# V640\. Code's operational logic does not correspond with its formatting\.

The analyzer has detected a potential error: code formatting following a conditional operator doesn't correspond to the program execution logic\. It's highly probable that opening and closing curly brackets are missing\.

Consider the following sample of incorrect code:

```cpp
if (a == 1)
  b = c; d = b;
```

In this case, the 'd \= b;' assignment will be executed all the time regardless of the 'a \=\= 1' condition\.

If the code contains a mistake, it can be fixed through adding curly brackets\. This is the fixed code:

```cpp
if (a == 1)
{ b = c; d = b; }
```



Another example of incorrect code:

```cpp
if (a == 1)
  b = c;
  d = b;
```

To fix the error here, we should use curly brackets too\. This is the fixed code:

```cpp
if (a == 1)
{
  b = c;
  d = b;
}
```

If the code is correct, it should be formatted in the following way, for the V640 warning not to be generated:

```cpp
if (a == 1)
  b = c;
d = b;
```



This type of errors can be often seen in programs that actively use macros\. Consider the following error found in one real application:

```cpp
#define DisposeSocket(a) shutdown(a, 2); closesocket(a)
...
if (sockfd > 0)
  (void) DisposeSocket(sockfd);
```

The call of the 'closesocket\(a\);' function will be executed all the time\. This will lead to a fault if the 'sockfd' variable is <\= 0\.

The error can be fixed by using curly brackets in the macro\. But you'd better create a full\-fledged function: code without macros is safer and more convenient to debug\.

This is what the correct code may look like:

```cpp
inline void DisposeSocket(int a) {
  shutdown(a, 2);
  closesocket(a);
}
...
if (sockfd > 0)
  DisposeSocket(sockfd);
```