Question

How could insert text by argument and automatically transform it to hex?

I tried with:

unsigned char aesKey[32] = argv[1];

but get errors

The output would be like this:

unsigned char aesKey[32] = {
    0x53, 0x28, 0x40, 0x6e, 0x2f, 0x64, 0x63, 0x5d, 0x2d, 0x61, 0x77, 0x40, 0x76, 0x71, 0x77, 0x28, 
    0x74, 0x61, 0x7d, 0x66, 0x61, 0x73, 0x3b, 0x5d, 0x66, 0x6d, 0x3c, 0x3f, 0x7b, 0x66, 0x72, 0x36
};

unsigned char *buf;

aes256_context ctx;
aes256_init(&ctx, aesKey);

for (unsigned long i = 0; i < lSize/16; i++) {
    buf = text + (i * 16);
    aes256_encrypt_ecb(&ctx, buf);
}

aes256_done(&ctx);

Thanks in advance

Answer 1

In C and C++, when you have code like

char name[]="John Smith";

The compiler knows at compile time what the size of that char array, and all the values will be. So it can allocate it on the stack frame and assign it the value.

When you have code like
char * strptr = foo();
char str[] = strptr;

The compiler doesn’t know what the size and value of the string pointed by strptr is. That is why this is not allowed in C/C++.

In other words, only string literals can be assigned to char arrays, and that too only at the time of declaration.

So

char name[] = "John Smith";

is allowed.

char name[32];
name = "John Smith";

is not allowed.

Use memcpy

So you could use memcpy. (Or use c++ alternative that others have alluded to)

unsigned char *aesKey;
size_t len = (strlen(argv[1])+1)*sizeof(unsigned char);
aesKey = malloc(len);
memcpy(aesKey, argv[1], len);

The old solution

(here is my previous answer, the answer above is better)
So you need to use strncpy.

unsigned char aesKey[32];
strncpy((char *) aesKey, argv[1], 32);

Notice the routine is strncpy not strcpy. strcpy is unsafe. (Thanks PRouleau for the arg fix)

If strncpy is not available in Visual Studio then you may have to try strcpy_s (Thanks Google: user:427390)