有人可以帮我解释这个算法吗？ C ++

i want to know how this code is working,

我想知道这段代码是如何工作的,

i can't understand the algorithm here

我无法理解这里的算法

#include <stdio.h>

void Draw(int length) {

int cons = 3,L,l;
length -= length % 2;
L = length + cons;

for (int i = 0; i < L; i++){

    l = cons + i * 2 - 2 * (i == L / 2) - 4 *(i - L / 2) * (i > L / 2);
    for (int j = 0; j < (L-l) / 2; j++) {
        printf(" ");
    }
    for (int j = 0; j < l; j++) {
        if (j == l/2)
            printf("|");
        else if (i == L/2)
            printf("=");
        else if (j == 0 && i < L/2 || (j == l - 1 && i > L / 2))
            printf("/");
        else if (j == l - 1 && i < L / 2 || (j == 0 && i > L / 2))
            printf("\\");
        else
            printf("*");
    }
    printf("\n");
}
}

int main(int argc, char **argv) {

Draw(5);
printf("\n");
Draw(6);
return 0;
}

I want clear clarification to this particular line, please :

我想清楚澄清这一特定的一行,请:

l = cons + i * 2 - 2 * (i == L / 2) - 4 *(i - L / 2) * (i > L / 2);

this code should print this shape

此代码应打印此形状

click to see the image

点击查看图片

or you can test the code here https://ideone.com/49mtT4

或者你可以在这里测试代码https://ideone.com/49mtT4

and i apologize for any inconvenience

对于给您带来的任何不便,我深表歉意

1 个解决方案

#1

since you you defined all your variables as int and you are making a calculation on boolean, it automatically convert that boolean into int. (i == L / 2) and (i > L / 2) these two values output either 1 or 0. consider true as 1 and false as 0.then you can understand the function

因为你将所有变量定义为int并且你在布尔值上进行计算,所以它会自动将该boolean转换为int。 (i == L / 2)和(i> L / 2)这两个值输出1或0.将true视为1,将false视为0.然后您可以理解函数

l = cons + i * 2 - 2 * (i == L / 2) - 4 *(i - L / 2) * (i > L / 2);

try follwing

#include <stdio.h>
#include <iostream>

using namespace std;


void Draw(int length) {

    int cons = 3,L,l,test;
    length -= length % 2;
    L = length + cons;

    for (int i = 0; i < L; i++){

        l = cons + i * 2 - 2 * (i == L / 2) - 4 *(i - L / 2) * (i > L / 2);
        test = (i == L / 2);
        cout << "(i == L / 2):" << test<< endl;
        test = (i > L / 2);
        cout << "(i > L / 2):" << test<< endl;
    }
}

int main(int argc, char **argv) {

    Draw(5);
    return 0;
}

Which will output

哪个会输出

(i == L / 2):0
(i > L / 2):0
(i == L / 2):0
(i > L / 2):0
(i == L / 2):0
(i > L / 2):0
(i == L / 2):1
(i > L / 2):0
(i == L / 2):0
(i > L / 2):1
(i == L / 2):0
(i > L / 2):1
(i == L / 2):0
(i > L / 2):1

https://ideone.com/PPwvRV