rupert wrote:
Quote:
i've got the following code:
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#include <iostream>
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#include <string>
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#include <vector>
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#include <iomanip>
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using namespace std;
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int main(double argc, char* argv[ ]) {
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>
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double r = 0.01;
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double s = 1.04;
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double t = 2.23;
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double u = 2.23;
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>
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vector<doublev;
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v.push_back(r);
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v.push_back(s);
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v.push_back(t);
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v.push_back(u);
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>
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streamsize prec = cout.precision();
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cout << "precision of doubles:..." << prec << endl;
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double sum = v[0]+v[1]+v[2]+v[3];
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cout << "actual number:..." << sum << endl;
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cout <<"presision set at 2:..."<< setprecision(2) << sum << endl
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<< "precision set at "<< prec << ":..." << setprecision(prec)
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<< " "<< sum << endl;
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>
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return EXIT_SUCCESS;
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}
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and the value of precision is 6 when there's only 3 sigificant figures.
I would have exptected precision() to return an integer of value 3, any
ideas?
When you say
double r = 1.04;
this is just a double to the compiler. Numbers do not have an intrinsic
precision. If you need that, you need to use interval arithmetic.
The precision that you print is not a property of the doubles but of the
stream that you are writing to. It just happens to default to 6. It would
do so regardless of which numbers (if any) you decide to write there.
Best
Kai-Uwe Bux
