std::cout deal with uint8_t as a character [duplicate]

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• uint8_t can't be printed with cout 8 answers

If I run this code:

std::cout << static_cast<uint8_t>(65);

It will output:

A

Which is the ASCII equivalent of the number 65. This is because uint8_t is simply defined as:

typedef unsigned char uint8_t;

Is this behavior a standard? I can not understand that if I want to print the value of a uint8_t variable, it will be printed 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.

P.S. I am using MSVS 2013.