C++ inline if fails in switch - c++

I just found out the hard way an inline if (A?B:C) does not work as expected in a switch statement.
where A a boolean, B and C both integer unequal to 0. The result of this statement is 0 when placed inside a switch.
I found a stackoverflow post [1] where this behaviour was mentioned but I can not find any explanation why this doesn't work as I would expect. What is causing this?
For example:
int foo = 6;
switch(foo)
{
case 6:
return 10 + true ? 2 : 4;
}
[1] Benefits of inline functions in C++?

This is nothing to do with switch.
10 + true ? 2 : 4
is equivalent to:
(10 + true) ? 2 : 4.
If you want it to act like:
10 + (true ? 2 : 4)
then you will need to write it like that.

Related

Can I prevent GNU bc from spewing out zeros?

Suppose I want to solve the equation x + 3 = 40 using GNU bc. One way I could do this would be to start by checking to see if 0 is a solution, then checking 1, and so on, until I get to the right answer. (Obviously not the best way to do algebra, but oh well.) So I enter the following code into GNU bc:
int solver(int x);
define solver(x){
if(x + 3 == 40) return x;
x = x + 1;
solver(x)
}
solver(0)
It produces 37 - the right answer, of course - but the 37 is then followed by 37 zeros. Based on some experimentation, it seems like each zero comes from an instance of the if statement being false, but how do I prevent the zeros from showing up? I'm using GNU bc to solve more complicated functions and create more complex lists of numbers, so it really isn't practical for me to sort through all the zeros. Any help would be appreciated, since I haven't yet figured anything out.
For each operation that isn't an assignment, bc prints an exit status. One way to suppress that is to assign to the dummy value . (which is just the value of the last result anyway), another way is to make sure you explicitly print exactly what you need.
I would have written your function like this:
#!/usr/bin/bc -q
define solver(x) {
if (x + 3 == 40) return x
return solver(x+1)
}
print solver(0), "\n"
quit
A few remarks for your attempt:
I don't understand what your first line is supposed to do, I just dropped it
I've indented the code, added some whitespace and removed the semicolons – mostly a matter of taste and readability
I've simplified the recursive call to avoid the solver(x) line stand on its own, as this produces the spurious 0
As for your suspicion that the if statement produces the zeroes: try, in an interactive session, the following:
1 == 2 # Equality test on its own produces output
0
1 == 1 # ... for both true and false statements
1
if (1 == 2) print "yes\n" # No output from false if condition
if (1 == 1) print "yes\n" # If statement is true, print string
yes

Postfix operators in while loops

I was looking through some C++ code today when I stumbled upon this:
while (c--) {
a = (a + 1) % n;
while(arr[a]) a = (a + 1) % n;
}
c was an integer. Don't while loops take boolean expressions to be evaluated? I know 1 and 0 are fine because they represent true and false, respectively, but in this case c was taking on values other than 0 and 1. What does this do?
Anything that isn't 0 is considered true, so this would loop until C=0.
Could just be a simple case of "using less code"

simple program without using "if condition"?

Is there any way of creating a simple java(or c,c ++, python) program that prints 3 (outputs the 3) when given input=6 and it gives output=6 when given input=3 without using "if conditions" ?
Assuming you're happy for it to produce other outputs on inputs that aren't 6 or 3, then you can just compute 9-x.
You can always just use a switch-case statement. Also, if you only want those two answers, you could also take the input as an int and do 9-[your int] and print that answer.
You can use the XOR bit operation. It compares pairs of bits and returns 0 if bits are equals and 1 if bits are different.
We have 3 = 011b and 6 = 110b. This numbers differ by 1 and 3 digit (bit), so XOR mask will be 101b = 5.
Code example:
public static int testMethod(int value){
return System.out.println(value ^ 5);
}
without if or without control flow statement/condition statement ?
you could use switch statement
private void tes(int i) {
switch (i) {
///give output 6 where input is 3
case 3:
System.out.println(6);
break;
///give output 3 where input is 6
case 6:
System.out.println(3);
break;
}
}

