Question

main()
{
char *s[] = { "apple","mango","berry","cherry"};
printf("%c\n",*(s+2));
printf("%c\n",( *s+2));
}

i tried this program and i got some un xpected results....
u ppl can find an xplanation for this?

Answer 1

well 1 not too sure what the * is doing there, also when dealing with an array you want to use the s[] not s... FOr example s[0] is apple, s[1] is mango etc.....

Answer 2

The first printf looks to dereference s+2, the second looks like it dereferences s and adds 2 to it.

Answer 3

@rsmith6559 sir please try compiling it ....it was not that easy

Answer 4

Since you have defined s as a pointer to a character array (char **), dereferencing s will give you the pointer to a string (like e.g. 0x400624). Since you said you want to print this value as a character (since you use '%c' in printf), these addresses are somehow translated into characters. I don't know how that translation is done, but it could result in printing some weird characters on your screen.

Answer 5

no i dont think so....coz we all get the same result... i mean we all got a star....so it cant be a random adrees in RAM

Answer 6

I did not get a star, but a 0 and an &. During compilation, the compiler will put those constant strings somewhere in the executable (if you're on Linux, try string <executable> to find such strings). Those addresses are therefor fixed once you compiled the code, so you will see the same symbol each time you run that executable.

Answer 7

ohh ok wait lemme run on linux an @slotema plz do wait i have another query

Answer 8

ok fine my next one

Answer 9

#include<iostream>
int main()
{
int x[3][5]={ {18,20,13,24,35}, {7,8,6,9,10},{19,22,30,21,15}} ;
int *n=&x[0][0]; std::cout<<"endl:- "<<(*(n+3)+1) <<":"<<*(*x+2)+5<<"\n";
std::cout<<"endl:- "<<*(*(x+2)+1) <<":"<<*(x[1]+2)+5;
}

u have amy idea how that *(n+3)+1 works?

Answer 10

*any

Answer 11

int main()
{
int x[3][5]={ {18,20,13,24,35}, {7,8,6,9,10},{19,22,30,21,15}} ;
int *n=&x[0][0];
std::cout<<"endl:- "<<(*(n+3)+1) <<":"<<*(*x+2)+5<<"\n";
std::cout<<"endl:- "<<*(*(x+2)+1) <<":"<<*(x[1]+2)+5;
}

Answer 12

So n points to the first element from x (which is 18). *(n+3) take the fourth (n, n+1, n+2, n+3) element in the array (which is 24). That 24 is than incremented by 1.

Answer 13

i thot it's a 2-d array so incrementin n by 3 shud make it point to the 3-row instead?

Answer 14

if you want it to act as a 2-D array, you should change the definition of n. Right now, n is just an int *, so it acts as a 1-D array.

Answer 15

so if i make it like int ** it's gonna change it's behavior?

Answer 16

so what if n is an int **
i got smthn like 12

Answer 17

It a bit more complicated than that. If you want to use a pointer as a 2-D array, the compiler should know the size of the second dimension. Only if that size is known, the compiler will know what do with a statement like n+1.

Therefor, you should define n as `int (*n)[5]` (post #6 in http://devmaster.net/forums/topic/12966-how-to-create-a-pointer-to-2d-array-in-c/page__view__findpost__p__69890).

Answer 18

it's about making an array of pointers r8?
i was trying to work on pointer arithmetic

Answer 19

what's if i declare a double pointer say t
int **t=&x[0][0];
and now i d like to use (*(t+3)+1)
i expected a segmentation fault but i dint get it
*(*(t+3)+1) gives me a segmentation fault....i convinced my self saying that ther's no x[3][1];
am i right?

Answer 20

There is indeed no x[3][*]. Try with 0, 1 or 2. I can't get the int ** pointer to compile properly without any errors. The solution in my previous post works for me.

Answer 21

did u try this (*(t+3)+1)
i got a warning but i got a value
39

Answer 22

My problem is that my compiler (GCC-4.7) won't accept `int **t = &x[0][0];` But as long as it works for you, that's the most important thing.

Answer 23

it wont? :-/
but i want an xplanation :(

Answer 24

i got a 4.6.3

Answer 25

Did you already try changing the 3 into 2 for `*(*(t+3)+1)`?

Answer 26

no a min

Answer 27

ops sory i got a 39 for (*(t+2)+1) abd 12 for (*(t+3)+1)

Answer 28

The thing with `(*(t+2)+1)` is that you're derefencing t once. Whatever you print is not the value of the array, but again a pointer to the array. Try dereferencing the whole thing.

Answer 29

that gives a segmentation fault

Answer 30

may be it's linearly accesing the array on memory now taking sizeof(t) and traversing over?

Answer 31

My guess is that it's somehow due to the fact that the compiler does not know what to do with a t+1. How many bytes would it skip? Try the approach I mentioned earlier (int (*t)[5]).

Answer 32

i kno that would work
it's simply creating an array of pointer and storing the address of beginning of row in that array
i jus wanna jus more on pointer arithmetic
it kinda confuses me everytime