Can anybody plz explain this code,specifically the 5th line..

abc=input('Enter image name or path');

a=imread(abc);

imshow(a);

b=size(a);

c=b(1)*b(2)/1024;

disp(fprintf('Name of image: %s',abc))

disp(fprintf('Size of the image is %f kB',c))

Answer 1

[OCDER]

An image is often represented as a MxNx3 matrix, where each element of a matrix is a uint8 value ranging from 0 to 255 (this is 1 byte of data. 1 byte = 8 bits = value from 0 to 255).

Thus, each pixel of an image is of size 1x1x3 matrix for the Red, Green, and Blue values. With that said, that code line 5 is incorrect if dealing with color images, as each pixel actually requires 3 bytes of data (Ex. White = [255, 255, 255] = 3 bytes)

image size should be:

c = prod(b) /1024 ; %for kB of data

If you have the liberty of editing the code, change the variable names to be more descriptive.

ImageSizeKB = prod(b) / 1024; %prod(b) will do b(1)*b(2)*b(3) = MxNx3 = total bytes in image

Another way to get image size is by using iminfo FileSize

ImageInfo = iminfo(abc);
ImageSizeKB = ImageInfo.FileSize / 1024;

Answer 2

[madhan ravi]

a small example to illustrate

a = rand(4,3)

b = size(a)

b(1)

b(2)

[madhan ravi]

size() returns dimensions of a matrix as row number and column number, the output is b here so b(1) is the a’s row number and b(2) is the columns number. So the fifth lines simple explanation is a’s Row number and it’s column number is multiplied and divided by 1024.

[madhan ravi]

imread() reads an image and outputs a matrix form of an image.

[madhan ravi]

imshow() Displays an image by reading the matrix

Answer 3

[Image Analyst]

This is BAD code and has several errors. What line 5 does is to try to compute the number of pixels and then kilobytes. b(1) is the number of rows. b(2) is the number of columns multiplied by the number of color channels. Then if you multiply them together you get the number of pixels, and if you assume they're 1 byte pixels and divide by 1024 you get kilobytes. Here is some less bad code:

fileName = input('Enter image name or path : ', 's')
theImage = imread(fileName);
imshow(theImage);
axis('on', 'image');
sizeDimensions = size(theImage);
sizeInBytes = sizeDimensions(1)*sizeDimensions(2)/1024
fileInfo = dir(fileName)
fprintf('Name of image: %s\n', fileName)
fprintf('Size of the image is %d bytes, or %.1f kB\n', fileInfo.bytes, fileInfo.bytes / 1024)

but it's still not that robust or user friendly. Please review these links:

http://www.mathworks.com/matlabcentral/answers/13205#answer_18099

http://blogs.mathworks.com/steve/2011/03/22/too-much-information-about-the-size-function/

I would not look to that programmer as a paragon of MATLAB code writing ability.

[Sumeet Badgujar]

But it does not give the perfect size... the image used is 11.2 kb,

the imread output gives [264,191,3]..calculated according this gives wrong file size

[Walter Roberson]

b(2) is the number of columns multiplied by the number of color channels.

Not in this case. Single output for size was used, so b will be as long as needed to represent the dimensions individually.

[Walter Roberson]

Image file sizes on disk represent sizes including any headers and comments (such as EXIF information), and take into account any compression (which might be lossy) that has been used. The size of the disk file can end up being larger than the amount of memory required to store the image in memory, but the file size could also end up being much less than is required to store the image in main memory if the compression is strong.