Unexpected result regarding comparing two pointer-related integer values in C++

I have a BST of three elements {1, 2, 3}. Its structure looks like
2
/ \
1 3
Now I try to calculate the height for each node using BSTHeight() defined below and have some problem with calculating the height of '2', which value is supposed to be 1 as the heights of '1' and '3' are defined as 0. My problem is that with direct use of heights from '2's two children (see part 2 highlighted below), its height is ALWAYS 0. However, its value is correct if I use two temporary integer variables (see part 1 highlighted below). I couldn't see any difference between the two approaches in terms of functionality. Can anyone help explain why?
void BSTHeight(bst_node *p_node)
{
if (!p_node)
return;
if (!p_node->p_lchild && !p_node->p_rchild) {
p_node->height = 0;
} else if (p_node->p_lchild && p_node->p_rchild) {
BSTHeight(p_node->p_lchild);
BSTHeight(p_node->p_rchild);
#if 0 // part 1
int lchild_height = p_node->p_lchild->height;
int rchild_height = p_node->p_rchild->height;
p_node->height = 1 + ((lchild_height > rchild_height) ? lchild_height : rchild_height);
#else // part 2
p_node->height = 1 + ((p_node->p_lchild->height) > (p_node->p_rchild->height)) ? (p_node->p_lchild->height) : (p_node->p_rchild->height);
#endif
} else if (!p_node->p_lchild) {
BSTHeight(p_node->p_rchild);
p_node->height = 1 + p_node->p_rchild->height;
} else {
BSTHeight(p_node->p_lchild);
p_node->height = 1 + p_node->p_lchild->height;
}
}
Problem lies in operator precedence. Addition binds stronger than ternary operator, hence you must surround ternary operator (?:) with brackets.
Below is the corrected version. Note that all brackets you used were superflous and I've removed them. I've added the only needed pair instead:
1 + (p_node->p_lchild->height > p_node->p_rchild->height ?
p_node->p_lchild->height : p_node->p_rchild->height);
Even better would be to use std::max (from <algorithm>) instead:
1 + std::max(p_node->p_lchild->height, p_node->p_rchild->height)

What is the "-->" operator in C++?

