realloc returns a pointer to the resized buffer; this pointer value may be different from the original pointer value, so you need to save that return value somewhere. realloc may return NULL if the request cannot be satsified (in which case the original buffer is left in place). For that reason, you want to save … Read more
No, you won’t have a memory leak. realloc will simply mark the rest “available” for future malloc operations. But you still have to free myPointer later on. As an aside, if you use 0 as the size in realloc, it will have the same effect as free on some implementations. As Steve Jessop and R.. … Read more
From: http://www.sgi.com/tech/stl/alloc.html This is probably the most questionable design decision. It would have probably been a bit more useful to provide a version of reallocate that either changed the size of the existing object without copying or returned NULL. This would have made it directly useful for objects with copy constructors. It would also have … Read more
The standard technique is to introduce a new variable to hold the return from realloc. You then only overwrite your input variable if it succeeds: tmp = realloc(orig, newsize); if (tmp == NULL) { // could not realloc, but orig still valid } else { orig = tmp; }
When std::vector<T> runs out of capacity it has to allocate a new block. You have correctly covered the reasons. IMO it would make sense to augment the allocator interface. Two of us tried to for C++11 and we were unable to gain support for it: [1] [2] I became convinced that in order to make … Read more
Others have answered how malloc(0) works. I will answer one of the questions that you asked that hasn’t been answered yet (I think). The question is about realloc(malloc(0), 0): What does malloc(0) return? Would the answer be same for realloc(malloc(0),0)? The standard says this about realloc(ptr, size): if ptr is NULL, it behaves like malloc(size), … Read more
Just don’t call free() on your original ptr in the happy path. Essentially realloc() has done that for you. ptr = malloc(sizeof(int)); ptr1 = realloc(ptr, count * sizeof(int)); if (ptr1 == NULL) // reallocated pointer ptr1 { printf(“\nExiting!!”); free(ptr); exit(0); } else { ptr = ptr1; // the reallocation succeeded, we can overwrite our original … Read more
Use ::std::vector! Type* t = (Type*)malloc(sizeof(Type)*n) memset(t, 0, sizeof(Type)*m) becomes ::std::vector<Type> t(n, 0); Then t = (Type*)realloc(t, sizeof(Type) * n2); becomes t.resize(n2); If you want to pass pointer into function, instead of Foo(t) use Foo(&t[0]) It is absolutely correct C++ code, because vector is a smart C-array.