c program output

Question

1.  #include <stdio.h>

2.     main()

3.     {

4.         char *p = 0;

5.         *p = 'a';

6.         printf("value in pointer p is %c\n", *p);

7.     }

Answer 1

It will throw an error because you didn't point to a valid memory location(or a variable name) when you made the assignment char *p = 0. Hence, undefined behaviour.

Answer 2

*p=<anything other than a variable>

This line is executed in the following way.

<anything other than a variable or a pointer> is a literal and is stored in the read-only part of the memory and the pointer is pointing to that memory. We can write *p as p[0]. So, when p[0]='a' is being executed, it wants to modify a data inside the read-only part and this is illegal. Hence it will give runtime error.

Comments

@sampatghosh i am having doubt... you meant since already *p is assigned '0' and '0' being a literal is stored in the read only part of memory..so when *p is reassigned to 'a' it is trying to modify the previous value i.e '0' but it is illegal as it is stored in read only part of memory...?? pls pls pls pls pls pls clear my doubt..  

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@sampatghosh

error is line 4 or 5 or both??