使用 c 问题的简单计算器

Question

我是 C 新手，我正在尝试制作一个简单的计算器

#include <stdio.h>
#include <stdlib.h>

int main() {
    int firstNumber, secondNumber, answer;
    char mathSign[20];

    printf("Enter first number \n");
    scanf("%d", &firstNumber);

    printf("Enter sign \n a - Add \n s = Subtract \n m - Multiply \n d - Divide \n");
    scanf("%s", mathSign);

    printf("Enter second number \n");
    scanf("%d", &secondNumber);

    if (mathSign == 'a') {
        answer = firstNumber + secondNumber;
    } else if (mathSign == 's') {
        answer = firstNumber - secondNumber;
    } else if (mathSign == 'm') {
        answer = firstNumber * secondNumber;
    } else if (mathSign == 'd') {
        answer = firstNumber / secondNumber;
    }

    printf("Your answer is %d", answer);

    return 0;
}

由于某种原因，答案总是以52 结尾。我应该让整数加倍吗？

Answer 1

试试这个代码。这应该有效。此外，这将是优化的。

#include <stdio.h>
#include <stdlib.h>

int main() {
    int firstNumber, secondNumber, answer;
    char mathSign;

    printf("1.Enter first number \n");
    scanf("%d", &firstNumber);

    printf("Enter sign \n a - Add \n s = Subtract \n m - Multiply \n d - Divide \n");
    scanf(" %c", &mathSign);

    printf("Enter second number \n");
    scanf("%d", &secondNumber);

    switch (tolower(mathSign)) {
    case 'a':
        answer = firstNumber + secondNumber;
        break;
    case 's':
        answer = firstNumber - secondNumber;
        break;
    case 'm':
        answer = firstNumber * secondNumber;
        break;
    case 'd':
        answer = firstNumber / secondNumber;
        break;
    default:
        printf("Invalid Choice");
        break;
    }

    printf("Your answer is %d", answer);
    return 0;
}

Answer 2

你测试了mathSign == 'a'，但mathSign是一个数组，所以你应该写mathSign[0] == 'a'。编译器应该产生警告。您不应忽略这些警告，它们表明问题通常是真正的错误。您可以通过提高警告级别来获得更多有用的诊断信息：gcc -Wall -W -Werror 或 clang -Weverything -Werror...

注意scanf("%s", mathSign); 如果用户键入的单词超过 19 个字符，将导致缓冲区溢出。您应该使用以下方法保护目标数组：

scanf("%19s", mathSign);

此外，您应该检查转换是否成功，否则如果输入无法转换并且您使用仍未初始化的值，您可能会出现未定义的行为。

您可以使用" %c" 来读取单个字符并使用经典运算符来提高可读性。注意%c 之前的空格：它让scanf() 跳过待处理的空白字符，例如上一个响应中的换行符。

这是修改后的版本：

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    int firstNumber, secondNumber, answer;
    char mathSign;

    printf("Enter first number: ");
    if (scanf("%d", &firstNumber) != 1)
        return 1;

    printf("Enter operation\n + - Add\n - = Subtract\n * - Multiply\n / - Divide\n");
    if (scanf(" %c", &mathSign) != 1)
        return 1;

    printf("Enter second number: ");
    if (scanf("%d", &secondNumber) != 1)
        return 1;

    if (mathSign == '+') {
        answer = firstNumber + secondNumber;
    } else if (mathSign == '-') {
        answer = firstNumber - secondNumber;
    } else if (mathSign == '*') {
        answer = firstNumber * secondNumber;
    } else if (mathSign == '/') {
        answer = firstNumber / secondNumber;
    } else {
        printf("unsupported operation: %c\n", mathSign);
        return 1;
    }

    printf("Your answer is %d\n", answer);

    return 0;
}

您可以使用switch 语句进一步简化程序：

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    int firstNumber, secondNumber, answer;
    char mathSign;

    printf("Enter a simple expression: ");
    if (scanf("%i %c%i", &firstNumber, &mathSign, &secondNumber) != 3)
        return 1;

    switch (mathSign) {
      case '+': answer = firstNumber + secondNumber; break;
      case '-': answer = firstNumber - secondNumber; break;
      case '*': answer = firstNumber * secondNumber; break;
      case '/': answer = firstNumber / secondNumber; break;
      case '%': answer = firstNumber % secondNumber; break;
      default: 
        printf("unsupported operation: %c\n", mathSign);
        return 1;
    }
    printf("the result is %d\n", answer);
    return 0;
}

Answer 3

当您只使用mathSign 时，它是一个指向数组的指针。 要访问第一个元素，您必须使用 mathSign[0]。

另外，由于您只对符号的一个字符感兴趣，您可以声明 char mathSign 而不是 char mathSign[20] 然后使用 scanf(" %c", mathSign); 读取符号字符 在这种情况下，比较 mathSign == 'a 是正确的。