@programminginc - страница 1844

Hi! this is a question not directly related to cpp but to the observer programming pattern. I have a Player that wants to build a Tower in the Board. What I would've done is: inside the okButton event, subtract money from player, and add a tower to the board, meaning this event knows the interfaces of player and board so it has some level of coupling. If I use the observer pattern, I create an observable buildTower, to which Board and Player were subscribed to in some initialization procedure. Player and Board each receive the notification, and the logic for discounting player resources is in the player, and for placing the tower in the board is in the board. Is this a good use of the observer pattern? If before I want to check if the player has enough resources to build a tower how should I do that using the observer pattern?

Al

hello i try to initilize an array of strings *char[3] = {"One" , "Two" , "Four"}; ... if i try to printf ("%s\n",char[0]); i havfe seg fault ... not sure why

Al

char *num [3] = {"One" , "Two" , "Four"}; thats the arrya sorry

Al

when i try to printf ("%s",num[0]);

Al

i receive seg fault..

Al

im not sure why

Artöm

const

olli

when i try to printf ("%s",num[0]);

is this the only time you access the array? #include <stdio.h> int main() { char *num [3] = {"One" , "Two" , "Four"}; printf("%s\n", num[0]); } Does not segfault

Anonymous

Can we use array elements for condition checking in loops EG: for(AR[0];AR[I]!='/0';AR[0]++)

Harmanjit

I am a beginner so i might be wrong Shobin Try if else condition n check ur array comparison, u will knw that we can use it or not

Harmanjit

I think it will work. M not sure . Do it practically

Sid Sun

is this the only time you access the array? #include <stdio.h> int main() { char *num [3] = {"One" , "Two" , "Four"}; printf("%s\n", num[0]); } Does not segfault

num is an array of pointers to characters, you will need to derefernce it when you access the value and you cannot store strings over there, a char is one character I might be wrong

Sid Sun

I probably am.. 🤦‍♂

olli

num is an array of pointers to characters, you will need to derefernce it when you access the value and you cannot store strings over there, a char is one character I might be wrong

the code works as expected. Just wanted to point out that the given information by @Nutellink are not sufficient to reproduce the issue.

Sid Sun

the code works as expected. Just wanted to point out that the given information by @Nutellink are not sufficient to reproduce the issue.

Hypothetically, if I did want an array of pointers to char.. how would I define them?

olli

Hypothetically, if I did want an array of pointers to char.. how would I define them?

The general solution would be char ** array, if you know the size you can also use char * array[SIZE]

olli

The second one allocates the space, for the first one you would need to allocate the space manually (e.g malloc )

Sid Sun

The general solution would be char ** array, if you know the size you can also use char * array[SIZE]

Well, you used the second one to store array of strings (char array), I don't understand how it is equivalent of pointers to an array of pointers to single chars

olli

Well, you used the second one to store array of strings (char array), I don't understand how it is equivalent of pointers to an array of pointers to single chars

char * array[3] is not a char array, it's an array of pointers to char. Since an array can decay to a pointer you can assign it to a pointer to pointer to char (char ** ) e.g. const char * ar1[3] = { "one", "two", "three" }; const char ** ar2 = ar1; does this answer your question?

Sid Sun

char * array[3] is not a char array, it's an array of pointers to char. Since an array can decay to a pointer you can assign it to a pointer to pointer to char (char ** ) e.g. const char * ar1[3] = { "one", "two", "three" }; const char ** ar2 = ar1; does this answer your question?

Well, if it's an array of pointers to char, how can we store "one"?

olli

Well, if it's an array of pointers to char, how can we store "one"?

a single pointer to char? Like this char *charPointer;

Sid Sun

a single pointer to char? Like this char *charPointer;

I mean you declared: const char * ar1[3] = { "one", "two", "three" }; Since it is an array of constant pointers to chars, how come we can store "one" in the location pointed by the first pointer or "two" in the next one and so on..