C Programming - Pointers - Discussion

@All.

int i, n;
char *x="Alice";
n = strlen(x); // n=5
*x = x[n]; //*x='\0'(the base adress' element will be replaced by null character)
for(i=0; i<=n; i++)
{
printf("%s ", x); //in 1st iteration,it will print '\0lice' then 'ice' and so on..
x++;
}
printf("\n", x);
return 0;

Output :lice ice ce e

Yes, it's a Segmentation fault (core dumped).

The 5th character of the string is '\0',therefore *x=*(x[0])==x[5]='\0'.Now the string x become x="lice".

For every x++ it will leave one character behind and print the rest for n=5 times. So, the output will be alice lice ice ce e.

In gcc platform shows an segmentation fault, why? I don't understand can you explain anyone of these.

We cannot change 'A' to '\0'. Since "Alice" is stored in read-only shared memory. So 'A' will not get replaced.

Only we can make the pointer to point something else,
x = x + n //can be done.
*x = x[n] //cannot be done. //gives error.

n=strlen(x);
gives n=5
*x=x[5]; can be written as *(x+0)=x[5]
and
x[5] contains '/0' so 'A' will replcd by '/0' since

Now our string will be like this:
x[0]='\0'
x[1]='l'
x[2]='i'
x[3]='c'
x[4]='e'
x[5]='\0'

Now look carefully :-

string become x[]={'\0','l','i','c','e','\0'};
for loop will be execute 6 times from i=0 to i=5

1st : at first position it find end char ('\0') so it will print nothing and x is incremented by 1 byte
2nd : x => l so it will print rest of string "lice" ,x++
3rd: x=>i so print "ice" and x++
4th: x=> c so print "ce" and x++
5th: x=> e so print "e" and x++
6th:x=> '\0' so nothing will be printed
at last printf("%s",x) will print nothing bz x =>'\0'
so result will be => lice ice ce e

Am getting option C while compiling. Please help me with right answer.

The question itself wrong or you need to add one more option for answer as "Undefined behaviour".

As I said above, In Linux, it throws "Segmentation fault" because of "*x = x[n];". It is correct since we are trying to change the string stored in read only memory. So what we have to do is create the memory which is not read only (In heap or in stack frame) as shown below

Solution: To avoid "Segmentation fault"

1.In heap:
char *x = (char *)malloc(10*sizeof(char));
strcpy(x,"Alice"); // copy in allocated mem
n = strlen(x);
*x = x[n];

2. In stack frame: Create char array & make the pointer to point the char array.
char *x;
char s[]="Alice";
x = s;
n = strlen(x);
*x = x[n];

In Linux, it throws "Segmentation fault" because of "*x = x[n];". It is correct since we are trying to change string constant.

Guys tell me in a simple way.