问题
I am trying to create an array of 1 byte values, using unsigned chars to hold the values.
my code looks like this:
unsigned char state[4][4] = { {0xd4, 0xe0, 0xb8, 0x1e},
{0xbf, 0xb4, 0x41, 0x27},
{0x5d, 0x52, 0x11, 0x98},
{0x30, 0xae, 0xf1, 0xe5}};
Where each value is 2 hex digits which makes a byte. However, when I attempt to print the array out using
cout << "\nInitial State \n------------------------------------------------------\n";
for(int x = 0; x < 4; x++)
{
for(int y = 0; y < 4; y++)
cout << hex << setw(2) << setfill('0') << state[x][y] << " ";
cout << "\n";
}
I don't get the hex value, but rather weird symbols
Initial State
---------------------
0Ô 0à 0¸ 0
0¿ 0´ 0A 0'
0] 0R 0 0˜
00 0® 0ñ 0å
I understand the 0s come from set(w) and setfill('0') but I don't understand why it isn't showing the proper values. However, if I make the state an unsigned long, I then get the proper values.
So I'm confused. If a char holds exactly one byte, why can't the char display the byte as a hex value?
回答1:
cout is specialized to output the char type as a character, not a number. That goes for unsigned chars too. You get around it by casting to an unsigned int.
cout << hex << setw(2) << setfill('0') << static_cast<unsigned int>(state[x][y]) << " ";
来源：https://stackoverflow.com/questions/33764309/why-cant-an-unsigned-char-properly-display-a-hex-value