Tukukkk

I am new to C++ and sometimes it is difficult for me to understand complex if statements, could anyone help explain following code to me? Thanks in advance.
Given:

int vis[25], g[25][25], Ty[25][25];



for (int i = 1; i < 30; i++)

{ 

if(!vis[i] && g[x][i] == 1 && Ty[f][i] != n) {...}

}

Vis is an array and only initialized and there is no any assigned value in it at the moment.
So what does !vis[i] mean?
Does it mean vis[i] !=0 or vis[i]==1 or something else ?

int vis[25], g[25][25], Ty[25][25];

Declares 3 arrays. "vis" is 1D array of size 25, "g" is 2D array of size 25x25, and same with "Ty".

    if(!vis[i] && g[x][i] == 1 && Ty[f][i] != n) {...}

In C++, integers evaluate to "false" in boolean expressions if the value is 0, and "true" for all other values. So in the if statement, the first expression "!vis[i]" will evaluate to true when vis[i] == 0.

The second expression will evaluate true when the value at index [x][i] in 'g' is equal to 1.

The third statement will evaluate to true when the value at index [f][i] in 'Ty' is not equal to some variable 'n', which is presumably defined somewhere in your program like 'x'.

*Note - as drescherjm noted in a comment, the for-loop should only go up to 24, as the size of your arrays is 25 in either direction, so accessing an element outside the range 0-24 (inclusive) is undefined behavior.

• A int is only false if it's 0. This means with any other number the string is true.


• A ! in front of a logical statement negates it.

So the if check:

1. 
   

   the array vis on position i has to be 0.

   
   
   

   !vis[i] == !(vis[i] != 0) == vis[i] == 0

   
   



2. The array value of g on x, i must be one



3. The array of Ty on position f, i cannot be equal to the value in n