python:Return concept

Question

I have a simple code:

def string_times(str, n):
  if n==0:
    return ""
  else:
    for i in range(n):
        return str
        i=i+1
      
print(string_times('Hi',2))

The expected output is HiHi but my output is HI. Why is this happening?

``return`` immediately ends the function. Did you intend to ``print`` instead? — MisterMiyagi, Aug 20 '20 at 10:55
Because you are doing a "return" in the for loop therefore it returns the string and quit the function — ShayHa, Aug 20 '20 at 10:56

score 1 · Answer 1 · answered Aug 20 '20 at 10:57

1

You can directly use like:

print("Hi"*2)

Output: HiHi

answered Aug 20 '20 at 10:57

Akshay Jain

score -1 · Accepted Answer · answered Aug 20 '20 at 11:01

I'd write the function like this:


def string_times(str_, n):
    if n != 0:
        for i in range(n):
            print(str_, end='')
    else:
        print(' ',end='')

I'd not use 'str' as my variable and print within the function.

Then you can call the function python string_times('Hi',2) and it will get you the result you want.

If you want each 'Hi' in a different row just remove the end param

2 Answers2