Because the expression num1++ evaluates to num1. You may want to do: ++num1 which evaluates to num1 + 1. Note however that both expressions increment num1 by one. Evaluating num1 in the next statement evalutes to the incremented value. In short In C, why doesn’t num1++ increment in the printf()? num1++ does increment num1 but … Read more
–> is not an operator. It is in fact two separate operators, — and >. The conditional’s code decrements x, while returning x‘s original (not decremented) value, and then compares the original value with 0 using the > operator. To better understand, the statement could be written as follows: while( (x–) > 0 ) 4 … Read more
fopen, calloc, and realloc can all fail, returning NULL. In the case of realloc, the original allocation will remain untouched, so immediately overwriting the original pointer with the newly returned value will leak memory and throw away potentially useful data. Reallocating every iteration is rather costly. A better strategy is to reallocate when the buffer … Read more
Your code exhibits undefined behaviour because you passed a non-static, uninitialised pointer to realloc whose contents were indeterminate. From C11, Memory Management Functions: The realloc function deallocates the old object pointed to by ptr and returns a pointer to a new object that has the size specified by size. ….if ptr does not match a … Read more
When processing regular characters (final else clause) you read additional input instead of operating over text, i.e. instead of: scanf(“%c”, &document[parano][sentno][wordno][charno]); printf(“%c\n”, document[parano][sentno][wordno][charno]); charno++; you want to do: document[parano][sentno][wordno][charno++] = text[i]; By the time we hit text[i] == ‘ ‘ for the first time the value of ***document is “3\n1 2\n2\n” but you wanted it … Read more
You want to modify the value of an int* (your array) so need to pass a pointer to it into your increase function: void increase(int** data) { *data = realloc(*data, 5 * sizeof int); } Calling code would then look like: int *data = malloc(4 * sizeof *data); /* do stuff with data */ increase(&data); … Read more
Invalid assumption: void *my_realloc(void *ptr, size_t size) { unsigned char *old_ptr = (unsigned char *)ptr; You’re breaking the realloc() specification here. realloc() doesn’t assume that the ptr would always be pointing to a string. It is type-agnostic. And the cast is redundant. There’s an implicit conversion from void * to any other pointer type. Undefined … Read more
OP’s mostly has it. It is just printing the first element each time. // printf(“%d”, *arr); printf(“%d”, arr[n-1]); is it possible to do it with an array? No. In C an array cannot change size once it is defined. The memory size allocated and referenced by a pointer can change though. solved How to grow … Read more
What you’re doing will work. Each time you increment contents, you also increment counter, so contents – counter gives you the original pointer that you can free. Of course, a better way of doing this would be to use a temporary pointer to increment through the allocated memory so you can use the original to … Read more
This answer is based on the assumption that you do proper handling of your pointers between the part where you allocate memory for 1 element each and where you try to reallocate the memory. This part // realloc to correct size for(int a = 0; a < word_nb; a++) { *(word_list + (word_nb-1)) = (char … Read more