Drawing Arrows

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• Download demo project - 13.6 Kb

Excuse

Basically, there was a request for some code to draw arrows and I had never seen any. Plus, I'm not really in the mood to hang drapes or vacuum.

Using the code

This is a simple API for drawing lines with arrowheads. It looks like this:

// ARROWSTRUCT

//

// Defines the attributes of an arrow.

typedef struct tARROWSTRUCT {
    int nWidth;     // width (in pixels) of the full base of the arrowhead

    float fTheta;   // angle (in radians) at the arrow tip between the two

                    //  sides of the arrowhead

    bool bFill;     // flag indicating whether or not the arrowhead should be

                    //  filled

} ARROWSTRUCT;

// ArrowTo()

//

// Draws an arrow, using the current pen and brush, from the current position

//  to the passed point using the attributes defined in the ARROWSTRUCT.

void ArrowTo(HDC hDC, int x, int y, ARROWSTRUCT *pArrow);
void ArrowTo(HDC hDC, const POINT *lpTo, ARROWSTRUCT *pArrow);

Simply fill an ARROWSTRUCT with the desired attributes, make sure the current DC position is correct (MoveTo(), etc.), and call one of the two ArrowTo() functions. The size parameters (nWidth and fTheta) are defined as follows:

Technique

This goes back to high-school algebra and trigonometry. The ArrowTo() function first builds a vector of the full line. Then it calculates the points for the sides of the arrowhead based on the nWidth and fTheta attributes you pass. Badda-boom-badda-bing, you got your arrowhead.

Here's some pseudo-pseudocode:

lineVector = toPoint - fromPoint
lineLength = length of lineVector

// calculate point at base of arrowhead

tPointOnLine = nWidth / (2 * (tanf(fTheta) / 2) * lineLength);
pointOnLine = toPoint + -tPointOnLine * lineVector

// calculate left and right points of arrowhead

normalVector = (-lineVector.y, lineVector.x)
tNormal = nWidth / (2 * lineLength)
leftPoint = pointOnLine + tNormal * normalVector
rightPoint = pointOnLine + -tNormal * normalVector

History

• December 1 2002 - Created.