Thread: 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.

>>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++.