Can someone help me figure out why this simple function isn´t working?

Question

I´m trying to copy a string that is in the comand line in argv to "op"

This is the function in C that i have been trying to fix and just can´t. I know it´s something pretty basic but it´s going over my head:

  int main(int argc, char* argv[])
{
   char op[90];
   strcpy(op,*argv[1]);
   printf("\n %s",op);
   return 0;
}

It might help to explain specifically and exactly what you are trying to do. — ScottyBlades, Jan 14 '21 at 17:11
Note that `*argv[1]`is the first character of `argv[1]`.I guess you need `strcpy(op, argv[1]);` — Damien, Jan 14 '21 at 17:15
The compiler can help you. Enable warnings, read what they are saying, try to fix them. — , Jan 14 '21 at 17:15

score 1 · Answer 1 · answered Jan 14 '21 at 17:18

1

You don't need to specify the pointer sign when using strcpy(). Change:

strcpy(op, *argv[1]);

into:

strcpy(op, argv[1]);

But still you get segfault if no argument is passed. You should create a condition to check if argc != 1, if it's true then instantly stop the program before anything unusual happens.

answered Jan 14 '21 at 17:18

Rohan Bari

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And to be really safe, you be using strncpy() instead of strcpy(). – Lee Daniel Crocker Jan 14 '21 at 19:40

Can someone help me figure out why this simple function isn´t working?

1 Answers1