Why this dies

[Tubs]

I have no idea what code that is.

[wtd]

It's Common Lisp. It's my way of just posting the answer.

[Tubs]

Weird.

This function is still eluding me. (Recursion is not my fort'e) Any tips at all? Currently if a power of 2 is inserted it returns n.

code:

#include <stdio.h> int A (int n); int main(int argc, char *argv[]) {   int n, x;   printf ("n = ");   scanf ("%d", &n);   printf ("\n");   x = A (n);   printf ("A (n) = %d\n", x);   system ("pause");   return 0; } int A (int n) {   int ans;   if (n == 1)     ans = 1;   else     ans = A(n / 2) + A(n / 2);   return ans; }

[wtd]

code:

int A (int n) {   int ans;   if (n == 1)     ans = 1;   else     ans = A(n / 2) + A(n / 2);   return ans; }

Your use of a variable here is not a good habit to get into.

code:

int A (int n) {   if (n == 1)     return 1;   else     return A(n / 2) + A(n / 2); }

You're returning 1 here if n equals 1. Is that what the problem says should happen?

[Tubs]

Yeah, I thought the A(1) = 3 was just an example to show that you were on the right track. Also, I followed the format provided in my text book on how to do a recursive fibonacci sequency and it used an 'ans' variable.

Thanks again

[Martin]

As I have recently discovered

code:

if (1 == n)

is far superior to

code:

if (n == 1)

Which of the following will crash?

code:

if (n == 1)

code:

if (n = 1)

code:

if (1 == n)

code:

if (1 = n)

[wtd]

Unfortunately this is sacrificing expressivity of the code to compensate for a [potential ambiguity in C++'s syntax.

It is vastly more natural to think "n == 1". Here "n" is emphasized because it is the subject. We're not interested in whether 1 equals "n". We're interested in whether "n" equals 1. Technically speaking, they're entirely equivalent. However, there is a linguistic factor in writing good code.

[Martin]

Well the easy solution is to start thinking in Japanese.

[Hikaru79]

Martin wrote:

Well the easy solution is to start thinking in Japanese.

That'd be closer to "n 1 =="

[Tubs]

This code is telling me that data.name and next are not part of a structure or union, which is almost definately because I have some incorrect pointer usage somewhere, but I get all confuzzeled when it comes to double pointers. Any tips? (I would post the program in its entirety but it is a heirchy of many functions)

code:

#include "coaster_node_t.h" #include <string.h> #include <stdlib.h> #include <stdio.h> coaster_node_t *remove_duplicate_coasters( coaster_node_t **list, char namedelete[] ) {   if (*list != NULL)   {     while (*list != NULL)     {       if (strcmp(*list->data.name, namedelete) != 0)       {         coaster_list_delete (&*list, namedelete);       }             *list->next;     }     return (*list);   }   else   {     printf ("List is empty.\n");     return (*list);   } }

[wtd]

And without seeing the definition of these data structures, I'm supposed to tell you what about this code?

[Tubs]

Just wondering if there was any obvious misuse of pointer syntax. Here is the whole file:

[Tubs]

Misuse of pointer syntax such as not putting brackets around *list
Fixed linker problem with having extraneous functions, now it dumps core after running, is'nt that great.

[wtd]

I've gotten spoiled. That level of pointer indirection is making my head hurt.

[wtd]

You have:

code:

*list->next;

In the function you pasted here.

This is in void context and does nothing. Why do you have it there?