CODEWITHSUNDEEP
c
Ben Tomasini
Ben Tomasini

Reputation: 33

In C, how to set an unsigned char array from argv

In the code below, I would like to have the value of myvar be provided by a program argument.

#include <stdio.h>

int main(int argc, const char **argv)
{
    const unsigned char myvar[] = "myvalue";
    return 0;
}

How would I get myvar to contain the value of the string from argv[1]?

Upvotes: 0

Views: 1590

Answers (2)

dbush
dbush

Reputation: 223972

An array cannot be initialized by a pointer or by another array. You can only initialize it with an initializer list or (in the char of a char array) a string constant.

What you can do it copy the contents of another string with strcpy. And since you'll be using this array as a parameter to an encryption function, it will probably need to be a fixed size.

char myvar[8] = { 0 };        // initialize all values to 0
strncpy(myvar, argv[1], 8);   // copy the first 8 bytes

Upvotes: 0

Marco
Marco

Reputation: 7271

If you are only reading, then you can simply copy the address of argv[1] like this:

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

int main(int argc, const char **argv) {
    const unsigned char *myvar = NULL;

    // Be sure to check argc first
    if (argc < 2) {
        fprintf(stderr, "Not enough arguments.\n");

        return EXIT_FAILURE;
    }

    myvar = (const unsigned char *)argv[1];

    printf("myvar = %s\n", myvar);
}

If you want to change myvar then you should copy the string with strncpy or alike.

Upvotes: 3

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