如何将 integer 转换为常量 void 指针？

Question

int getNumber(){
   int number;
   cin >> number;
   return number;
}

const void * getPointer(){
   const void *p = (const void *)getNumber();
   return p;
}

Gives me an error:

warning: cast to pointer from integer of different size

and the console crashes

Answer 1

How to cast an integer to a constant void pointer?

Start with an integer type that is wide enough to represent all pointer values. std::intptr_t is such type. If you want to convert from a narrower type, then you can convert it to a wider type first. Then use reinterpret_cast.

and the console crashes

The shown program is not the cause of that crash.

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Note that converting an integer to a pointer is an advanced technique that is rarely needed. If you are a beginner, then you likely are facing the XY-problem .

Answer 2

getNumber() returns a copy. You cast the copy to a pointer. (There is a difference between the copy and the address of the copy, which is what the pointer should be)

int nb = getNumber();
const void *cp_void = (const void *)nb; // this will not yield the expected result.
int *p_nb = &nb; // memory adress to variable .. but variable will go out of scope

As the comments suggest just dynamically allocating the int is not the answer and there is a very likely a design flaw (in the code not shown).

The two snippets shown getNumber() and getPointer() are not enough for us to comment or improve on the rest of the program.