ECE 3331, Dr. Hebert, Summer-3, 2016 HW 11 Hardcopy HW due Tues 07/19 Program due Sunday 07/17. Problem 1. Section 10.6, Exercise 3.

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1 ECE 3331, Dr. Hebert, Summer-3, 2016 HW 11 Hardcopy HW due Tues 07/19 Program due Sunday 07/17 Problem 1. Section 10.6, Exercise 3. Problem 2. Section 10.6, Exercise 5. Problem 3. Section 10.6, Exercise 6 Problem 4. Section 10.8, Exercise 2. Problem 5. Section 10.8, Exercise 5. Problem 6. Section 10.9, Exercise 1. Problem 7. Section 10.9, Exercise 3. Problem 8. Programming project. The BMP (bitmap) Image-file Format Read this whole section before starting. The BMP (bitmap) format is a standard image-file format developed by Microsoft and used by Windows Operating systems and others. This document describes how to construct a 24-bit BMP file. A 24-bit BMP file uses 24 bits to denote the color of each pixel in an image: an 8-bit value (0-255) for the blue component, an 8-bit value (0-255) for the green component, and an 8-bit value (0-255) for the red component of a pixel. Using BGR to denote the blue, green, and red components of a pixel, if the pixel is bright blue, it would have (B,G,R)=(255,0,0). If a pixel is bright green it would have (B,G,R)=(0,255,0). If the pixel is bright red, it would have (B,G,R)=(0,0,255). Black is formed by the absence of color; so if a pixel is black it would have (B,G,R)=(0,0,0). White is formed by the superposition of the brightest blue, green, and red; so a white pixel would have (B,G,R)=(255,255,255). Your 24-bit BMP file will have the following 3 parts. 1) A bitmap file header (14 bytes) is the 1st 14 bytes in a BMP image-file. The bitmap-file header, called a BITMAPFILEHEADER, contains information about the type, size, and layout of the bitmap file. (More info below) 2) A bitmap information header (40 bytes) called a BITMAPINFOHEADER follows the BITMAPFILEHEADER in the *.bmp image file. The bitmap-information header specifies the dimensions, compression type, and color format for the bitmap file. (More info below) 3) Image-data, a sequence of bytes that define the BGR value at each pixel in the image. The image-data stores the BGR values of each pixel in order: beginning at the bottom row of the image, going across the row, then the 1st pixel of the next higher row of the image, etc. Let's consider creating a BMP (bitmap) file that will contain a 256x256 pixel color image as shown below. The image will be black on the left and bright red on the right.

2 For this image, the BGR values in each row of the image will be the same. As your eye travels across the image from left-to-right, the pixels become brighter and brighter red, as follows. Consider that the image is 256 pixels in width and 256 pixels in height. The BMP file will have the width and height of the image in the file header, each will be stored as an unsigned int (4 bytes) least significant byte first. The decimal value 256 is represented with a 4-byte unsigned integer as the 4 bytes (least significant byte 1st) 0, 1, 0, 0 which denotes the decimal non-negative value 0+1*256+0*256*256+0*256*256*256=256. The decimal value 503, for examples, would be stored as 247, 1, 0, 0 which denotes the decimal non-negative value 247+1*256+0*256*256+0*256*256*256=503. Additionally in the header of the BMP file, we will need to store the number of bytes of image-data; i.e. the total number of bytes required to store the BGR values for each pixel in the image. Since we are storing the blue-green-red value of each pixel as 3 consecutive bytes, we will need 3*256*256=196,608 bytes for the image data. The value 196,608 will also have to be stored in the header of the BMP file as a 4-byte unsigned int value, least significant byte first. The decimal value 196,608 is represented as a 4-byte integer value as the 4 bytes (least significant byte 1st) 0, 0, 3, 0 which denotes the decimal value 0+0*256+3*256*256+0*256*256*256=196,608. The total number of bytes in the BMP file will be *256*256=196,662 via (bitmap file header =14 bytes)+(bitmap information header =40 bytes)+(image data BGR =196,608bytes). The decimal value 196,662 is represented as a 4-byte integer value as the 4 bytes (least significant byte 1st) 54, 0, 3, 0. Now let's examine the first 14 bytes in the 256x256 pixel BMP file shown above, i.e. the BITMAPFILEHEADER. BYTE locations Description Decimal Byte Value Comments bytes 1-2 (2) BMP file 66, 77 (Type) designation bytes 3-6 (4) (Size) total # bytes in BMP imagefile 54, 0, 3, 0 LeastSigByte 1st; MostSigByte last bytes 7-8 (2) reserved bytes 0, 0 must be 0's (Reserved1) bytes 9-10 (2) reserved bytes 0, 0 must be 0's (Reserved2) bytes (4) (OffsetBytes) offset (in bytes) from start of file to image-data =54. using 4 bytes: 54, 0, 0, 0 Least Signif Byte 1st; Most Signif Byte last Figure 5.1 The 14-byte BITMAPFILEHEADER that begins your BMP file.

