number in hexadecimal

[ZimCa]

I need help i wont to try something like this

int a = 5;

string hex = a << hex;

[mbozzi]

std::cout << std::hex << my_number << "\n";

[ZimCa]

Yes but i wont to do this

int a = 5;

address = a in hexidecimal ?

ReadProcessMemory(hProc, address, &nVal, (DWORD)sizeof(nVal), NULL);

cout << nVal << endl;

[mbozzi]

A number's value does not change when it is represented using a different base.

Edit: please use code tags.

[ZimCa]

yes but how to int addres = a in hexidecimal ?

[ZimCa]

for(NOS;NOS <= trazido;NOS+=4)
    {
        ofstream provera("SveProvereneAdrese\\ML1.txt");
        provera << hex  << NOS << endl;
        provera.close();
        fstream MemoryAddress("SveProvereneAdrese\\ML1.txt");
        LPVOID linee;
        MemoryAddress >> linee;
        cout << linee << endl;
        ReadValue(linee,app.c_str());
        MemoryAddress.close();
    }

here is my code
do you see i use ofstream and fstream to convert int nos value in hexedecimal

[mbozzi]

int address = a
Because 5 in decimal is the same value as 5 in hexadecimal, which is the same value as 0000 0101 in binary, which is the same value as 12 in ternary, and so on.

You don't need to care how the number is represented. The number is the same value no matter how you choose to represent it.

[ZimCa]

here is my problem i trying this

int a = 10 // 10 in hexadecimal = c
how to do
this LPVOID addres = a in hexadecimal

[mbozzi]

Use code tags, please.

Since `a' is an integer, you need to type-cast it to a void pointer.

Number bases have nothing to do with it.

LPVOID address = reinterpret_cast <void*> (a);

[ZimCa]

Man its WORKS :DDDDDDDDDDD
ty <3