Larger than and less than in C switch statement

Question

I'm trying to write a code that has a lot of comparison

Write a program in “QUANT.C” which “quantifies” numbers. Read an integer “x” and test it, producing the following output:

x greater than or equal to 1000 print “hugely positive”
x from 999 to 100 (including 100) print “very positive”
x between 100 and 0 print “positive”
x exactly 0 print “zero”
x between 0 and -100 print “negative”
x from -100 to -999 (including -100) print “very negative”
x less than or equal to -1000 print “hugely negative”

Thus -10 would print “negative”, -100 “very negative” and 458 “very positive”.

Then I tried to solve it using a switch statement, but it didn't work. Do I have to solve it using an if statement or there is a method to solve it using a switch statement?

#include <stdio.h>

int main(void)
{
    int a=0;
    printf("please enter a number : \n");

    scanf("%i",&a);

    switch(a)
    {
        case (a>1000):
            printf("hugely positive");
            break;

        case (a>=100 && a<999):
            printf("very positive");
            break;

        case (a>=0 && a<100):
            printf("positive");
            break;

        case 0:
            printf("zero");
            break;

        case (a>-100 && a<0):
            printf("negative");
            break;

        case (a<-100 && a>-999):
            printf("very negative");
            break;

        case (a<=-1000):
            printf("hugely negative");
            break;

    return 0;
}

Solution 1

There is no clean way to solve this with switch, as cases need to be integral types. Have a look at if-else if-else.

Solution 2

A switch-less and if-else-less method:

#include <stdio.h>

int main(void)
{
    int a=0, i;
    struct {
        int value;
        const char *description;
    } list[] = {
        { -999, "hugely negative" },
        { -99, "very negative" },
        { 0, "negative" },
        { 1, "zero" },
        { 100, "positive" },
        { 1000, "very positive" },
        { 1001, "hugely positive" }
    };

    printf("please enter a number : \n");
    scanf("%i",&a);

    for (i=0; i<6 && a>=list[i].value; i++) ;
    printf ("%s\n", list[i].description);

    return 0;
}

The for-loop contains no code (there is just an empty statement ;) but it still runs over the array with values and exits when the entered value a is equal to or larger than the value element in the array. At that point, i holds the index value for the description to print.

Solution 3

If you are using gcc, you have "luck" because it supports exactly what you want by using a language extension:

#include <limits.h>
...

switch(a)
{
case 1000 ... INT_MAX: // note: cannot omit the space between 1000 and ...
    printf("hugely positive");
   break;
case 100 ... 999:
    printf("very positive");
   break;
...
}

This is non-standard though, and other compilers will not understand your code. It's often mentioned that you should write your programs only using standard features ("portability").

So consider using the "streamlined" if-elseif-else construct:

if (a >= 1000)
{
    printf("hugely positive");
}
else if (a >= 100)
{
    printf("very positive");
}
else if ...
...
else // might put a helpful comment here, like "a <= -1000"
{
    printf("hugely negative");
}

Solution 4

(a>1000) evaluates to either 1 [true] or 0 [false].

Compile and you will get the error:

test_15.c:12: error: case label does not reduce to an integer constant

This means, you have to use an integer constant value for the case labels. An If-else if-else loop should work just fine for this case.

Solution 5

Use:

switch (option(a)) {
    case (0): ...
    case (1): ...
    case (2): ...
    case (n): ...

Where the option() function is simply a function with if else.

It lets you keep the clean look of a switch and the logic part is elsewhere.

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