In C since C99, even dynamic multidimensional arrays can be easily allocated in one go with malloc and freed with free: double (*A)[n] = malloc(sizeof(double[n][n])); for (size_t i = 0; i < n; ++i) for (size_t j = 0; j < n; ++j) A[i][j] = someinvolvedfunction(i, j); free(A);
Just about every modern operating system will recover all the allocated memory space after a program exits. The only exception I can think of might be something like Palm OS where the program’s static storage and runtime memory are pretty much the same thing, so not freeing might cause the program to take up more … Read more
OK some answers about malloc were already posted. The more interesting part is how free works (and in this direction, malloc too can be understood better). In many malloc/free implementations, free does normally not return the memory to the operating system (or at least only in rare cases). The reason is that you will get … Read more
Unless you are forced to use C, you should never use malloc. Always use new. If you need a big chunk of data just do something like: char *pBuffer = new char[1024]; Be careful though this is not correct: //This is incorrect – may delete only one element, may corrupt the heap, or worse… delete … Read more
TL;DR int *sieve = (int *) malloc(sizeof(int) * length); has two problems. The cast and that you’re using the type instead of variable as argument for sizeof. Instead, do like this: int *sieve = malloc(sizeof *sieve * length); Long version No; you don’t cast the result, since: It is unnecessary, as void * is automatically … Read more
In C++, it would be achieved with this. std::vector<node *> problem(length); The code you show is effectively emulating a small subset of the features of vector. Namely, an array-like container that is aware of its size.
What yo actually do is called “undefined behavior”. You allocate 4 bytes and then store into that buffer integers. It works fine for the first integer (or maybe two if sizeof(int)==2 on your machine) but for the next integer the behavior becomes undefined. It may segfault immediately and it may take more similar stores until … Read more
Just do char* only_valid_data = new char[data_size]; to allocate. To free you do delete[] only_valid_data; Important note: When you allocate memory with new it will allocate data_size elements, not data_size bytes (like malloc does). The size of an element is the size of the non-pointer base type, in your case e.g. sizeof(*only_valid_data). In this case … 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
I would recommend you to replace if(c!=’ ‘ && c!=’\t’ && c!=’\n’) with if(!isspace(c)) // Iam easier and more readable from ctype.h. Which detects all these characters ‘ ‘ space ‘\t’ horizontal tab ‘\n’ newline ‘\v’ vertical tab ‘\f’ feed ‘\r’ carriage return Also you should change char*** word_array; (three star general) to pointer to … Read more