Taking advantage of short-circuiting in boolean expressions:
int max(int a, int b, int c)
{
int m = a;
(m < b) && (m = b); //these are not conditional statements.
(m < c) && (m = c); //these are just boolean expressions.
return m;
}
Explanation:
In boolean AND operation such as x && y, y is evaluated if and only if x is true. If x is false, then y is not evaluated, because the whole expression would be false which can be deduced without even evaluating y. This is called short-circuiting when the value of a boolean expression can be deduced without evaluating all operands in it.
Apply this principle to the above code. Initially m is a. Now if (m < b) is true, then that means, b is greater than m (which is actually a), so the second subexpression (m = b) is evaluated and m is set to b. If however (m < b) is false, then second subexpression will not be evaluated and m will remain a (which is greater than b). In a similar way, second expression is evaluated (on the next line).
In short, you can read the expression (m < x) && (m = x) as follows : set m to x if and only if m is less than x i.e (m < x) is true. Hope this helps you understanding the code.
Test code:
int main() {
printf("%d\n", max(1,2,3));
printf("%d\n", max(2,3,1));
printf("%d\n", max(3,1,2));
return 0;
}
Output:
3
3
3
Note the implementation of max gives warnings because evaluated expressions are not used:
prog.c:6: warning: value computed is not used
prog.c:7: warning: value computed is not used
To avoid these (harmless) warnings, you can implement max as:
int max(int a, int b, int c)
{
int m = a;
(void)((m < b) && (m = b)); //these are not conditional statements.
(void)((m < c) && (m = c)); //these are just boolean expressions.
return m;
}
The trick is that now we’re casting the boolean expressions to void, which causes suppression of the warnings: