read argv[1] into say "key" using atoi. Should only accept int. If a non int like 3g is entered how do I reject is as non int

Question

How do I fix this. My program must not accept any alphabet.

Int must accept only valid numbers and reject the string if it contains a alphabet

int main(int argc, char *argv[])
{
    char str[1000];
    int i, ch, key;
    {
        if (argc != 2)
        {
             printf("key\n");
             return 1; // exit prog
        }
        // read argv value into key after argc valid
        //to prevent sementation error
        key = atoi(argv[1]);
        // do multi wrap arounds use leftover digit
        key = (key % 26);
            //check for positive number
        if (key < 1)
        {
        printf("key\n");
        return 1; // exit prog
        }
        // ask for text if pos number
        else if (key >= 0)
        {
        printf("text:");
        // check string and read in
        fgets(str, sizeof str, stdin);
        i = strlen(str);
}

If I type

./name (no input, alphabet, two strings returns 1 as it should.)
./name (number should work and return 0)
./name a1 or b5 return 1 but 1a or 5b return 0 and should not.

In the debugger I can see it only reads the value of the number into argv as 1 or 5 not as 1a or 5b

First, don't use `atoi()` on unchecked input - it has absolutely no error checking. It returns an `int` no matter what you pass into it, and it has no way of returning an error if the input wasn't a numeric value. — Andrew Henle, Aug 03 '20 at 12:37
Does this answer your question? [How to check if a string is a number?](https://stackoverflow.com/questions/16644906/how-to-check-if-a-string-is-a-number) — Cid, Aug 03 '20 at 12:47
The "key" must read in the value of argv for an up to int only if the int is valid (2^31). — Johan Maritz, Aug 04 '20 at 05:38

score 1 · Answer 1 · edited Aug 03 '20 at 16:54

1

Instead of the function atoi use another standard function strtol. For example

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

int main(void) 
{
    const char *s1 = "3g";
    const char *s2 = "32";
    
    char *endptr;

    errno = 0;
    
    long key = strtol( s1, &endptr, 10 );
    if ( *endptr != '\0' || errno != 0 )
    {
        puts( "invalid value supplied" );
    }
    else
    {
        printf( "key = %ld\n", key );
        
    }

    endptr = NULL;
    errno = 0;
    
    key = strtol( s2, &endptr, 10 );
    if ( *endptr != '\0' || errno != 0 )
    {
        puts( "invalid value supplied" );
    }
    else
    {
        printf( "key = %ld\n", key );
        
    }
    
    return 0;
}

The program output is

invalid value supplied
key = 32

edited Aug 03 '20 at 16:54

chux - Reinstate Monica

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answered Aug 03 '20 at 13:02

Vlad from Moscow

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Corner: `if ( *endptr != '\0' || errno != 0 )` does not detect `s1` as `""`, yet that is a little tricky to pass into `main()` anyways. Perhaps `if (s1 == endptr || *endptr != '\0' || errno != 0 ) `? – chux - Reinstate Monica Aug 03 '20 at 15:08
@chux-ReinstateMonica It is the way how atoi is defined. atoi: (int)strtol(nptr, (char **)NULL, 10) – Vlad from Moscow Aug 03 '20 at 16:20
@chux-ReinstateMonica I have updated the code substituting int key; for long key; – Vlad from Moscow Aug 03 '20 at 16:38

MikeCAT · Answer 2 · 2020-08-03T13:32:29.340

0

You can use strtol() to convert strings to integers and obtain the last position of converted string.

#include <limits.h>
#include <stdlib.h>
#include <errno.h>

// ...

int key;
// ...
char* end;
long key_temp;
// ...
errno = 0;
key_temp = strtol(argv[1], &end, 10);
if (argv[1][0] == '\0') {
    // no character exists
} else if (*end != '\0') {
    // invalid character exists
} else if ((errno == ERANGE && (key_temp == LONG_MIN || key_temp == LONG_MAX)) ||
key_temp < INT_MIN || INT_MAX < key_temp) {
    // too small or too big
} else if (key_temp == 0 && errno != 0) {
    // other error
} else {
    // looks OK
    key = (int)key_temp;
}

edited Aug 03 '20 at 13:32

answered Aug 03 '20 at 12:52

MikeCAT

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The expression `key_temp < LONG_MIN || LONG_MAX < key_temp` is always false (given the type of `key_temp` is `long`) – M. Nejat Aydin Aug 03 '20 at 13:12
@M.NejatAydin Thank you, fixed. – MikeCAT Aug 03 '20 at 13:13
Still not guaranteed to work as intended: It might be that `LONG_MIN == INT_MIN` and `LONG_MAX == INT_MAX`. There are some platforms that these are true. The expression will be always false on such platforms. – M. Nejat Aydin Aug 03 '20 at 13:18
@M.NejatAydin fixed. – MikeCAT Aug 03 '20 at 13:33
`(errno == ERANGE && (key_temp == LONG_MIN || key_temp == LONG_MAX))` is unclear. Why not use `(errno == ERANGE)` instead and skip the `key_temp == ...` tests? – chux - Reinstate Monica Aug 03 '20 at 14:59
`else if (key_temp == 0 && errno != 0)` is also unclear. Why not simplify to `else if (errno != 0)`. I see no C spec requiring the return value as 0 on some other error. – chux - Reinstate Monica Aug 03 '20 at 15:01
//I did not want to rewrite my code from start so: I added the following into my code just before the "key = atoi(argv[1]);" line else if(strspn(argv[1], "0123456789") == strlen(argv[1])) { } else if(strspn(argv[1], "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ./") == strlen(argv[1])) { printf("key\n"); return 1; } – Johan Maritz Aug 04 '20 at 08:56

score 0 · Answer 3 · answered Aug 04 '20 at 09:04

   //I did not want to rewrite my code from start so: I added the following into my 
   //code just before the "key = atoi(argv[1]);" line

                                                                 
     else if(strspn(argv[1], "0123456789") == strlen(argv[1]))
     {
     }   
     else if(strspn(argv[1], "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ./") == strlen(argv[1]))
    {  printf("key\n");
        return 1; }

read argv[1] into say "key" using atoi. Should only accept int. If a non int like 3g is entered how do I reject is as non int

3 Answers3