动态函数调用实现下列操作，输入2个数以及操作符计算结果。@求最大公约数 $求最小公倍数 -求差 +求和

int sumValue(int x, int y) {
    return x + y;
}
int subValue(int x, int y) {
    return x - y;
}
int temp = 0;
int gcdValue(int x, int y) {//greatest common divisor 最大公约数
    if (x > y) {//使得x存储小值, y存储大值
        temp = x;
        x = y;
        y = temp;
    }
    int gcd = 0;
    for (int i = x; i > 1; i--) {
        if ((x % i == 0) && (y % i == 0)) {
            gcd = i;
            break;
        }
    }
    return gcd;
}
int lcmValue(int x, int y) {//lowest common multiple 求最小公倍数
    if (x > y) {//使得x存储小值, y存储大值
        temp = x;
        x = y;
        y = temp;
    }
    int lcm = 0;
    for (int i = y; i <= x * y; i++) {
        if ((i % x == 0) && (i % y == 0)) {
            lcm = i;
            break;
        }
    }
    return lcm;
}
typedef int (*PFUN) (int, int);
typedef struct operatorFctPair {
    char operator;//存储运算符
    PFUN function;//存储函数地址
}OperatorFctPair;

//operator是用户输入的运算符, p用来接收匹配表, count是匹配表中成员的个数
PFUN getFctByopetator(char operator, OperatorFctPair *p, int count) {
    for (int i = 0; i < count; i++) {
        if ((p + i)->operator == operator) {
            //如果找到对应的函数, 将函数地址返回
            return (p + i)->function;
        }
    }
    return NULL;
}

int computer(int x, int y, PFUN pFct, char operator, OperatorFctPair *pair, int count) {//动态调用
    PFUN function = getFctByopetator(operator, pair, count);
    return function(x, y);
}

int main(int argc, const char * argv[])
{

    // 动态函数调用实现下列操作，输入2个数以及操作符计算结果。@求最大公约数 $求最小公倍数 -求差 +求和  等等
    OperatorFctPair pair[4] = {
        {'@', gcdValue},
        {'$', lcmValue},
        {'+', sumValue},
        {'-', subValue}
    };
    char operatorName = 0;
    int a = 0, b = 0;
    printf("请输入要运算的两个整数和运算符:(@求最大公约数, $求最小公倍数, -求差, +求和)\n");
    scanf("%d%c%d", &a, &operatorName, &b);
    if (getFctByopetator(operatorName, pair, 4) != NULL) {
        int result = computer(a, b, gcdValue, operatorName, pair, 4);
        printf("%d %c %d = %d\n", a, operatorName, b, result);
    } else {
        printf("%c运算没有实现, 请按照提示进行输入\n", operatorName);
    }
    return 0;
}