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anonymous
 3 years 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?
anonymous
 3 years 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?

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KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0well 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.....

rsmith6559
 3 years ago
Best ResponseYou've already chosen the best response.0The first printf looks to dereference s+2, the second looks like it dereferences s and adds 2 to it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@rsmith6559 sir please try compiling it ....it was not that easy

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Since 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ohh ok wait lemme run on linux an @slotema plz do wait i have another query

anonymous
 3 years ago
Best 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0int 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; }

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so if i make it like int ** it's gonna change it's behavior?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so what if n is an int ** i got smthn like 12

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It 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).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it's about making an array of pointers r8? i was trying to work on pointer arithmetic

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what'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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There 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.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0My 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it wont? :/ but i want an xplanation :(

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that gives a segmentation fault

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0may be it's linearly accesing the array on memory now taking sizeof(t) and traversing over?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0My 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]).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
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