char array not properly initialized

Question

I have am array of chars that I'm using for bytecode. Printing them out one by one should yield the same hex values that you see here:

char toWrite[] = {'\x50','\x48','\xB8','\x00','\x00','\x00','\x00','\x00','\x00','\x00','\x00','\xFF','\xE0' };

When I try to print these values out in a loop, however, they are mangled. What I see instead is:

50 48 ffffffb8 00 00 00 00 00 00 00 ffffffff ffffffe0

Why are these chars printing wrong? I am iterating in a foreach loop, and every single element is passed to

cout << hex << (int)currentChar << endl;

One option is to cast differently, (int)(unsigned char)currentChar — M.M
– M.M, Commented Feb 20, 2017 at 1:52
Please don't advocate c-style casts, static_cast should always be preferred to tell the compiler the intention. — zett42
– zett42, Commented Feb 20, 2017 at 2:10

Dannyu NDos · Accepted Answer · 2017-02-20 01:34:36Z

3

For most systems, char is 8 bits wide and is a signed integer type. Storing \xB8 will make its most significant bit 1, which will make it negative. And casting it to int will also result in a negative value, resulting in 0xffffffb8 if int is 32 bits wide.

You should use unsigned char:

unsigned char toWrite[] = {/*...*/};

Also, static_cast is more preferable than C-style cast:

cout << hex << static_cast<int>(currentChar) << endl;

answered Feb 20, 2017 at 1:34

Dannyu NDos

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