C ++，用于检查软件是否在32位或64位系统上运行的函数

Question

As I explained in many questions, I'm trying to move a software from a 32-bit system to a 64-bit system. I had some problem with malloc() function , but now I solved it by correcting a parameter.

In that part of my code, if I run on a 32-bit system, I can use:

(int**) malloc (const * sizeof(int))

But, on a 64-bit system, I have to use:

(int**) malloc (const * sizeof(int64_t))

I'd like to manage these crossroads with an if() condition, so I need a boolean isIt64system() function that behaves this way:

if(isIt64system()) then [64-bit code]

else [32-bit code]

Does this function exist in C++? Is there any function that tells me if software's running on a 32-bit system or 64-bit system?

Answer 1

Rather than writing two size-dependent branches, just write one correct, portable code path. In your case:

(int**)malloc(count*sizeof(int*));

This will work correctly regardless of the sizeof of int* on your system.

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Postscript: As you can see from this literal answer to your question, you are better off not having an if:

 if(sizeof(int*) == sizeof(int)) x = (int**)malloc(count*sizeof(int)); else if (sizeof(int*) == sizeof(int64_t)) x = (int**)malloc(count*sizeof(int64_t)); 

Hopefully you can see how absurdly redundant that code is, and how it should be replaced by a single well-constructed malloc call.

Answer 2

您的编译器将具有预处理器定义，这些定义使您可以检查32位和64位。

Answer 3

The best would be to use something like this

#ifdef __LP64__
    <64bit code>
#else
    <32bit code>
#endif

But if you really need a function for it then this should work.

bool is64bit() {
    if (sizeof(int*) == 4) {
        return false;
    }
    return true;
}