I am currently experimenting with pointers, in my program I am trying to display the pointer variable's value in the standard output. Only a blank line appears, does someone know how to fix this and display 42?
Code:#include <stdio.h> void point_funct(int *********nbr) { *********nbr = 42; } int main() { int *********i; point_funct(i); printf("%d", *********i); return (0); }
I suggest you make it work with just one star first, before going all galactic.
FYI, anything more than three stars is rare in the real world.Code:#include <stdio.h> #include <stdlib.h> void foo ( int **p ) { **p = 42; // The answer } void bar ( int *p ) { *p = 88; // MPH } int main ( ) { int speed; bar(&speed); printf("How fast - %d\n", speed); int answer; int *p_answer = &answer; foo(&p_answer); printf("The question - %d\n", answer); return 0; }
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
You are attempting to dereference i 9 times, but there are no addresses set up to allow this.
How about:
orCode:#include <stdio.h> int main() { int a = 42; int *b = &a; int **c = &b; int ***d = &c; int ****e = &d; int *****f = &e; printf("%d\n", *****f); return 0; }
Code:#include <stdio.h> void f(int ******x) { printf("%d\n", ******x); } void e(int *****x) { f(&x); } void d(int ****x) { e(&x); } void c(int ***x) { d(&x); } void b(int **x) { c(&x); } void a(int *x) { b(&x); } int main() { int x = 42; a(&x); return 0; }
A little inaccuracy saves tons of explanation. - H.H. Munro
I found this helpful, thank you!
