Cout not printing number

Question

Issue

I'm getting no output from a simple cout, whereas a printf will always print the number:

std::cout << variableuint8;  // prints nothing
printf("%u", variableuint8); // prints the number

I've never run into this behavior before, and while I have a work around I would like to understand it.

Context:

Yay, pointers. So it may be a little more involved than as stated. Here's the context - I don't think it should matter, by the time cout and printf get the info it's dereferenced to a simple unsigned char. This is probably more information than needed to resolve the issue, but I wanted to be complete on the off chance that it was relevant.

typedef unsigned char   UInt8;
typedef unsigned short  UInt16;

typedef struct{
   UInt8   len;
   // some other definitions. sizeof(DeviceNotification) <= 8
}DeviceNotification;

UInt8 somerandomdata[8];
DeviceNotification * notification;

notification = (DeviceNotification *) somerandomdata;

std::cout << "Len: (" << notification->len << ")\n";
printf("Len: (%u)\n", notification->len);

The random data is initialized elsewhere. For this output, the first byte in the array is 0x08:

Output:

Len: ()
Len: (8)

Environment

• Development machine:
  • OS X 10.6.8
  • Compiler LLVM 1.7
  • Xcode 3.2.6
• Test machine:
  • OS X 10.6.8

Answer 1

It's printing the char as a character.

#include <iostream>

int main() {
        unsigned char c = 0x41;
        std::cout << c << std::endl;
        return 0;
}

This prints 'A' to the stdout.

Although in that case, your code should print Len: (*) , and I did have verified it prints Len: (*) .

---

Edit: Since in character encodings like ASCII or UTF-8 your console is likely using, the character corresponding to 8 (Backspace) is not visible, it will look like it's not printed.

(In some cases it may cause the previous character (the ( ) to disappear because it is a backspace character. In DOS it may show ◘)

Answer 2

Have you tried to cast it into int, to avoid depending on the char type that is used by the implementation:

std::cout << (int)variableuint8;

Here is an explanation of what is going on: What is an unsigned char?

To get an idea, your implementation is using unsigned char as char . The ASCII code for 0x08 is the backspace control character, and of course it is not a printable character, that is why you don't see an output in the case of std::cout .

Answer 3

Would std::cout << "Len: (" << (unsigned short)notification->len << ")\\n"; print the correct value? I guess casting to an integer is what might be looking for.