Reading from argv in C

Question

I want to read the path of three files (for example, "../c/..file"), and a long from argv[] and bind them to three already created char* 's, and also a long value called num.

Here's my main function:

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

    char* file1 = NULL;
    char* file2 = NULL;
    char* file3 = NULL;
    long num = 0;

    //copy the file paths to the char*'s
    strcpy(file1, argv[1]);
    strcpy(file2, argv[2]);
    strcpy(file3, argv[3]);
    // convert from string to long int
    num = strtol(argv[4],NULL,0);

}

However this does not work, and the filename of the files and the value of the long, don't end up on the variables, as they're supposed to.

How can i fix this ?

In my program i check the argc values to make sure i'm not passing wrong stuff, but here i wrote the function this way just to illustrate my problem.

Answer 1

Don't strcpy into pointers that aren't pointing at allocated memory. Instead, just set them equal to the pointers in the argv array, which already do point at allocated memory. Like so:

int main(int argc, char *argv[])
{
    if (argc < 5)
        fprintf(stderr, "usage: %s <file1> <file2> <file3> <num>\n", argv[0]), exit(1);

    char *file1 = argv[1];
    char *file2 = argv[2];
    char *file3 = argv[3];
    long  num   = strtol(argv[4], NULL, 0);

    /* Do stuff */

    return 0;
}

Answer 2

If you want to copy strings from argv then you have to allocate memory for these strings in the program. For example

int main(int argc, char* argv[])
{
    char *file1, *file2, *file3;
    long num = 0;

    if ( argc > 1 )
    {
        file1 = malloc( strlen( argv[1] ) + 1 ); 
        strcpy( file1, argv[1] );
    }

    if ( argc > 2 )
    {
        file2 = malloc( strlen( argv[2] ) + 1 ); 
        strcpy( file2, argv[2] );
    }

    if ( argc > 3 )
    {
        file3 = malloc( strlen( argv[3] ) + 1 ); 
        strcpy( file3, argv[3] );
    }

    if ( argc > 4 )
    {
        num = strtol( argv[4], NULL, 0 );
    }

    //...