用memcpy_s用未签名的char替换memcpy
假设我们有一个执行此操作的旧代码：
unsigned char* dest = new unsigned char[length];
memcpy(dest, source, length);
指针source作为该方法的输入参数传递。 长度是一个无符号的长变量。
现在，我想用它的安全版本替换被认为不安全的纪念呼叫，因此使用 memcpy_s 。 在其文档的基础上，此方法采用三个参数，即
- 目标
- 目标缓冲区的
- 大小，in memcpy_s的字节和WMEMCPY_S的宽字符（WCHAR_T）。来源
- 要复制的字符数。
我对第四参数有一些担忧。应该是这样的：
err = memcpy_s(dest, sizeof(dest), a2, length * sizeof (unsigned char));
是正确的吗？谢谢
Let's suppose we have a legacy code that performs this operation:
unsigned char* dest = new unsigned char[length];
memcpy(dest, source, length);
where the pointer source is passed as input parameter of that method. length is an unsigned long variable.
Now I want to replace the memcpy call, considered not secure, with the secure version of it, so with memcpy_s.
In base of its documentation, this method takes three parameters,
- destination
- Size of the destination buffer, in bytes for memcpy_s and wide characters (wchar_t) for wmemcpy_s.
- the source
- the number of characters to copy.
I'have some concern regarding the fourth parameter. Shall it be something like that:
err = memcpy_s(dest, sizeof(dest), a2, length * sizeof (unsigned char));
Is that correct? Thanks
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memcpy_s（）不是从根本上“更安全”。它只是执行A 很少的理智检查。就您而言，其中一些甚至是多余的。因此，如果要从无效的参数中“捍卫”您的函数实现，则可以确保source不是nullptr;所有其他“安全”检查都可以通过任何其他“安全性”检查：nullptr- 您只是成功分配了它。长度，则不能超过rsize_max。就是这样，无需使用
memcpy_s（）。另外，
sizeof（unsigned char）是1，一定要。memcpy_s()is not fundamentally "more secure". It just performs a few sanity checks. In your case, some of these are even redundant. So, if you want to "defend" your function implementation from invalid arguments, you could make suresourceis notnullptr; all the other "security" checks are guaranteed to pass anyway:nullptr- you just successfully allocated it.length, then it can't be more thanRSIZE_MAX.That's it, no need to use
memcpy_s().Also,
sizeof(unsigned char)is 1, necessarily.