jpo38 - 1 year ago 50

C++ Question

I'm writting a function that will convert a bitset to a int/uint value considering that the bitset could have fewer bits than target type.

Here is the function I wrote:

`template <typename T,size_t count> static T convertBitSetToNumber( const std::bitset<count>& bitset )`

{

T result;

#define targetSize (sizeof( T )*CHAR_BIT)

if ( targetSize > count )

{

// if bitset is 0xF00, converting it as 0x0F00 will lose sign information (0xF00 is negative, while 0x0F00 is positive)

// This is because sign bit is on the left.

// then, we need to add a zero (4bits) on the right and then convert 0xF000, later, we will divide by 16 (2^4) to preserve sign and value

size_t missingbits = targetSize - count;

std::bitset<targetSize> extended;

extended.reset(); // set all to 0

for ( size_t i = 0; i != count; ++i )

{

if ( i < count )

extended[i+missingbits] = bitset[i];

}

result = static_cast<T>( extended.to_ullong() );

result = result >> missingbits;

return result;

}

else

{

return static_cast<T>( bitset.to_ullong() );

}

}

And the "test program":

`uint16_t val1 = Base::BitsetUtl::convertBitSetToNumber<uint16_t,12>( std::bitset<12>( "100010011010" ) );`

// val1 is 0x089A

int16_t val2 = Base::BitsetUtl::convertBitSetToNumber<int16_t,12>( std::bitset<12>( "100010011010" ) );

// val2 is 0xF89A

See the program works when using

`uint16`

`uint16_t val = 0x89A0; // 1000100110100000`

val = val >> 4; // 0000100010011010

However, it fails when using

`int16_t`

`0x89A0 >> 4`

`0xF89A`

`0x089A`

`int16_t val = 0x89A0; // 1000100110100000`

val = val >> 4; // 1111100010011010

I don't understand why >> operator sometimes insert 0 and sometimes 1. And I can't find out how to safely do the final operation of my function (

`result = result >> missingbits;`

Answer Source

It's because shifting is an arithmetic operation, and that *promotes* the operands to `int`

, which will do sign extension.

I.e. promoting the signed 16-bit integer (`int16_t`

) `0x89a0`

to a 32-bit signed integer (`int`

) causes the value to become `0xffff89a0`

, which is the value that is shifted.

See e.g. this arithmetic operation conversion reference for more information.

You should cast the variable (or value) to an unsigned integer (i.e. `uint16_t`

in your case):

```
val = static_cast<uint16_t>(val) >> 4;
```

If the type is not really know, like if it's a template argument, then you can use `std::make_unsigned`

:

```
val = static_cast<typename std::make_unsigned<T>::type>(val) >> 4;
```