难以理解 C 中的函数

Question

我正在完成一项编程任务。我在 main 中进行了密钥验证，但决定尝试使其成为一个单独的函数。我还不太了解函数，所以我看不出哪里出错了。每当我运行程序时，即使我知道它不是，我总是只会得到“密钥有效”。正如我所说，程序在 main 中运行良好。

#include <cs50.h>
#include <ctype.h>
#include <stdio.h>
#include <string.h>

int validate (int c, string v[]); //prototpe validate function

int main (int argc, string argv[])
{
    int validate (int argc, string argv[]); //run validate for argc and argv

    printf("Key is valid\n"); //if valid, print message

}

int validate (int c, string v[])
{
    //Validate that only one Command Line Argument was entered
    if (c != 2) //Check the number if entered commands at execution
    {
        //If more than one, print error message
        printf("Key must be the only Command Line Argument\n");
        return 1; //Return false
    }

    //Validate that Key length is 26 characters
    if (strlen(v[1]) != 26) //Don't forget null 0
    {
        printf("Key must contain exactly 26 characters\n");
        return 1; //Return false
    }

    //Validate that all Key characters are alphabetic
    for (int i = 0, n = strlen(v[1]); i < n; i++)
    {
        //Check each element of the array
        if (isalpha(v[1][i]))
        {
            continue; //Continue if alphabetic
        }
        else
        {
            //if non-alphabetic, print error code
            printf("Key must contain only alphabetic characters\n");
            return 1; //Return false
        }
    }

    //Validate that each Key character is unique
    for (int x = 0, z = strlen(v[1]) - 1; x < z; x++)
    {
        //Create a second loop to compare every element to the first before incrementing the first
        for (int y = x + 1; y < z; y++)
        {
            //Cast array elements to int, check if each element equals next element in array
            if (v[1][x] == v[1][y])
            {
                printf("Key must contain exactly 26 unique characters\n");
                return 1;
            }
        }
    }

    return 0; //Key is valid, so return true
}

Answer 1

您只是声明函数validate，而不是运行该函数并无条件打印Key is valid。

要运行函数validate并仅在返回0时打印Key is valid，main函数应该是这样的：

int main (int argc, string argv[])
{
    if (validate (argc, argv) == 0) //run validate for argc and argv and check its response
    {
        // put printing inside if statement so that it runs only if the condition is true
        printf("Key is valid\n"); //if valid, print message
    }
}

Answer 2

函数声明提示编译器我将返回一个特定的数据类型，并且我将接受参数部分中的给定数据类型。例如，

int check(int a, int b); -> return type is int and the function will accept 2 integer parameters.
int mul(int a, float b); -> return type is int and the function will accept 1 integer parameter and 1 float.
void check(); -> returns nothing, accepts nothing

函数调用就像你的代码正在调用要执行的函数。

int c = check(2, 3);
int b = mul(3, 0.12);
check();

您的函数正在返回一些值。您必须获取该值并根据以下值执行其余代码。

if (validate (argc, argv) == 0) {
    printf("key is valid");
} else {
    printf("key is not valid");
}