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");
else
printf("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");
else
printf("the answer is %d\n",answer);
}
return 0;
}
You also need to check if the scanf did not fail.
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