Finding target coordinates from given coordinates

user_10306507 0
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Hi their I have an excercise and below is my solution to it. Is their a better solution?

A figure stands in the 2-dimensional space on coordinate (Fx/Fy) and can move with one step in 8 directions to a neighboring field: (E SE S SW W NW N NE) The figure should move as fast as possible to a target point (Zx/Zy). First read the coordinates of the figure and the target point. Then move the figure along the shortest possible way to the destination, outputting the coordinates of each step and at the end outputting the number of steps required.

What I don't like on the code is how many lines of code I need.

#include <stdio.h>

int fx;
int fy;
int zx;
int zy;
int xgleich = 0;
int ygleich = 0;
int cnt = 1;


// Diagonale Funktion
int diagonalsteps(int fx,int zx)
{
    while (xgleich == 0 && ygleich == 0)
    {
        if (fx < zx)
        {
            fx++;
        }
        else
        {
            fx--;
        }

        if (fy < zy)
        {
            fy++;
        }
        else 
        {
            fy--;
        }
        printf("\n%dter Schritt: X:%d, Y=%d\n", cnt, fx,fy);
        cnt++;

        if (fx == zx)
        {
            xgleich = 1;
        }
        if (fy == zy)
        {
            ygleich = 1;
        }
    }
}

/* gerader Weg */
int geradesteps(int fy, int zy)
{
    if (xgleich == 1)
    {
        if(fy < zy)
        {
            while (fy < zy)
            {
                fy++;
                printf("\n%dter Schritt: X=%d, Y=%d\n", cnt, fx, fy);
                cnt++;
            }
        }
        if (ygleich == 1)
        {
            if (fx < zx)
            {
                while (fx < zx)
                {
                    fx++;
                    printf("\n%dter Schritt: X=%d, Y=%d\n", cnt, fx, fy);
                    cnt++;
                }
            }
            if (fx > zx)
            {
                while (fx > zx)
                {
                    fx--;
                    printf("\n%dter Schritt: X=%d, Y=%d\n", cnt, fx, fy);
                    cnt++;
                }
            }
        }
    }
}

int main()
{
    printf("Bitte geben Sie den Koordinaten Fx und Fy Punkt ein: ");
    scanf("%d %d", &fx, &fy);
    printf("%d und %d\n", fx, fy);

    printf("Bitte geben Sie die Ziel Koordinaten Zx und Zy Punkt ein: ");
    scanf("%d %d", &zx, &zy);
    printf("%d und %d\n", zx, zy);

    if (fx == zx)
    {
        xgleich = 1;
    }

    if (fy == zy)
    {
        ygleich = 1;
    }
    diagonalsteps(fx,zx);
    geradesteps(fy,zy);

    return 0;
}
coordinates c
1 Answers

If the number of lines is what concerns you, you can replace:

int fx;
int fy;
int zx;
int zy;
int xgleich = 0;
int ygleich = 0;
int cnt = 1;

with

int fx, fy, zx, zy, xgleich = 0, ygleich = 0, cnt = 1;

and

if (fx < zx)
{
    fx++;
}
else
{
    fx--;
}

with

fx += (fx < zx) ? +1 : -1;

The if condition is redundant here:

if(fy < zy)
{
    while (fy < zy)

just use:

while (fy < zy)

you can also replace

if (fx < zx)
{
    while (fx < zx)
    {
        fx++;
        printf("\n%dter Schritt: X=%d, Y=%d\n", cnt, fx, fy);
        cnt++;
    }
}
if (fx > zx)
{
    while (fx > zx)
    {
        fx--;
        printf("\n%dter Schritt: X=%d, Y=%d\n", cnt, fx, fy);
        cnt++;
    }
}

with

int inc = (fx < zx) ? +1 : -1;

while (fx != zx)
{
    fx += inc;
    printf("\n%dter Schritt: X=%d, Y=%d\n", cnt++, fx, fy);
}

Finally, compile with warnings, you promise to return something from your functions:

demo.c: In function ‘diagonalsteps’:
demo.c:46:1: warning: control reaches end of non-void function [-Wreturn-type]
   46 | }
      | ^
demo.c: In function ‘geradesteps’:
demo.c:84:1: warning: control reaches end of non-void function [-Wreturn-type]
   84 | }
      | ^

In this case:

void diagonalsteps(int fx,int zx)

and

void geradesteps(int fy, int zy)

is what you want.

user_16063450
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