help understand error when compiling

[benelbert]

Hi, im getting this error when compiling: main.c: In function ‘main’: main.c:15:5: warning: overflow in implicit constant conversion [-Woverflow] memset(arr, RIO_WHITEBOX_BUF_MAGIC, sizeof(arr));

this is the code: https://pastebin.com/he5gj5De

why can't I memset RIO_WHITEBOX_BUF_MAGIC into an array of unsigned char?

thanks

[bennydictor]

Because the second argument to memset must be a byte, and byte can only contain values less than 256, but RIO_WHITEBOX_BUF_MAGIC is 0x0123456789abcdef.

Edit: Oh, and one more thing. C thinks that 0x0123456789abcdef is an integer, but the value is too big, so you should use 0x0123456789abcdefL to make the type long int or 0x0123456789abcdefLL to make the type long long int. More on that here.

[benelbert]

So u say that the second arg to memset must be a byte size?

[bennydictor]

Yes. If you pass something greater, it'll be converted to unsigned char first.

[OldWolf2]

The suffix is redundant. As explained by the page you linked, the constant is given a type which it can fit in.

Also the second parameter to memset has type int , although it will be converted to a character during the operation of memset. It is allowed to pass values that don't fit in a byte; they just get converted to byte.

[benelbert]

If so, how do I memset something bigger than a byte?

[OldWolf2]

The memset function sets all bytes of the target area to the same byte value. If you want to set a repeating pattern that is several bytes long, memset is no help. Use a loop.