Question

我有这个功能，一切都有效，直到a和b都等于1.当发生这种情况时，当它应该只有0或1时，进位位将变为2.我的算法对于进位位是错误的 carry_bits | =（（a＆amp; mask）＆amp;（b＆amp; mask））＆lt;＆lt; 1U;

bool add7bits( unsigned int a, unsigned int b, unsigned int *carry_out, unsigned int     *overflow_out, unsigned int *sum_out)
{
if ( a > ( ( 1u << NUM_BITS ) - 1 ) )
{
    cout << "a value: " << a << " is too large" << endl;
    return false;
}
else if ( b > ( ( 1u << NUM_BITS ) - 1 ) )
{
    cout << "b value: " << b << " is too large" << endl;
    return false;
}
else
{
    unsigned int sum_bits = 0;
    unsigned int carry_bits = 0;
    // Use a mask to access the specific bits of interest
    unsigned int mask = 1u;

    // Handle rightmost bit as a half-adder (no carry input)
    // sum = a ^ b
    // c_out = a & b
    sum_bits |= ( a & mask) ^ ( b & mask);

    // The carry _out_ from this stage sets the carry in for the next,
    // that is, the next higher bit in the carry_bits value

    carry_bits |= ((a & mask) & (b & mask)) << 1u;

    // The remaining bits must be handled with the full adder logic. The last
    // adder in the chain's carry out becomes the carry output return
    // value of this function.
            mask = 1u << 1;

            sum_bits |= ((a & mask) ^ (b & mask) ^ carry_bits);
            carry_bits |= ((a & mask) & (b & mask)) << 1u; 

            mask = 1u << 2;
            sum_bits |= ((a & mask) ^ (b & mask) ^ carry_bits);
            carry_bits |= ((a & mask) & (b & mask)) << 1u;

            mask = 1u << 3;
            sum_bits |= ((a & mask) ^ (b & mask) ^ carry_bits);
            carry_bits |= ((a & mask) & (b & mask)) << 1u;

            mask = 1u << 4;
            sum_bits |= ((a & mask) ^ (b & mask) ^ carry_bits);
            carry_bits |= ((a & mask) & (b & mask)) << 1u;          

            mask = 1u << 5;
            sum_bits |= ((a & mask) ^ (b & mask) ^ carry_bits);
            carry_bits |= ((a & mask) & (b & mask)) << 1u;  
    // Handle bit 6 separately.
    mask = 1u << 6;
    sum_bits = ((a ^b) ^ carry_bits);
    carry_bits |= ((a & mask) & (b & mask)) << 1u;


// Determine the overflow by checking if a and b are both 1.
    bool overflow = false;
    if ((a & mask) & (b & mask))
        overflow = true;
    // ...
    //
 *sum_out = sum_bits;
 *carry_out = carry_bits;
 *overflow_out = overflow;
 cout << a << " + " << b << " = " << *sum_out << endl;;
 cout << "Carry: " << *carry_out << endl;
 cout << "Overflow: " << *overflow_out << endl;
    return true;
}
}

Answer 1

进位为1，它位于第二位，因此carry_bits为...0010或2。

Answer 2

您将所有先前的进位位置入每级的和位，而不仅仅是当前级的进位位。你想要

sum_bits |= ((a & mask) ^ (b & mask) ^ (carry_bits & mask));

在每个阶段，或等同，更简单：

sum_bits |= (a ^ b ^ carry_bits) & mask;

您的执行计算不正确 - 它不考虑进位。你想要：

carry_bits |= ((a & b | a & carry_bits | b & carry_bits) & mask) << 1;

您的最终执行位是进位位的第8位，因此您需要：

*carry_out = carry_bits >> NUM_BITS;

只写那一点。

“溢出”在这里没有多大意义，因为你似乎在进行无符号加法（因此“溢出”到它存在的程度，与执行相同）。如果你正在进行2s-complement signed add，你需要最后两个进位的xor：

*overflow_out = ((carry_out >> NUM_BITS) ^ (carry_out >> (NUM_BITS-1))) & 1;

在我的7位加法器中，进位是不正确的，我不知道为什么

2 个答案: