这个简单的程序有问题

Question

该程序只是简单地通过从当前年份中减去他们的出生日期来计算用户年龄。当我运行程序时，它编译成功，但我得到一个很长的数字，例如 -215863352。添加 if 和 else 条件只是为了测试它们，我正在使用它们编写各种程序以确保我理解 c 中的语法。我想我错过了一些简单但无法弄清楚的东西。

#include <stdio.h>
int main()
{
    int year;
    int cyear;
    int age = cyear - year;

    printf("Please enter the year you were born: \n");
    scanf("%i", &year);
    printf("Now enter the current year: \n");
    scanf("%i", &cyear);

    if (1+1 == 2){
        printf("You must be %i", age);
    }
    else {
        printf("Cannot compute age, GOODBYE:\n");
    }
    return 0;
}

Answer 1

您正在计算从用户获取输入之前的年龄。所以age 变量存储了一个垃圾值。

解决方案：

在使用 scanf 获取cyear 的输入之后，定位从用户获取输入之后的年龄计算。正确的代码如下 #include

int main()
{
    int year;
    int cyear;
    int age =0;     //initialise with 0

    printf("Please enter the year you were born: \n");
    scanf("%i", &year);
    printf("Now enter the current year: \n");
    scanf("%i", &cyear);
    
    age = cyear - year;     //note the change here

    if (1+1 == 2){
        printf("You must be %i", age);
    }
    else {
        printf("Cannot compute age, GOODBYE:\n");
    }
    return 0;
}

Answer 2

enter code here
   #include <stdio.h>
   int main()
   {
  long long int year;
  printf("Please enter the year you were born: \n");
scanf("%lld",&year);
long long int cyear;
  printf("Now enter the current year: \n");
scanf("%lld",&cyear);

long long  int age = cyear-year;

if (1){
    printf("You must be %lld", age);
}
else {  printf("Now enter the current year: \n");
scanf("%lld",&cyear);

    printf("Cannot compute age, GOODBYE:\n");
}
return 0;

}