// strings is a 2D array (each string is 11 bytes long)
char strings[][11] = {"0123456789", "2222244444", "3333366666"};   

printf("String 3 Character 2 is %c\n", strings[2][1]);

How can I code this print statement using pointer arithmetic instead of the strings[2][1] ?

回答1:

In C, a[b] is the same as *(a+b) (and since addition is commutative, that implies that it's also equivalent to b[a]. People writing for the International Obfuscated C Code Contest frequently rely on this, using things like x["string"];. Needless to say, it's best to avoid that sort of thing unless you're intentionally being evil though...

Edit:For anybody who's sure their understanding of the subject is up to snuff should feel free to analyze the following and correctly predict its output before running it:

#include <stdio.h>

char *c[] = { "ENTER", "NEW", "POINT", "FIRST" };
char **cp[] = { c+3, c+2, c+1, c };
char ***cpp = cp;

main()
{
    printf("%s", **++cpp);
    printf("%s ", *--*++cpp+3);
    printf("%s", *cpp[-2]+3);
    printf("%s\n", cpp[-1][-1]+1);
    return 0;
}

If memory serves, the credit (blame?) for that particular code goes to Thad Smith.

回答2:

How did I do with this?

char strings[][11] = { "0123456789", "2222244444", "3333366666" }; 
printf("String 3 Character 2 is %c\n", *(*(strings + 2) + 1));

回答3:

 *(*(strings+2)+1)