C Programming - Pointers - Discussion

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

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];

I will explain u guys for linux platform for this the answer is option 'c'.
first see this prog...Actually it is like above prog but with more abstract
#include<string.h>

int main()
{
int i, n;
char *x="Alice";
n = strlen(x);
printf("%d\n",n);
*x = x[n];
for(i=0; i<=n; i++)
{
printf("%s ", x);
printf(" ");
x++;
}
printf("\n", x);
return 0;
}

for this o/p is5
Alice lice ice ce e
so first printf in for loop it prints Alice next space shows next iteration and x++ shows lice and x++ then lice and loop continues and atlast x is e.....
hope u understand ....
sorry for my english...

Actual code should be this, otherwise, an internal error will occur.

#include<stdio.h>
#include<string.h>
int main()
{
int i, n;
char *x = (char *)malloc(10*sizeof(char));
strcpy(x,"Alice");
n = strlen(x);//value of the n is 5
*x = x[n];//it copied the x[5] value('\0') to its base location so now string becomes " lice"
for(i=0; i<=n; i++)
{
printf("%s ", x);
x++;
}
printf("\n", x);
return 0;
}
now for i =0 printf will print from base location lets say 2000->" lice"
now for i=1 printf will print from 2001 upto null occur and so on->" lice ice ce e.

#include<stdio.h>
#include<string.h>

int main()
{
int i, n;
char *x="Alice";
n = strlen(x);
// *x = x[n];// this line introduce run time error, if you run on linux plateform.without this output will be{Alice lice ice ce e};
for(i=0; i<=n; i++)
{
printf("%s ", x);
x++;
}
printf("\n", x);
return 0;
}

o/p={Alice lice ice ce e};

@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

See dude *x=*x[] one and only same for char type declaration
here string length is five at the begining .
now intially *x=x[5]; so x represents the base address of array x[5] i.e x pointing to first element of array x[5] ie A , so its prints Alice ,next
x++ , so x is pointing to l , then it prints lice ,
at last x comes out of loop pointing to e , so it prints e.
thats it

@ALL.

The code should be:

#include<stdio.h>
#include<string.h>

int main()
{
int i, n;
char x[20] = "Alice";
n = strlen(x);
char *const p=x;
*p = p[n];
char *q = p;
for(i=0; i<=n; i++)
{
printf("%s ", q);
q++;
}
printf("\n", p);
return 0;
}

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'

So in 1st iteration i print \0lice
2nd : ice
3rd : ce
4th : e
5th : \0
i.e lice ice ce e

I run this program in Linux based system and output will be segmentation fault.

Cause x is pointing to string constant "Alice". So when we try to modify the string constant through x it is a illegal access and throw a runtime error.

*x = x[n]; This line is causing the error.