setprecision()

if u want to setprecision you need to do it like this:

cout<<setprecision(3)<<..........

is there a way that you only need to say it once so it stay in memory?????

Use the stream member precision().

Code:

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std::cout.precision(3);

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EDIT: You can find other interesting iostream function members here:
http://www.cplusplus.com/ref/iostream/ostream/
and
http://www.cppreference.com/cppio/index.html

Code:

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      if(d==0)std::setprecision(0);
      if(d==1)std::setprecision(1);
      if(d==2)std::setprecision(2);
      if(d==3)std::setprecision(3);
      if(d==4)std::setprecision(4);
      if(d==5)std::setprecision(5);
      if(d==6)std::setprecision(6);
      if(d==7)std::setprecision(7);
      if(d==8)std::setprecision(8);
      if(d==9)std::setprecision(6);

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still not doing it if i do 1.225*1.252245 with 1precision i still not get that precision:s

I don't think you understood:

Code:

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double value = 23.4567;

std::cout.precision(3);

std::cout << value << std::endl; // outputs 23.4

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ow , i missread it

but now it works

but , i want that the setprecision only is true behind the , so:
precision(1)
99.11*1=99.1

how do you that?

The number of digits of an integer can be calculated by adding one to the common logarithm of that number. In C++ the common logarithm is given by log10() defined in <cmath>.

So... one way is:

Code:

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#include <cmath>
#include <iostream>

using namespace std;

int main {
    double value = 99.11;

    int numdigits = log10(value) + 1;  // returns 2. The number of whole digits in value

    int myprecision = numdigits + 1;  // You want one more for one decimal place

    cout.precision(myprecision);

    cout << value << endl;  // outputs 99.1
}

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EDIT: removed an std:: ... force of habit.