I have a problem with my code. I would like to know what the "variation" value is. But it always gives me extremely high or negative values. So when I type in terminal for example ./NAME 3 I end up with another number. How can i fix it?
#include <stdio.h>
#include <cs50.h>
int main(int argc, string argv[])
{
int variation = (int)argv[1];
if (argc == 2)
{
printf("%i\n", variation);
return 0;
}
else
{
return 1;
}
}
[Note: C
does not have builtin string
type. What C
has is character array
]
This is a very easy to solve problem. Here's the code given, it uses stdlib
's builtin atoi
function to convert string
to int
:
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]) {
if (argc == 2) {
char *var_str = argv[1];
int variation = atoi(var_str);
printf("variation: %d\n", variation);
}
else if (argc < 2) {
printf("no agrument provided.\n");
} else {
printf("only one argument was expected. %d provided.\n", argc);
}
return 0;
}
Update (A more secure version using strtol
):
As strtol
is a better function to convert string
to long int
, than atoi
, because of it's error detecting utility
, it's better to use strtol
in this regard. This practice is heavily enforced by user @klutt
. I would like to thank her/him for advising me to include the code using strtol
too. It is given below:
// a more secure version
#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
#include <limits.h>
int main(int argc, char *argv[]) {
if (argc == 2) {
char *var_str = argv[1];
char *text = NULL;
int base = 10; // default base
errno = 0; // intializing errno
long int variation = 0;
variation = strtol(var_str, &text, base);
// now on to error detection
// full error trace
if (var_str == text) {
printf ("error: no number found\n");
} else if (errno == ERANGE && variation == LONG_MIN) {
printf ("error: underflow\n");
} else if (errno == ERANGE && variation == LONG_MAX) {
printf ("error: overflow\n");
} else if (errno != 0 && variation == 0) {
printf ("error: unknown error\n");
} else if (errno == 0 && var_str && !*text) {
printf ("variation: %ld\n", variation);
} else if (errno == 0 && var_str && *text != 0) {
printf ("variation: %d, but some text included: %s\n", variation, text);
}
} else if (argc < 2) {
printf("no agrument provided.\n");
} else {
printf("only one argument was expected. %d provided.\n", argc);
}
return 0;
}
If you have any questions, ask me in comment...