Why Doesn\'t `cout` Print Unsigned Characters Correctly in C ?

cout Failing to Print Unsigned Character: Resolving the Issue

In C , the issue of cout not printing unsigned char is often encountered. To understand this, let's analyze the code example provided:

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

using namespace std;

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

In this code, the unsigned char variable a is assigned the value 1. When printing a using printf, the result is "1." However, the output using cout << a displays a seemingly random character.

The reason for this discrepancy is that unsigned char can store values from 0 to 255. When a is 1, it corresponds to the non-printable ASCII character 'SOH' (start of heading). printf handles non-printable characters differently than cout.

To determine if a character is printable, use the std::isprint function:

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

This will print "0," indicating that 'SOH' is non-printable.

To force cout to print 1, cast a to an unsigned integer:

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

This will successfully print "1."

Understanding the fundamental difference between printf and cout in handling non-printable characters is crucial to resolving this issue. Additionally, std::isprint can help determine whether a character should be printed in a human-readable form.

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