why only the succeeding value adds up???

yeah, but why does that mean more power?

you could replace the whole switch with a small array (no more conditionals):

Code:

  int total[]={0/*unused*/,25,55,45,...};
  printf(...)
  scanf("%u",&order);
  sum+=total[order%the_appropriate_value];

Originally Posted by root4

you could replace the whole switch with a small array (no more conditionals):

Code:

  int total[]={0/*unused*/,25,55,45,...};
  printf(...)
  scanf("%u",&order);
  sum+=total[order%the_appropriate_value];

Although it probably is a good idea to range-check the values.

--
Mats

You CAN of course use if/else instead of switch/case - but consider what the compiler does if you have a value 8 for this:

Code:

if(order==1){
       total=25.00;
       }
else if (order==2){
	total=55.00;
	}
else if(order==3){
	total=45.00;
	}
else if (order==4){
	total=35.00;
	}
else if(order==5){
	total=50.00;
	}
else if (order==6){
	total=70.00;
	}
else if(order==7){
	total=80.00;
	}
else if(order==8){
	total=15.00;
	}

Unless it figures out that it's a "switch in disguise", it will compare down the chain until it finds the one that matches. In a switch/case block, the compiler will (most of the time) generate a jump-table, which means that no matter how many variants it is, there will be a O(1) type, rather than O(n) that the if/else version does.

Note that reasonably modern/clever compilers, such as gcc 3.4 can not figure out that the above is a "switch/case in disguise". I just tried out the above code as a if/else, and as a switch/case, and for the average case, the code is simply shorter and better for the switch/case statement.

But if it's part of your assignment to use if-else, then do so.

--
Mats