为什么我会出现段错误？

Question

I'm very new to C, I am attempting to read the contents of one file character by character and output them to the stream. But even with my fopen() command commented out I receive segfault (core dumped).

I must run a command: ./a.out < testWords.in > myOut.txt to execute my file properly.

Here is what I have so far:

#include  <stdio.h>
void main(char *fileName[]) 
{
  printf("filename is %s.\n",fileName[0]);
  //Get file based on a string inputed
  FILE *fp=fopen(fileName[0],"r"); //Fetches our file as read only
  char ch;
  int lineCount = 0;
  int wordCount = 0;
  int charCount = 0;

  //Failed to find/open file. NULL character.
  if (fp == 0) printf("Woops! Couldn't open file!\n");

  //While not at end of file, grab next char.
  else while( (ch=fgetc(fp)) != EOF) 
    {
      if (ch == '\n') //on newline
      {
    //Prints (charCount,wordCount)\n lineCount:
    printf("(%d,%d)%c%d:",charCount,wordCount,ch,lineCount);
    charCount = 0;
    lineCount += 1;
      }
      else printf("%c",ch); //mirrors char.
    }
  fclose(fp); //Closes file (gotta be tidy!)

}

Answer 1

You can't just invent a way to call main . You need to use one of the standard ways, like this:

int main(int argc, char *argv[])
{
    if (argc < 2) {
        fprintf(stderr, "Missing filename\n");
        return -1;
    }
    FILE *fp = fopen(argv[1], "r");
    // ...
}

And note that argv[0] contains the program name (if available; if not it contains an empty string).

Your program segfaulted because you received the int argc argument into your char *filename[] parameter. If you ran the program with a single command line parameter, the value passed in as the first argument would have been 2, which is not a valid pointer value. The expression filename[0] dereferences that address and causes a segfault.

Answer 2

Any time you get a segfault in C, you should smell a bad pointer or address in an argument list. In this particular case., the signature of main is always int main(int argc, char** argv) . Yours isn't.

What you want is

int main(int argc, char ** argv) {
  ...
  FILE * fp = fopen(argv[1]); // Quiz: why argv[1]? What's argv[0]?

You're getting away with it in the compiler because, basically, luck.

I also notice in your example call, there's actually no argument in the argument list, because you're using redirection.

Answer 3

Use:

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

And use argv[1] as fileName.

Main function must receive always that two parameters.