Code in case if the image not is visible.
#include <stdio.h>
int fun(int n)
{
if(n=4)
return n;
else
return 2*fun(n+1);
}
int main()
{
printf("%d", fun(2));
}
This is the code snippet and the output is given as 4 by the professor.
How is the output correct?
Is it possible that n=4 is assigned in the 'if-else' statement as the assignment operator is correct, but the "if" condition will not work as the syntax is wrong and the output will be directly given as 4.
The answer is correct, and there aren't any syntax errors.
= is an assignment operator, and in C/C++, (n = 4) is a valid expression that evaluates to true as long as the expression is not (n = 0), because n will then be considered as false by C. Note that in C/C++, 0 is false and everything else is true.
Hence, if (n = 4) is perfectly valid and always evaluates to true. Of course, in the process, there will also be an assignment involved.
Thus, what happens in the code above is that
the integer n is assigned the value 4 in n = 4
(n=4) as an expression returns true.
return n (4).
So the answer is 4.
Assignment in a if-else statement is valid syntax and the branching will depend on the value of n.
Example:
#include <iostream>
int main() {
int n = 2;
if (n = 0) {
std::cout << "Never printed\n";
}
if (n = 4) {
std::cout << "Always printed\n";
}
return 0;
}
Compiler explorer: https://godbolt.org/z/fEYPcq
You should use == operator to make a confront between two compatible values.
if(n = 4) assign 4 to n and then the if statement is always true. So the return value will always be 4.
"Is it possible that n = 4 is assigned in the if-else statement as the assignment operator is correct, but the if condition will not work as the syntax is wrong....?"
The syntax is not wrong and the if condition does work. It is perfectly valid. And yes, the assignment is also valid/correct.
With if (n = 4) you assign the value of 4 to the function-local variable n, although this makes less sense since n is a parameter and is meant to be feed with different values at each call to the function fun().
But I guess the intention of your professor is exactly to demonstrate this trickery, so it makes sense.
So the value of n is not 2 anymore; It is 4.
This is a valid expression for the if condition and evaluates to 1/true since the value/expression to be assigned is or does not evaluate not 0.
Usually a compiler will warn you about doing so nonetheless to avoid any undesired result here by suggesting optional parentheses around the assignment like: if ((n = 4)).
Clang:
warning: using the result of an assignment as a condition without parentheses [-Wparentheses]
GCC:
warning: suggest parentheses around assignment used as truth value [-Wparentheses]
If you explicitly want to remove these warnings, use the -Wno-parentheses flag. But it is recommended not to do so.
Since the if condition is true, it doesn't get to the recursive part in the else condition and the function fun immediately returns n which is 4.
So is the return value of 4 displayed by the call to printf() in the caller.
My favorite to remember that is this:
int war = false;
if (war = true) { launch nuke; }
Credits go to WTP.
Maybe you'll catch the joke. ;-)
Is it possible to do the following without getting a stackoverflow?
int foo(int i = 0)
{
return foo(--i) || i;
}
When foo(--i) evaluates to 0, it should return false, therefore returning i and ending the execution.
You want return !i || foo(--i).
Note that || is short-circutted. This means that evaluation (which is performed from left to right) will continue for only as long as the result of the expression is not known. So the way I've written it, the case i being zero will block the recursion.
(Moving on to more advanced issues, you need to be very careful when evaluating expressions when the same variable appears in more than one sub-expression and it's value is changed in some of them. My having !i and --i could get me in trouble: I'm not too far away from undefined behaviour here. In fact, it turns out that my code is completely safe since || is a sequence point and the order of evaluation with || is well-defined. But do be careful.)
From what you have coded foo never returns because it will continuosly go on decreasing i and recurse. You need a check inside foo which returns whatever you want when you get i as 0.
