Segmentation Fault when attempting to print value in C

Question

The following C code returns a \"segmentation fault\" error. I do not understand why it does not return the value 20. What is my error?

#include 

        

Answer 1

You haven't allocated memory to n, so

*n = 20;

attempts to write unspecified memory.

Try

#include <stdlib.h>

int *n = malloc(sizeof *n);
/* use n */
free(n);

Answer 2

You haven't allocated space for your int, you've only declared a pointer to an int.

The pointer is uninitialized, and so writing to that unknown space in memory is undefined behavior and causes problems. This typically causes a segfault.

You can allocate a slot for an integer using malloc:

n = malloc(sizeof(int));

And use a corresponding call to free to free up the memory later:

free(n);

But allocating a single slot for an integer is pretty unusual, typically you would allocate the int on the stack:

int n;
n = 20;

Answer 3

You are trying to write 20 in garbage value. You must allocate space for it by using one of *alloc() functions or creating an int on stack and getting the andress of it(as Richard J. Ross III mentioned on comments).

dynamic allocation:

int n*; 
n = malloc(sizeof(int));  /* allocate space for an int */
if(n != NULL) {
 /* do something.. */ 
 free(n); /* free 'n' */
} else {
  /*No space available. */
}

or on the stack:

int int_on_stack;
int *n = &int_on_stack;
*n = 20;
printf("%i\n", *n); // 20