Does specifying array size for a user input string in C matter?

Question

I am writing a code to take a user's input from the terminal as a string. I've read online that the correct way to instantiate a string in C is to use an array of characters. My question is if I instantiate an array of size [10], is that 10 indexes? 10 bits? 10 bytes? See the code below:

#include <stdio.h>

int main(int argc, char **argv){

    char str[10] = "Jessica";
    scanf("%s", &str);
    printf("%c\n", str[15]);

}

In this example "str" is initialized to size 10 and I am able to to print out str[15] assuming that when the user inputs a a string it goes up to that index.

My questions are:

1. Does the size of the "str" array increase after taking a value from scanf?
2. At what amount of string characters will my original array have overflow? .

Answer 1

When you declare an array of char as you have done:

char str[10] = "Jessica";

then you are telling the compiler that the array will hold up to 10 values of the type char (generally - maybe even always - this is an 8-bit character). When you then try to access a 'member' of that array with an index that goes beyond the allocated size, you will get what is known as Undefined Behaviour, which means that absolutely anything may happen: your program may crash; you may get what looks like a 'sensible' value; you may find that your hard disk is entirely erased! The behaviour is undefined. So, make sure you stick within the limits you set in the declaration: for str[n] in your case, the behaviour is undefined if n < 0 or n > 9 (array indexes start at ZERO). Your code:

printf("%c\n", str[15]);

does just what I have described - it goes beyond the 'bounds' of your str array and, thus, will cause the described undefined behaviour (UB).

Also, your scanf("%s", &str); may also cause such UB, if the user enters a string of characters longer than 9 (one must be reserved for a terminating nul character)! You can prevent this by telling the scanf function to accept a maximum number of characters:

scanf("%9s", str);

where the integer given after the % is the maximum input length allowed (anything after this will be ignored). Also, as str is defined as an array, then you don't need the explicit "address of" operator (&) in scanf - it is already there, as an array reference decays to a pointer!

Hope this helps! Feel free to ask for further clarification and/or explanation.

Answer 2

One of C's funny little foibles is that in almost all cases it does not check to make sure you are not overflowing your arrays.

It's your job to make sure you don't access outside the bounds of your arrays, and if you accidentally do, almost anything can happen. (Formally, it's undefined behavior.)

About the only thing that can't happen is that you get a nice error message

Error: array out-of-bounds access at line 23

(Well, theoretically that could happen, but in practice, virtually no C implementation checks for array bounds violations or issues messages like that.)

See also this answer to a similar question.

Answer 3

An array declares the given number of whatever you are declaring. So in the case of:

char str[10]

You are declaring an array of ten chars.

1. Does the size of the "str" array increase after taking a value from scanf?

No, the size does not change.

2. At what amount of string characters will my original array have overflow?

An array of 10 chars will hold nine characters and the null terminator. So, technically, it limits the string to nine characters.

printf("%c\n", str[15]);

This code references the 16th character in your array. Because your array only holds ten characters, you are accessing memory outside of the array. It's anyone's guess as to if your program even owns that memory and, if it does, you are referencing memory that is part of another variable. This is a recipe for disaster.