I was writing a console application that would try to "guess" a number by trial and error, it worked fine and all but it left me wondering about a certain part that I wrote absentmindedly,
The code is:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int x,i,a,cc;
for(;;){
scanf("%d",&x);
a=50;
i=100/a;
for(cc=0;;cc++)
{
if(x<a)
{
printf("%d was too big\n",a);
a=a-((100/(i<<=1))?:1);
}
else if (x>a)
{
printf("%d was too small\n",a);
a=a+((100/(i<<=1))?:1);
}
else
{
printf("%d was the right number\n-----------------%d---------------------\n",a,cc);
break;
}
}
}
return 0;
}
More specifically the part that confused me is
a=a+((100/(i<<=1))?:1);
//Code, code
a=a-((100/(i<<=1))?:1);
I used ((100/(i<<=1))?:1) to make sure that if 100/(i<<=1) returned 0 (or false) the whole expression would evaluate to 1 ((100/(i<<=1))?:***1***), and I left the part of the conditional that would work if it was true empty ((100/(i<<=1))? _this space_ :1), it seems to work correctly but is there any risk in leaving that part of the conditional empty?
This is a GNU C extension (see ?: wikipedia entry), so for portability you should explicitly state the second operand.
In the 'true' case, it is returning the result of the conditional.
The following statements are almost equivalent:
a = x ?: y;
a = x ? x : y;
The only difference is in the first statement, x is always evaluated once, whereas in the second, x will be evaluated twice if it is true. So the only difference is when evaluating x has side effects.
Either way, I'd consider this a subtle use of the syntax... and if you have any empathy for those maintaining your code, you should explicitly state the operand. :)
On the other hand, it's a nice little trick for a common use case.
This is a GCC extension to the C language. When nothing appears between ?:, then the value of the comparison is used in the true case.
The middle operand in a conditional expression may be omitted. Then if the first operand is nonzero, its value is the value of the conditional expression.
Therefore, the expression
x ? : y
has the value of x if that is nonzero; otherwise, the value of y.
This example is perfectly equivalent to
x ? x : y
In this simple case, the ability to omit the middle operand is not especially useful. When it becomes useful is when the first operand does, or may (if it is a macro argument), contain a side effect. Then repeating the operand in the middle would perform the side effect twice. Omitting the middle operand uses the value already computed without the undesirable effects of recomputing it.
I cannot figure out what the difference between the following pieces of code is:
int t = __double2int_rd(pos.x/params.cellSize.x*2.0)&1;
if( t ==0) {...}
and
if(__double2int_rd(pos.x/params.cellSize.x*2.0)&1 == 0) {...}
The second option never returns true, while the first behaves as expected.
Does anyone have any ideas?
The second expression first evaluates (1==0) whose result is always false. Then ANDs it with the result of the function __double2int_rd.
Therefore it actually evaluates:
if(__double2int_rd(pos.x/params.cellSize.x*2.0) & 0)
Which would always be false.
The equivalent of the first expression would be:
if((__double2int_rd(pos.x/params.cellSize.x*2.0) & 1) == 0)
Mind the brackets.
Its a good programming practice to add brackets if you are not sure about the order of evaluation of expressions.
Is there any reason not to use the bitwise operators &, |, and ^ for "bool" values in C++?
I sometimes run into situations where I want exactly one of two conditions to be true (XOR), so I just throw the ^ operator into a conditional expression. I also sometimes want all parts of a condition to be evaluated whether the result is true or not (rather than short-circuiting), so I use & and |. I also need to accumulate Boolean values sometimes, and &= and |= can be quite useful.
I've gotten a few raised eyebrows when doing this, but the code is still meaningful and cleaner than it would be otherwise. Is there any reason NOT to use these for bools? Are there any modern compilers that give bad results for this?
|| and && are boolean operators and the built-in ones are guaranteed to return either true or false. Nothing else.
|, & and ^ are bitwise operators. When the domain of numbers you operate on is just 1 and 0, then they are exactly the same, but in cases where your booleans are not strictly 1 and 0 – as is the case with the C language – you may end up with some behavior you didn't want. For instance:
BOOL two = 2;
BOOL one = 1;
BOOL and = two & one; //and = 0
BOOL cand = two && one; //cand = 1
In C++, however, the bool type is guaranteed to be only either a true or a false (which convert implicitly to respectively 1 and 0), so it's less of a worry from this stance, but the fact that people aren't used to seeing such things in code makes a good argument for not doing it. Just say b = b && x and be done with it.
