我可以将一个未签名的char*转换为unsigned int*吗?

error: invalid static_cast from type ‘unsigned char*’ to type ‘uint32_t* {aka unsigned int*}’
     uint32_t *starti = static_cast<uint32_t*>(&memory[164]);

I've allocated an array of chars, and I want to read 4 bytes as a 32bit int, but I get a compiler error. I know that I can bit shift, like this:

我已经分配了一个字符数组，我想把4个字节读为32位int，但是我得到了一个编译器错误。我知道我可以改变，像这样:

(start[0] << 24) + (start[1] << 16) + (start[2] << 8) + start[3];

And it will do the same thing, but this is a lot of extra work.

它会做同样的事情，但这是很多额外的工作。

Is it possible to just cast those four bytes as an int somehow?

是否可以将这四个字节转换成int类型?

4 个解决方案

#1

static_cast is meant to be used for "well-behaved" casts, such as double -> int. You must use reinterpret_cast:

static_cast应该用于“行为良好的”类型转换，比如double -> int。

uint32_t *starti = reinterpret_cast<uint32_t*>(&memory[164]);

Or, if you are up to it, C-style casts:

或者，如果你能胜任，c风格的演员:

uint32_t *starti = (uint32_t*)&memory[164];

#2

Yes, you can convert an unsigned char* pointer value to uint32_t* (using either a C-style cast or a reinterpret_cast) -- but that doesn't mean you can necessarily use the result.

是的，您可以将未签名的char*指针值转换为uint32_t*(使用c样式的转换或重新解释的cast)——但这并不意味着您一定可以使用结果。

The result of such a conversion might not point to an address that's properly aligned to hold a uint32_t object. For example, an unsigned char* might point to an odd address; if uint32_t requires even alignment, you'll have undefined behavior when you try to dereference the result.

这样一个转换的结果可能不会指向一个正确对齐的地址以持有uint32_t对象。例如，一个未签名的char*可能指向一个奇数地址;如果uint32_t需要对齐，那么当您试图取消结果时，就会有未定义的行为。

If you can guarantee somehow that the unsigned char* does point to a properly aligned address, you should be ok.

如果您可以保证没有签名的char*确实指向一个正确对齐的地址，那么应该可以。

#3

I am used to BDS2006 C++ but anyway this should work fine on other compilers too

我习惯了BDS2006 c++，但不管怎样，这也适用于其他编译器。

char memory[164];
int *p0,*p1,*p2;
p0=((int*)((void*)(memory)));    // p0 starts from start
p1=((int*)((void*)(memory+64))); // p1 starts from 64th char
p2=((int*)((void*)(&memory[64]))); // p2 starts from 64th char

#4

You can use reinterpret_cast as suggested by faranwath but please understand the risk of going that route.

您可以按照faranwath的建议使用reviewt_cast，但是请理解走这条路线的风险。

The value of what you get back will be radically different in a little endian system vs a big endian system. Your method will work in both cases.

在一个小的endian系统和一个大的endian系统中，你得到的东西的价值将会完全不同。你的方法在这两种情况下都适用。

#1