C语言中第二个switch语句后面为什么要加return 0 ？作用是什么

Question

#include<stdio.h>
int main()
{
int num, unit, decade;
printf("Enter a two-digit number:");
scanf_s("%d", &num);
printf("You entered the number ");
decade = num / 10;
unit = num % 10;
switch (decade) {
case 9:printf("ninety"); break;
case 8:printf("eighty"); break;
case 7:printf("seventy"); break;
case 6:printf("sixty"); break;
case 5:printf("fifty"); break;
case 4:printf("forty"); break;
case 3:printf("thirty"); break;
case 2:printf("twenty"); break;
case 1:
switch (unit) {
case 9:printf("nineteen"); break;
case 8:printf("eighteen"); break;
case 7:printf("seventeen"); break;
case 6:printf("sixteen"); break;
case 5:printf("fifteen"); break;
case 4:printf("fourteen"); break;
case 3:printf("thirteen"); break;
case 2:printf("twelve"); break;
case 1:printf("eleven"); break;
}
return 0;//为什么要加return 0?
}
printf("-");
switch (unit) {
case 9:printf("nine"); break;
case 8:printf("eight"); break;
case 7:printf("seven"); break;
case 6:printf("six"); break;
case 5:printf("five"); break;
case 4:printf("four"); break;
case 3:printf("three"); break;
case 2:printf("two"); break;
case 1:printf("one"); break;
}
}

Answer 1

return 0;就是退出 main()主函数
当switch (decade) 中 decade等于1时执行case 1:分支中switch (unit) {.......}后遇到return 0;就退出 main()主函数了,后面的printf("-");就不执行了