python - 类型错误 : 'float' object is not callable error with bhaskara program

Question

#entrada de dados
a = int(input("valor de a "))
b = int(input("valor de b "))
c = int(input("valor de c "))
#
import math
#
d =b*b - 4 * a * c
#
if d < 0:
        print('nao tem raizes reais')
elif d ==0:
        raiz = (-1*b + math.sqrt(d))/(2 * a)
        print('o valor e '), print(raiz)
elif d > 0:
        raiz = (-1*b + math.sqrt(d))(2 * a)
        raiz1 = (-1*b - math.sqrt(d))(2 * a)
        print('as raizes sao'), print(raiz), print(raiz1)
#fim

我试着输入这段代码，但我得到了这个错误:

Traceback (most recent call last): File "C:/Python34/test podre.py", line 16, in raiz = (-1*b + math.sqrt(d))(2 * a) TypeError: 'float' object is not callable

怎么了？

Answer 1

python中不能用纯括号将两个对象相乘，需要用*分隔。

raiz = (-1*b + math.sqrt(d))(2 * a)

成为

raiz = (-1*b + math.sqrt(d))*(2 * a)

这是您编辑的代码:

a = int(input("valor de a "))
b = int(input("valor de b "))
c = int(input("valor de c "))
#
import math
#
d =b*b - 4 * a * c
#
if d < 0:
        print('nao tem raizes reais')
elif d ==0:
        raiz = (-1*b + math.sqrt(d))/(2 * a)
        print('o valor e '), print(raiz)
elif d > 0:
        raiz = (-1*b + math.sqrt(d))*(2 * a)
        raiz1 = (-1*b - math.sqrt(d))*(2 * a)
        print('as raizes sao'), print(raiz), print(raiz1)