returning value without breaking a loop-CodePudding

I intend to make a while loop inside a defined function. In addition, I want to return a value on every iteration. Yet it doesn't allow me to iterate over the loop. Here is the plan:

def func(x):
    n=3
    while(n>0): 
        x = x 1  
        return x

print(func(6))

I know the reason to such issue-return function breaks the loop. Yet, I insist to use a defined function. Therefore, is there a way to somehow iterate over returning a value, given that such script is inside a defined function?

CodePudding user response：

When you want to return a value and continue the function in the next call at the point where you returned, use yield instead of return.

Technically this produces a so called generator, which gives you the return values value by value. With next() you can iterate over the values. You can also convert it into a list or some other data structure.

Your original function would like this:

def foo(n):
  for i in range(n):
    yield i

And to use it:

gen = foo(100)
print(next(gen))

gen = foo(100)
l = list(gen)
print(l)

Keep in mind that the generator calculates the results 'on demand', so it does not allocate too much memory to store results. When converting this into a list, all results are caclculated and stored in the memory, which causes problems for large n.

CodePudding user response：

Depending on your use case, you may simply use print(x) inside the loop and then return the final value.

If you actually need to return intermediate values to a caller function, you can use yield.

CodePudding user response：

You can create a generator for that, so you could yield values from your generator.

Example:

def func(x):
    n=3
    while(n>0): 
        x = x 1  
        yield x
func_call = func(6)  # create generator
print(next(func_call)) # 7
print(next(func_call)) # 8