您如何在C ++中“重新分配"? [英] How do you 'realloc' in C++?

问题描述

如何在C ++中realloc?该语言似乎缺少了-有new和delete，但没有resize！

How can I realloc in C++? It seems to be missing from the language - there is new and delete but not resize!

我需要它，因为随着程序读取更多数据，我需要重新分配缓冲区以容纳它.我认为delete选择旧指针并new选择新的更大指针不是正确的选择.

I need it because as my program reads more data, I need to reallocate the buffer to hold it. I don't think deleteing the old pointer and newing a new, bigger one, is the right option.

推荐答案

使用:: std :: vector！

Use ::std::vector!

Type* t = (Type*)malloc(sizeof(Type)*n) 
memset(t, 0, sizeof(Type)*m)

成为

::std::vector<Type> t(n, 0);

然后

t = (Type*)realloc(t, sizeof(Type) * n2);

成为

t.resize(n2);

如果要将指针传递给函数，而不是

If you want to pass pointer into function, instead of

Foo(t)

使用

Foo(&t[0])

这是绝对正确的C ++代码，因为向量是一个智能C数组.

It is absolutely correct C++ code, because vector is a smart C-array.

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