#include <stdio.h> #include <limits.h> // int comprising limit #include <float.h> // limit float comprising void main () { the printf ( "% of The D bytes of int IS, The bytes of a float IS% D \ n-", the sizeof (int), the sizeof (a float)); the printf (" of The max of int IS% D, The min of int IS% D \ n-", INT_MAX, INT_MIN); the printf ( "the max of Float is% f , the min of Float is% .100f \ n", FLT_MAX, FLT_MIN); //%.100f display bit decimal 100% f default 6 decimal places getchar (); / * the Output of the bytes of int. 4 iS, iS the bytes of a float. 4 of the iS 2147483647 max of int, int the iS min of -2147483648 The max of Float is 340282346638528860000000000000000000000.000000, the min of Float is 0.0000000000000000000000000000000000000117549435082228750000000000000000000000000000000000000000000000 for the above analysis of the results, and flaot int why the same 4 bytes, will not represent the return Like it? Since each bit of the data bit int, 4 bytes, 32-bit, as a sign bit, the other data bit is valid, the limited 32-bit only indicates 2147483647. The float 4 bytes, 32-bit, a sign bit, the other 31 bits, a portion of valid data bits, exponent part, the base part, due to the index range, will represent larger FLT_MAX is The maximum positive and negative numbers, FLT_MIN is the minimum negative number c is a double default language is generally represented by floating-point numbers. This expression to express the real number using scientific notation, i.e., using a mantissa (Mantissa), a base (Base), an index (Exponent,) and a symbol representing the positive and negative to express the real fixed point comprising integer and fractional fixed point (0.123443 no integer part) * / }