传递 argv 和 argc 以能够读取任何文件类型的任何文件名

Question

我正在尝试将我的程序从能够读取硬编码 name.txt 更改为任何名称或可读类型。我在正确传递 argc 和 argv 时遇到问题，并给出诸如 "too few arguments in main" 之类的错误我试图在使用的函数中初始化这两个，但未确定。我将如何正确使用这两个工具？（箭头指向 arg 行）这是我的代码：

#include <stdio.h>
#include <stdlib.h>

// Pointer to store numbers
int *pointerNUM;

// Read in the parts file and returns the length
int readFile(int argc, char *argv[])      //<-----------------------------
{
    // File pointer
    FILE *fptr;
    // numberOfNUM for number of numbers
    // cntVAR for counter variable
    int numberOfNUM, cntVAR;
    // Open the file for reading
    fptr = fopen(argv[1], "r"); //<-----------------------------------------

    // Check that it opened properly
    if (fptr == NULL)
    {
        printf("Cannot open file \n");
        exit(0);
    }// End of if condition

    // Reads number of numbers in the file
    fscanf(fptr, "%d", &numberOfNUM);

    // Dynamically allocates memory to pointer pointerNUM
    pointerNUM = (int *) calloc(numberOfNUM, sizeof(int));
    // Loops numberOfNUM times
    for(cntVAR = 0; cntVAR < numberOfNUM; cntVAR++)
        // Reads each number and stores it in array
        fscanf(fptr, "%d", &pointerNUM[cntVAR]);
    // Returns the length of the numbers
    return numberOfNUM;
}// End of function

// Function to display numbers
void display(int numberOfNUM)
{
    int cntVAR;

    // Loops numberOfNUM times
    for(cntVAR = 0; cntVAR < numberOfNUM; cntVAR++)
        // Displays each number
        printf("%4d, ", pointerNUM[cntVAR]);
}// End of function

// Function for insertion sort
void insertionSort(int numberOfNUM)
{
    int x, key, y;

    // Loops numberOfNUM times
    for (x = 1; x < numberOfNUM; x++)
    {
        // Stores i index position data in key
        key = pointerNUM[x];
        // Stores x minus one as y value
        y = x - 1;

        /*
        Move elements of pointerNUM[0..x - 1], that are greater than key,
        to one position ahead of their current position
        */
        while (y >= 0 && pointerNUM[y] > key)
        {
            // Stores pointerNUM y index position value at pointerNUM y next index position
            pointerNUM[y + 1] = pointerNUM[y];
            // Decrease the y value by one
            y = y - 1;
        }// End of while
        // Stores the key value at pointerNUM y plus one index position
        pointerNUM[y + 1] = key;
    }// End of for loop
}// End of function

// main function
int main()
{
    // To store the numbers of number in file
    int numberOfNUM;

    // Calls the function to read numbers and stores the length
    numberOfNUM = readFile(); <-----------------------------------------
    // Calls the function to displays the numbers before sorting
    printf("\n Before sort : ");
    display(numberOfNUM);
    // Calls the function for sorting
    insertionSort(numberOfNUM);
    // Calls the function to displays the numbers after sorting
    printf("\n After sort: ");
    display(numberOfNUM);
}

Answer 1

您的 main() 函数原型是错误的。如果你想在 C 中使用 argc 和 argv，你需要使用支持它们的 main 版本。

int main(int argc, char *argv[])
{
   // Check argc, and argv for correctness

   numberOfNum = readFile(argc, argv);
   ...
}

然后运行它会做：

./myProgram arg1 arg2 ... argN

Answer 2

我注意到有四个即时错误：

int readFile(int argc, char *argv[])
...
numberOfNUM = readFile();

您的函数 readFile 有两个参数，但是当您调用它时根本不传递任何参数：

第二个是你好像误解了argc和argv。这些只是我们为 main() 的参数提供的常用名称。实际上，您可以随意命名它们，即：

int main(int snapple, char **horses)

由此导致：

int main()

你实际上并没有接受 argc 或 argv，你需要将它们添加到你的 main。这样做将解决您的 main 问题，但请确保将它们提供给您的 readFile 电话。

第三，你从来没有fclose你打开的文件。你需要这样做。

我注意到的第四件事有点迂腐，但你将main定义为returning an int，但你没有return.

传递 argv 和 argc 以能够读取任何文件类型的任何文件名

Passing argv and argc to be able to read any file name of any file type

c

parameter-passing

argv

argc