C++ switch statement count output [closed] - c++

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Closed 11 years ago.
I am having trouble forming the output statement for this program. The correct output should print the same input number that its numerical value holds. .

Answer 1

It looks like you want something similar to this:
#include <iostream>
using namespace std;
int main()
{
int standard, count1, count2, count3, count4, count5;
// Initialize your int's
standard = count1 = count2 = count3 = count4 = count5 = 0;
do {
cout<< "Enter value: ";
cin >> standard;
// Switch through your cases
switch(standard)
{
case 1:
count1++;
break;
case 2:
count2++;
break;
case 3:
count3++;
break;
case 4:
count4++;
break;
case 5:
count5++;
break;
}
} while(standard != 0);
// Do this type of loop for all your count vars
for(int i=0;i<count1;++i)
cout<< "1";
for(int i=0;i<count2;++i)
cout<< "2";
// And so on...
return 0;
}
For the for loops, note that you are going from 0 to < the total number. If you went to <= you would go passed that number. The range [0,4] = 5 (or in our case [0,5) for the set of integers) and so does [1,5] so remember to mind your offsets. Good luck!
EDIT: Also recognize that the position of the switch statement has been moved. By placing it in the loop, you actually count every time it runs through the loop. For instance, the loop "blocks" (in other words, it waits for input) at your cin line. When it receives an input (though you have no error-checking, so a non-int value may cause a segfault), it will then check it against the switch statement and increment accordingly.
By placing the switch statement outside the loop (as you initially had it), please realize that nothing happens. You have no default condition in your switch statement, nor a case for handling 0. If you check your switch after the loop, standard == 0 since that is how you exited the loop in the first place (therefore, standard's last value will be retained).
Furthermore, revisit the for loop syntax. This can be done in several ways, there are many good articles here on SO which can help you further by using the search function, but it basically works like this:
for(STARTING_POINT;SOME_CONDITION;SOME_CHANGE);
STARTING_POINT = The value where you should start your loop
SOME_CONDITION = When this condition is true, then the loop will terminate
SOME_CHANGE = how to change your variable (i.e. from starting point) until it reaches the condition
So a loop which looks like this:
for(int i=1;i<=5;++i){ ... }
means to initialize int i to value 1
Until i <= 5 run through your loop
Then, on every iteration, increment i by one (which is ++i or i++)
Finally, as you can see, I added an initialization line to my int's. What this does is it gives your variables some starting value rather than garbage value.
I hope this helps
Regards,
Dennis M.

Answer 2

You definitely need to move the switch statement within the while loop (Edit: this was based on the OP's initial posting of the code).

Answer 3

I really don't understand what you're trying to accomplish, but this part here:
for (count1=0; count1<=5; count1++)
cout << "1";
Is going to print out '1' 6 times in a row, regardless of the input. Could you explain a bit more what you're trying to do?

Answer 4

Here's my interpretation: You want to input a sequence of numbers (ending with zero), then output those exact same numbers, using loops.
Here's some pseudocode to give you a hint.
Declare an array of int's large enough to hold all your input numbers.
Declare a counter c = 0.
Begin loop while input is not 0
Input number from user and store it in array[c].
Increment counter c.
End loop
Begin loop i from 0 to c-1
Output number that is in array[i].
End loop
No need to use a switch statement and a counter for each possible number. Just store each input in it's own spot in an array.