How to cast/convert an unsigned char array to a const unsigned char array in C?

Question

I know that this is a silly question. How do you convert an unsigned char array (with values) to a const unsigned char array?

Below is the code that I have tried. It does not make sense, but a const char does not allow you to change the values (as meant by the word const) but I need a const unsigned char array to pass to a function.

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

int main()
{
    uint8_t a[] = {0, 10, 0, 0, 35, 45, 99, 100,88,67,34,23,56};
    unsigned char k[13];
    for(int i=0; i<sizeof(k)/sizeof(unsigned char); i++)
    {
        k[i] = a[i];
        printf("%X\t%d\n",k[i],k[i]);
    }

    // k = (const)k[]; -- cannot cast char array to const char array 

    return 0;
}

Please let me know what can be done, thanks. Below is the function declaration that k (const unsigned char *k) is to be passed.

int crypto_aead_encrypt(
    unsigned char *c, unsigned long long *clen,
    const unsigned char *m, unsigned long long mlen,
    const unsigned char *ad, unsigned long long adlen,
    const unsigned char *nsec,
    const unsigned char *npub,
    const unsigned char *k
)

Edit the question to show the function declaration. A function that is said to take a “const unsigned char array” as an argument probably actually has a parameter whose type is const unsigned char *, a pointer to a const unsigned char. To pass your array a to it, you merely use a as an argument. An array argument will be automatically converted to a pointer to its first element, yielding an unsigned char *, and, for a properly declared function, that will be automatically converted to a const unsigned char *. — Eric Postpischil
– Eric Postpischil, Commented Aug 2, 2022 at 20:02
As suspected, the parameter is const unsigned char *, not an array. So simply pass a. What problem are you actually having? — Eric Postpischil
– Eric Postpischil, Commented Aug 2, 2022 at 20:46
@Barmar, you can absolutely cast a non-const to a const in C — Elzaidir
– Elzaidir, Commented Aug 2, 2022 at 20:58
@Elzaidir But you can't assign to an array, so you can't convert the arrayk from non-const to const. — Barmar
– Barmar, Commented Aug 2, 2022 at 20:59

Schwern · Accepted Answer · 2022-08-02 21:00:29Z

const here is a qualifier meaning "this function will not alter this variable". It's a guarantee about the function's behavior, not about what you're passing in. It's ok to pass a non-const variable to a function taking a const.

For example, the following is fine. It's ok to pass a char * to const char *. main is allowed to alter the string, but foo is not.

void foo(const char *a) {
    puts(a);
}

int main()
{
    char *str = strdup("testing");
    foo(str);
}

However, the following is an error. It doesn't matter that str was not declared with const, foo promised not to modify its argument.

void foo(const char *a) {
    // error: assignment of read-only location '*a'
    a[0] = 'r';
    puts(a);
}

int main()
{
    char *str = strdup("testing");
    foo(str);
}

You should not go the other way. Passing a const to a non-const function is undefined. It might work, it might not.

// note: expected 'char *' but argument is of type 'const char *
void foo(char *a) {
    a[0] = 'r';
    puts(a);
}

int main()
{
    const char str[] = "testing";

    // warning: passing argument 1 of 'foo' discards 'const' qualifier from pointer target type [-Wdiscarded-qualifiers]
    foo(str);
}

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