try:
num=float(num)
except:
print "Invalid input"
continue
this part of my code seems to be bugging, but when i remove the try and except everything works smoothly, so this seems to be the problem.
i want to convert the input inside a while loop to an integer, if the input isn't an integer it'll display an error and just continue with the loop and ask again. however, it doesn't continue the loop and just keeps printing "Invalid input" forever. how come it isn't continuing the loop?
here is the entire code, in case something else might be wrong:
c=0
num2=0
num=raw_input("Enter a number.")
while num!=str("done"):
try:
num=float(num)
except:
print "Invalid input"
continue
c=c+1
num2=num+num2
num=raw_input("Enter a number.")
avg=num2/c
print num2, "\t", c, "\t", avg
You can solve the problem by moving the variable assignments into the try block. That way, the stuff you want to skip is automatically avoided when an exception is raised. Now there is no reason to continue
and the next prompt will be displayed.
c=0
num2=0
num=raw_input("Enter a number.")
while num!=str("done"):
try:
num=float(num)
c=c+1
num2=num+num2
except:
print "Invalid input"
num=raw_input("Enter a number.")
avg=num2/c
print num2, "\t", c, "\t", avg
You can tighten this a bit further by removing the need to duplicate the prompt
c=0
num2=0
while True:
num=raw_input("Enter a number. ")
if num == "done":
break
try:
num2+=float(num)
c=c+1
except:
print "Invalid input"
avg=num2/c
print num2, "\t", c, "\t", avg
continue means return back to while , and as num never changes, you will be stuck in an infinite loop.
If you want to escape the loop when that exception occurs then use the term break instead.
# this function will not stop untill no exception trown in the try block , it will stop when no exception thrown and return the value
def get() :
i = 0
while (i == 0 ) :
try:
print("enter a digit str : ")
a = raw_input()
d = int(a)
except:
print 'Some error.... :( '
else:
print 'Everything OK'
i = 1
return d
print(get())
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