I am implementing a 32 bit CLA Adder like how a 16 bit adder is implemented in Wikipedia

Problem is how do I determine if the block overflows? I will need the carry into bit 32 (which is now in the last 4 bit CLA Adder) and XOR with carry out of bit 32(Check difference in MSB)?


Is my overflow logic correct

where Cin is the carry into the whole 16 bit block, P* is the Block Propagate, G* the block generate, and carry into bit 32 (typo, bit 16 MSB actually)


3 Answers 3


I thought about seeing what my brain could regurgitate of long ago explnatins, as opposed to commin sense observations, BUT here's a page which does it all very well indeed.

They explain a "nice" 'trick'.

  • Two's compelment overflow detection can be achieved by XORing the carry-in and the carry-out bo=its of the leftmost full adder.

Their diagram. See above ref for detailed comment:

enter image description here

  • 2
    \$\begingroup\$ Thats a 1 bit adder, the problem with using a >1 bit adder is that the carry into the MSB is hidden somewhere in the adder itself. Hope you got me? I could have an output for my block adder for Cin to MSB but that seems off place? Alternatively, I could have evey block output an Overflow bit, is this the usual implementation? It will give quite abit redundant logic \$\endgroup\$
    – Jiew Meng
    Commented Oct 23, 2011 at 13:22
  • \$\begingroup\$ The last stage output is the XOR of the two inputs and the carry in; thus the carry in is the xor of the two inputs and the output. \$\endgroup\$
    – supercat
    Commented Sep 15, 2015 at 16:46

If you are designing the CLA (carry lookahead adder) block, you could have the block output the carry from the bit 2 digit (which must already be calculated in order to form the correct value for bit 3 of the sum).

If the CLA block interface is a given, you can derive overflow from the sign bits of the two adder inputs, A(31) and B(31), and the sign bit of the output sum. If A and B are the same sign, the sum sign must match them iff there is no overflow. For more detail, see this page or this page.


You can retrace C(n-1) from P, G and S

If you use: X = XOR( P, G)

or you can also use XOR( !P, !G); P=Propagate, G=Generate

Ci(n-1) = XOR( X, S) ; S = sum result of highest bit

Now you have Ci(n-1) and Ci(n)

V = XOR ( Ci(n-1), Ci(n) ) ; see other theories above

G= P= X= Ci= Ci
AxB =Y xCi =S A&B A+B GxP XxS check
00 0 0 0 0 0 0 0 ok
11 0 1 1 1 1 0 1 ok
01 1 0 1 0 1 1 0 ok
10 1 1 0 0 1 1 1 ok

x = XOR; + = OR; & = AND; ! = NOT enter image description here


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