Is this array increasing or what did it do?

Question

I tried to make program where you insert first a digit for how much your name have letters. Then I like to get your name (using arrays). The problem in this program is that if I insert 1 for N and then insert name who had more letters than that we entered on start. Why this works and I get the full name?

And the second question is how to convert a char into int so if the user insert number for how much letter he have in their name, and later insert a name with more letters, I can print "Error you insert more letters for your name then expected"?

#include <stdio.h>
int main()
{
    int N;
    printf("Insert how much letter does your name have");
    scanf("%d",&N);
    char name[N];
    printf("Insert your name ");
    scanf("%s",&name);
    printf("You insert %s",name);

    return 0;
}

You have forgot that you need one element to store the `\0`-terminator for the string. If you make it like you have now, you will easily get into buffer overflow. Also `scanf()` should have a length modifier to avoid exactly the same thing. — RobertS supports Monica Cellio, Dec 01 '19 at 14:18
An array of N characters does not have enough elements for a name of N characters. +1 is required for the terminating null. Writing out of bounds has undefined behaviour, one of the acceptable behaviours is ignoring the situation completely with unpredictable results. To read the name you can allocate an array that allows a name *one* larger than expected and then fault if the name was indeed that long... — Antti Haapala -- Слава Україні, Dec 01 '19 at 14:19
Your `scanf()` call uses `%s` without a maximum length modifier, and will thus happily write out of bounds of your array, which is undefined behavior. — Shawn, Dec 01 '19 at 14:21

RobertS supports Monica Cellio · Accepted Answer · 2019-12-04T15:46:28.623

The problem in this program is that if I insert 1 for N and then insert name who had more letters than that we entered on start. Why this works and I get the full name?

The reason why it (in your case) works is called "Undefined Behavior". You can´t expect/guarantee anything if you insert such issues in your program. It is possible that the right values appear but high-probably on another implementation you will get completely different results. Always and ever avoid the likes of those and program in the right way with a clean-focus on what you actually do.

Here a link for more informations about what UB is: Undefined, unspecified and implementation-defined behavior

how to convert a char into int so if the user insert number for how much letter he have in his name, and later insert a name with more letters, I can print "Error you insert more letters for your name then expected"?

I do not understand, why a char to int conversion should help in that case. Rather you should scan name with the use of strlen() of how many characters the string is contained of and make a if/else-statement thereafter based on the return value of strlen().

Also as i already mentioned in my comment, but twice is better:

You have forgot that you need one element to store the \0-terminator for the string, which is automatically inserted at the end of the string if you use scanf("%s",&name);, so name should be declared as char name[N + 1];. If you make it like you have now, you will easily get into buffer overflow. Also if possible, scanf() should have a width modifier to avoid exactly the same thing, even if the width is not fixed at compilation and generated at run-time. A trick to use a variable (not fixed) width modifier, as you need it if you want to incorporate one, is written down in one answer on that question:

Width Specifier for scanf() - Length of characters to consume is not fixed at compilation and only determined at run-time. How to make it variable?

Always remember: Make your program as safe as possible.

Is this array increasing or what did it do?

1 Answers1