CODEWITHSUNDEEP
cmallocargc
Carterini
Carterini

Reputation: 165

Placing an unspecified amount of command line inputs into an array in C

I essentially want to read values from the command line as typed in by the user and place them into an array. My attempt at this so far is as follows

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

#define     w   1.0


int main(int argc, char argv[])
{
    int     tmp;

    double  *x, *v, *m, *k, *R;

    x = malloc((argc-1)*sizeof(double));

    printf("%lf\n", argc);

    for(tmp=0; tmp<argc-1; tmp++)
    {
        x[tmp] = argv[tmp+1];
        printf("%lf\n", x[tmp]);
    }
}

The print of the value of argc is equal to 0 yet the for loop will repeat three times which doesn't make sense and the values it gives are just completely wrong. Sorry i'm a bit of an amateur at this. Thanks in advance!

Upvotes: 0

Views: 122

Answers (4)

user3151973
user3151973

Reputation: 86

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

//calloc = malloc for N elemnts of given size
double* myArray = (double*)calloc(argc, sizeof(double));

//Check out arg 0!
for (i = 0; i<argc; i++)
{
    printf("arg [%d] = %s\n",i,argv[i]);
}
for (i = 1; i<argc; i++)
{
    sscanf(argv[i], "%lf", &myArray[i]);
    printf("myArray[%d] = %f\n",i, myArray[i]);
}
//avoid memory leak
free(myArray);
}

I see that my answer came a bit to late, anyways I just wanted to point out the arg[0] is not input.

Upvotes: 0

barak manos
barak manos

Reputation: 30136

printf("%lf\n", argc) will cause a memory access violation, because:

  1. argc is an integer type (4 bytes).

  2. printf("%lf... attempts to read 8 bytes.

You must change it to printf("%d\n", argc).

Upvotes: 0

Shahbaz
Shahbaz

Reputation: 47533

There are many problems with your code, some already mentioned by others such as argv that needs to have a type of char *[].

First of all, if argc is 1, then your malloc will fail because allocating memory of size 0 doesn't make sense.

Second, printf("%lf\n", argc); is incorrect. argc is int and you need to use %d to print it. The reason you see 0, which is wrong, is this.

Third, x[tmp] = argv[tmp+1]; is incorrect too. argv[tmp+1] is a char *, which means it's a string. While x[tmp] is a double. You can't just assign a string to a double. What you need to do is to convert the string to a double, for example:

sscanf(argv[tmp+1], "%lf", &x[tmp]);

or

x[tmp] = strtod(argv[tmp+1]);

Hint: always compile your code with common warnings. With gcc, that would be -Wall command line argument. The compiler can tell you about many of these mistakes.

Upvotes: 2

Ed Heal
Ed Heal

Reputation: 60007

For a start the signature of main should be

int main(int argc, char *argv[])

Then you are passed strings. You need to convert them to doubles sscanf will do this - see http://linux.die.net/man/3/sscanf

Upvotes: 1

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