Segmentation fault in caesar

Question

I am returning a segmentation fault with this code. The argc portion of the code is function correctly if i enter too many arguments returning "1". If the correct number of arguments are entered I get a segmentation fault. I am pretty new to this and could use some help figuring out why.

Here is output from testing when i added a print f to the beginning.

./caesar your key is (null) segmentation fault

./caesar 2 your key is 2 segmentation fault

./caesar 2 3 your key is 2 segmentation fault

#include <stdio.h>
#include <cs50.h>
#include <string.h>

int main (int argc, string argv[])
{
    int counter = 0;

    printf("your key is: %s \n", argv[1]);


    for (int i =0, n=strlen(argv[1]); i<n; i++)//tries to read argv[1] to determine if all of the elements are digits
    {
        if (argv[1][i]<=9)
        {
            counter ++;//adds to counter when decimal digits are found
        }
        else
        {
            counter--;//subtracts from counter if non decimals are found
        }
    }
    if (argc==2 && counter == strlen(argv[1]))//if arguments are entered properly prompt user for message to encode.
    {
       string k = get_string("Plaintext: \n");
    }
    else
    {
        printf("1 \n");
    }
}

I know argv is taking in the second argument stored in argv[1]. I am just not sure how to correct this outcome. Any help would be greatly appreciated.

Thanks

Answer 1

The seg fault happens when the program is run without any parameters. It is looking for a value in argv[1]. If there are no parameters given, then there argv[1] is null. Then, when it calls strlen(argv[1]), it throws a seg fault. When a function like strlen (and a lot of related functions) is called with null input, it seg faults.

When I run the code with a key, it executes to the end without any seg faults. Perhaps you haven't recompiled your code since you updated it?

There are a number of other issues in the code, but this question related only to the seg faults. New problems deserve a new question.

Happy coding.

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