【算法Ⅰ~Ⅳ(C++实现)】习题3.1 寻找int float double能表示的最大最小值

cout.setf( ios::showbase|ios::uppercase );

cout << "int min = " << numeric_limits<int>::min() << "---" << hex << numeric_limits<int>::min() << endl;
cout.unsetf( ios::hex );
cout << "int max = " << numeric_limits<int>::max() << "---" << hex << numeric_limits<int>::max() << endl;
cout << endl;
cout.unsetf( ios::hex );

cout << "lont int min = " << numeric_limits< long int>::min() << "---" << hex << numeric_limits<long int>::min() << endl;
cout.unsetf( ios::hex );
cout << "long int max = " << numeric_limits<long int>::max() << "---" << hex << numeric_limits<long int>::max() << endl;
cout << endl;
cout.unsetf( ios::hex );

cout << "short int min = " << numeric_limits<short int>::min() << "---" << hex
<< numeric_limits<short int>::min() << endl;
cout.unsetf( ios::hex );
cout << "short int max = " << numeric_limits<short int>::max() << "---" << hex
<< numeric_limits<short int>::max() << endl;
cout << endl;
cout.unsetf( ios::hex );

cout << "float min = " << numeric_limits<float>::min() << "---" << hex << numeric_limits<float>::min() << endl;
cout << "float max = " << numeric_limits<float>::max() << "---" << hex << numeric_limits<float>::max() << endl;
cout << endl;

cout << "double min = " << numeric_limits<double>::min() << "---" << hex << numeric_limits<double>::min() << endl;
cout << "double max = " << numeric_limits<double>::max() << "---" << hex << numeric_limits<double>::max() << endl;
cout << endl;

int min = -2147483648---0X80000000
int max = 2147483647---0X7FFFFFFF

lont int min = -2147483648---0X80000000
long int max = 2147483647---0X7FFFFFFF

short int min = -32768---0X8000
short int max = 32767---0X7FFF

float min = 1.17549E-38---1.17549E-38
float max = 3.40282E+38---3.40282E+38

double min = 2.22507E-308---2.22507E-308
double max = 1.79769E+308---1.79769E+308