Char automatically converts to int (I guess)

Question

I have following code

char temp[] = { 0xAE, 0xFF };
printf("%X\n", temp[0]);

Why output is FFFFFFAE, not just AE?

I tried

printf("%X\n", 0b10101110);

And output is correct: AE.

Suggestions?

Answer 1

The answer you're getting, FFFFFFAE, is a result of the char data type being signed. If you check the value, you'll notice that it's equal to -82, where -82 + 256 = 174, or 0xAE in hexadecimal.

The reason you get the correct output when you print 0b10101110 or even 174 is because you're using the literal values directly, whereas in your example you're first putting the 0xAE value in a signed char where the value is then being sort of "reinterpreted modulo 128", if you wanna think of it that way.

So in other words:

    0   =    0 = 0x00 
    127 =  127 = 0x7F
    128 = -128 = 0xFFFFFF80
    129 = -127 = 0xFFFFFF81
    174 =  -82 = 0xFFFFFFAE
    255 =   -1 = 0xFFFFFFFF
    256 =    0 = 0x00

To fix this "problem", you could declare the same array you initially did, just make sure to use an unsigned char type array and your values should print as you expect.

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


int main()
{
    unsigned char temp[] = { 0xAE, 0xFF };

    printf("%X\n", temp[0]);
    printf("%d\n\n", temp[0]);

    printf("%X\n", temp[1]);
    printf("%d\n\n", temp[1]);

    return EXIT_SUCCESS;
}

Output:

AE
174

FF
255

Answer 2

https://linux.die.net/man/3/printf

According to the man page, %x or %X accept an unsigned integer. Thus it will read 4 bytes from the stack.

In any case, under most architectures you can't pass a parameter that is less then a word (i.e. int or long) in size, and in your case it will be converted to int.

In the first case, you're passing a char, so it will be casted to int. Both are signed, so a signed cast is performed, thus you see preceding FFs.

In your second example, you're actually passing an int all the way, so no cast is performed.

If you'd try:

printf("%X\n", (char) 0b10101110);

You'd see that FFFFFFAE will be printed.

Answer 3

When you pass a smaller than int data type (as char is) to a variadic function (as printf(3) is) the parameter is converted to int in case the parameter is signed and to unsigned int in the case it is unsigned. What is being done and you observe is a sign extension, as the most significative bit of the char variable is active, it is replicated to the thre bytes needed to complete an int.

To solve this and to have the data in 8 bits, you have two possibilities:

• Allow your signed char to convert to an int (with sign extension) then mask the bits 8 and above.

  printf("%X\n", (int) my_char & 0xff);
  

• Declare your variable as unsigned, so it is promoted to an unsigned int.

  unsigned char my_char;
  ...
  printf("%X\n", my_char);
  

Answer 4

This code causes undefined behaviour. The argument to %X must have type unsigned int, but you supply char.

Undefined behaviour means that anything can happen; including, but not limited to, extra F's appearing in the output.