Reference no: EM132101280
PLS USE C++
Your program will generate a pseudo-random color BMP image of size and location as specified by the user.
Your program will use a maximum size of 512x512.
Ask the user to enter the horizontal and vertical size (e.g. M and N) notifying the user of the maximum size allowable.
Prompt the user for the output filename with a bmp extension where the image is to be stored (in the directory from which your program is run).
Generate the correct size image and store a valid pseudo-random color BMP image file at the specified file location.
Use your program from the programming assignment on the previous HW as a starting point.
Use the rand() function to generate the random BGR values 0-255.
Seed the rand() function using the srand() function with the argument of srand( ) being the time-of-day obtained from your computer using the following syntax. The statement is
srand((unsigned int)time(NULL));
You will need to include in your preprocessor commands the line
#include<time.h> in order to use the function call time(NULL).
The new piece of information about the BMP file format that you will need to implement in this project is the BMP requirement that the image data for each row of image pixels must occupy an integer number of 4-bytes. For example, if the image is (width,height)=(3,4), each row of the image is 3 pixels, whose BGR values occupy (3)(3)=9 bytes.
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bytes 3-6 (4) (Size)
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total # bytes in
BMP image-
file=102
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64, 0, 0, 0
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LeastSigByte 1st; MostSigByte last
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bytes 5-8 (4) (Width)
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Image-width in pixels=3
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3, 0, 0, 0
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LSByte 1st
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bytes 9-12 (4) (Height)
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Image-height in pixels=4
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4, 0, 0, 0
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LSByte 1st
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bytes 21-24 (4) (ImageDatasize)
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Total # bytes in
imagedata =48
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48, 0, 0, 0
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LSByte 1st
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But since the image data for each image row of a BMP image must occupy an integer number of 4-byte words, you will have to pad 12-9=3 bytes of zeroes to the end of the image data for each and every row. Therefore, the size of the BMP file is 14+40+(4)(12)=102 bytes.
The BITMAPFILEHEADER for the 3x4 BMP image would therefore have the value 102 in bytes 3-6 (Size). The BITMASPINFOHEADER would have Width=3 and Height=4
and ImageDataSize=(4)(12)=48
Let's, for example, use the integer width for the number of pixels in each row and the integer rowsize for the number of bytes in each image row.
If width=1, 1x3=3 bytes for BGR but rowsize=4.
If width=2, 2x3=6 bytes for BGR but rowsize=8.
If width=3, 3x3=9 bytes for BGR but rowsize=12.
If width=4, 4x3=12 bytes for BGR and rowsize=12.
So how do you compute rowsize given width?
If (3*width) %4=0, no zero-padding at the end of the 3*width bytes in an image row.
If (3*width) %4=1, zero-padding 3 additional bytes at the end of the 3*width bytes in the image row.
If (3*width) %4=2, zero-padding 2 additional bytes at the end of the 3*width bytes in the image row. If (3*width) %4=3, zero-padding 1 additional bytes at the end of the 3*width bytes in the image row.
Again,
IMPORTANT:
Recall, if you fprintf() the ascii character code (10) for newline '\n' into a file, fprintf(fp,"%c",10); the operating system will put 2-bytes into the file: carriage return (asci code 13) then linefeed (10). This is a problem, because the value for the red component of the 10th pixel in the image data is exactly value 10. No other value in the BITMAPFILEHEADER or BITMAPINFOHEADER has the value 10. To fix this, instead of setting the red component of the 10th pixel to 10, set it equal to 11.
If your program is written correctly, after you run your program, you can use your mouse to doubleclick on your output file and the default application for displaying images on your computer will display a pseudo-random color image.
PLS USE C++