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对于学习C语言的小白来说，经常会遇到求解分段函数的问题，下面是小编写的一段求解分段函数的代码，希望对于初学的你们有所帮助
代码求解的分段函数为：
y=f(x)=4x/3(x<=15)
y=f(x)=2.5x-10.5(x>15)
#include<stdio.h> int main(void) { double x,y; printf("enter x(x>0):\n");
scanf("%lf",&x); if(x<=15){ y=4*x/3; } else{ y=2.5*x-10.5; }
printf("y=f(%.2f)=%.2f\n",x,y); return 0; }
上面分为两个部分来看代码，小编用到了一个if-else原来判断条件，因为原题当中是以输入的x的值来作为判断依据的。
小知识：
if-else语句的写法格式：
if（表达式）
语句1；
else
语句2；
在这个判断当中，如果if里面的条件满足就执行if里面的语句块，如果条件不满足，则执行else语句块里面的内容。这个是按照实际条件来说的。
当然，除了这一种判断之外，我们以后还会遇到do-while来判断的，这里小编先不做引申，小编会在后面专门写一篇关于这个的专栏的。
除了上面的if-else之外，我们还会遇到多重判断，即if-else if-...-else这样的情况，遇到这种问题的时候我们就需要看我们的要求了，
