invalid initializer error for argv[0] and argv[1]?-CodePudding

Whenever I compile my c program, I get this error for both of my argv[]'s. error: invalid initializer char baseDir[] = argv[0]; error: invalid initializer char pattern[] = argv[1];

This is a snippet of the main method in my program.

void walkDir(char *baseddir, char *pattern);

int main(int argc, char *argv[]){
  char baseDir[] = argv[0];
  char pattern[] = argv[1];
  walkDir(baseDir, pattern);
  if(count==0){
     printf("No match found \n");
  }
printf("\n Done \n");
return 0;
}

CodePudding user response：

argv[0] and argv[1] are char*. They point at strings of unknown length. argv[0] usually holds a pointer to the name of the program itself while argv[1] holds a pointer to the first argument supplied to the program.

To initialize a char[] you need something like this:

char baseDir[] = "a string literal";

In your case, it's not a string literal so you can't initialize a char[] with it, so you probably should just copy the pointers:

const char *baseDir = argv[1]; // a pointer to the first argument
const char *pattern = argv[2]; // a pointer to the second argument

I added const since you shouldn't change the strings they are pointing at. I suppose you are not planning to change the strings in your walkDir function either, so then change it accordingly:

void walkDir(const char *baseddir, const char *pattern)