I am looking at some legacy C code and got confused, it is something like:
UINT A, B = 1;
if((A = B) == 1){
return(TRUE);
} else {
return(FALSE);
}
We all know there will be a compiler warning if we do if(A = B), but here it looks like the 'if' is checking A against 1, am I correct?
5 Answers
First, it assigns the value of B to A (A = B), then it checks if the result of this assignment, which is A and evaluates to 1, is equal to 1.
So technically you are correct: On the way it checks A against 1.
To make things easier to read, the code is equivalent to:
UINT A, B = 1;
A = B;
if(A == 1){
return(TRUE);
} else {
return(FALSE);
}
Rather, your code is always assigning B to A, and it is moreover checking whether the value of B (and thus also A) is equal to 1.
There's nothing "legacy" about this, this is generally a pretty handy idiom if you need the result of an operation but also want to check for errors:
int result;
if ((result = foo()) != -1)
{
printf("The result is: %i\n", result);
}
else
{
// panic
}
If you want to keep it on 1 line:
if ((A = B), A == 1)
does the same thing.
We are trying to avoid if statements to make code more readable.
UINT A, B = 1;
bool fResult = false;
fResult = (A == B);
return(fResult);
And if there must be an condition to act on (not) equality, see this example.
UINT A, B = 1;
bool fResult = false;
fResult = (A == B);
if(fResult)
{
doThis();
}
else
{
doThat();
}
return(fResult);
Correct. The value A has after the assignment will be compared to 1.
This code sample is equivalent to just:
return (TRUE);
