The * is part of the type. book is of type Book *, that is a pointer to a Book. class Book { }; Book * book1 = new Book(); Book * book2 = book1; //now book1 and book2 point to the same Book 4 solved What is the significance of * in * ? … Read more
From the documentation pbParamInfo – Pointer to a null-terminated string of bytes specifying the types of the parameters following pbParamInfo. … – Variable list of parameters, of types specified in pbParamInfo. You are passing NULL for pbParamInfo, which I assume means that no data will be stored in the returned variant, so of course the … Read more
Not quite sure what output you want, but probably you want to output the strings in the array. You can’t detect the end of an array directly. You either need to put a sentinel value at the end of the array, for example NULL: #include <stdio.h> int main(void) { char *t[10] = { “Program”, “hjl”, … Read more
How is value getting retained even after function completing execution? Undefined behavior (UB). It might appear to “work”, but that is not specified to be so by C. fun(&ptr); runs fine, yet printf(“%p”,ptr); is UB as the value ptr is no longer valid. Many systems will tolerate this UB. De-referencing ptr, with printf(“%d\n”,*ptr); is even … Read more
The difference is: names1 is declared simply as an array of string pointers without other defined characteristics. Using string literals here will put the string literals into a section in your executable file which is read-only, because the compiler can this way re-use them. For example, when you use char* a = “abc”; char* b … Read more
.NET runtime is a managed execution runtime which (among other things) provides garbage collection. .NET garbage collector (GC) not only manages the allocation and release of memory, but also transparently moves the objects around the “managed heap”, blocking the rest of your code while doing it. It also compacts (defragments) the memory by moving longer … Read more
*((int*)2000) is not “perfectly valid code”. The result of converting an integer to a pointer is implementation-defined. It’s likely that in your implementation, (int*)2000 results in an invalid pointer. Trying to dereference an invalid pointer produces undefined behavior, which means that anything can happen. When you ran the program with the printf line uncommented, it … Read more
As pointed out by others, this statement suffers from undefined behavior: *(&x+5) += 7; Memory address &x+5 is outside the bounds of variable x. Writing to that address is a very bad idea. OP’s code sample exploits certain C compiler implementation details. Probably educational; it can be used to demonstrate how hackers can exploit missing … Read more
The simple answer that won’t help you much is that this is done for backwards compatibility with C, that has exactly the same syntax to define variables. The rationale for that design would have to come from the creators of C, and I don’t really know it. What I do know is that most coding … Read more
You just need to delete the array referenced by del: delete[] del; You are already deleting the arrays referenced by hello and bye: delete[] * del; delete[] * (del + 1); Though this would be more idiomatic: delete[] del[0]; delete[] del[1]; Or even better, avoid using new and delete altogether, taking advantage of modern C++ … Read more
This code int∗ p = &a[2]; is equivalent to this: int∗ p = a + 2; so even mathematically you can see that p – a is equal to a + 2 – a so I think result should be obvious that it is equal to 2. Note name of array can decay to a … Read more
It is certainly undefined behavior, but I am strong proponent of understanding symptoms of undefined behavior for the benefit of spotting one. The results observed can be explained in following manner: const int a = 5 defined integer constant. Compiler now assumes that value will never be modified for the duration of the whole function, … Read more
The while condition should probably read as: while (abs(nextValue – currValue) > 0.00000000001 && iter < 100000) Note that There is no semicolon at the end. The entire condition must be in parentheses. and is replaced by && – this is not strictly necessary because and is valid in C++ as far as I know, … Read more
You have two separate questions here. A pointer is sometimes used to point to an array or buffer in memory. Therefore it supports the [] syntax. In this case, using pi[x] where x is not 0 is invalid as you are not pointing to an array or buffer. Streams have an overload for char pointers … Read more
Your code (corrected): public static unsafe bool ExistInURL(params object[] ob) { byte* voidPtr = stackalloc byte[5]; *((int*)voidPtr) = 0; while (*(((int*)voidPtr)) < ob.Length) { if ((ob[*((int*)voidPtr)] == null) || (ob[*((int*)voidPtr)].ToString().Trim() == “”)) { *((sbyte*)(voidPtr + 4)) = 0; goto Label_004B; } (*(((int*)voidPtr)))++; } *((sbyte*)(voidPtr + 4)) = 1; Label_004B: return *(((bool*)(voidPtr + 4))); } instead … Read more