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When do you use the " * " and not on pointers?
Author
Well i get how pointers work, but wheni put them in code, i get some errors, so when i take out the" * " in front of the pointer it workds.
ie.
int somenum, *psomenum;
psomenum = &somenum //i dont use the * why?
somefunc(psomenum) //again i dont have to use it why?
but some times i do have to use the " * ". So im just asking when do you use the * and when not, and why?
I''ll try to help you.
when declaring a pointer:
int num; //normal int
int *pNum; //pointer to int
and when changing the value that is being pointed to by the pointer:
pNum = # //point to num,
*pNum = 10; //change value of num to 10
I think thats all..
...go on and live with no regrets, you only have one life...
when declaring a pointer:
int num; //normal int
int *pNum; //pointer to int
and when changing the value that is being pointed to by the pointer:
pNum = # //point to num,
*pNum = 10; //change value of num to 10
I think thats all..
...go on and live with no regrets, you only have one life...
Look at it this way. The * and & operators are opposite of each other. The & operator tells the compiler that you have a variable, not a pointer, and you want a pointer to that address returned.(Actually it isnt quite a pointer, it''s the actual mem address, but forget that). THe * says that you have a mem address, like a pointer, and you want the variable being pointed to.
int somenum, *psomenum;
somenum = 5;
psomenum = NULL; //Doesnt point to anywhere
psomenum = &somenum //psomenum points to somenum
cout << somenum; //Prints out "5"
cout << &somenum //prints out some hexadecimal junk
cout << *psomenum; //prints out 5
cout << psomenum; //same hex crap as before
-----------------------------
The sad thing about artificial intelligence is that it lacks artifice and therefore intelligence.
int somenum, *psomenum;
somenum = 5;
psomenum = NULL; //Doesnt point to anywhere
psomenum = &somenum //psomenum points to somenum
cout << somenum; //Prints out "5"
cout << &somenum //prints out some hexadecimal junk
cout << *psomenum; //prints out 5
cout << psomenum; //same hex crap as before
-----------------------------
The sad thing about artificial intelligence is that it lacks artifice and therefore intelligence.
Hi, When you declare a pointer:
int *psomething;
you are saying that you want a variable of type (int *) which is a pointer to int. This pointer is nothing more than a variable that holds an address.
when you say:
psomething = &something
you are saying I want the variable psomething(which holds an address) to hold the address of something.
when you say:
*psomething = 5;
you are saying, look at the address that psomething holds and go to that location(which is something), and put a 5 in it.
hope this helps.
- Free Your Mind -
int *psomething;
you are saying that you want a variable of type (int *) which is a pointer to int. This pointer is nothing more than a variable that holds an address.
when you say:
psomething = &something
you are saying I want the variable psomething(which holds an address) to hold the address of something.
when you say:
*psomething = 5;
you are saying, look at the address that psomething holds and go to that location(which is something), and put a 5 in it.
hope this helps.
- Free Your Mind -
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