CODEWITHSUNDEEP
c++castingtype-conversion
Alex Brooks
Alex Brooks

Reputation: 1151

Data loss between conversion

Why is it that I loose data between the conversions below even though both types take up the same amount of space? If the conversion was done bitwise, it should be true that x = z unless data is being stripped during the conversion, right? Is there a way to do the two conversions without losing data (i.e. so that x = z)?

main.cpp:

#include <stdio.h>
#include <stdint.h>

int main() {
    double   x = 5.5;
    uint64_t y = static_cast<uint64_t>(x);
    double   z = static_cast<double>(y) // Desire : z = 5.5;

    printf("Size of double: %lu\nSize of uint64_t: %lu\n", sizeof(double), sizeof(uint64_t));
    printf("%f\n%lu\n%f\n", x, y, z);
}

Results:

Size of double: 8
Size of uint64_t: 8
5.500000
5
5.000000

Upvotes: 3

Views: 2418

Answers (1)

Reed Copsey
Reed Copsey

Reputation: 564441

The conversion is not bitwise.

The first conversion converts the value to an unsigned integer:

uint64_t y = static_cast<uint64_t>(x);  // y == 5

The second takes that integer, and converts it to a double of the same value

double   z = static_cast<double>(y) // Convert 5 to 5.0

The fact that the types use the same amount of memory is irrelevant, as the static_cast changes the type. You would see the same behavior if you converted to then from an uint32_t, as well. It does not merely perform a bitwise conversion.

If you want to perform a bitwise conversion, you could do this via pointer manipulation:

double   x = 5.5;
uint64_t y = *((uint64_t*)&x);  // Will effectively be a "garbage" value
double   z = *((double*)(&y)); // Will be 5.5

Upvotes: 10

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