为什么我的python代码不执行，而是给我一个“无”错误，而只执行其中的一部分？

Question

I made a program to solve a quadratic equation by inputting a, b, and c and then it would return the solutions, or if it had any and it would return that it had no real solutions if not.

It gave me a couple of syntax errors that I fixed, and I tried to import the time module to delay it, but it still doesn't work. I think it is because I didn't give it like a keyword to start running or something.

import math

print('What quadratic do you want to solve?')        

a = input(print("a= "))
b = input(print("b= "))
c = input(print("c= "))

discriminant = b**2 - 4*a*c

if discriminant < 0:
    print('There are no solutions to this quadratic!')

x_1 = (-b + math.sqrt(discriminant))/2*a
x_2 = (-b - math.sqrt(discriminant))/2*a

if x_1 == x_2:
    print('The solution is ' + x_1)

print('The solutions are ' + x_1 + ' and ' + x_2)

I expected the output to ask the values of a, b, and c and then give me a result, but it just gave me

What quadratic do you want to solve?
a= 
None

Answer 1

It is asking you for the value of a , but the code is a bit weird so the display gets messed up. The program will not proceed until you enter the value for a ; it is waiting patiently for you to do that.

Normally, input function is expecting a single argument, which is the prompt it would display. print function is used for printing, and returns None . What happens is, print("a=") prints that, and returns None ; then you get input(None) which understands None to be the prompt to be printed, and you get this sort of messed up output. The code should be either

print("a=", end="")
a = input()

or just

a = input("a=")

There's more errors in your code once you get past this; eg b**2 - 4*a*c is calculated, but never stored anywhere, and thus ends up discarded. Meanwhile, discriminant variable was never defined, and you'll hear Python complain about it. Also, something / 2*a means (something / 2) * a , but correct formula needs something / (2 * a) (which you can also write as something / 2 / a ).

EDIT with more details from comments, as I missed them (and thanks to TrebledJ for being more observant): since Python3 input returns strings, you will need to cast them to floats (using float(...) function) for them to be usable in calculation; and similarly, the results, being numbers, will not cleanly concatenate with strings using + (you would have to either cast them to strings using str(...) function, or use .format string method or % formatting operator to get everything to mesh).

Answer 2

Python's print statement always returns None . So in input(print("a= ")) , you're essentially evaluating input(None) . Another problem in your code is in the line

print('The solutions are ' + x_1 + ' and ' + x_2)

You can't concatenate int and str objects so you must turn the number into a str object before printing it.

print('The solutions are ' + str(x_1) + ' and ' + str(x_2))

import math

print('What quadratic do you want to solve?')        

a = input("a= ")
b = input("b= ")
c = input("c= ")

discriminant = b**2 - 4*a*c

if discriminant < 0:
    print('There are no solutions to this quadratic!')

x_1 = (-b + math.sqrt(discriminant))/2*a
x_2 = (-b - math.sqrt(discriminant))/2*a

if x_1 == x_2:
    print('The solution is ' + x_1)

print('The solutions are ' + str(x_1) + ' and ' + str(x_2))

Answer 3

Why your program is returning None is perfectly explained in the above two answers. So, you can also write the print and input statement in a single line as

a = int(input(f"a = " ))

And before in the input statement, you need to specify int or float which will be the type of a, b and c. Without it, they will be of type str which will again raise a TypeError in declaring the discriminant variable.

And finally in the if statement, after checking if the discriminant is < 0 and printing the statement you need to quit the program using exit() function. Else the rest of the program will be executed regardless and we don't want that.

Also, check the above two answers, they have rightly pointed out other small mistakes in the code.