Question

Why are the two following code segments not equivalent?

void print (char* s) {
  if (*s == '\0')
     return;
    print(s+1);
   cout << *s;
}

void print (char* s) {
  if (*s == '\0')
     return;
    print(++s);
   cout << *s;
}

Answer 1

Since it looks like the OP changed print(s++) to print(++s), which is hugely different, here’s an explanation for this new version.

In the first example, you have:

print(s+1);
cout << *s;

s+1 does not modify s. So if s is 4, and you print(s+1), afterwards s will still be 4.

print(++s);
cout << *s;

In this case, ++s modifies the local value of s. It increments it by 1. So if it was 4 before print(++s), it will be 5 afterwards.

In both cases, a value equivalent to s+1 would be passed to the print function, causing it to print the next character.

So the difference between the 2 functions is that the first one will recursively print character #0, then 1, 2, 3, …, while the second function prints 1, 2, 3, 4, … (it skips the first character and prints the “\0” afterwards).

Example:
For the s+1 version, print("hello") will result in h e l l o
For the ++s version, print("hello") will result in e l l o \0