Two main reasons. In short, consider carefully; there could be a good reason for it, but if there is be VERY explicit in your comments because it can be brittle and, as you say yourself, people aren't generally used to seeing code like this.
Bitwise xor != Logical xor (except for 0 and 1)
Firstly, if you are operating on values other than false and true (or 0 and 1, as integers), the ^ operator can introduce behavior not equivalent to a logical xor. For example:
int one = 1;
int two = 2;
// bitwise xor
if (one ^ two)
{
// executes because expression = 3 and any non-zero integer evaluates to true
}
// logical xor; more correctly would be coded as
// if (bool(one) != bool(two))
// but spelled out to be explicit in the context of the problem
if ((one && !two) || (!one && two))
{
// does not execute b/c expression = ((true && false) || (false && true))
// which evaluates to false
}
Credit to user #Patrick for expressing this first.
Order of operations
Second, |, &, and ^, as bitwise operators, do not short-circuit. In addition, multiple bitwise operators chained together in a single statement -- even with explicit parentheses -- can be reordered by optimizing compilers, because all 3 operations are normally commutative. This is important if the order of the operations matters.
In other words
bool result = true;
result = result && a() && b();
// will not call a() if result false, will not call b() if result or a() false
will not always give the same result (or end state) as
bool result = true;
result &= (a() & b());
// a() and b() both will be called, but not necessarily in that order in an
// optimizing compiler
This is especially important because you may not control methods a() and b(), or somebody else may come along and change them later not understanding the dependency, and cause a nasty (and often release-build only) bug.
The raised eyebrows should tell you enough to stop doing it. You don't write the code for the compiler, you write it for your fellow programmers first and then for the compiler. Even if the compilers work, surprising other people is not what you want - bitwise operators are for bit operations not for bools.
I suppose you also eat apples with a fork? It works but it surprises people so it's better not to do it.
I think
a != b
is what you want
Disadvantages of the bitlevel operators.
You ask:
“Is there any reason not to use the bitwise operators &, |, and ^ for "bool" values in C++? ”
Yes, the logical operators, that is the built-in high level boolean operators !, && and ||, offer the following advantages:
Guaranteed conversion of arguments to bool, i.e. to 0 and 1 ordinal value.
Guaranteed short circuit evaluation where expression evaluation stops as soon as the final result is known.
This can be interpreted as a tree-value logic, with True, False and Indeterminate.
Readable textual equivalents not, and and or, even if I don't use them myself.
As reader Antimony notes in a comment also the bitlevel operators have alternative tokens, namely bitand, bitor, xor and compl, but in my opinion these are less readable than and, or and not.
Simply put, each such advantage of the high level operators is a disadvantage of the bitlevel operators.
In particular, since the bitwise operators lack argument conversion to 0/1 you get e.g. 1 & 2 → 0, while 1 && 2 → true. Also ^, bitwise exclusive or, can misbehave in this way. Regarded as boolean values 1 and 2 are the same, namely true, but regarded as bitpatterns they're different.
How to express logical either/or in C++.
You then provide a bit of background for the question,
“I sometimes run into situations where I want exactly one of two conditions to be true (XOR), so I just throw the ^ operator into a conditional expression.”
Well, the bitwise operators have higher precedence than the logical operators. This means in particular that in a mixed expression such as
a && b ^ c
you get the perhaps unexpected result a && (b ^ c).
Instead write just
(a && b) != c
expressing more concisely what you mean.
