`new' operator

[slippy]

#include "cstrf.hh"

int main() {

  const char* s = "The quick brown fox";
  const char* t = "jumped over the lazy dogs";
  char* u = new char; // This works, but I don't know why

  strcp(s, u);

  std::cout<< "Length: " << strl(s) << '\n';
  std::cout<< "Copy: (" << u << ")\n";
}

int strl(const char* s) {

  int l = 0; //do local variables declared in functions have default initializers?

  while(*s++ != 0) { l++; }
  return l;

}

void strcp(const char* from, char* to) {

  while(*to++ = *from++);

}

int strcmp(const char* s1, const char* s2) {

  // -1 if s1 < s2, 0 if s1 == s2, +1 if s1 > s2

  int c1;
  int c2;

  while(c1 += static_cast<int>(*s1++));
  while(c2 += static_cast<int>(*s2++));

  switch(c1 - c2) {
  case 0:
    return 0;
  default:
    if(c1 < c2) return -1;
    else return 1;
  }
}

header file :

#include <iostream>

int strl(const char*);
void strcp(const char*, char*);
int strcmp(const char*, const char*);

Does `new' instantiate an object and allocate memory for it? I was having trouble with things like

char* u = 0;

and

char u[] = "";

My guess is that the size of `u' is unknown in either of these cases and so the program will crash when it tries to do anything with `u' at all (it segfaulted on my end).

Thanks.

[Elysia]

For new: In short yes. Read http://www.cprogramming.com/tutorial/lesson6.html.
char* u = 0 <--- no memory allocated so you can't write or read from it.
charu [] = "" <-- Size == 1. If you try to read (or write) from/to element 1 or higher, you can get an error.

[CornedBee]

The tricky thing about pointers is that they let you do horrible things without ever complaining. It may seem to work. Then you add a line of code, and suddenly it crashes.

Thus, you can never learn pointers just by trying out stuff. You actually have to understand them.