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Parse multidimensional arrays to functions [c++] |
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So i was wondering if theres a easy and practical way to parse a multidimensional arrays to a function without changing the function, lets say i have two arrays with different dimensions, and i want to use them in the same fuction, how?
example:
Code:
int array[2][4] = {{1,2,3,4},{4,5,6,7}}
int array[3][2] = {{1,2},{3,4},{5,6}}
//How to use both of those arrays in the same function without changing anything?
function coolStoryBro(int arrayf[][],sizeX,sizeY)
{
//etc
}
of course this is just an example and doesn't work :P
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trye see if pointers work... |
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rudi: sorry, i am a complete noob in c++, how?! D: |
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bah. first of all that should not compile. you cant have declare and define two arrays with the same name, second of all im too drubk to say more and bedtime |
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it was an example, oh don't tell me i have to read my 500 pages c++ book again? pl0x halp D:
p0int3rz r h4rdz
(leet language makes everything better right?) |
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it was an example, oh don't tell me i have to read my 500 pages c++ book again? pl0x halp D:
p0int3rz r h4rdz
(leet language makes everything better right?) |
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Here's a program to illustrate:
Code:
#include <stdio.h>
int array1[2][4] = {{1,2,3,4},{4,5,6,7}};
int array2[3][2] = {{1,2},{3,4},{5,6}};
void printArray(int* array, int sizeX, int sizeY) {
printf("{");
for (int x = 0; x < sizeX; x++) {
printf("{");
for (int y = 0; y < sizeY; y++) {
printf("%i", array[x + y*sizeX]);
if (y != sizeY - 1)
printf(",");
}
printf("}");
if (x != sizeX - 1)
printf(",");
}
printf("}\n");
}
int main(void) {
printArray((int*)array1, 2, 4);
printArray((int*)array2, 3, 2);
return 0;
}
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Whoops, change the line that prints the value to this:
Code:printf("%i", array[x*sizeY + y]);
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GO MAEK A DEMO BUOT IT HARHARHARHARHARHAR |
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is this a homework :O? |
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pointers are easy. |
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blacksheep: im sorry to break it to you but if you want to learn how to code graphics you need to first learn how to code. and when it comes to c++ if you dont know pointers properly you'll have plenty of headaches trying to figure stuff out. |
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odd question: is it possible to cast a linear array as an n-dimensional array, or is xpansive's way the only way to do it?
something like
Code:
if (sizeof(long) != 4) throw err("not long enough or too long");
void* stuff = malloc((480*800) << 2);
*long[480][800] thing = null
thing = (*(long[480][800]))stuff;
violate_and_mess_with(thing);
thing = null;
free(stuff); //pun intended
or even
Code:
void coolStoryBro(int* arrayf,int sizeX, int sizeY)
{ int[sizeX,sizeY] myway = *int[sizeX,sizeY](arrayf);
//etc...}
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Quintix, you'll have a great problem doing that, because the C syntax when specifying a size for a local variable will always allocate memory.
Quote:
int array[5][9]; // this is not a type declaration but an actual allocation request
This is why the [][] are on the variable side. |
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Actually I found a C99 conforming version:
Quote:
void printItC99 (int sizeX, int sizeY, int array[sizeX][sizeY]) {
printf("{");
for (int x = 0; x < sizeX; x++) {
printf("{");
for (int y = 0; y < sizeY; y++) {
printf("%i", array[x][y]);
if (y != sizeY - 1)
printf(",");
}
printf("}");
if (x != sizeX - 1)
printf(",");
}
printf("}\n");
}
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It actually works in c90 mode with gcc
Quote:
void printItC90 (int sizeX, int sizeY, int array[sizeX][sizeY]) {
int x, y;
printf("{");
for (x = 0; x < sizeX; x++) {
printf("{");
for (y = 0; y < sizeY; y++) {
printf("%i", array[x][y]);
if (y != sizeY - 1)
printf(",");
}
printf("}");
if (x != sizeX - 1)
printf(",");
}
printf("}\n");
}
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sidenote: I still learn new things about C in 2012? :) |
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