Thread: program using char and string is working weirdly

> int i=0;
> char str1[i];
> char str2[i];

How much space is allocated to your arrays - none.

> printf("Please enter shift [0--25]: %d %c \n", input, str1[i]);
What was str1 initialised to before printing it - nothing.

You probably want something like str1[100] and %s for reading in a string.

Thank you for that. I have made a few adjustments, however, it still isn't working as planned. Because the following error message won't let me compile it:
invalid operands to binary & (have 'int *' and 'int') //reference to line 12

Code:

#include <stdio.h>
#include <string.h>
int main (int argc, char *argv[]) {    

    int i=100;
    int input=0;
    char str1[i];
    char str2[i];
    int n;

    printf("Please enter shift [0--25]: %d %c \n", input, str1[i]);
    n=scanf("%d %c", &input &str1[i]);

    i=0;
    while (str1[i] != '\0') {
        str2[i] = str1[i];
        i=i+input;
    }
    str2[i] = '\0';
    printf("%c", str2[i]);

    return n;
}

Why do you insist on putting [i] in every place after str1?

In a declaration the number in the square brackets specifies the size of the array,
whereas elsewhere it means that you want to access the item with the number specified.

Given an array of size 100 only has valid indexes 0 to 99, you're going out of bounds. Subsequent code goes way out of bounds in fact.
You need to understand and learn the part in bold.