如何使用malloc()将C语句转换为C ++? [英] How to convert a C statement using malloc() to C++?

问题描述

我的C代码中有一个缓冲区

I have a buffer in C with this code

    buffer = malloc(sizeof(uint32_t)*CACHE_LEN*2);

如何将这一行更改为C ++?

How can I change this line to C++?

使用malloc或new[]更好吗?
我不明白这个sizeof(uint32_t)的含义.

Is it better with malloc or with new[]?
I cannot understand the meaning of this sizeof(uint32_t).

推荐答案

您提到的行将在C ++中很好地编译，除了以下事实:如果buffer不是void *，则可能需要强制转换在C ++中返回malloc的值，您不需要在C中执行(也许不应该这样做).

The line you mention will compile fine in C++, save for that fact that if buffer is not a void *, then you may need to cast the return value of malloc in C++, which you don't need to do in C (and probably should not do).

EG:

uint32_t *buffer = (uint32_t *) malloc(sizeof(uint32_t)*CACHE_LEN*2);

但是，您可能希望转换为new/delete范例；这不仅需要更改该行.

However, you may wish to convert to a new / delete paradigm; that will require more than just changing that line.

EG:

uint32_t *buffer = new uint32_t[CACHE_LEN*2];
...
delete[] buffer;

如果您仍然想更像C ++，请使用std::vector或类似的方法:

If you want to be more C++-like still, then use std::vector or similar:

std::vector<uint32_t> buffer(CACHE_LEN*2);

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