x² + y² = r<\/p><\/blockquote>\n
Variables x<\/em> and y<\/em> are the circle’s coordinates. Variable r<\/em> represents the radius or, more precisely, the square root of r<\/em> is the radius. Figure 1 illustrates a basic circle plotted using the Grapher program on the Macintosh.<\/p>\n Figure 1. The circle formula.<\/p><\/div>\n If you’re going to plot coordinates from left to right on the Cartesian plane, you need to solve the circle equation for y<\/em>. By my math memory, that would be:<\/p>\n y = √r-x²<\/span><\/p><\/blockquote>\n A mathematician would cringe at that solution, because it’s kinda wrong: If you plot that equation, you get only the top half of the circle. To get the bottom half you also have to take into account negative values of y<\/em>. That’s okay, because all you need to plot a circle on a computer is the top half anyway, as the following code demonstrates.<\/p>\n The circle()<\/em> function plots the circle by using the equation y = √r-x²<\/span><\/em>. Here’s how that breaks down, starting at Line 35:<\/p>\n The for<\/em> loop marches along the x<\/em> axis based on the Lines 37 and 38 obtain the square of the I didn’t have to use the Line 39 calculates the y<\/em> position by using the formula y = √r-x²<\/span><\/em>:<\/p>\n The abs()<\/em> function calculates the absolute value (positive numbers only) of And the circle is drawn. Sample output is shown in Figure 2. Yes, it’s an oval, which is due to the proportions of single characters in text mode graphics. (See if you can figure out how to compensate for that in the code!)<\/p>\n Figure 2. The circle is plotted by using text mode graphics.<\/p><\/div>\n The problem?<\/p>\n Oh, yeah: Come on, you knew<\/em> that there would be a problem!<\/p>\n The current rendition of the circle()<\/em> function assumes a focus of 0,0. If you want to plot the center of the circle at location 3,8, you need to modify the circle()<\/em> function, specifically Lines 38, 40, and 41. Here’s the revised circle()<\/em> function:<\/p>\n Or you can click here<\/a> to download the full source code file.<\/p>\n","protected":false},"excerpt":{"rendered":" Here’s how to graph a circle in text mode. Crazy? Of course! Continue reading ![\"Figure](\"http:\/\/c-for-dummies.com\/blog\/wp-content\/uploads\/2014\/07\/0719-figure11.png\")
\r\n#include <stdio.h>\r\n#include <math.h>\r\n#include <stdlib.h>\r\n\r\n#define WIDTH 60\r\n#define HEIGHT 20\r\n#define X WIDTH\/2\r\n#define Y HEIGHT\/2\r\n#define XMAX WIDTH-X-1\r\n#define XMIN -(WIDTH-X)\r\n#define YMAX HEIGHT-Y\r\n#define YMIN -(HEIGHT-Y)+1\r\n\r\nchar grid[HEIGHT][WIDTH];\r\n\r\nint circle(int x, int y, int radius);\r\nint plot(int x, int y);\r\nvoid init_grid(void);\r\nvoid show_grid(void);\r\n\r\nint main()\r\n{\r\n init_grid();\r\n circle(0,0,8);\r\n show_grid();\r\n\r\n return(0);\r\n}\r\n\r\n\/* Plot a circle *\/<\/span>\r\nint circle(int x, int y, int radius)\r\n{\r\n float xpos,ypos,radsqr,xsqr;\r\n\r\n for(xpos=x-radius;xpos<=x+radius;xpos+=0.1)\r\n {\r\n radsqr = pow(radius,2);\r\n xsqr = pow(xpos,2);\r\n ypos = sqrt(abs(radsqr - xsqr));\r\n plot(rintf(xpos),rintf(ypos));\r\n plot(rintf(xpos),rintf(-ypos));\r\n }\r\n\r\n return(1);\r\n}\r\n\r\n\/* Set \"pixel\" at specific coordinates *\/<\/span>\r\nint plot(int x, int y)\r\n{\r\n if( x > XMAX || x < XMIN || y > YMAX || y < YMIN )\r\n return(-1);\r\n\r\n grid[Y-y][X+x] = '*';\r\n return(1);\r\n}\r\n\r\n\/* Initialize grid *\/<\/span>\r\nvoid init_grid(void)\r\n{\r\n int x,y;\r\n\r\n for(y=0;y<HEIGHT;y++)\r\n for(x=0;x<WIDTH;x++)\r\n grid[y][x] = ' ';\r\n \/* draw the axis *\/<\/span>\r\n for(y=0;y<HEIGHT;y++)\r\n grid[y][X] = '|';\r\n for(x=0;x<WIDTH;x++)\r\n grid[Y][x] = '-';\r\n grid[Y][X] = '+';\r\n}\r\n\r\n\/* display grid *\/<\/span>\r\nvoid show_grid(void)\r\n{\r\n int x,y;\r\n\r\n for(y=0;y<HEIGHT;y++)\r\n {\r\n for(x=0;x<WIDTH;x++)\r\n putchar(grid[y][x]);\r\n putchar('\\n');\r\n }\r\n}<\/pre>\nfor(xpos=x-radius;xpos<=x+radius;xpos+=0.1)<\/code><\/p>\nradius<\/code> argument. I use the full value of radius<\/code> here, so I’m not exactly following the formula. That’s okay; the circle()<\/em> function asks for a radius value and not the square root of that value. Also, the stepping value is 0.1, which ensures that something more-or-less resembling a circle is drawn.<\/p>\nradius<\/code> and xpos<\/code> variables by using the pow()<\/em> function:<\/p>\nradsqr = pow(radius,2);
\nxsqr = pow(xpos,2);<\/code><\/p>\nradsqr<\/code> or xsqr<\/code> variables; these could have been calculated on the next line, but doing so makes that statement rather messy. Also, you could use radius*radius<\/code> instead of pow(radius,2)<\/code>, but I prefer using the pow()<\/em> function. Remember: Unlike other programming languages, C lacks a power operator.<\/p>\nypos = sqrt(abs(radsqr - xsqr));<\/code><\/p>\nradsqr<\/code> minus xsqr<\/code>. It’s necessary because that value could be negative, which would cause the sqrt()<\/em> function to squirt<\/em> all over the compiler. And because this formula returns only positive values (the top of the circle), the next two plot()<\/em> statements (Lines 40 and 41) must plot both positive and negative values of ypos<\/code>:<\/p>\nplot(rintf(xpos),rintf(ypos));
\nplot(rintf(xpos),rintf(-ypos));<\/code><\/p>\n![\"Figure](\"http:\/\/c-for-dummies.com\/blog\/wp-content\/uploads\/2014\/07\/0726-figure2.png\")
\r\n\/* Plot a circle *\/\r\nint circle(int x, int y, int radius)\r\n{\r\n float xpos,ypos,radsqr,xsqr;\r\n\r\n for(xpos=x-radius;xpos<=x+radius;xpos+=0.1)\r\n {\r\n radsqr = pow(radius,2);\r\n xsqr = pow(xpos-x<\/span>,2);\r\n ypos = sqrt(abs(radsqr - xsqr));\r\n plot(rintf(xpos),rintf(ypos)+y<\/span>);\r\n plot(rintf(xpos),rintf(-ypos)+y<\/span>);\r\n }\r\n\r\n return(1);\r\n}<\/code><\/pre>\n