122 lines
3.5 KiB
C++
122 lines
3.5 KiB
C++
/*
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Problem Statement: a) Write C++ program to draw 2-D object and perform following basic transformations:
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Scaling, Translation, Rotation. Apply the concept of operator overloading
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Code from Computer Graphics (SPPU - Second Year - Computer Engineering - Content) repository on KSKA Git: https://git.kska.io/sppu-se-comp-codes/CG
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*/
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// BEGINNING OF CODE
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#include<iostream>
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#include<graphics.h>
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#include<cmath>
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using namespace std;
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int main()
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{
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int gd = DETECT,gm;
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initgraph(&gd, &gm, NULL);
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initwindow(1000,800);
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float a[3][3];
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int x1,y1,x2,y2,choice;
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cout<<"Enter x1 and y1 coordinates: ";
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cin>>a[0][0]>>a[1][0];
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cout<<"\nEnter x2 and y2 coordinates: ";
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cin>>a[0][1]>>a[1][1];
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cout<<"\nEnter x3 and y3 coordinates: ";
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cin>>a[0][2]>>a[1][2];
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a[2][0] =a[2][1] = a[2][2] = 1;
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line(a[0][0],a[1][0],a[0][1],a[1][1]);
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line(a[0][1],a[1][1],a[0][2],a[1][2]);
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line(a[0][2],a[1][2],a[0][0],a[1][0]);
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delay(10000);
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cout<<"\n1. Translation \n2. Scaling \n3. Rotation";
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cout<<"\nEnter choice: ";
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cin>>choice;
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switch(choice)
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{
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case 1:
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{
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float b[3][3],c[3][3];
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b[0][0] = b[1][1] = b[2][2] = 1;
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b[0][1] = b[1][0] = b[2][0] = b[2][1] = 0;
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cout<<"\nEnter tx and ty: ";
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cin>>b[0][2]>>b[1][2];
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for(int i = 0; i<3; i++)
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{
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for(int j = 0; j<3; j++)
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{
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c[i][j] = 0;
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for(int k = 0; k<3; k++)
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{
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c[i][j] += b[i][k]*a[k][j];
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}
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}
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}
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setcolor(BLUE);
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line(c[0][0],c[1][0],c[0][1],c[1][1]);
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line(c[0][1],c[1][1],c[0][2],c[1][2]);
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line(c[0][2],c[1][2],c[0][0],c[1][0]);
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delay(7000);
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}
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break;
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case 2:
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{
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float b[3][3],c[3][3];
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b[0][1] = b[0][2] = b[1][0] = b[1][2] = b[2][0] = b[2][1] = 0;
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b[2][2] = 1;
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cout<<"\nEnter sx and sy: ";
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cin>>b[0][0]>>b[1][1];
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for(int i = 0; i<3; i++)
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{
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for(int j = 0; j<3; j++)
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{
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c[i][j] = 0;
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for(int k = 0; k<3; k++)
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{
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c[i][j] += b[i][k]*a[k][j];
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}
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}
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}
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setcolor(BLUE);
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line(c[0][0],c[1][0],c[0][1],c[1][1]);
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line(c[0][1],c[1][1],c[0][2],c[1][2]);
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line(c[0][2],c[1][2],c[0][0],c[1][0]);
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delay(7000);
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}
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break;
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case 3:
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{
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float b[3][3],c[3][3];
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float z;
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cout<<"\nEnter angle: ";
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cin>>z;
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float ang = (z*3.14)/180;
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b[0][2] = b[1][2] = b[2][0] = b[2][1] = 0;
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b[2][2] = 1;
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b[0][0] = cos(ang);
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b[0][1] = sin(ang);
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b[1][0] = -(sin(ang));
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b[1][1] = cos(ang);
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for(int i = 0; i<3; i++)
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{
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for(int j = 0; j<3; j++)
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{
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c[i][j] = 0;
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for(int k = 0; k<3; k++)
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{
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c[i][j] += b[i][k]*a[k][j];
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}
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}
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}
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setcolor(BLUE);
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line(c[0][0],c[1][0],c[0][1],c[1][1]);
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line(c[0][1],c[1][1],c[0][2],c[1][2]);
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line(c[0][2],c[1][2],c[0][0],c[1][0]);
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delay(7000);
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}
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}
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closegraph();
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return 0;
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}
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// END OF CODE
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