具有0个命令行参数的简单程序分段错误

Question

so I have the following program which outputs the parameter with the maximum length. I want to make an exception so when I give it 0 parameters I get an error telling me that I need at lest one parameter.

// Program which given an number of program parameters
// (command-line parameters, calulates the length of each one
// and writes the longest to standard output.

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


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

    int i;
    int maxLength = strlen(argv[1]);    
    char* maxString = argv[1]; 

   if (argc<=0)
    {
    printf("You cannot have 0 parameters\n");
    exit(0);
    }

    for(i = 2; i < argc; ++i) 
    {
    // find out the length of the current argument
        int length = strlen(argv[i]);

        if (maxLength < length)
        {
            maxLength = length;
        maxString = argv[i];    
        } // if

    } // for

printf("Largest string is %s\n", maxString);

} // main

This is my code but for some reason I am getting a segmentation error when I'm giving it 0 arguments, instead of the message.

Any thoughts?

Answer 1

Edit after editing the question: Also you're using argv[1] before you checked argc . This is an error.

If you pass no arguments, argc is going to be 1 (because argv[0] is usually the executable name).

So

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

int main(int argc, char *argv[])
{
    if(argc<=1)
    {
        printf("You cannot have 0 parameters\n");
        exit(255);
    } else
    {
        int i;
        int maxLength = 0;
        const char* maxString = 0;

        for(i = 1; i < argc; ++i)
        {
            // find out the length of the current argument
            int length = strlen(argv[i]);
            if(maxLength <= length)
            {
                maxLength = length;
                maxString = argv[i];
            }
        }
        printf("Largest string is %s\n", maxString);
    }
} // main

Answer 2

如果在命令行中给零参数，则argc将为1而不是0。可执行文件的名称将为第一个参数。