Why Does `cout` Misbehave When Printing Unsigned Char Values?

Misbehaving cout: Why Unsigned Char Output Causes Confusion

In programming, using the cout function to print unsigned char values can sometimes lead to unexpected results. Let's delve into the issue and explore why this happens.

Problem Description

Consider the following code snippet:

#include<iostream>
#include<stdio.h>

using namespace std;

main() {
    unsigned char a;
    a=1;
    printf("%d", a);
    cout<<a;
}

When executed, this code prints 1 followed by some seemingly random characters. But why does the output behave this way?

Answer

The confusion stems from the character corresponding to ASCII value 1. This character is a non-printable control character and is typically not visible when displayed as text. As a result, cout prints it as garbage. To confirm this, we can use the isprint function to check if the character is printable:

std::cout << std::isprint(a) << std::endl;

This will print 0 (false), indicating that the character is non-printable.

To rectify the issue and display the value 1 in both printf and cout, we can cast the unsigned char to an unsigned integer:

cout << static_cast<unsigned>(a) << std::endl;

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