does adding 1 in my code doesn't make any change??

Code:

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#include<iostream>
using namespace std;
int main()
{
    using namespace std;
    float a=2e+6;
    cout<<a<<endl;
    float b=a+1;
    cout<<b<<endl;
    cout<<b-a<<endl;
    return 0;
}

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output is a shown in attachment..
if b and a are same then why their subtraction gives 1??
however i have assigned b=a+1.. if it is so then why both are shown same??

Quote:

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Originally Posted by vkash

if it is so then why both are shown same?

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An approximation of the value in scientific notation was printed. You could try to avoid this by placing:

Code:

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cout.precision(7);

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before you start printing.

Quote:

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Originally Posted by vkash

if b and a are same then why their subtraction gives 1??

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There is a difference in what those variables are internally and how you are displaying them. Obviously b and a are not the same internally. If you use the debugger, you would see that the values are different.

Also, what is that stuff at the bottom of the console display ("Process Returned xxx" and the time)? Are you using Visual C++?

Regards,

Paul McKenzie

Quote:

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Originally Posted by laserlight

An approximation of the value in scientific notation was printed. You could try to avoid this by placing:

Code:

---

cout.precision(7);

---

before you start printing.

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thanks laser light for your cout.precision....

Quote:

---

There is a difference in what those variables are internally and how you are displaying them. Obviously b and a are not the same internally. If you use the debugger, you would see that the values are different.

Also, what is that stuff at the bottom of the console display ("Process Returned xxx" and the time)? Are you using Visual C++?

Regards,

Paul McKenzie

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errrrrr....
sorry it's not VC++.. recently my windows xp corrupted so i am on linux(no VC++ here) so it's codeblocks..
after all thanks for your answer.thanks for telling me that internally both a and b are not same... :)