分段错误（核心转储）C程序

Question

I am trying to run this program with the command

./box2 5

/*
 * box2.c
 *
 *  Created on: Mar 19, 2014
 *      Author: Ian
 */
 #include <stdio.h>
 void printchars(char c, int n);
 int main( int argc, char*argv)
 {
    int n = argv[1];
    printchars('*', n);
    return 0;
 }

 void printchars(char c, int n)
 {
    int x;
    for (x = n + 2 ; x > 0; x--)
    {

        if (x != 1 && x != n)
        {
            printf("%c", c);
            int count = n;
            while (count - 2 != 0)
            {
                printf(" ");
                 count--;
            }
        }
        else
        {
            int num = n;
            while (num != 0)
            {
                printf("%c", c);
                num--;
            }
        }
        printf("\n");
     }
 }

I always get the error Segmentation Fault(core dumped) each time i try it.

    *****
    *   *
    *   *
    *   *
    *   *
    *   *
    *****

This is what I should get.

I have no idea what to do to fix this. I get no errors when I compile and that error is the only thing that comes up when I try to run the program.

Answer 1

argv[1] is a char * . You need to convert that to an int . You could do that by calling atoi() as follows:

int n = atoi(argv[1]);

Note that this will not handle argument errors. Note also that your definition of argv is wrong: it should be char * argv[] .

Answer 2

将int main( int argc, char*argv)更改为int main( int argc, char**argv) 。

Answer 3

This is wrong:

 int main( int argc, char*argv)

since argv must be declared as a ptr-to-ptr-to-char. Likewise, this:

 int n = argv[1];

can't work. You need something like

 int main(int argc, char **argv)
 ...
 int n = atoi(argv[1]);