为什么添加字符（来自argv）导致C无效

Question

first post (as you would know) and the form tells me that my title is bad and will get downvoted but I can't do better :) I've worked on this a bunch already.

I'm trying to cipher text by adding a number given in the command line to a string.

Why does

include <stdio.h>
int main(int argc, char * argv[])
{
    printf("%i", argc);
    char k = argv[1][0];
    printf("%c", k);
    char * s = "a";
    printf("%c", s[0] + 0);
}

correctly print "a" (besides for printing argc and k) but when the last line is

    printf("%c", s[0] + k);

it just prints argc and k and then nothing. I wanted it to print 20a (when running ./question 0 ).

( I tried

char k = 0;
char * s = "a";
printf("%c", s[0] + k);

and that does work :( ) It seems like the fact that k is coming from argv is the issue but I don't have any idea why (not that I really understand any of this) Thanks!

Answer 1

argv and argc and "a" are all red herrings. They have nothing to do with the problem. The problem lies in not understanding the difference between two character values, '0' and 0 (the latter is aso known as '\\0' ).

In order to understand the difference I suggest experimenting with this program:

#include <stdio.h>
int main(void) {
    char a = 'a', b = '0';
    printf ("as char: a=%c b=%c sum=%c\n", a, b, a+b);
    printf ("as int: a=%d b=%d sum=%d\n", a, b, a+b);
}

Use different values for b , including 0 , 1 , '0' and '1' and see what happens.

Answer 2

The problem is that the expression s[0] + k is that adding a letter and a number together (conceptually speaking, ignoring the data types for a minute) can easily result in a value that is past the end of the alphabet, and printing it out as if it was a letter isn't necessarily going to work.

Bonus problem: reading the command line parameter into a single character isn't giving you a value of 0, but '0' which has a numerical value of 48. You can't just pull out the first character from the string. What if the user runs the program and passes 10 as a parameter?

You need to make sure that the result of your calculation is within the bounds of the alphabet.

As suggested by commenters above, take a look at the ASCII table, and try using some character constants sich as 'a' 'z' 'A' and 'Z' to modify the result

Answer 3

The answer is very simple.

The ASCII code of 'a' is 97.

Assuming only the printable characters in the arguments, the lowest printable ASCII code from the argv is 32.

97+32 = integer overflow which is UB - on ideone.com system it is -127

you can try yourself https://ideone.com/D8DLcy

#include <stdio.h>

int main(void) {
    for(char x = 32; x < 127; x++)
    {
        char m = 'a' + x;

        printf("'a' +  %c = char value in decimal is:%hhd - and the char is%c\n", x, m, m);
    }
    return 0;
}

Answer 4

From the information you provided, it looks like you're implementing a simple shift cipher. Note the use of modulo for cases where the input for k is greater than 25.

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

#define     MAX_STRING  16

int main(int argc, char *argv[])
{
    char        num[MAX_STRING];

    char * s = "a";
    int k;

    if (argc == 2)
    {
        snprintf(num,MAX_STRING, argv[1], 10);

        k = (int)strtol(num, NULL, 10) % 26;

        printf("%c\n", s[0] + k);

    }
    else
    {
        fprintf(stderr,"Invalid number of arguments\n");
        return(EXIT_FAILURE);
    }
    return(EXIT_SUCCESS);
}