For the multiple argument either/or there is no C++ operator that does the job. For example, if you write a ^ b ^ c than that is not an expression that says “either a, b or c is true“. Instead it says, “An odd number of a, b and c are true“, which might be 1 of them or all 3…
To express the general either/or when a, b and c are of type bool, just write
(a + b + c) == 1
or, with non-bool arguments, convert them to bool:
(!!a + !!b + !!c) == 1
Using &= to accumulate boolean results.
You further elaborate,
“I also need to accumulate Boolean values sometimes, and &= and |=? can be quite useful.”
Well, this corresponds to checking whether respectively all or any condition is satisfied, and de Morgan’s law tells you how to go from one to the other. I.e. you only need one of them. You could in principle use *= as a &&=-operator (for as good old George Boole discovered, logical AND can very easily be expressed as multiplication), but I think that that would perplex and perhaps mislead maintainers of the code.
Consider also:
struct Bool
{
bool value;
void operator&=( bool const v ) { value = value && v; }
operator bool() const { return value; }
};
#include <iostream>
int main()
{
using namespace std;
Bool a = {true};
a &= true || false;
a &= 1234;
cout << boolalpha << a << endl;
bool b = {true};
b &= true || false;
b &= 1234;
cout << boolalpha << b << endl;
}
Output with Visual C++ 11.0 and g++ 4.7.1:
true
false
The reason for the difference in results is that the bitlevel &= does not provide a conversion to bool of its right hand side argument.
So, which of these results do you desire for your use of &=?
If the former, true, then better define an operator (e.g. as above) or named function, or use an explicit conversion of the right hand side expression, or write the update in full.
Contrary to Patrick's answer, C++ has no ^^ operator for performing a short-circuiting exclusive or. If you think about it for a second, having a ^^ operator wouldn't make sense anyway: with exclusive or, the result always depends on both operands. However, Patrick's warning about non-bool "Boolean" types holds equally well when comparing 1 & 2 to 1 && 2. One classic example of this is the Windows GetMessage() function, which returns a tri-state BOOL: nonzero, 0, or -1.
Using & instead of && and | instead of || is not an uncommon typo, so if you are deliberately doing it, it deserves a comment saying why.
Patrick made good points, and I'm not going to repeat them. However might I suggest reducing 'if' statements to readable english wherever possible by using well-named boolean vars.For example, and this is using boolean operators but you could equally use bitwise and name the bools appropriately:
bool onlyAIsTrue = (a && !b); // you could use bitwise XOR here
bool onlyBIsTrue = (b && !a); // and not need this second line
if (onlyAIsTrue || onlyBIsTrue)
{
.. stuff ..
}
You might think that using a boolean seems unnecessary, but it helps with two main things:
Your code is easier to understand because the intermediate boolean for the 'if' condition makes the intention of the condition more explicit.
If you are using non-standard or unexpected code, such as bitwise operators on boolean values, people can much more easily see why you've done this.
EDIT: You didnt explicitly say you wanted the conditionals for 'if' statements (although this seems most likely), that was my assumption. But my suggestion of an intermediate boolean value still stands.
Using bitwise operations for bool helps save unnecessary branch prediction logic by the processor, resulting from a 'cmp' instruction brought in by logical operations.
Replacing the logical with bitwise operations (where all operands are bool) generates more efficient code offering the same result. The efficiency ideally should outweigh all the short-circuit benefits that can be leveraged in the ordering using logical operations.
This can make code a bit un-readable albeit the programmer should comment it with reasons why it was done so.
IIRC, many C++ compilers will warn when attempting to cast the result of a bitwise operation as a bool. You would have to use a type cast to make the compiler happy.
Using a bitwise operation in an if expression would serve the same criticism, though perhaps not by the compiler. Any non-zero value is considered true, so something like "if (7 & 3)" will be true. This behavior may be acceptable in Perl, but C/C++ are very explicit languages. I think the Spock eyebrow is due diligence. :) I would append "== 0" or "!= 0" to make it perfectly clear what your objective was.
But anyway, it sounds like a personal preference. I would run the code through lint or similar tool and see if it also thinks it's an unwise strategy. Personally, it reads like a coding mistake.