How to store entered 26 character values against argv[1] to string type array key

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[dm_code_snippet background=”yes” background-mobile=”yes” slim=”no” line-numbers=”no” bg-color=”#abb8c3″ theme=”dark” language=”clike” wrapped=”yes” height=”” copy-text=”Copy Code” copy-confirmed=”Copied”]

#include
#include
#include
#include
#include
int main(int argc, string argv[])
{
int counter = 0;
    if (argc != 2)
    {
    return 1;
    }
    else
    {
    printf("enter plainkey: ");
    }
string key = argv[1][];
int t = strlen(key);
    if(t != 26)
    {
    return 1;
    }
    while(counter < t)
    {
    isdigit (argv[1][counter]);
    return 1;
    counter = counter + 1;
    }
string s = get_string("plaintext: ");
int countstring = strlen(s);
    for(int i = 0; i <= countstring; i++)
        if (isalpha s[i])
        {
        int s[i]= int t;
        t = s[i];
        }
        else
        {
        s[i] = s[i];
        }
    printf("ciphertext: %s",s);
 }

[/dm_code_snippet]

On compiling,

 

 

There is a problem with:

string key = argv[1][ ];

How do I denote 26 characters that are supposed to be entered by the user against argv[1]. Need to store entered 26 character values to key.

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Reply

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You should note that agrv[1] is a string while argv[1][ ] is a character. You are declaring key as string. So you should be assigning a value of string to it and not character. Use string key = argv[1];[learn_press_profile]

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