cout not printing unsigned char

Question

I am working on below code:

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

using namespace std;

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

It is printing 1 and some garbage.

Why cout is behaving so?

Answer 1

cout << a is printing a value which appears to be garbage to you. It is not garbage actually. It is just a non-printable ASCII character which is getting printed anyway. Note that ASCII character corresponding to 1 is non-printable. You can check whether a is printable or not using, std::isprint as:

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

It will print 0 (read: false) indicating the character is non-printable

--

Anyway, if you want your cout to print 1 also, then cast a to this:

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

Answer 2

I had a similar issue here that I've long forgotten about. The resolution to this problem with iostream's cout can be done like this:

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

main() {
    unsigned char a;
    a=1;
    printf("%d", a);
    std::cout<< +a << std::endl;

    return 0;
}

instead of casting it back to another type if you want cout to print the unsigned char value as opposed to the ascii character. You need to promote it.

If you noticed all I did was add a + before the unsigned char. This is unary addition that will promote the unsigned char to give you the actual number representation.

User Baum mit Augen is responsible for reminding me of this solution.

Answer 3

You need to typecast a as integer as cout<< (int)(a);. With this you will observe 1 on the output. With cout << a;, the print will be SOH (Start of Heading) corresponding to ascii value of 1 which can't be printed and hence, some special character is observed.

EDIT:

To be more accurate, the cout statement should be cout << static_cast<unsigned>(a) as Nawaz has mentioned.

Answer 4

The C compiler has its own way of defining the type of the printed output, because you can specify the type of the output.

Ex:

uint8_t c = 100;
printf("%d",c);

so you can also print c as an int by %d, or char %c, string %s or a hex value %x.

Where C++ has its own way too, the cout prints the 8-bit values as a char by default. So, you have to use specifiers with the output argument.

You can either use:

1. a + before the name of the output argument

   uint8_t data_byte = 100;
   cout << "val: " << +data_byte << endl;
   

2. use a function cast unsigned(var); like,

    uint8_t data_byte = 100;
    cout << "val: " << unsigned(data_byte) << endl;
   

Answer 5

A char will have a numerical value associated with it. The character represented by 1 is the SOH or Start of header character which is usually not printed (similar to printing the NUL character \0).

You can have a look at an ASCII table to see which character will be printed, but an example with a value such as 65 might help. Where 65 is the A characater.

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


int main() {
    // initialise with an integer
    char a = 65;

    // print the integer representation
    printf("%d\n", a);

    // print as a character
    std::cout << a << std::endl;

    // initialise with a character 
    char b = 'B';

    // print the integer representation
    printf("%d\n", b);
    std::cout << static_cast<int>(b) << std::endl;
    
    // print as a character
    std::cout << b << std::endl;
}

Outputs:

65
A
66
66
B

Answer 6

printf("%u",a); its so simple try it