CODEWITHSUNDEEP
carguments
B.B10
B.B10

Reputation: 171

C argument conversion

First,I have to say that I am new to C programming. What I'm trying to do is write a program that takes an argument input and converts it into an integer and then returns its value. My code looks like this:

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

int main (int argc, char *argv[])
{
   int fromArgv = NULL;  /* holds value from argv[1] */
   fromArgv = atoi (argv[1]);    /* convert argv[1] to int */

   /* if incorrect no. of arguments entered */
   if (argc != 2) {
   fprintf (stderr, "error: wrong number of arguments\n");
   exit (EXIT_FAILURE);
   } 

   return fromArgv;
}

I get the following error when trying to compile: 

ex1.c: In function ‘main’:
ex1.c:6:18: error: initialization makes integer from pointer without a cast [-Werror]
cc1: all warnings being treated as errors

Upvotes: 0

Views: 148

Answers (5)

ShinTakezou
ShinTakezou

Reputation: 9661

As mentioned by others, you are initializing an integer with a "null pointer" (that from a type point of view is a pointer), which is wrong. You wanted to initialize it to zero,

       int fromArgv = 0;

Then, you should pick the argument and convert it to int after and not before checking the number of arguments. So, you should move the "atoi" line after the if (argc != 2) part.

Consider also the possibility to use strtol instead of atoi. The former allows you to check if the conversion succeeded.

Upvotes: 1

md5
md5

Reputation: 23699

NULL is a null pointer constant , and you attempt to assign it to an integer value (fromArgv), therefore your compiler reports a warning.

The following initialization works fine:

fromArgv = 0;

Although, it is useless, since you are modifying the value of your variable on the next line:

int fromArgv = atoi(argv[1]);

By the way, it is a bad idea to return this kind of value from main (it is often reserved to error codes), and you are cbecking the number of arguments too late.

Upvotes: 3

user3693
user3693

Reputation: 95

Why do you ever have to initialize fromArgv ?

just

int fromArgv = atoi(arg[1]);

should be working fine

Upvotes: 0

1&#39;&#39;
1&#39;&#39;

Reputation: 27095

Your problem is the line

int fromArgv = NULL;

I believe NULL is usually defined as (void *) 0, which is a pointer. You're assigning this pointer to an int variable, which gives you the error.

However, there's no need to initialize fromArgv in this case. You can just do:

int fromArgv;  /* holds value from argv[1] */
fromArgv = atoi (argv[1]);    /* convert argv[1] to int */

or even just

int fromArgv = atoi (argv[1]);    /* convert argv[1] to int */

What you do have to do, however, is make sure argv[1] exists before you access it so you don't get a segfault if the user does not enter any command-line arguments. You should move your if (argc != 2) test before the assignment of fromArgv.

Upvotes: 5

prateeksarda
prateeksarda

Reputation: 947

u cant initialize fromArgv to NULL because NULL is a pointer and u are assigning it to int just change the line as int fromArgv ;

Upvotes: 0

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