Question

The code snippet

char c1;
char c2;
c1 = 0X7F;
c2 = 0X80;
printf("c1 = %X\nc2 = %X\n", c1, c2);

outputs

c1 = 7F
c2 = FFFFFF80

why is it that c2 does not print as 80? (i’m guessing it’s something to do with the fact that the most significant bit of 0X80 is 1 while the MSB of 0x7F is 0). how can i get c2 to print as simply 80?

Answer 1

Because printf is a variadic function. Arguments to variadic functions undergo the default promotions; one effect of this is that char arguments are converted to int. On your platform, char is signed, so c2 is actually negative. printf is therefore printing all the leading ones that occur in the int (these are due to two’s-complement representation).

There are (at least) 3 methods to work around this:

• Use %02X rather than %X as your printf format specifier.
• Use unsigned char rather than char (this gets promoted to unsigned int, so no leading ones)
• Explicitly convert each of your chars to int, and then mask: (int)c2 & 0xFF.

[Note: The corollary of this is that the behaviour of char c2 = 0x80 is actually implementation-defined; if char is signed then this will overflow.]