Uint8_T Can't Be Printed With Cout

uint8_t can't be printed with cout

It doesn't really print a blank, but most probably the ASCII character with value 5, which is non-printable (or invisible). There's a number of invisible ASCII character codes, most of them below value 32, which is the blank actually.

You have to convert aa to unsigned int to output the numeric value, since ostream& operator<<(ostream&, unsigned char) tries to output the visible character value.

uint8_t aa=5;

cout << "value is " << unsigned(aa) << endl;

std::cout deal with uint8_t as a character

Is this behavior a standard

The behavior is standard in that if uint8_t is a typedef of unsigned char then it will always print a character as std::ostream has an overload for unsigned char and prints out the contents of the variable as a character.

Should not be a better way to define uint8_t that guaranteed to be dealt with as a number not a character?

In order to do this the C++ committee would have had to introduce a new fundamental type. Currently the only types that has a sizeof() that is equal to 1 is char, signed char, and unsigned char. It is possible they could use a bool but bool does not have to have a size of 1 and then you are still in the same boat since

int main()
{
    bool foo = 42;
    std::cout << foo << '\n';
}

will print 1, not 42 as any non zero is true and true is printed as 1 but default.

I'm not saying it can't be done but it is a lot of work for something that can be handled with a cast or a function

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C++17 introduces std::byte which is defined as enum class byte : unsigned char {};. So it will be one byte wide but it is not a character type. Unfortunately, since it is an enum class it comes with it's own limitations. The bit-wise operators have been defined for it but there is no built in stream operators for it so you would need to define your own to input and output it. That means you are still converting it but at least you wont conflict with the built in operators for unsigned char. That gives you something like

std::ostream& operator <<(std::ostream& os, std::byte b)
{
    return os << std::to_integer<unsigned int>(b);
}

std::istream& operator <<(std::istream& is, std::byte& b)
{
    unsigned int temp;
    is >> temp;
    b = std::byte{b};
    return is;
}

int main()
{
    std::byte foo{10};
    std::cout << foo;
}

Improper printing of uint8_t variable [duplicate]

uint8_t is not a separate data type. On systems that provide it the actual type is aliased to some standard data type, most commonly, an unsigned char.

Operator << provides an overload that takes unsigned char, and prints it as a character. When you are printing your uint8_t variable as an int, cast it to an int for printing:

std::cout << "My ID is "<< int(myID) <<std::endl;
//                         ^^^^^

Demo.

uint8_t values issues with vectors

I was able to find the solution. the old logic was too complex and not working with different resolutions. So I made a simpler one that does need to take care of rounding up issues.
first took each point and tried to go replace values index by index.

for(auto i: landmarks)
{
    for(int m = (i[0]/grid_resolution)-2; m <= (i[0]/grid_resolution)+2; m++)
    {
        for(int n = (i[1]/grid_resolution)-2; n <= (i[1]/grid_resolution)+2; n++)
        {
            grid[m][n] = 255;
        }
    }
}

I can't print out sentence with 'cout' in c++

Change AddingTasks by:

int AddingTasks()
{
    cout << "type your task here: ";
    cin >> task;
    tasksList.push_back(task);
    cout << tasksList[tasksList.size() - 1] << " added " << endl;
    return 0;
}

What is happening is, you're trying to access an object that does not exist in the vector.

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