CODEWITHSUNDEEP
c++
notaorb
notaorb

Reputation: 2190

Conversion from signed int to signed char?

I'm following an example from Stroustrups C_++ 4th Ed. Page 143, where there is Errata. -160, should be -140.

Specifically the conversion of a signed int to a signed char.

#include <iostream>
using namespace std;

int main(int argc, char *argv[])
{  
    signed char sc = -140;
    cout << "signed int = " <<
    int((signed int) (0b11111111111111111111111101110100))
          << endl; // -140, 4 bytes
    cout << "signed int & 0xFF = " << int((signed int) -140 & 0xFF)
          << endl; // 116, narrow 2 bytes    
    unsigned char uc = sc; // 256 - 140 = 116
    return 0;
}

I understand how the conversion is a narrowing of 4 bytes, to 2 bytes which result in -140 being converted to 116. I am confused on the comment Stroustrup adds for the line unsigned char uc = sc where it is // 256 - 140 = 116. I see it results in the correct answer 116, but am unsure how that conversion is done. I'm aware that an unsigned char max limit completely full of 1's is 255 or 2^8 - 1. Does anyone know why this math in the comment works?

UPDATE: The solution is -140 mod 256 = 116.

Thanks

Upvotes: 1

Views: 1420

Answers (2)

Mark Ransom
Mark Ransom

Reputation: 308520

Two byte int hasn't been common for a long time now; most int are 4 bytes.

Regardless, it's the top bit that matters for a signed number. The expression -140 & 0xff results in 0b01110100 which is 116, because all those sign bits get chopped off.

To make it even simpler, let's look at the full bit patterns.

-140 = 0b11111111111111111111111101110100 (first sign bit is set)
0xff = 0b00000000000000000000000011111111 (bottom 8 bits set, all others zero)
&    : 0b00000000000000000000000001110100 = 116

Upvotes: 1

HolyBlackCat
HolyBlackCat

Reputation: 96886

When converting a value to an unsigned type, the value is taken modulo N,
where N = 2sizeof(T)*CHAR_BIT.

In simple terms, it means that N*i is added to the value to put it in the representable range, where i is an integer (possibly negative).

You can always determine a single specific value of i that puts the value into the representable range. In your case, i = 1, so, 256 * 1 is added to the value.

Upvotes: 2

Related Questions