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c++cturbo-c

binary representation in turbo c

I have a very basic query. For the following code:

int i = -1;
unsigned int j = (unsigned int) i;
printf("%d",j); // I get 65535 the int limit in turbo C 16 bit compiler

Shouldn't it simply drop the negative sign or does it wrap around the limits ?

Solution

  • The scheme most commonly used to represent negative numbers is called Two's Complement. In this scheme -1 has the bit pattern of all 1 bits: 1111 1111 1111 1111. When interpreted as an unsigned integer, which is what your cast means, that's 65535.

    In other words, it does "simply drop the negative sign", except that doing so has a different effect than what you expect.