When the realloc function can’t get more memory to expand the block to the size you asked for. This is either because there is literally no memory available, or the virtual space is exhausted (e.g. a 32-bit application on a 64-bit machine with 16GB of RAM still can’t use more than 4GB in the process … Read more
Error was in v_push_back implementation : v->e_array = check_a(memcpy((void*)((char*)v->e_array + (v->no_e * v->e_sz)) ,new_val, v->e_sz)); //pointer has not to be saved check_a(memcpy((void*)((char*)v->e_array + (v->no_e * v->e_sz)) ,new_val, v->e_sz)); Because memcpy returns a pointer , in this case, to the last element of the array so we would have lost access to all previous memory solved … Read more
This sounds like it is homework. What does your text book can course material say? If you come here for just the answers without the understanding, then you may pass the assignment, but you won’t learn from the course. Assuming that you will use this information to learn, here is some information to get you … Read more
if (newsize == 0) { free(ptr); return; } if (ptr == NULL) return malloc(size); // otherwise do a true realloc As for What if there is not a contigious size of the size wanted. Then realloc returns NULL and sets errno to indicate the error. 9 solved using realloc in C with malloc [closed]
From the man page: The realloc() function changes the size of the memory block pointed to by ptr to size bytes. So instead of: int *newbuffer = (int*) realloc (stacky, size); you probably want int *newbuffer = (int*) realloc (stacky, size * sizeof(int)); BTW: No need for cast when using malloc and friends. See Do … Read more
%%.2f needs an extra % symbol to make the final .2f into a format rather than the string literal that’s being displayed. Secondly realloc is intended to resize a previously calloc-ed array in memory. solved Source code using a structure pointer