easy question about passing by reference
Author
Must have done this a million times but I cant remember how to now or I have something wrong somewhere else...
I have a 2D array (c++) called array[800][600] which I want to pass as a pointer to another function. How do I
a) call the function say if the function is called function()
b) define the function in the header library and
c) reference the first and second parts of the array in the function
Hope that makes sense.
Cheers
Planetblaze.com - www.planetblaze.com - As METAL as it gets!
DISCLAIMER: Please note I am no expert but the following is as much as I can remember, and I hope it helps.
You pass multidimensional arrays by reference, and in all but the leftmost dimension you must specify a size.
To call:
function(array); // note no & is required as arrays are automatically passed by ref
Not sure about c, soz.
pan narrans
Study + Hard Work + Loud Profanity = Good Code
[edited by - pan narrans on September 30, 2002 2:39:44 PM]
You pass multidimensional arrays by reference, and in all but the leftmost dimension you must specify a size.
So in the header:void function(int a[][600]);orvoid function(int a[800][600]);will work. To call:
function(array); // note no & is required as arrays are automatically passed by ref
Not sure about c, soz.
pan narrans
Study + Hard Work + Loud Profanity = Good Code
[edited by - pan narrans on September 30, 2002 2:39:44 PM]
function( type **array);
function( type *array[]);
both is possible please don t ask me now how to do it with an 3d array i don t know it and i don t think it is possible though
please correct me when i am wrong
but the function parameters are quite common you can find them in nearly every main function of console applications
function( type *array[]);
both is possible please don t ask me now how to do it with an 3d array i don t know it and i don t think it is possible though
please correct me when i am wrong
but the function parameters are quite common you can find them in nearly every main function of console applications
Author
quote:Original post by Basiror
function( type **array);
function( type *array[]);
both is possible please don t ask me now how to do it with an 3d array i don t know it and i don t think it is possible though
please correct me when i am wrong
but the function parameters are quite common you can find them in nearly every main function of console applications
error C2664: ''CheckForCollisionWithPlayer'' : cannot convert parameter 5 from ''unsigned long [800][600]'' to ''unsigned long ** ''
The first one doesnt seem to work either. it compiles but the array doesnt seem to be passed.
Planetblaze.com - www.planetblaze.com - As METAL as it gets!
Author
quote:Original post by pan narrans
DISCLAIMER: Please note I am no expert but the following is as much as I can remember, and I hope it helps.
You pass multidimensional arrays by reference, and in all but the leftmost dimension you must specify a size.So in the header:void function(int a[][600]);orvoid function(int a[800][600]);will work.
To call:
function(array); // note no & is required as arrays are automatically passed by ref
Not sure about c, soz.
pan narrans
Study + Hard Work + Loud Profanity = Good Code
[edited by - pan narrans on September 30, 2002 2:39:44 PM]
the [][600] method seems to work cheers
Planetblaze.com - www.planetblaze.com - As METAL as it gets!
allocating an array like this array[100][100] isn''t bad, but internally the compiler uses this kind of math to access the element : index=i*100+j , this is how a compiler translates the
a[j] access , so i think that *array = new int [100*100] ,is better , just because you can write an optimized accessing function and do all sort of pointer arithmetic trick like shifting by a power of 2 when reading/writing
i hope it helps, sorry for my english ( this will be my new signature )
a[j] access , so i think that *array = new int [100*100] ,is better , just because you can write an optimized accessing function and do all sort of pointer arithmetic trick like shifting by a power of 2 when reading/writing
i hope it helps, sorry for my english ( this will be my new signature )
quote:Original post by Basiror
function( type **array);
function( type *array[]);
This won''t work. type **array is a different type than type var[600][800].
pan narrans'' response was 100% correct, except for the fact that his suggestion _was_ C, so it was even more correct than he thought.
Please ignore other insane responses.
if you want to use void function(type **a) the array should be allocated dynamically that is declare the array as:
type **array = new type[800][600]
or whatever dimensions you want. then you can just pass in array as the parameter for the function.
type **array = new type[800][600]
or whatever dimensions you want. then you can just pass in array as the parameter for the function.
OK, maybe it's just the latent road rage since I sat gridlocked at a light through 3 cycles before getting home, but you people are freakin' morons.
error C2440: 'initializing' : cannot convert from 'int (*)[600]' to 'int ** '
You know, a sign of real intellect is keeping your mouth shut when you don't know what you're talking about.
Again, the very first freakin' response was the absolute, 100% correct answer. The rest of you need to take a class and shut your pie-holes.
[edited by - Stoffel on September 30, 2002 10:00:33 PM]
int **array = new int[800][600]; error C2440: 'initializing' : cannot convert from 'int (*)[600]' to 'int ** '
You know, a sign of real intellect is keeping your mouth shut when you don't know what you're talking about.
Again, the very first freakin' response was the absolute, 100% correct answer. The rest of you need to take a class and shut your pie-holes.
[edited by - Stoffel on September 30, 2002 10:00:33 PM]
This topic is closed to new replies.
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