[英]Trouble understanding Functions in C
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,所以我看不出我哪里出错了。 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
}
您只是声明了函数validate
而不是运行它,并且打印Key is valid
无条件Key is valid
。
要运行函数validate
和 print Key is valid
仅在返回0
时才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
}
}
函数声明提示编译器我将返回一个特定的数据类型,我将在参数部分接受给定的数据类型。 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
函数调用就像您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该值执行其余代码,如下所示。
if (validate (argc, argv) == 0) {
printf("key is valid");
} else {
printf("key is not valid");
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.