英文:
why do i get 6 from default statement if my input was another character other than (+, -, *, /)
问题
我每次执行默认语句时都会得到6，无论输入是什么这是我的输入(输入2个数字第一个数字 4第二个数字 8输入操作符操作符 H)输出结果(错误答案是 6)
英文:
#include <stdio.h>int calculator(int num1,int num2, char p){switch(p){case '+':return num1+num2;case '-':return num1-num2;case '*':return num1*num2;case '/':return num1/num2;default:printf("Error\n");break;}}int main(){int num1,num2,answer=0;char p;while(1){answer=0;printf("Enter 2 numbers\n");scanf("%d%d",&num1,&num2);printf("Enter the operator\n");scanf(" %c",&p);answer=calculator(num1,num2,p);printf("the answer is %d\n",answer);}return 0;}
I get 6 every time the default statement is executed no matter the input
this is my input
(
Enter 2 numbers
first number 4
second number 8
Enter the operator
operator H
)
the output
(
Error
the answer is 6
)
答案1
得分: 4
你的 calculator 函数在 break 的情况下没有执行 return。严格来说，这是未定义行为。请参考 https://stackoverflow.com/questions/7280877/why-and-how-does-gcc-compile-a-function-with-a-missing-return-statement
英文:
Your calculator function doesn't execute a return for the break fallthrough case. Strictly speaking this is undefined behavior. See https://stackoverflow.com/questions/7280877/why-and-how-does-gcc-compile-a-function-with-a-missing-return-statement
答案2
得分: 1
以下是您要翻译的代码部分：
这是因为在您的 `default:` 情况下没有返回任何内容，它是未定义的行为。您还需要通知调用者函数失败了。您可以通过设置 [哨兵值][1] 或引入另一个参数来指示错误来实现这一点。#include <stdio.h>int calculator(int num1, int num2, char p, int *error){int result = 0;*error = 0;switch (p){case '+':result = num1 + num2;break;case '-':result = num1 - num2;break;case '*':result = num1 * num2;break;case '/':result = num1 / num2;break;default:*error = 1;break;}return result;}int main(void){int num1, num2, answer = 0, error;char p;while (1){answer = 0;printf("输入2个数字\n");if (scanf("%d %d", &num1, &num2) != 2){printf("您输入了错误的数字\n");continue;}printf("输入操作符\n");if (scanf(" %c", &p) != 1) { /* 处理错误 */ };answer = calculator(num1, num2, p, &error);if (error)printf("错误\n");elseprintf("答案是 %d\n", answer);}return 0;}您还需要检查 `scanf` 是否失败。[1]: https://en.wikipedia.org/wiki/Sentinel_value
请注意，这是代码的翻译版本，没有其他内容。
英文:
It is undefined behaviour as you do not return anything in your default: case. Also you need to let the caller that your function failed. You can do this by setting a sentinel value or introducing another parameter which will indicate the error.
#include <stdio.h>int calculator(int num1,int num2, char p, int *error){int result = 0;*error = 0;switch(p){case '+':result = num1+num2;break;case '-':result = num1-num2;break;case '*':result = num1*num2;break;case '/':result = num1/num2;break;default:*error = 1;break;}return result;}int main(void){int num1,num2,answer=0, error;char p;while(1){answer=0;printf("Enter 2 numbers\n");if(scanf("%d %d",&num1,&num2) != 2){printf("you have entered wrong numbers\n");continue;}printf("Enter the operator\n");if(scanf(" %c",&p) != 1) { /* handle arror */};answer=calculator(num1,num2,p, &error);if(error)printf("Error\n");elseprintf("the answer is %d\n",answer);}return 0;}
You also need to check if the scanf did not fail.
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