3 The first 14 bytes of the BMP file will be 66, 77, 54, 0, 3, 0, 0, 0, 0, 0, 54, 0, 0, 0 Now, let's examine the next 40 bytes in the BMP file, i.e. the BITMAPINFOHEADER. BYTE locations Description Decimal Byte Comments Values bytes 1-4 (4) # bytes in 40, 0, 0, 0 (Size) BITMAPINFOHEADER bytes 5-8 (4) Image-width in 0, 1, 0, 0 (Width) pixels=256 bytes 9-12 (4) Image-height in 0, 1, 0, 0 (Height) pixels=256 bytes (2) (Planes) # planes in image=1 1, 0 typically = 1 bytes (2) # of bits per pixel=24 24, 0 (BitCount) bytes (4) (Compression Type) Type of Compression 0, 0, 0, 0 0 means no compression bytes (4) (ImageDatasize) bytes (4) (PixelsPerMeter_X) bytes (4) (PixelsPerMeter_Y) bytes (4) (Colors_Used) bytes (4) (Colors_Important) Total # bytes in imagedata 192,608 using 4 bytes is 0, 0, 3, 0 # horiz pixels per meter=0 0, 0, 0, 0 (Not used). # vert pixels per meter=0 0, 0, 0, 0 (Not used). # of entries in color 0, 0, 0, 0 not used for this table=0 image. Set equal to 0 # of significant colors=0 0, 0, 0, 0 not used for this image. Set equal to 0 Figure 5.2 The 40-byte BITMAPINFOHEADER that follows the BITMAPFILEHEADER in a BMP file. The next 40 bytes in the BMP file will therefore be 40, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 24, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0. To generate the BGR pixels in each row of the image above, you could use a double "for" loop (a nested for-loop) to increment across each image row and across each pixel in that row. for(i=0; i<256; i++){\\ iterate across the rows in the image for(j=0; j<256; j++){ \\ iterate across the pixels in the ith row.... } }. As you visit each pixel, you would generate the 3-byte BGR value for that pixel. The blue B and green G components would be 0 for every pixel in the image. The red R value for each pixel would be its j value from the inner for-loop.

4 For example, if i=14 (you are visiting the 14th row) and j=23 (you are visiting the 23rd pixel in the 14th row) the BGR value for that pixel is BGR=(0, 0, 23)= (0,0,j). There was two additional pieces of information about the BMP format. A) The BMP format requires that each image row consist of an integer multiple of 4-bytes. If each image row needs for example 15 bytes to store the BGR pixel values, then an additional 1-byte ( a "padding byte") needs to be stored after the BGR pixel values for each row, so that the row occupies 16-bytes, an integer multiple (4) of 4-bytes. B) The BMP format writes out the pixels in the bottom row of the image to the file first, and it writes the pixels of the top row of the image last. For the reddish image above, each image row contains 256 BGR values or (256)*(3)=768 bytes. Of course, 768 is a whole multiple of 4, since 768=4*192. Therefore, no "padding bytes" need to be written to the file after each image row. To calloc( ) memory for the BGR values of the N pixels in an image row, we could do unsigned char *irow,n; N=99; \\ 99 pixels per image row. irow=(unsigned char)calloc(1,3*n); But 3*99=297 is not a multiple of 4 bytes, so "padding bytes" would need to be written to the file so that the bytes for the row would occupy a whole multiple of 4 bytes. Since 300=4*75, we need to add =3 "padding bytes" to the data in the BMP file for each image row. Note that modulus 297%4=1. Also 4-1=3 bytes, and 3 bytes is the number of padding bytes that we would need. So the way to calloc( ) memory for each image row is unsigned char *irow,n; N=99; \\ 99 pixels per image row. M=3*N+(4-(3*N)%4); irow=(unsigned char)calloc(1,3*n); To read-in or write-out a BMP image, you should open the file as unformatted "rb" or "wb". In this project, you will write a program to read-in a 24-bit BMP color image and write out a color "negative" of the image to a BMP file. A color "negative" simply changes each color component B,G,R to 255-B, 255-G,255- R. Your program will prompt the user for the path/filename of the input BMP file, and write out a file called neagative.bmp to the same folder in which the input file is stored. 1. Write a structures for the BITMAPFILEHEADER and for the BITMAPINFOHEADER. Since the byte-order (least significant byte first) of many of the fields of the BITMAPFILEHEADER and BITMAPINFOHEADER are important, the fields, for example, for filesize should not be simply an unsigned int. If we were to fwrite( ) the structure to a file, the byte order written to the file depends on the operating system of the computer. Most, but not all, of the operating systems would write the least-significant byte of the unsigned int; but some operating systems write out the most-significant byte first. Your program should be able to e compiled and run on any operating system and still produce a valid BMP file. Therefore, filesize is most easily stored as a 4-byte vector unsigned char filesize[4], struct bmpfileheader{ unsigned char bmp[2]; unsigned char filesize[4]; unsigned int reserved;

5 unsigned char dataoffset[4]; }; Write a similar structure for the BITMAPINFOHEADER. 2. Now declare a structure for the BMP header made up of the BITMAPFILEHEADER structure and the BITMAPINFOHEADER structure struct bmpheader{ struct bmpfileheader fileheader; struct bmpinfoheader infoheader; } 3. Declare a bmpheader variable struct bmpheader header; 4. Write a function whereby you pass a 4-byte unsigned int value and the function fills a vector unsigned char v[4] with the 4 bytes ordered with least-significant byte first in v[0]. We worked this out in lecture as void ui24bytes(unsigned int itmp, unsigned char * v){ v[0]= (unsigned char)(itmp%256); v[3]= (unsigned char)(itmp/pow(256,3)); v[2]= (unsigned char)( (itmp-v[3]*pow(256,3))/pow(256,2)); v[1]= (unsigned char)( (itmp-v[3]*pow(256,3)-v[2]*pow(256,2))/256); } For example, ui24bytes(23917, header.fileheader.filesize); would will in the four bytes denoting a filesize of Fill in all the fields of the bmp header using fread( ) from the input BMP file. 6. If the 1st two bytes of the input file are not 66 and 77, print a message to the user and exit. 7. If the file is indeed BMP, write out the header to the output BMP file. 8. Now read-in the imagedata byte -by - byte, invert each byte (255-byte) and write it to the output file. Each byte value v of the input image data, however, will be changed to 255-v in the output image data. In the final version of your program, do not change the "padding bytes" p in each image row to 255-p.