realloc

In C the standard memory handling functions are malloc() , realloc() and free() . However, C++ stdlib allocators only parallel two of them: there is no reallocation function. Of course, it would not be possible to do exactly the same as realloc() , because simply copying memory is not appropriate for non-aggregate types. But would there be a problem with, say, this function: bool reallocate (pointer ptr, size_type num_now, size_type num_requested); where ptr is previously allocated with the same allocator for num_now objects; num_requested >= num_now ; and semantics as follows: if allocator

char* p = malloc(1024); char* q = realloc(p,2048); 现在的问题是我们应该如何处理指针 p。 刚开始按照我最直观的理解，如果就是直接将 p = NULL;。 到最后只需要释放 q的空间就可以了。 因为最近在做个封装。结果在做单元测试的时候发现。有时候我在 free(q); 的时候会出错。这样我就郁闷了。 后来仔细一跟踪，发现 realloc 完以后 q 和 p 的指针地址是一样。不过有时候又不一样。 仔细查了下资料。得到如下信息： 1.如果 当前连续内存块足够 realloc 的话，只是将p所指向的空间扩大，并返回p的指针地址。 这个时候 q 和 p 指向的地址是一样的。 2.如果 当前连续内存块不够长度，再找一个足够长的地方，分配一块新的内存，q，并将 p指向的内容 copy到 q，返回 q。并将p所指向的内存空间删除。 这样也就是说 realloc 有时候会产生一个新的内存地址 有的时候不会。所以在分配完成后。我们需要判断下 p 是否等于 q。并做相应的处理。 这里有点要注意的是要避免 p = realloc(p,2048); 这种写法。有可能会造成 realloc 分配失败后，p原先所指向的内存地址丢失。 来源： https://www.cnblogs.com/god-of-death/p/11334846.html

A different question inspired the following thought: Does std::vector<T> have to move all the elements when it increases its capacity? As far as I understand, the standard behaviour is for the underlying allocator to request an entire chunk of the new size, then move all the old elements over, then destroy the old elements and then deallocate the old memory. This behaviour appears to be the only possible correct solution given the standard allocator interface. But I was wondering, would it make sense to amend the allocator to offer a reallocate(std::size_t) function which would return a pair

How can I realloc in C++? It seems to be missing from the language - there is new and delete but not resize ! I need it because as my program reads more data, I need to reallocate the buffer to hold it. I don't think delete ing the old pointer and new ing a new, bigger one, is the right option. f0b0s 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++

This question already has an answer here: what's the point in malloc(0)? 17 answers What does malloc(0) returns? Would the answer be same for realloc(malloc(0),0) ? #include<stdio.h> #include<malloc.h> int main() { printf("%p\n", malloc(0)); printf("%p\n", realloc(malloc(0), 0)); return 0; } Output from linux gcc: manav@manav-workstation:~$ gcc -Wall mal.c manav@manav-workstation:~$ ./a.out 0x9363008 (nil) manav@manav-workstation:~$ The output keep changing everytime for malloc(0) . Is this a standard answer? And why would anyone be interested in getting such a pointer, other than academic