Question

对于我的任务，我的程序需要询问用户他想要执行的五个函数中的哪一个。五个功能包括：

数字的总和
数字因素
第n个术语的斐波纳契值。
gcd of two numbers
a b的力量。

将重复提示用户，直到他希望退出。我所有的功能都很好。但是，我认为我搞砸了其中一个循环，因为一旦我输入了我想要执行的函数并输入一个值，它就会在无限循环中继续显示答案。

#include <stdio.h>
#include <math.h>

// function to find summation of a number
int summation(int k) {
  int i;

  for(i = k; i >= 0; i--) {
    k = i + (i-1);
  }
  return k;
}

// function to find the factorail of a number
int factorial(int num) {
  int i;

  for(i = num - 1; i > 0; i--) {
    num = num * i;
  }
  return num;
}

// dunxtion to find the fibonacci of the nth term
int fibonacci(int n){
  int i, t1 = 0, t2 = 1, nextTerm;

  for(i = 1; i <= n; i++) {
    if(i == 1) {
        printf("%d, ", t1);
        continue;
    }
    if(i == 2) {
        printf("%d, ", t2);
        continue;
    }
    nextTerm = t1 + t2;
    t1 = t2;
    t2 = nextTerm;
    printf("%d, ", nextTerm);
  }
  return nextTerm;
}

// function to find the gcd of two numbers
int gcd(int n, int m) {
  int i, gcd;

  for(i=1; i <= n && i <= m; i++) {
    // Checks if i is a factor of both integers
    if(n % i == 0 && m % i == 0)
        gcd = i;
  }
  return gcd;
}

// function to find value of n to the power of m
int power(int n, int m) {
  return pow(n, m);
}

int main(void) {;
  int option ,n, m;

//Asks user for what they want to find
  printf("If you would like to find the summation of a number, enter 1 \n");
  printf("If you would like to find the factorial of a number, enter 2 \n");
  printf("If you would like to find the fibonacci sequence of a number, enter 3 \n");
  printf("If you would like to find the gcd of two numbers, enter 4 \n");
  printf("If you would like to find the power of a number a to b, enter 5 \n");
  printf("If you would like to exit, enter 0 \n");
  scanf("%d", &option);

// Enables the program to prompt the user until they wish to exit
  while(option != 0) {  
    switch(option) { //If user wishes to find the summation
        case 1: if(option == 1) {   
                    printf("Enter a number: ");
                    scanf("%d", &n);
                    while(n > 0) {
                        if(n < 1) { //message displayed if an invalid value is entered
                            printf("invalid value");
                        }
                        else {
                            printf("Summation of %d is %d", n, summation(n));
                        }
                    }
                }
        case 2: if(option == 2) { //if user wishes to find factorial of a number
                    printf("Enter a number: ");
                    scanf("%d", &n);
                    while(n >= 0) {//message displayed if an invalid value is entered
                        if(n < 0) {
                            printf("invalid value");
                        }
                        else {
                            printf("factorial of %d is %d", n, factorial(n));
                        }
                    }
                }
        case 3: if(option == 3) { //if user wishes to find the fibonacci value of the nth term
                    printf("Enter a number: ");
                    scanf("%d", &n);
                    while(n >= 0) {//message displayed if an invalid value is entered
                        if(n < 0) {
                            printf("invalid value");
                        }
                        else {
                            printf("fibonacci of %d is %d", n, fibonacci(n));
                        }
                    }
                }
        case 4: if(option == 4) { 
                    printf("Enter a number: ");
                    scanf("%d %d", &n, &m);
                    while(n >= 0 && m >= 0) {
                        if(n < 0 || m < 0) {//message displayed if an invalid value is entered
                            printf("invalid value");
                        }
                        else {
                            printf("GCD of %d and %d is %d", n, m, gcd(n, m));
                        }
                    }
                }
        case 5: if(option == 5) {
                    printf("Enter a number: ");
                    scanf("%d %d", &n, &m);
                    while(n >= 0 && m >= 0) {
                        if(n <= 0 || m < 0) {
                            printf("invalid value");
                        }
                        else {
                            printf("%d to the power of %d is %d", n, m, power(n, m));
                        }
                    }
                }
        default: if(option == 0) {
            break;
        }
    }
    scanf("%d", &option);
}
}

Answer 1

首先，C具有非结构化switch语句。

您需要在每个break;正文后添加case语句，以便将特定案例的执行限制到case下提到的正文

否则，默认情况下，（缺少break语句）所有case语句都以直通方式运行。您可以阅读更多相关信息here。

也就是说，关于单个函数的重复执行，大多数（如果不是全部）逻辑都存在严重缺陷。例如，让我们来看看

   while(n > 0) {
     if(n < 1) { //message displayed if an invalid value is entered
         printf("invalid value");
     }
     else {
         printf("Summation of %d is %d", n, summation(n));
     }
   }

在这里，你回复n在某个时刻变为0以突破循环，但你根本没有修改n。

详细说明，C使用pass-by-value进行参数传递，因此对于函数内部的调用summation(n)，对接收n值的参数所做的任何更改都是没有反映在调用者中，因此调用者中的n保持不变。

Answer 2

在每个案例结束时，您只需要break语句像：

 case 1: if(option == 1) { 
                printf("Enter a number: ");
                scanf("%d", &n);
                while(n > 0) {
                    if(n < 1) { //message displayed if an invalid value is entered
                        printf("invalid value");
                    }
                    else {
                        printf("Summation of %d is %d", n, summation(n));
                    }
                }
            } 
break;

如果没有break语句，控件将落到下一个案例中。

使用几种不同的功能

2 个答案: