Comparing strings?

Hello,
I have a question about comparing strings.
First, I will post the code I've written:

Code:

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

void make_word(char* s, char c)
{
        int len = strlen(s);
        s[len] = c;
        s[len + 1] = '\0';
}

int main()
{
        char grid [3][4] = {{'a', 'b', 'c'},
                            {'d', 'e', 'f'},
                            {'g', 'h', 'i'}};

        char word[] = "abc";

        char found_word[] = "";       //room for found_word

        int i;

        for(i = 0; i != 3; i++)
        {
            make_word(found_word, grid[i]);
        }




        if(word == grid[3])
        {
            printf("yes");
        }




        return 0;
}

So what I am trying to do is to compare two strings.
In the code, there is an array grid and a string found_word.
The string found_word is made from the array grid.
The function allows the grid to become a string.
So what I have done is make the 'a' 'b' 'c' into a string "abc", which is same as "abc" from the string word.
So now the word contains "abc" and found_word also contains "abc"
However, when I try the if loop, they are not the same.
How do I make the array and the string the same so that when the if loop runs, it prints the "yes"?

Please help

1.Allocate memory for char*s before using it....
2.grid[i] is not a (char) ..it is a (char*)
3.Your

Code:

void make_word(char*,char)

function appends the character in the arguments after the string(as you meant it)..but found_word contains only a '\o' and grid[i] isn't a character..

First of all... it's int main (void)

Second, your make word function is not a text search. All it does is add a 0 onto the end of a string.

Third ... char found_word[] = ""; ... is NOT how you create text buffers. The result will be an array of 1 character containing a 0.

Finally ... you cannot compare strings across the = sign in C. It doesn't know how to do that so you will need to find a library function (or write your own) to do that for you.

Originally Posted by CommonTater

First of all... it's int main (void)

Does not matter here: like void, an empty parameter list in a function definition specifies that the function has no parameters.

>>Does not matter here: like void, an empty parameter list in a function definition specifies that the function has no parameters.

In C++. But not in C
For C, it means what parameter (number, type) is not specified.

Code:

void foo() 
{

}


...
 foo()
 foo(3,2);
 foo("hello", 2.32,3);

This will compile cleanly in C. But not in C++.

Originally Posted by Bayint Naung

In C++. But not in C
For C, it means what parameter (number, type) is not specified.

We're talking about a function definition, not a function declaration that is not a function definition:

Originally Posted by C99 Clause 6.7.5.3 Paragraph 14

An identifier list declares only the identifiers of the parameters of the function. An empty list in a function declarator that is part of a definition of that function specifies that the function has no parameters. The empty list in a function declarator that is not part of a definition of that function specifies that no information about the number or types of the parameters is supplied.

Originally Posted by Bayint Naung

This will compile cleanly in C.

I have observed this before in gcc, but based on my understanding of the C standard, I believe that "compile cleanly" is a compiler bug. That said, as a rule of thumb explicit is better than implicit, and C99 does say that the use of empty parentheses in this way is an "obsolescent feature" (though whether it will really be removed from the language is another matter).