How to convert std::string representation of 64-bit address to uint64_t?

c++

Solution

#include <sstream>
#include <string>
#include <iostream>
#include <iomanip>
#include <cstdint>

int main()
{
    std::string s("0x12345");
    std::stringstream strm(s);
    std::uint64_t n;
    strm >> std::hex >> n;
    std::cout << std::hex << n << std::endl;
    return 0;
}

This prints `12345`, as expected.

Edit: If you want to convert from little-endian to big-endian, that's possible too:

#include <sstream>
#include <string>
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstdint>

int main()
{
    std::string s("0x12345");
    std::stringstream strm(s);

    union {
        std::uint64_t n;
        std::uint8_t a[8];
    } u;

    strm >> std::hex >> u.n;
    std::reverse(u.a, u.a + 8);

    std::cout << std::hex << std::setfill('0') << std::setw(16) << u.n << std::endl;
    return 0;
}

Problem

I have a `std::string` representing a 64-bit memory address in little-endian, hexadecimal form. How to convert this to a `uint64_t` representation?

Original source

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