C Program

Question

#include‬<stdio.h>
int main()
{
char *s[] = { "knowledge","is","power"};
char **p;
p = s;
printf("%s ", ++*p);
printf("%s ", *p++);
printf("%s ", ++*p);

return 0;
}

Comments

nowledge
nowledge
s

Answer 1

s is array of three character pointers and p is capabale of holding a address of a pointer variable which in turn points to a character.

up to p = s statement situation is like below : 

Here, few assumptions : size of char variable = 1 Byte, size of pointer varibale = 4 Byte.

[addresses are assumed]

Addresses of first,second and third element of s are 4000,4004,4008. Address of p is 5000.

First string : starting address = 1000

Second string : starting address = 2000

Third string : starting address = 3000

[ These starting locations are linker/compiler dependent, not a C language standard http://stackoverflow.com/questions/38997119/storing-of-string-literals-in-consecutive-memory-locations?noredirect=1#comment65347237_38997119]

[I have just assumed for easy explanation purpose]

and values of s[0] = 1000,s[1] = 2000,s[2] = 3000;

---

printf("%s ",++*p);

*p gives the 1st element of s, and value of 1st element of s, ( s[0] ) is = 1000. [ *p is an lvalue ,so, no issue with ++ operator ]

We pre-increment this first element and s[0] becomes 1001. 1001 points to 'n' of the first string. By using %s specifier printf() prints the entire string starting from n.

=> O/P = nowledge

---

printf("%s ",*p++);

Here we have a little bit of precedence comparison, (post increment ) ++ wins. So, exp. evaluation : *(p++)

Since it is a post-increment dereference will take place on the old_value of p which is 4000. after dereferencing resulting address points to 1001 again.

=> O/P = nowledge.

value of p is incremented to 4004.

---

printf("%s ",++*p);

*p gives the 2nd element of s, and value of 2nd element of s, ( s[1] ) is = 2000. [ *p is an lvalue ,so, no issue with ++ operator ]

We pre-increment this second element and s[1] becomes 2001. 2001 points to 's' of the second string. By using %s specifier printf() prints the entire string starting from s.

=> O/P = s

Comments

#include <stdio.h>
#include <string.h>
int main() {

	char *s[] = {"knowledge","is","power"};
	char **p;
	p = s;

	int i=0;

	printf(" String are: \n\n");
	printf("  %s \n", *(s));
	printf("  %s \n", *(s+1));
	printf("  %s \n", *(s+2));
	printf("  \n\n");

	printf("  Starting locations of the string are: \n\n");

	printf("  %d\n",*(s));
	printf("  %d\n",*(s+1));
	printf("  %d\n",*(s+2));
	printf("  \n\n");
 
return 0;
}

O/P:

d@D:~/Desktop/Gate_C_programs$ ./a.out 
 String are: 

  knowledge 
  is 
  power 
  

  Starting locations of the string are: 

  134514112
  134514122
  134514125
  

d@D:~/Desktop/Gate_C_programs$ 

Here string are stored in continuous locations, without any gap.

[at least what I can tell by obseving these virtual addresses]

but I am not sure whether is it true always.

I understand your concern. I have idependently chosen starting location of second string, and you are asking that it should depend on previous string location. Am I right ? . By looking at the above O/P it seems like second string indeed starts just after the null character of the first string.

I posted this issue in stackoverflow and found some usefull insights of storage of string literals in c.

http://stackoverflow.com/questions/38997119/storing-of-string-literals-in-consecutive-memory-locations/38997175?noredirect=1#comment65346141_38997175

---

Thanks @Debashish... :)

---

@Debashish :) Nice Pictorial explanation.

Can u plz give ans to a similar type  of pictorial explanation for below question in link.

https://gateoverflow.in/66452/modified-gate-question-self-doubt

Answer 2

1. #include‬<stdio.h>
2. int main()
3. {
4. char *s[] = { "knowledge","is","power"};
5. char **p;                  // p si double pointer i.e. p contain address which contain address at which something stores
6. p = s;                      // p contain address of s 
7. printf("%s ", ++(*p));   // p pointing to knowledge then increment  then print i.e. nowledge [here p contain address of nowledge].
8. printf("%s ", *(p++));   // here first p is print then increment is done i.e. nowledge print then increament to is  string
9. printf("%s ", ++(*p));   // increament pointer for string is i.e. point to single "s"  and print it
10. return 0;
11. }

Note:1) Precedence of prefix ++ and * is same. Associativity of both is right to left.
        2) Precedence of postfix ++ is higher than both * and prefix ++. Associativity of postfix ++ is left to right.

Answer 3

O/P:-

nowledge

nowledge

s

 

Explanation:-

1.First printf ++*p equals to

 *p then ++

*p means s[0]

S[0] ++ means 

Now  *p contains to 'n's address of string "knowledge"

 so.o/p:- nowledge

 2. printf *p++

Here both unary operator * and ++ have same precedence (see precedence table image below) and it associate Right to left

So here effect is

*(p++)//but due to postfix it increment at next statement

So overall result of 2nd printf is *p only So it gives same result.

Means o/p:- nowledge

But after printing it increments p

So p now after 2nd print points to s[1].

 

3. Print ++*p equals to
*p then ++

*p means s[1]

S[1] ++ means  *p contains Adress of 's' of string "is"

So o/p is s

 

References By C progtamming Language By Dennis Ritchie

1.Clue of 2nd printf

2..precedency Table

Comments

When I want to change the "knowledge" to "lnowledge"

And tried code like this it is giving runtime error

        #include <stdio.h>
     
        int main(void) {
        	char s[]={"knowledge"};
            char *p;
            p = s;
            printf("%s",++*p);
            //printf("%s",*p++);
            //printf("%s",++*p);
        	return 0;
        }
     

---

@srestha,

replace %s with %d ...

---

int main(void) {
        	char s[]={"knowledge"};
            char *p;
            p = s;
            *p='l';
            printf("%s", s);

        }

---

@Kapil that is to see ASCII code rt?

@Manoj , good but if I want it with increment operator?

then where is the problem?

---

 int main(void) {
        	char s[]={"knowledge"};
            char *p;
            p = s;
            ++*p;
            printf("%s",p);

        	return 0;
        }

 ++*p;

operation modify the content of array which is single char if you supplied it to %s then of course there will be runtime issue.

If still problem check this everything will be clear

int main(void) {

        	char s[]={"knowledge"};
            char *p;
            p = s;
            printf("%c",++*p);
            return 0;
        }

---

@srestha

for ur objective..only one printf is enough

printf("%s",++*p); //Here is Ur err

++*p here not returning any adress

it just changing k to l

While  printf expects address(due to %s format specifier)

 

So correct code is

++*p

then

printf("%s",p);

Answer 4

Here You can think p as 2-D array.

P[0] = "knowledge"

p[1] ="is"

p[2]= "power"

++*p can be written as ++(*p)

Here *p will give k's address and ++ gives n's address ... So it prints nowledge...

*p++ here ++ same priority as * by "denies richie" book and associativity is from right to left...

p points to p[0] means base address of first row and associativity is from right to left so first ++ has been evaluated...Now we have done ++  we got p[1] but before value has been updated *p has been evaluated...

Previously *p at the "n" so it points nowledge.. After that we get p[1]..

PS: when we doing p++ or ++p pointer will increment value by current row size and jump to next row...

When.  (*p)++ or ++(*p) than it will increments value by one cell only where *p points...