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Using % in C++

Posted on 2004-09-25
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If I wanted to use a % sign to indicate percentage as opposed to mod .........Is this possible? what would I need to know do to do this....
I am writting a code with 12%,15%,and 18% increments.

12% rate  if gross wage <=200.00   ////    or should It be //// .12 rate if gross wage <=200.00
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Question by:bananaamy
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by:Jaime Olivares
Jaime Olivares earned 400 total points
ID: 12152251
I think it can't be done.
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Sys_Prog earned 1600 total points
ID: 12152804
I do not really understand your question

If u are using % sign in your calculation, then for sure it will be interpreted as mod

You will have to explicitly put 0.12 in your caculation if u need it to be interpreted as percentage

This is the case with built in data types

However, for user defned datatepes/classes, you could use operator overloading to treat this as an percentage OR a mod depending on your arguments.

Hoevere, its not recommended to change the meaning of a buit in operator

Amit
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by:Feldspar
ID: 12152820
"x = 0.12 * gross_wage" is the way to go.

Note that if you want to use an integer variable to represent the percentage, i.e. "int rate = 12" then you can use
"x = (rate/100.0) * gross_wage" - the 100.0 is to make sure the compiler uses floating point math, if you use (rate/100) it will use integer math and round that off to 0.
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by:Sys_Prog
ID: 12152900
That's OK bananaamy,

We all keep learning for the whole of life

So....best of luck for your learning adventure

Amit
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