Difference between C array and python array [duplicate]

Question

a=input("type sentence: ") for i in range(1, 100): if a[i] == 'a': print (a[i]) elif a[i] == 'o': break 

According to my experience, in C, we can use a[i] with for to check each index of a string array from a[0] to a[i]. However, whenever I use the same method in Python I get the error "IndexError: string index out of range". What is the problem in this case?

My goal is to check if the sentence I wrote has vowels and if it does, print the first vowel. I tried listing, but problem got worse.

Answer 1

Use

for i in range(len(a)): ... 

to avoid the IndexError when i is after the end of the string a.

Even better:

for c in a: if c == 'a': ... 

Answer 2

I suspect the input you gave isn't 100 characters long.

The problem lies in the fact that the input function in the first line creates the string object in a. This string object will be as long as the input. So by checking for a[100] on a string that is only 5 characters long will result in an IndexError.

How about trying

for character in a: if(...): ... 

edit: Checking for vowels can be easier too.

list_of_vowels = [a, e, i, o, u] if (character in list_of_vowels): do_something(character) 

Answer 3

It is very easy in Python, we do not need to compare each index. Just need one comparison using in operator.

a=input("type sentence: ") for i in a: if (i in ["a","e","i","o","u"]): print (i) break 

Answer 4

I guess you might be doing this in C:

char a[100]; scanf("%s", a); 

In C you need to pre-allocate a fixed-length "container" to later save the input string, and the length should be larger than that of expected input to be able to put the string in a.

However, in Python you don't need to do so. Python helps you do this extra work. In Python, executing

a = input() 

and you will get a str variable a which contains your input string. The length of a is automatically determined by the length of your input string, and you can get the length by using the len function.

So the version of your code that works should be

for i in range(1, len(a)): if a[i] == 'a': print (a[i]) elif a[i] == 'o': break