Question

The code is self-explanatory, but it gives me segmentation fault, why? :\

#include <stdio.h>
int main(void)
{
    char *c = "Hella";
    *(c+4) = 'o';
    printf("%s\n",c);
}

Answer 1

How to avoid it?

Don’t modify a string literal!

char *c = "Hella";

Declares a pointer c to a string literal “Hella” stored in implementation defined read only memory.
You are not allowed to modify this literal. An attempt to do so results in Undefined Behavior.

You are lucky that your program crashes, an Undefined Behavior does not always result in a crash but may cause your program to behave weirdly in any possible way.

What you need here is an array:

char c[] = "Hella";

Good Read:
What is the difference between char a[] = ?string?; and char *p = ?string?;?