hello i have this function
which finds the *size from a txt fileCode:void learn_size(int *size,char *point)
then im declaring a function like this
as im not able to declare read function with a real number in that xarray array because i do not know it from the beginning and keeps telling me that it does not recognize the *size how can i fix that problemCode:void read(int xarray[*size][*size])
Well no, because you could not have declared your array at run-time with a variable size.
Using a C99 compiler, you can do this.
Unless you can guarantee that size will be small, you're better off using malloc to allocate the space. A VLA can blow your stack allocation with zero warning, and no recovery.Code:#include <stdio.h> #include <stdlib.h> #include <math.h> void foo ( int size, int arr[size][size] ) { for ( int i = 0 ; i < size ; i++ ) { printf("%d\n",arr[i][i]); } } int main ( ) { int size = 3; int arr[size][size]; arr[0][0] = 0; arr[1][1] = 1; arr[2][2] = 2; foo(size,arr); return 0; }
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
the other approach is to pass a pointer foo(int size,int *arr) and compute the array indices yourself. for a two dimension array
you do arr[row * size + col]
Code:read(int size,int *arr) { int i; // row (first) index int j; // column (second) index // arr[0][0] i = 0; j = 0; arr[i * size + j] = 0; // array[0][1] i = 0; j = 1; arr[i * size + j] = 1; // array[1][0] i = 1; j = 0; arr[i * size + j] = 2;
