关系运算符的链接给出了错误的输出

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谁能给我解释一下吗？我做错什么了吗？当我运行该程序时，它没有显示正确的答案。

例如:当我输入体重 = 50 公斤和高度 = 130 厘米时，答案应该是

"Your BMI is 29.58. You are overweight2.You will have a chance to cause high blood pressure and diabetes need to control diet. And fitness."

但显示的答案是

"Your BMI is 29.58. You are normal......"

<小时/>

#include<stdio.h>
#include<conio.h>

int main() {
  float weight, height, bmi;

  printf("\nWelcome to program");

  printf("\nPlease enter your weight(kg) :");
  scanf("%f", &weight);

  printf("\nPlease enter your height(cm) :");
  scanf("%f", &height);

  bmi=weight/((height/100)*(height/100));

  if(bmi<18.50) { 
    printf("Your bmi is : %.2f",bmi);
    printf("You are Underweight.You should eat quality food and a sufficient amount of energy and exercise proper.");
  } else if(18.5<=bmi<23) {
    printf("Your bmi is : %.2f \nYou are normal.You should eat quality food and exercise proper.",bmi);
  } else if(23<=bmi<25) {
    printf("Your bmi is : %.2f \nYou are overweight1 if you have diabetes or high cholesterol,You should lose weight body mass index less than 23. ",bmi);
  } else if(25<=bmi<30) {
    printf("Your bmi is : %.2f \nYou are overweight2.You will have a chance to cause high blood pressure and diabetes need to control diet. And fitness.",bmi);
  } else if(bmi>=30) {
    printf("Your bmi is : %.2f \nYou are Obesity.Your risk of diseases that accompany obesity.you run the risk of highly pathogenic. You have to control food And serious fitness.",bmi);
  } else {
    printf(" Please try again! ");
  }

  return 0;
  getch();
}

Answer 1

在您的代码中，您尝试过的内容

 else if(18.5<=bmi<23)

不，这种链接关系运算符在 C 中是不可能的。您应该编写

 else if((18.5<=bmi) &&  (bmi < 23))

检查bmi值 [18.5, 23)其他情况依此类推。

<小时/>

编辑:

只是为了详细说明问题，像这样的表达式

18.5<=bmi<23

是完全有效的 C 语法。但本质上是一样的

((18.5<=bmi) < 23 )

由于 operator associativity .

所以，首先是 (18.5<=bmi)被评估并且结果(0或1)得到与23进行comapred ，这当然不是您想要的。