cout 不打印无符号字符

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

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