That is, the sum of the number of integers and decimals should not exceed 6 bits, if the fractional part is too long, then the system will be rounded to approximate 6 bits, if the integer part is too long, then also approximate to become 6 bit
For example:
cout<<123.4567 The result is 123.457,
cout<<123.4564 The result is 123.456,
cout<<123456.1 The result is 123456,
cout<<123456.5 the result is 123457.
For the constant string, theoretically infinite, but received the limitations of VC + + compiler, if the constant string length to the point of causing the line, it will lead to compilation error, that is,cout< <后接常串的话串的长度限制为编译器中的1行。>
For cout< <后加变量的情况：>
First of all, the case with decimals, with a slightly different number of constants, its output is related to the type of the variable, int type by prototype output, double type 6 bits (including 6 bits) according to the prototype output, more than 6 in the form of scientific and technological law output, other types are similar. By the way, the range of int is -214783648~214780647, a total of 2^32 the number of times, in the assignment even if the cross-border, the compiler will not error, nor warning, so be careful, there is a common sense, if the number of global population (about 6 billion), should be used double or long Double type, other types are not possible; for variable strings, that is, char arrays, I don't see any restrictions because I write:
Char v[50000];
memset (v,1,49999);
cout<<>
, there are countless small round faces on the screen (ASCII code 1 characters), even if you use memset cross-border assignment, there is no problem.
The above describes the Paragon NTFS for Mac PHP cout< <的一点看法，包括了paragon ntfs="" for="">
