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copied!<pre><code> if(expA==expB) { if(debug) { printf("We got into the if exponents are the same block \n"); printf("Exp a: %x, Exp b: %x \n",expA,expB); } // exponents are the same // Add Mantissa B to A mantA = ADD(mantB,mantA); if(debug) { printf("Addition of mantissa's %x\n",mantA); } // store into the return value the logical and of the mantissa with the existing exponent and sign // might want to do an OR() not an AND() returnValue = OR(signExpA,mantA); } // end if the eponents are the same </code></pre> <p>This is wrong.</p> <p>You're failing to account for the addition of the two implicit ones. When you're adding 2 + 3 you're adding 1.0 x 2^1 + 1.1 x 2^1 and you're ignoring everything before the decimal point... So you end up with 0.0 + 0.1 = 0.1 and just stick a 1 at the front. You need to add the two implicit ones as well.</p> <p>Try something like this:</p> <pre><code> if(expA==expB) { if(debug) { printf("We got into the if exponents are the same block \n"); printf("Exp a: %x, Exp b: %x \n",expA,expB); } // exponents are the same // Add Mantissa B to A mantA = OR(implicitOne,mantA); mantB = OR(implicitOne,mantB); mantA = ADD(mantB,mantA); // You need to normalize this now. But shifting to the right by 1 will suffice. mantA >>= 1; ++expA; // ... add the sign and you're done... if(debug) { printf("Addition of mantissa's %x\n",mantA); } // store into the return value the logical and of the mantissa with the existing exponent and sign // might want to do an OR() not an AND() returnValue = OR(signExpA,mantA); } // end if the eponents are the same </code></pre>

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