I have the following code:
int **ptr = (int **)malloc(sizeof(int*)*N); for(int i=0;i<N;i++) ptr[i]=(int*)malloc(sizeof(int)*N)); How can I free ptr using free? Should I loop over ptr and free ptr[i] or should I just do
free(ptr) and ptr will be freed?
6 Answers
Just the opposite of allocation:
for(int i = 0; i < N; i++) free(ptr[i]); free(ptr); You will have to loop over ptr[i], freeing each int* that you traverse, as you first suggest. For example:
for (int i = 0; i < N; i++) { int* currentIntPtr = ptr[i]; free(currentIntPtr); } Yes, you must loop over ptr and free each ptr[i]. To avoid memory leaks, the general rule is this: for each malloc(), there must be exactly one corresponding free().
Simple
while (N) free(ptr[--N]); free(ptr); Handsome
#define FALSE 0 #define TRUE 1 typedef int BOOL; void freev(void **ptr, int len, BOOL free_seg) { if (len < 0) while (*ptr) {free(*ptr); *ptr++ = NULL;} else while (len) {free(ptr[len]); ptr[len--] = NULL;} if (free_seg) free(ptr); } freev(ptr, N, TRUE); /* if known length */ freev(ptr, -1, TRUE); /* if NULL-terminated */ freev(ptr, -1, FALSE); /* to keep array */ Patrician
GLib functions:
g_ptr_array_free()for freeing arrays of pointers,g_strfreev()for freeing arrays of strings.
I find it hard to do any serious C programming without GLib. It introduces things such as dynamic strings and lays foundations for functional programming. It should really be part of the standard C run-time library. It would give C a breath of fresh air. It would make C a reasonable and competitive language again for the year 2019. But because it isn’t, it will add 1 MB to your application (either in DLL size or in executable size). Also the Windows distribution is maintained by sadists.
for(int i=0;i<N;i++) free(ptr[i]); free(ptr); you are not checking for malloc failure to allocate. You should always check.
void freeMatrix(int **matrix ,int row) { for(int i=0;i<row;i++) { free(matrix[i]); } free(matrix); }
