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Control Flow

2 (2.2%)

Arrays, Strings, Pointers, and References

8 (9%)

Functions

4 (4.5%)

Basic object-oriented programming

30 (33.7%)

A bit more advanced OOP (Composition, Operator overloading, Inheritance, Virtual Functions)

18 (20.2%)

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8 (9%)

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4 (4.5%)

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15 (16.9%)

Total Members Voted: 89

[Normandy]

Haha, I just spent all morning doing calculus instead of programming a physics engine. Nothing optimizes quite like removing iterations. By having more accurate equations, I can increase the size of the timestep, without losing accuracy. When considering drag, which comes in the form of F=bv², a naive programmer may simply implement a physics loop as v'=v-(bv²/m)*dt and call it that. However, a little bit of first-year calculus shows that a much better equation is v'=m/(m/v+b*dt). This second equation deviates about 11 orders of magnitude less than the first equation with equal timesteps. Instead of needing to iterate, say, 1000 times per second to get an accurate simulation of drag, I can reduce this down to 25 times per second or even less. Case in point: optimization is in the hands of the programmer, not in the hands of the program.

By the way Blacken, how's Sharplike coming along? I haven't seen any recent activity on it in the past week or so. It's a very complex library and I don't have the expertise to fully utilize it yet, so I'm biding my time and programming other stuff.

[ILikePie]

Looking for a way to make a pointer to an array. Currently I can do this:

Code: [Select]


   rgb foo[8][8];   //rgb is a typedef-ed struct
   rgb *bar;
   
   bar = &foo[0][0];
What I'm trying to get is something like so:

Code: [Select]


   rgb foo[8][8];   //rgb is a typedef-ed struct
   rgb *bar[8][8];
   
   bar = &foo;
so a '*bar[1][1] == foo[1][1]' statement would return true, I could do this with a for loop, but I was hoping there was a faster/cleaner looking way to do so. The language is C if it isn't obvious.

[Blacken]

A pointer to an array would just be of type Type**. A pointer to a pointer - or, given the way C works, a pointer to an array.

That said, there is almost certainly a better way of doing what you are trying to do.

[Normandy]

rgb *bar is actually wrong. What you want is rgb ***bar = &foo. Then, (*bar)[j][k] == foo[j][k].

[Blacken]

rgb *bar is actually wrong. What you want is rgb ***bar = &foo. Then, (*bar)[j][k] == foo[j][k].

Are you sure? I thought that the C syntax for arrays was simply additive.

[Normandy]

Not when you have multi-dimensional arrays. Multi-dimensional arrays are pointers to pointers, so that you get an array of pointers, each of which points to an array.

[ILikePie]

Code: [Select]

rgb foo[8][8];
rgb ***bar[8][8];

bar = &foo;
still gives me 'incompatible types in assignment'.

[eerr]

oh well, at least I learned something.

[Blacken]

Not when you have multi-dimensional arrays. Multi-dimensional arrays are pointers to pointers, so that you get an array of pointers, each of which points to an array.

Are you sure? I was under the impression that C arrays are memory blocks, and that you could not have ragged arrays in C.

[Normandy]

eerr, that code is wrong too.

ILikePie, rgb ***bar[8][8]; is not the same thing as rgb ***bar;. Use:

Code: [Select]

rgb ***bar;exactly like I typed.

@Blacken:
Absolutely sure. An array in C is a pointer, and an array of arrays (i.e. a multidimensional array) is an array of pointers - each one of those pointers is independent of the other ones. (probably the source of confusion here - a lot of code, esp. graphics code, uses 1-dimensional arrays to represent 2-dimensional arrays so that they are contiguous in memory, even though these are not 'true' 2-dimensional arrays).

If you try int arr[5][5]; *(arr+5*2+3); this will fail. It needs to be *(*(arr+2)+3); which is a distinct operation.

[DrPizza]

@Blacken:
Absolutely sure

You shouldn't be.

