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The first call to f(b) doesn't assign the return value, the second does.
To expand on what rsmith6559 said:
f(b) is a call to the function f with the value of b. The function does the internal math in a separate namespace. Nothing has actually changed in the original b. Then f returns the input + 5, but f(b) is on a line by itself so that goes nowhere. Sort of like if a person tosses a ball and nobody is there to catch it then it is not caught. The ball falls to the ground and rolls away.
b = f(b) starts out the same way. A copy of b is put into f in a separate namespace. f does its work and tosses back 5. This time there is b= out there so b catches the address that points to 5. To do this it had to drop the address that pointed to 0. Now b is a representation of 5 and no longer 0.
Because the print statements happen after each of these steps, you get the unchanged b then the changed b.
All of this is wonderfully summarized by rsmith6559, but if you still wanted more: there it is!
you got a function f
function f takes an input and adds 5 to it, and returns that value...
then its printing, what you input and the new value with this command
print b, ",", <---- right now b is still 0
b = f(b) <----- now b became b=0+5=5
print b <---- this is added to the top print command because there is a comma at the end of it
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f(b) # nothing
print b, ",", # print 0
b = f(b) # b = f(0) = "a = 0 + 5" = "a = 5" = 5
print b #print 5
when f(b) is called the first time, the function returns 5 but it doesn't change the value of b. So it remains 0.
but the the next time you assign b=f(b) as the function returns 5 which is then assigned to the variable b. Hence this time it prints 5.
so the output is 0,5