PTA 6-40 简单表达式求值

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于 2024-11-06 14:29:35 发布

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文章标签： c++ c语言

本文链接：https://blog.youkuaiyun.com/YHCsuper0429/article/details/143569810

版权

int cal(char s[]) {
    int num1 = 0, num2 = 0;
    char op;
    int i = 0;
    while (s[i] >= '0' && s[i] <= '9') {
        num1 = num1 * 10 + (s[i] - '0');
        i++;
    }
    op = s[i];
    i++;
    while (s[i] >= '0' && s[i] <= '9') {
        num2 = num2 * 10 + (s[i] - '0');
        i++;
    }
    switch (op) {
        case '+':
            return num1 + num2;
        case '-':
            return num1 - num2;
        case '*':
            return num1 * num2;
        case '/':
            if (num2 == 0) {
                return 0;
            }
            return num1 / num2;
    }
    return 0;
}