c++ - 为什么我的程序不打印字符串？

Question

我的程序需要学生的姓名和年龄，这个函数应该找到打印最小的学生的名字、最大的学生的名字和平均年龄。但是它不打印最小的学生的名字谁能告诉我为什么？

void check(string *nameStudent, int *ageStudent, int num) {
    int i, young = 0, old = 0, sum = 0, mov = 0;
    string a, b;
    double average;
    for (i = 0; i < num; i++){
        if (*(ageStudent + mov) < young) {
            young = *(ageStudent + mov);
            a = *(nameStudent + mov);
        }
        if (*(ageStudent + mov) > old) {
            old = *(ageStudent + mov);
            b = *(nameStudent + mov);
        }
        sum += *(ageStudent + mov);
        mov++;
    }
    average = (double) sum / num;
    cout << a << " Is youngest student and " << b << " Is oldest student " << endl << average << " is average age of students";
}

Answer 1

您将 young 初始化为 0。如果有一个学生的年龄小于 0，它只会改变 young 和 a。

不是对所有这些变量的初始值进行硬编码，而是从数组的第一个元素中获取它们。

if (num > 0) {
    young = old = sum = *ageStudent;
    a = b = *nameStudent;
}

然后您可以将循环更改为从 1 而不是 0 开始，因为您已经处理了数组的第一个元素。

顺便说一句，不需要 mov 变量，因为它总是包含与 i 相同的内容。请改用 *(ageStudent + i)。

void check(string *nameStudent, int *ageStudent, int num) {
    int i, young = 0, old = 0, sum = 0;
    string a, b;
    double average;
    if (num > 0) {
        young = old = sum = *ageStudent;
        a = b = *nameStudent;
    }
    for (i = 1; i < num; i++){
        if (*(ageStudent + i) < young) {
            young = *(ageStudent + i);
            a = *(nameStudent + i);
        }
        if (*(ageStudent + i) > old) {
            old = *(ageStudent + i);
            b = *(nameStudent + i);
        }
        sum += *(ageStudent + i);
        mov++;
    }
    average = num ? (double) sum / num : 0; // prevent division by 0
    cout << a << " Is youngest student and " << b << " Is oldest student " << endl << average << " is average age of students";
}