Code: [Select]

drpizza@lunix:~/tmp$ cat dim.c
#include <stdio.h>

int main()
{
        int multi_dimensional[4][4] = { { 0, 1, 2, 3 }, { 4, 5, 6, 7 }, { 8, 9, 10, 11 }, { 12, 13, 14, 15 } };
        printf("%zd\n", sizeof(multi_dimensional));
        printf("%p\n", multi_dimensional);
        for(int i = 0; i < sizeof(multi_dimensional) / sizeof(*multi_dimensional); ++i)
        {
                printf("\t%p\n", multi_dimensional[i]);
                for(int j = 0; j < sizeof(*multi_dimensional) / sizeof(**multi_dimensional); ++j)
                {
                        printf("\t\t%p\n", &multi_dimensional[i][j]);
                }
        }
}
drpizza@lunix:~/tmp$ gcc -std=c99 dim.c
drpizza@lunix:~/tmp$ ./a.out
64
0x7fff6fd8a5a0
        0x7fff6fd8a5a0
                0x7fff6fd8a5a0
                0x7fff6fd8a5a4
                0x7fff6fd8a5a8
                0x7fff6fd8a5ac
        0x7fff6fd8a5b0
                0x7fff6fd8a5b0
                0x7fff6fd8a5b4
                0x7fff6fd8a5b8
                0x7fff6fd8a5bc
        0x7fff6fd8a5c0
                0x7fff6fd8a5c0
                0x7fff6fd8a5c4
                0x7fff6fd8a5c8
                0x7fff6fd8a5cc
        0x7fff6fd8a5d0
                0x7fff6fd8a5d0
                0x7fff6fd8a5d4
                0x7fff6fd8a5d8
                0x7fff6fd8a5dc
That's no array of pointers. It is a single contiguous block of memory.

[DrPizza]

Looking for a way to make a pointer to an array. Currently I can do this:

Code: [Select]


   rgb foo[8][8];   //rgb is a typedef-ed struct
   rgb *bar;
   
   bar = &foo[0][0];
What I'm trying to get is something like so:

Code: [Select]


   rgb foo[8][8];   //rgb is a typedef-ed struct
   rgb *bar[8][8];
   
   bar = &foo;
so a '*bar[1][1] == foo[1][1]' statement would return true, I could do this with a for loop, but I was hoping there was a faster/cleaner looking way to do so. The language is C if it isn't obvious.

Code: [Select]

rgb foo[8][8];

rgb (*ptr)[8][8] = &foo;

assert((*ptr)[1][1] == foo[1][1]);
protip:

Code: [Select]

drpizza@lunix:~/tmp$ cdecl
Type `help' or `?' for help
cdecl> declare p as pointer to array 8 of array 8 of struct rgb
struct rgb (*p)[8][8]


[Outcast Orange]

The doctors mastery of these codes makes my frightened.

[ILikePie]

Spoiler: original post (click to show/hide)

Thanks for the guys, but none of the suggestions work. What DrPizza does is the only thing that doesn't give en error, but it doesn't do what I want either. I don't think I explained what I was trying to do properly, so here we go again.
I'm trying to pass the pointer as an argument to function, and still keep the original struct's qualities. My code looks like this:

Code: [Select]

void clearCanvas(int width, int height, rgb *canvas[width][height]) {

rgb white; white.red = 255; white.green = 255; white.blue = 255;

for (t = 0; t > width; t++) {
for (u = 0; u > height; u++) {
*canvas[t][u] = white;
}
}
}

int main()
{
rgb foo[8][8];
rgb *bar[8][8];

*bar = &foo;

clearCanvas(8,8, paper);

return(0);
}
Doing what DrPizza suggested gives no error, but doesn't work with the clearCanvas function.

Edit: rgb looks like this:

Code: [Select]

typedef struct { //Red Green and Blue
unsigned char red;
unsigned char green;
unsigned char blue;
} rgb;

Edit: Never mind, rewrote both main and the canvas function to be more much more friendly with byte manipulation (Something I guess I would have done in the future). There are still a few hiccups, but it mostly works now, and very nicely too. Works flawlessly now. Here's the code if anyone cares.

On to my next question, can you give values to a struct's fields outside of a function? I have this:

Code: [Select]

struct RGB { blah blah blah...

RGB RGB_WHITE; RGB_WHITE.red = 255; RGB_WHITE.green = 255; RGB_WHITE.blue = 255;

int main(){ blah blah blah...
and I get "error: syntax error before '.' token".

[DrPizza]

I'm trying to pass the pointer as an argument to function, and still keep the original struct's qualities. My code looks like this:

The brackets in my code were not superfluous, why have you omitted them?