错误:二进制操作数无效^(有'double'和'double') - 我不是试图用^作为幂运算符

Question

我一直在寻找答案,但找不到.我不是试图使用^作为幂运算符.

这是我的问题.我正在尝试从数字电视标准(物理层协议)重现交织器.这是一个学校的项目.这是我到目前为止在C中的代码:

#include <sys/io.h>
#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <time.h>
#include <unistd.h>
#include <stdbool.h>
#include <string.h>
#include <complex.h>
#include <math.h>

int main()
{
    int Mmax = 32768; 
    int Nr = log10(Mmax)/log10(2);
    printf("Nr = %i \n",Nr);

    int Lfm = 42; 

    int i,j,k,l; 

    // Create array R'i 
    unsigned char Rli[Mmax][Nr];
    for (i = 0; i < Mmax; i++)
        memset(Rli[i], 0, sizeof(unsigned char)*(Nr));

    for (i = 0; i < Mmax; i++)
    {
        for(j = 1; j < Nr; j++)
        {
            if(i == 0 || i == 1)
            {
                Rli[i][j] = 0;
            }

            else if(i == 2)
            { 
                Rli[i][j] = 0;
                if(j == Nr - 2)
                    Rli[i][j] = 1;
            }

            else
            {
                Rli[i][j+1] = Rli[i-1][j]; 
            }
        }

        if(i > 2)
        {
            Rli[i][1] = (((Rli[i-1][14] ^ Rli[i-1][13] ) ^ Rli[i-1][12] ) ^ Rli[i-1][2]);
        }
        Rli[i][0] = 0;
    }

    // wire permutation for array R'i - creates array Ri

    unsigned char Ri[Mmax][Nr];
    for (i = 0; i < Mmax; i++)
        memset(Ri[i], 0, sizeof(unsigned char)*(Nr));   

    for (i = 0; i < Mmax; i++)
    {
        Ri[i][1] = Rli[i][8]; //Rli[i][7];
        Ri[i][2] = Rli[i][9]; //Rli[i][8];
        Ri[i][3] = Rli[i][14]; //Rli[i][13];
        Ri[i][4] = Rli[i][4]; //Rli[i][3];
        Ri[i][5] = Rli[i][6]; //Rli[i][5];
        Ri[i][6] = Rli[i][13]; //Rli[i][12];
        Ri[i][7] = Rli[i][3]; //Rli[i][2];
        Ri[i][8] = Rli[i][2]; //Rli[i][1];
        Ri[i][9] = Rli[i][12]; //Rli[i][11];
        Ri[i][10] = Rli[i][5]; //Rli[i][4];
        Ri[i][11] = Rli[i][10]; //Rli[i][9];
        Ri[i][12] = Rli[i][11]; //Rli[i][10];
        Ri[i][13] = Rli[i][1]; //Rli[i][0];
        Ri[i][14] = Rli[i][7]; //Rli[i][6];
    }

    // Offset generator Gk

    int Gmax = floor(Lfm/2);
    unsigned char Gk[Gmax][Nr];
    for (i = 0; i < Gmax; i++)
        memset(Gk[i], 0, sizeof(unsigned char)*Nr-1);

    for (k = 0; k < Gmax; k++)
    {
        for(j = 0; j < Nr; j++)
        {
            if(k == 0)
            { 
                Gk[k][j] = 1; 
            }

            else
            { 
                Gk[k][j+1] = Gk[k-1][j];
            }
        } 

        if(k > 0)
        { 
            Gk[k][0] = ( Gk[k-1][14] ^ Gk[k-1][13] );
        }

    }

    // interleaving sequence

    int Ndata = 26303; // numero de data 
    unsigned char Hl[Ndata];
    double H1[Ndata], H2[Ndata];
    memset(Hl, 0, sizeof(unsigned char)*Ndata);
    memset(H1, 0, sizeof(double)*Ndata);
    memset(H2, 0, sizeof(double)*Ndata);

    int p,indice;

    // loop from page 107 of the physical layer protocol.
    for (i = 0; i < Lfm; i++) 
    {
        for (k = 0; k < Mmax; k++) 
        {
            p = 0;
            for (j = 0; j < Nr; j++) // sum
            {
                if(j >= 0 && j <= Nr - 2)
                    H1[p] += Ri[k][j]*pow(2,j);
                else if(j >= 0 && j <= Nr - 1)
                    {
                    indice = floor(i/2);
                    H2[p] += Gk[indice][j]*pow(2,j);
                    }
            }
    /*****/ Hl[p] = ((i % 2)*pow(2,Nr-1) + H1[p] ) ^ H2[p]; /*****/
            if (Hl[p] < Ndata)
                p += 1;
        }
    }

当我尝试编译时,/*****/之间的行最后会产生错误.

freqint.c:146:43: error: invalid operands to binary ^ (have ‘double’ and ‘double’)

        Hl[p] = ((i % 2)*pow(2,Nr-1) + H1[p] ) ^ H2[p];

我想在那里进行XOR操作,但我无法做到正确.

我从这里拿走这个.

交错序列位于第107页的底部

如何编写代码以避免出现错误消息并仍然执行异或操作？

Answer 1

正如评论中所述,^操作员不能在a上使用double,但您不需要使用double.

你正在使用两个返回a的函数double,即pow和floor.

每次打电话pow,你都会通过2基地.通过将值移位1指数,可以更有效地提高2到幂.

同样,每次你打电话给floor你时,要将整数值除以2.由于整数除法会自动降低除法的余数(假设您只使用正值),调用floor不会给你带来任何东西.

因此,更改pow(2,x)to的所有实例(1 << x),并更改floor(x/2)to的所有实例(x/2).然后你可以声明H1和H2作为数组,unsigned int你将能够使用XOR运算符^.