从二进制文件读取uint8_t

Question

I have a file where are binary numbers and I have to read them. I use:

ifstream data("date.txt", ios_base::binary);

int count_16 = 0; //count how many uint16_s I have already read
uint16_t numbers_16; // Allocate storage for uint16_s
int count_8 = 0;
uint8_t numbers_8;
int count_char = 0;
char name[20];

data.seekg(0U, ios_base::beg); // Move the input position indicator to the beginning of the file for reading
data.read(reinterpret_cast<char*>(&numbers_16), sizeof(uint16_t)); // into numbers_16
cout << numbers_16; 
count_16++;

data.seekg(count_16 * sizeof(uint16_t) + count_8 * sizeof(uint8_t) + count_char * sizeof(name+1));
data.read(reinterpret_cast<char*>(&numbers_16), sizeof(numbers_16)); // Read the element into number
cout << numbers_16 << endl;
count_16++;

everything works untill that:

data.seekg(count_16 * sizeof(uint16_t) + count_8 * sizeof(uint8_t) + count_char * sizeof(name+1));
data.read(reinterpret_cast<char*>(&numbers_8), sizeof(numbers_8)); // Read the element into number
cout << numbers_8 << endl;
count_8++;

And here I don't get any number or anything readable. I don't know why this method works for uint16_t but not for uint8_t. Can someone explain why it is wrong and how to read uint8_t from file?

Answer 1

It turned out it is good why of reading. The only mistake was here:

cout << numbers_8 << endl;

I just wrote (int) before numbers_8 and now it looks fine.