Strings

1.What is output of below code in C ? Explain in Detail ?

Code Snippet:-
#include<stdio.h>
#include<string.h>
int main {
   char *p = "abc";
    strcpy(p,"def");
    printf("p = %s\n", p);
    return 0;
}

Answer :- Output of this code likely to be undefined behavior and it may result a segmentation fault or a crash. 

Let's analyze the code step by step:

char *p = "abc";

Here, a character pointer p is declared and initialized to point to the string literal "abc." String literals in C are typically stored in read-only memory (usually in a data segment), and attempting to modify them directly is undefined behavior.

strcpy(p, "def");

The strcpy function is used to copy a string from the source to the destination. In this case, the destination is the pointer p, which points to the string literal "abc". 

The source is the string literal "def" Since "abc" is a string literal and read-only, trying to copy "def" into the memory location pointed to by p will result in undefined behavior.

In practical scenarios, this code may produce a segmentation fault or crash at runtime due to the attempt to modify read-only memory. However, it is essential to note that the specific behavior of such code is undefined and can vary depending on the compiler and the system on which it runs. 

2. How to avoid the above (problem#1) crash or seg fault ? or alternatives approach ? 

Answer :- To avoid undefined behavior in such cases, we should use an array to store the string or allocate memory dynamically using malloc for the character buffer that we intend to modify:

char p[] = "abc"; // Using an array
strcpy(p, "def"); // This is now valid since p is an array with writable memory. Or using dynamic memory allocation:

OR 

char *p = malloc(strlen("abc") + 1); // +1 for null terminator
strcpy(p, "abc");
// Some code...
strcpy(p, "def"); // Now, this will work since memory is allocated dynamically.
// Don't forget to free the dynamically allocated memory when it's no longer needed.
free(p);