3 ms·
#define COUNT(array) (sizeof(array) / sizeof((array)[0])) int array[1024]; for (int *p = array; p < array + COUNT(array); ++p) { // ... }
by pwdisswordfish8 5y ago
#define COUNT(array) (sizeof(array) / sizeof((array)[0]))
int array[1024];
for (int *p = array; p < array + COUNT(array); ++p) {
// ...
}
For the loop to be well-behaved, such pointer formation must be well-defined.