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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

This Question is Closed

KonradZuseBest ResponseYou've already chosen the best response.0
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.....
 one year ago

rsmith6559Best ResponseYou've already chosen the best response.0
The first printf looks to dereference s+2, the second looks like it dereferences s and adds 2 to it.
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
@rsmith6559 sir please try compiling it ....it was not that easy
 one year ago

slotemaBest ResponseYou've already chosen the best response.0
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.
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
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
 one year ago

slotemaBest ResponseYou've already chosen the best response.0
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.
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
ohh ok wait lemme run on linux an @slotema plz do wait i have another query
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
ok fine my next one
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
#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?
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
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; }
 one year ago

slotemaBest ResponseYou've already chosen the best response.0
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.
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
i thot it's a 2d array so incrementin n by 3 shud make it point to the 3row instead?
 one year ago

slotemaBest ResponseYou've already chosen the best response.0
if you want it to act as a 2D array, you should change the definition of n. Right now, n is just an int *, so it acts as a 1D array.
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
so if i make it like int ** it's gonna change it's behavior?
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
so what if n is an int ** i got smthn like 12
 one year ago

slotemaBest ResponseYou've already chosen the best response.0
It a bit more complicated than that. If you want to use a pointer as a 2D 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/12966howtocreateapointerto2darrayinc/page__view__findpost__p__69890).
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
it's about making an array of pointers r8? i was trying to work on pointer arithmetic
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
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?
 one year ago

slotemaBest ResponseYou've already chosen the best response.0
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.
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
did u try this (*(t+3)+1) i got a warning but i got a value 39
 one year ago

slotemaBest ResponseYou've already chosen the best response.0
My problem is that my compiler (GCC4.7) won't accept `int **t = &x[0][0];` But as long as it works for you, that's the most important thing.
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
it wont? :/ but i want an xplanation :(
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
i got a 4.6.3
 one year ago

slotemaBest ResponseYou've already chosen the best response.0
Did you already try changing the 3 into 2 for `*(*(t+3)+1)`?
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
ops sory i got a 39 for (*(t+2)+1) abd 12 for (*(t+3)+1)
 one year ago

slotemaBest ResponseYou've already chosen the best response.0
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.
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
that gives a segmentation fault
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
may be it's linearly accesing the array on memory now taking sizeof(t) and traversing over?
 one year ago

slotemaBest ResponseYou've already chosen the best response.0
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]).
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
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
 one year ago
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