【演算法Ⅰ~Ⅳ(C++實現)】習題3.1 尋找int float double能表示的最大最小值

包含標頭檔limits.h

numeric_limits<類型>::max()   求某資料類型的最大值（min()則求最小值）

浮點數的最小值是能表示的最小正數

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;

結果：

int min = -2147483648---0X80000000int max = 2147483647---0X7FFFFFFFlont int min = -2147483648---0X80000000long int max = 2147483647---0X7FFFFFFFshort int min = -32768---0X8000short int max = 32767---0X7FFFfloat min = 1.17549E-38---1.17549E-38float max = 3.40282E+38---3.40282E+38double min = 2.22507E-308---2.22507E-308double max = 1.79769E+308---1.79769E+308