C++矩阵中的正弦函数

Question

C++矩阵中的正弦函数

3

我该如何用0和1在C++矩阵中表示正弦函数的图形？

char sinM[DN][DM];
memset(sinM,32, sizeof(sinM));

for(int i=1;i<DN;++i) {

    double val = sin(i*M_PI/180);
    int valY = int(val*DM);
    sinM[i][valY] = '1';
}

这似乎不太正常。

- Alessandro

5

处理 sin 函数的负返回值。 - Nico Schertler

为什么要将 sinM 的内容的1/4设置为32？这会断开矩阵的前1/4行。 - user3235832

DN和DM的值是多少？M_PI至少是双精度吗？你想如何处理负数值？你能提供一个期望输出的例子吗？ - Bob__

1个回答

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- Bob__ · Accepted Answer

如果您正在寻找类似于这样的内容：

0000000000000000000000000000000000000000000000000000000000000000
0000000000000111111100000000000000000000000000000000000000000000
0000000000111000000011100000000000000000000000000000000000000000
0000000011000000000000011000000000000000000000000000000000000000
0000001100000000000000000110000000000000000000000000000000000000
0000110000000000000000000001100000000000000000000000000000000000
0001000000000000000000000000010000000000000000000000000000000000
0110000000000000000000000000001100000000000000000000000000000000
1000000000000000000000000000000010000000000000000000000000000000
0000000000000000000000000000000001100000000000000000000000000011
0000000000000000000000000000000000010000000000000000000000000100
0000000000000000000000000000000000001100000000000000000000011000
0000000000000000000000000000000000000011000000000000000001100000
0000000000000000000000000000000000000000110000000000000110000000
0000000000000000000000000000000000000000001110000000111000000000
0000000000000000000000000000000000000000000001111111000000000000

那么你可以使用这个：

#include <iostream>
#include <cmath>

const size_t HEIGHT=16;
const size_t LENGTH=64;
const double pi = std::acos(-1);

int main() {
    char board[HEIGHT][LENGTH];
    double x_scale = 2.0 * pi / LENGTH;
    size_t y_scale = (HEIGHT - 1) / 2;
    size_t y;
                                        // initialize 
    for ( size_t i=0; i<HEIGHT; i++)
        for (size_t j=0; j<LENGTH; j++) board[i][j]='0';
                                        // "draw" the function in the matrix
    for ( size_t x=0; x<LENGTH; x++) {
        y = (size_t)(std::round(y_scale + 1 - std::sin(x*x_scale) * y_scale));
        board[y][x] = '1';
    }
                                        // print the matrix
    for ( size_t i=0; i<HEIGHT; i++) {
        for (size_t j=0; j<LENGTH; j++) {
            std::cout << board[i][j];
        }
        std::cout << std::endl;
    }
    return 0;
}