It sounds like you are rote learning rather than actually understanding the reason behind doing that.Originally Posted by transgalactic2
That would be true if arr is an int* and you wanted to change the pointer itself from within the function such that the change is reflected in the caller. Here you just want to pass a pointer to a pointer, so why take its address?
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i tried to write a program for it
but i get read accsses error
i dont know why??
Code:#include <stdio.h> #include <stdlib.h> int what1(int **arr, int m, int n); int what2 (int **arr, int row, int col, int m, int n); int main() { int ** arr=(int**)malloc(6*sizeof(int*)); *arr=(int*)malloc(5*sizeof(int)); arr[0][0]=0; arr[0][1]=1; arr[0][2]=1; arr[0][3]=0; arr[0][4]=1; arr[1][0]=0; arr[1][1]=0; arr[1][2]=1; arr[1][3]=0; arr[1][4]=0; arr[2][0]=0; arr[2][1]=0; arr[2][2]=1; arr[2][3]=1; arr[2][4]=1; arr[3][0]=0; arr[3][1]=1; arr[3][2]=0; arr[3][3]=0; arr[3][4]=0; arr[4][0]=1; arr[4][1]=0; arr[4][2]=0; arr[4][3]=1; arr[4][4]=1; arr[5][0]=1; arr[5][1]=0; arr[5][2]=1; arr[5][3]=1; arr[5][4]=0; what1(arr,6,5) ; return 0; } int what1(int **arr, int m, int n){ int i, j, tmp, stam=0; for(i=0; i<m; i++) for(j=0; j<n; j++){ tmp = what2(arr,i,j,m,n); if (tmp>stam) stam = tmp; } return stam; } int what2 (int **arr, int row, int col, int m, int n){ int i,j,tmp, stam=0; if (row < 0 || row >= m || col < 0 || col >= n) return 0; if (arr[row][col] == 0) return 0; arr[row][col] = 0; for(i=-1; i<2; i++) for(j=-1; j<2; j++){ if(!i && !j) continue; tmp = 1 + what2(arr, row+i, col+j, m, n); if (tmp > stam) stam = tmp; } arr[row][col] = 1; return stam; }
You only allocated space for one pointer element to point to in your dynamically allocated array of pointers.
It does not have to be so hard. For example:
Code:#include <stdio.h> #include <stdlib.h> #include <stddef.h> void print(int **numbers, size_t num_rows, size_t num_cols); int main(void) { const size_t num_rows = 4; const size_t num_cols = 3; /* For simplification, assume that malloc() never returns a null pointer. */ int **numbers = malloc(num_rows * sizeof(*numbers)); size_t n = 1; size_t i; for (i = 0; i < num_rows; ++i) { size_t j; numbers[i] = malloc(num_cols * sizeof(*numbers[i])); for (j = 0; j < num_cols; ++j) { numbers[i][j] = n++; } } print(numbers, num_rows, num_cols); for (i = 0; i < num_rows; ++i) { free(numbers[i]); } free(numbers); return 0; } void print(int **numbers, size_t num_rows, size_t num_cols) { size_t i; for (i = 0; i < num_rows; ++i) { size_t j; for (j = 0; j < num_cols; ++j) { printf("%d ", numbers[i][j]); } puts(""); } }
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yes we changed the head of the linked list
so we use &head
