Function that checks for overflow in addition of unsigned ints - c

Question

I have to make functions that check for overflow in integer addition, subtraction, and unsigned int addition(using only ! ~ | & ^ + >> <<). I have functions figured out for signed integer addition and subtraction, but I can't figure out how to do one for unsigned int addition.
How would I go about doing this?
Here is the code I have for the 2 functions I have completed:
int twosAddOk(int x, int y){
int z=x+y;
int a=x>>31;
int b=y>>31;
int c=z>>31;
return !!(a^b)|(!(a^c)&!(b^c));
}
int twosSubtractOK(int x, int y){
int z=x+~y+1;
return !(((x^y & x^z))>>31);
}

Answer 1

You can calculate the carry-out from the MSB the hard way:
int unsignedAddOk(unsigned int x, unsigned int y){
unsigned int x0=(~(1U<<31))&x; // MSB of x cleared
unsigned int y0=(~(1U<<31))&y; // MSB of y cleared
int c=(x0+y0)>>31; // Carry-in of MSB
int a=x>>31; // MSB of x
int b=y>>31; // MSB of y
return !((a&b)|(a&c)|(b&c));
}

Answer 2

Perhaps a solution without coding the magic number 31
// Return 1 on overflow
int unsigned_add_overflow_test(unsigned a, unsigned b) {
// Add all but the LSBits and then add 1 if both LSBits are 1
// When overflow would occur with a + b, sum's MSBit is set.
unsigned sum = (a >> 1) + (b >> 1) + (a&b&1);
// Test MSBit set
// vvv--------- All bits set
return !!(sum & ~(-1u >> 1));
// ^^^^^^^^^^ -- All bits set, except MSBit
// ^^^^^^^^^^^ -- MSBit set, rest are 0
}
Or as a one-liner
!!( ((a >> 1) + (b >> 1) + (a&b&1)) & ~(-1u >> 1)) )