大于和小于C switch语句 [英] Larger than and less than in C switch statement

问题描述

我正在尝试编写具有很多比较的代码

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

在 QUANT.C中编写一个程序，其中量化数字。读取一个整数 x并对其进行测试，并在输出后产生
：

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

x大于或等于1000打印非常正

x从999到100（包括100）打印非常积极

x在100和0之间打印积极

x恰好0打印零

x在0到-100之间打印负

x从-100到-999（包括-100）打印非常负

x小于或等于到-1000打印非常负面

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"

因此-10将打印负面，-100非常负面和458非常正面。

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

然后我尝试使用switch语句解决它，但是没有用。我是否必须使用 if 语句来解决它，还是有一种方法可以使用switch语句来解决它？

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;
}

推荐答案

没有干净的方法用case来解决这个问题，因为情况需要是整数类型。看看if-else if-else。

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.

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