Another newbie question! "Segmentation Fault on char *"

[fantasticamir]

Guys,

DO you know why I got segmentation fault on this piece of code?

  char p[]={68,69,98,112,105};
  char *p1;
  for (int i=0; i<5; i++)       p1[i]=p[i];
  p1[5]='\0';
 
  cout << p1<<endl;

Thanks,

Amir.

[ilikepython]

Guys,

DO you know why I got segmentation fault on this piece of code?

char p[]={68,69,98,112,105};
char *p1;
for (int i=0; i<5; i++) p1[i]=p[i];
p1[5]='\0';

cout << p1<<endl;

Thanks,

Amir.

It is probably because "p1" does not have a size declared. Try declaring it like "char p1[5]" or "char p1[6]" if you want the nul zero character.

[JosAH]

It is probably because "p1" does not have a size declared. Try declaring it like "char p1[5]" or "char p1[6]" if you want the nul zero character.

Variable p1 cannot have a size declared: it's a pointer. The error is that p1 doesn't
point to anything so storing a char at where p1 points to is shooting yourself
in the foot or worse.

The solution is to make p1 point to a bunch of chars, enough to hold all the
characters that were stored in the p array.

kind regards,

Jos

[Ganon11]

Mainly, youir problem comesbecause you have declared a char pointer p1, but you did not give it anything to point to. You can either make it a native chara array like ilikepython suggested, or use

char *p1 = new char[5]; // or the appropriate malloc() call in C