CODEWITHSUNDEEP
c++integer-division
user1544067
user1544067

Reputation: 1764

Get percent of number in c++

How can I calculate several percent of int?
for example I want to get 30% from number, if I will use this example of code, I will get wrong answer:

int number = 250;
int result = (number / 100) * 30;

result will be 60, and the real answer is 75, is there any way to calculate it?

Upvotes: 11

Views: 145760

Answers (11)

Welinton Fernandes
Welinton Fernandes

Reputation: 1

This is a First-degree equation, simpler than it looks!

float annualGrossSalary;
cout << "Please enter your annual salary: ";
cin >> annualGrossSalary;
float annualNetSalary = (annualGrossSalary - (annualGrossSalary / 100) * 40);
float monthyGrossSalary = annualGrossSalary / 12;
float monthyNetSalary = annualNetSalary / 12;
cout << "Your monthly gross salary is " << monthyGrossSalary << "\n";
cout << "Your monthly net salary is " << monthyNetSalary << "\n";
cout << "In 10 years you will earn " << annualNetSalary * 10 << "\n";

Upvotes: -1

Tim
Tim

Reputation: 11

Another possibility, good in Windows when working with 32 bit numbers.

int num = 750;

int result;

result = MulDiv( num, 30, 100 );

https://learn.microsoft.com/en-us/windows/win32/api/winbase/nf-winbase-muldiv

Upvotes: 1

marvin orsua
marvin orsua

Reputation: 1

score/items*30 ex;

15%

int score, items;
cout<<"input your score"<< score<<endl;
cin>>score;
cout<<"input number of items"<<items<<endl;
cin>>items;
float total=score/items*15
cout<<"you got "<<total<<""<<endl;

Upvotes: -1

user3506720
user3506720

Reputation: 1

The whole program is like:

double number , result;

cout << "Enter the number:";
cin >> number;

result = number * 30.0 / 100;
cout << "Result = " << result << endl;

and you will get your answer....

Upvotes: 0

Joni
Joni

Reputation: 111219

Multiply before dividing:

int result = number * 30 / 100;

The reason you get the result you get is that division with integer types produces an integer result: 250 / 100 is 2. If you multiply before dividing you still get an integer result, but at least you haven't lost data in intermediate steps. If you have to deal with really huge numbers there is a danger of overflowing the range permitted by int though.

Alternatively you can use floating point arithmetic, where division can produce fractions of integers:

int result = number * 0.30;

This may produce unexpected results though, so you're better off using integers only as above. Or write 3.0/10 instead of 0.30.

Upvotes: 18

user1544067
user1544067

Reputation: 1764

Now I think about some solution: 

int number = 250;
int result = number * 100;
result = (result / 100) * 30;
result /= 100;

Upvotes: 0

Sceptical Jule
Sceptical Jule

Reputation: 917

Switching the operands (as others suggested) would work too, but just in case you do not want to, there's another solution:

int number = 250;
int result = static_cast<double>(number) / 100 * 30;

Upvotes: 4

Ms. Nobody
Ms. Nobody

Reputation: 1219

Use float:

float number = 250;
float result = (number / 100.0) * 30

Also just putting there the 100.0 with deciaml point might be enough.

Because if you do it your way 250 / 100 in integer equals 20( you can put 100 in 200 only twice and integer doesn't care about the rest 50) and times 30 = 60.

Upvotes: 0

CodeCrusader
CodeCrusader

Reputation: 579

Try this,

int number = 250;
float result = number  * (float)(30/100.0);

Upvotes: 2

Emil Sit
Emil Sit

Reputation: 23512

You need to perform floating point math, otherwise the compiler does all the arithmetic here as integers.

You might try

int result = (number / 100.0) * 30;

Upvotes: 0

Mats Petersson
Mats Petersson

Reputation: 129314

Assuming the numbers are small(ish), you can just turn it around:

 int result = (number * 30) / 100;

(Parenthesis not required, but helps clarify).

This won't work if either of the numbers are several million, but should be fine for numbers smaller than that.

Upvotes: 3

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