how to round off the return value

Try this:

unsigned int CalculatePercentage(int TotalComponent,int PartialComponent){
return 100.0f*PartialComponent/TotalComponent;
}

You might want to read a C book, specifically a section on types and conversions between them. Otherwise prepare for more surprises like this one. Regards, - mike

Try this:

unsigned int CalculatePercentage(int TotalComponent,int PartialComponent){
return 100.0f*PartialComponent/TotalComponent;
}

You might want to read a C book, specifically a section on types and conversions between them. Otherwise prepare for more surprises like this one. Regards, - mike

hello mike im sori thats suppose to be

return per;

okie i'll try your code thanks for the reply

hello mike tried your code how will i know if it return 38.3% i transfered the code in c

 per=CalculatePercentage(235,90);
 printf("Percent %d",per);

the return value is 38% the .3 is disregarded

even if i change per to float

main(){
float per;
 per=CalculatePercentage(235,90);
 printf("Percent %f",per);
}

the return value is 38% the .3 is disregarded

Isn't it obvious?
You really really need to read a book on C. Getting answers to questions like these in forums only postpones the inevitable - reading the books.

- mike

 per = CalculatePercentage( 235, 90 );
 printf( "Percent %d", per );

the return value is 38% the .3 is disregarded

As Mike says, the reson for this should be obvious - it's textbook stuff.

main() {
 float per;
 per = CalculatePercentage( 235, 90 );
 printf( "Percent %f", per );
}

the return value is 38% the .3 is disregarded

How have you checked the return value?

Obviously, if the return value has been truncated before calling printf, then just changing the format specifier in the printf cannot magically re-generate the discarded information!

Or maybe you just changed your source code, and forgot to rebuild it...?

PS Note how the addition of a little whitespace makes your code so much easier to read!