char* to double and back to char* again ( 64 bit application)

I am trying to convert a char* to double and back to char* again. the following code works fine if the application you created is 32-bit but doesn't work for 64-bit application. The problem occurs when you try to convert back to char* from int. f开发者_如何学运维or example if the hello = 0x000000013fcf7888 then converted is 0x000000003fcf7888 only the last 32 bits are right.

#include <iostream>
#include <stdlib.h>
#include <tchar.h>
using namespace std;


int _tmain(int argc, _TCHAR* argv[]){

    char* hello = "hello";
    unsigned int hello_to_int = (unsigned int)hello;
    double hello_to_double = (double)hello_to_int;

    cout<<hello<<endl;
    cout<<hello_to_int<<"\n"<<hello_to_double<<endl;

    unsigned int converted_int = (unsigned int)hello_to_double;
    char* converted = reinterpret_cast<char*>(converted_int);

    cout<<converted_int<<"\n"<<converted<<endl;

    getchar();
    return 0;
}

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On 64-bit Windows pointers are 64-bit while int is 32-bit. This is why you're losing data in the upper 32-bits while casting. Instead of int use long long to hold the intermediate result.

char* hello = "hello";
unsigned long long hello_to_int = (unsigned long long)hello;

Make similar changes for the reverse conversion. But this is not guaranteed to make the conversions function correctly because a double can easily represent the entire 32-bit integer range without loss of precision but the same is not true for a 64-bit integer.

Also, this isn't going to work

unsigned int converted_int = (unsigned int)hello_to_double;

That conversion will simply truncate anything digits after the decimal point in the floating point representation. The problem exists even if you change the data type to unsigned long long. You'll need to reinterpret_cast<unsigned long long> to make it work.

Even after all that you may still run into trouble depending on the value of the pointer. The conversion to double may cause the value to be a signalling NaN for instance, in which cause your code might throw an exception.

Simple answer is, unless you're trying this out for fun, don't do conversions like these.

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You can't cast a char* to int on 64-bit Windows because an int is 32 bits, while a char* is 64 bits because it's a pointer. Since a double is always 64 bits, you might be able to get away with casting between a double and char*.

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A couple of issues with encoding any integer (specifically, a collection of bits) into a floating point value:

1. Conversions from 64-bit integers to doubles can be lossy. A double has 53-bits of actual precision, so integers above 2^52 (give or take an extra 2) will not necessarily be represented precisely.
2. If you decide to reinterpret the bits of a pointer as a double instead (via union or reinterpret_cast) you will still have issues if you happen to encode a pointer as set of bits that are not a valid double representation. Unless you can guarantee that the double value never gets written back by the FPU, the FPU can silently transform an invalid double into another invalid double (see NaN), i.e., a double value that represents the same value but has different bits. (See this for issues related to using floating point formats as bits.)

You can probably safely get away with encoding a 32-bit pointer in a double, as that will definitely fit within the 53-bit precision range.

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only the last 32 bits are right.

That's because an int in your platform is only 32 bits long. Note that reinterpret_cast only guarantees that you can convert a pointer to an int of sufficient size (not your case), and back.

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If it works in any system, anywhere, just all yourself lucky and move on. Converting a pointer to an integer is one thing (as long as the integer is large enough, you can get away with it), but a double is a floating point number - what you are doing simply doesn't make any sense, because a double is NOT necessarily capable of representing any random number. A double has range and precision limitations, and limits on how it represents things. It can represent numbers across a wide range of values, but it can't represent EVERY number in that range.

Remember that a double has two components: the mantissa and the exponent. Together, these allow you to represent either very big or very small numbers, but the mantissa has limited number of bits. If you run out of bits in the mantissa, you're going to lose some bits in the number you are trying to represent.

Apparently you got away with it under certain circumstances, but you're asking it to do something it wasn't made for, and for which it is manifestly inappropriate.

Just don't do that - it's not supposed to work.

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This is as expected.

Typically a char* is going to be 32 bits on a 32-bit system, 64 bits on a 64-bit system; double is typically 64 bits on both systems. (These sizes are typical, and probably correct for Windows; the language permits a lot more variations.)

Conversion from a pointer to a floating-point type is, as far as I know, undefined. That doesn't just mean that the result of the conversion is undefined; the behavior of a program that attempts to perform such a conversion is undefined. If you're lucky, the program will crash or fail to compile.

But you're converting from a pointer to an integer (which is permitted, but implementation-defined) and then from an integer to a double (which is permitted and meaningful for meaningful numeric values -- but converted pointer values are not numerically meaningful). You're losing information because not all of the 64 bits of a double are used to represent the magnitude of the number; typically 11 or so bits are used to represent the exponent.

What you're doing quite simply makes no sense.

What exactly are you trying to accomplish? Whatever it is, there's surely a better way to do it.