Question

#include<stdio.h>
int main()
{
    char *s[] = { "knowledge","is","power"};
    char **p;
    p = s;
    printf("%s ", ++*p);
    printf("%s ", *p++);
    printf("%s ", ++*p);

    return 0;
}

Output:

nowledge nowledge s

Please explain the output specially output from the 2nd printf() statement.I think that because ++ and * have same precedence therefore in *p++ p should be incremented first and then use *(associativity from right to left for unary operators).

Answer 1

According to C++ Operator Precedence:

1. “*” has the same precedence as prefix “++” but must be avaluated rigth to left.

   printf(“%s “, ++*p);

So first *p is evaluated, then ++(*p), leading to the second character in the first string.

1. “*” has less precedence than suffix “++”.

   printf(“%s “, *p++);

So first p is incremented, but it is a post-increment. The value returned from the operation is the original one. This way, the * operates over the original pointer, that pointed to the second char on the first string.

Note that, this time, ++ is operating over p, and not over *p.

1. Since “2”, p points to the second string. When you do ++*p you are now pointing to the second character of the second string (“s”). As you are again using a pre-increment, the value passed to printf is already changed.

   printf(“%s “, ++*p);

I may get clearer if you do a little change and print the pointer value aswell (ignore the warnings):

printf("%s [%p]\n", ++*p, p );
printf("%s [%p]\n ", *p++, p );
printf("%s [%p]\n ", ++*p, p );

nowledge [0x7fff6f5519e0]
nowledge [0x7fff6f5519e8]
 s [0x7fff6f5519e8]