How to Print Unsigned 8-bit Variables Correctly in C ?

Manipulating Unsigned 8-bit Variables in C

When dealing with unsigned 8-bit variables in C , you may encounter challenges when attempting to print them correctly using ostream. By default, ostream interprets these variables as char, leading to unintended output.

Understanding the Issue

Unsigned char and uint8_t are both data types representing 8-bit unsigned integers. However, in C , ostream automatically treats these values as characters, resulting in incorrect hex representation. The following code snippet demonstrates the issue:

unsigned char a = 0;
unsigned char b = 0xff;
cout << "a is " << hex << a << "; b is " << hex << b << endl;

The output will be:

a is ^@; b is 377

Instead of the desired output:

a is 0; b is ff

Resolving the Problem

To correctly print unsigned char variables as hex, you need to explicitly cast them to int before printing them:

cout << "a is " << hex << (int) a << "; b is " << hex << (int) b << endl;

This will produce the intended output:

a is 0; b is ff

If you desire leading zero padding in your output, you can use the following code:

#include <iomanip>

//...

cout << "a is " << setw(2) << setfill('0') << hex << (int) a ;

Alternative Solutions

MartinStettner's solution provides an alternative method to resolve the issue:

std::cout << std::hex << static_cast<unsigned int>(a);

This method avoids the need for explicit casting and provides a more concise solution.

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