Question

I have created the following program which allows a user to guess a word 3 times before ending the program. I’m using a function to read the users input. When I compile the program I get the error ‘expected expression before char’. Some feedback would be great thanks!

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

void get_user_input(char *guess[10]);

void get_user_input(char *guess[10])
{ 
     printf("Please guess the word: \n");
     scanf("%s", guess);
}

int main(void)
{
     const char secret[10] = "pink";
     char guess[10];
     int i;

     for (i=0; i < 3; i++)
     {
         get_user_input(char *guess[10]);

         if (strcmp(secret, guess)==0)
         {
             printf("Your guess was correct");
             return 0;
         }
         else
         {
             printf("Your guess was incorrect. Please try again\n");
         }
     } 
     return 0;
}

Answer 1

You have an extra char here:

for (i=0; i < 3; i++)
{
    get_user_input(char *guess[10]);

Just get rid of it. You just need to pass the variable in.

get_user_input(guess);

EDIT :

The other problem seems to be this function:

void get_user_input(char *guess[10]);

change it to this:

void get_user_input(char *guess)
{ 
 printf("Please guess the word: \n");
 scanf("%s", guess);
}

and it should work. However, be aware that you run the risk of overrunning your guess array.