Question

I have the hex value 0x5a800000000b and I’m trying to get a printf statement in C to print it to the console.

So far (because I’m useless in C) I’m able to get the ‘b’ to print, using the syntax:

printf("Hex value%x\n", value);

The value is stored in an integer type U32, but after trying all different combinations of %llx, %lx, I just keep getting compiler warnings.

I’m guessing that I’m getting the printf syntax wrong, but I can’t seem to find the right % option, can someone help me out?

Thanks

Answer 1

It’s not a problem with the printf.

The problem is that a 32 bits variable cannot hold the value 0x5a800000000b. 32 bits can hold only 8 hex digits: 0x0000000b. Hence the b on output.

To store such a large value, you should use a 64 bits variable.

long long value = 0x5a800000000bLL

Note also the double L at the end. It tells the compiler that the constant is also a long long.

Then you can use %llx in the printf format string.