After reading Hidden Features and Dark Corners of C++/STL on comp.lang.c++.moderated, I was completely surprised that the following snippet compiled and worked in both Visual Studio 2008 and G++ 4.4.
Here's the code:
#include <stdio.h>
int main()
{
int x = 10;
while (x --> 0) // x goes to 0
{
printf("%d ", x);
}
}
Output:
9 8 7 6 5 4 3 2 1 0
I'd assume this is C, since it works in GCC as well. Where is this defined in the standard, and where has it come from?
--> is not an operator. It is in fact two separate operators, -- and >.
The conditional's code decrements x, while returning x's original (not decremented) value, and then compares the original value with 0 using the > operator.
To better understand, the statement could be written as follows:
while( (x--) > 0 )
Or for something completely different... x slides to 0.
while (x --\
\
\
\
> 0)
printf("%d ", x);
Not so mathematical, but... every picture paints a thousand words...
That's a very complicated operator, so even ISO/IEC JTC1 (Joint Technical Committee 1) placed its description in two different parts of the C++ Standard.
Joking aside, they are two different operators: -- and > described respectively in §5.2.6/2 and §5.9 of the C++03 Standard.
x can go to zero even faster in the opposite direction in C++:
int x = 10;
while( 0 <---- x )
{
printf("%d ", x);
}
8 6 4 2
You can control speed with an arrow!
int x = 100;
while( 0 <-------------------- x )
{
printf("%d ", x);
}
90 80 70 60 50 40 30 20 10
;)
It's equivalent to
while (x-- > 0)
x-- (post decrement) is equivalent to x = x-1 (but returning the original value of x), so the code transforms to:
while(x > 0) {
x = x-1;
// logic
}
x--; // The post decrement done when x <= 0
It's
#include <stdio.h>
int main(void) {
int x = 10;
while (x-- > 0) { // x goes to 0
printf("%d ", x);
}
return 0;
}
Just the space makes the things look funny, -- decrements and > compares.
The usage of --> has historical relevance. Decrementing was (and still is in some cases), faster than incrementing on the x86 architecture. Using --> suggests that x is going to 0, and appeals to those with mathematical backgrounds.
Utterly geek, but I will be using this:
#define as ;while
int main(int argc, char* argv[])
{
int n = atoi(argv[1]);
do printf("n is %d\n", n) as ( n --> 0);
return 0;
}
while( x-- > 0 )
is how that's parsed.
One book I read (I don't remember correctly which book) stated: Compilers try to parse expressions to the biggest token by using the left right rule.
In this case, the expression:
x-->0
Parses to biggest tokens:
token 1: x
token 2: --
token 3: >
token 4: 0
conclude: x-- > 0
The same rule applies to this expression:
a-----b
After parse:
token 1: a
token 2: --
token 3: --
token 4: -
token 5: b
conclude: (a--)-- - b
This is exactly the same as
while (x--)
Anyway, we have a "goes to" operator now. "-->" is easy to be remembered as a direction, and "while x goes to zero" is meaning-straight.
Furthermore, it is a little more efficient than "for (x = 10; x > 0; x --)" on some platforms.
This code first compares x and 0 and then decrements x. (Also said in the first answer: You're post-decrementing x and then comparing x and 0 with the > operator.) See the output of this code:
9 8 7 6 5 4 3 2 1 0
We now first compare and then decrement by seeing 0 in the output.
If we want to first decrement and then compare, use this code:
#include <stdio.h>
int main(void)
{
int x = 10;
while( --x> 0 ) // x goes to 0
{
printf("%d ", x);
}
return 0;
}
That output is:
9 8 7 6 5 4 3 2 1
My compiler will print out 9876543210 when I run this code.
#include <iostream>
int main()
{
int x = 10;
while( x --> 0 ) // x goes to 0
{
std::cout << x;
}
}
As expected. The while( x-- > 0 ) actually means while( x > 0). The x-- post decrements x.
while( x > 0 )
{
x--;
std::cout << x;
}
is a different way of writing the same thing.
It is nice that the original looks like "while x goes to 0" though.
There is a space missing between -- and >. x is post decremented, that is, decremented after checking the condition x>0 ?.
-- is the decrement operator and > is the greater-than operator.
The two operators are applied as a single one like -->.
It's a combination of two operators. First -- is for decrementing the value, and > is for checking whether the value is greater than the right-hand operand.
#include<stdio.h>
int main()
{
int x = 10;
while (x-- > 0)
printf("%d ",x);
return 0;
}
The output will be:
9 8 7 6 5 4 3 2 1 0
C and C++ obey the "maximal munch" rule. The same way a---b is translated to (a--) - b, in your case x-->0 translates to (x--)>0.
What the rule says essentially is that going left to right, expressions are formed by taking the maximum of characters which will form a valid token.
Actually, x is post-decrementing and with that condition is being checked. It's not -->, it's (x--) > 0
Note: value of x is changed after the condition is checked, because it post-decrementing. Some similar cases can also occur, for example:
--> x-->0
++> x++>0
-->= x-->=0
++>= x++>=0
char sep = '\n' /1\
; int i = 68 /1 \
; while (i --- 1\
\
/1/1/1 /1\
/1\
/1\
/1\
/1\
/ 1\
/ 1 \
/ 1 \
/ 1 \
/1 /1 \
/1 /1 \
/1 /1 /1/1> 0) std::cout \
<<i<< sep;
For larger numbers, C++20 introduces some more advanced looping features.
First to catch i we can build an inverse loop-de-loop and deflect it onto the std::ostream. However, the speed of i is implementation-defined, so we can use the new C++20 speed operator <<i<< to speed it up. We must also catch it by building wall, if we don't, i leaves the scope and de referencing it causes undefined behavior. To specify the separator, we can use:
std::cout \
sep
and there we have a for loop from 67 to 1.
Instead of regular arrow operator (-->) you can use armor-piercing arrow operator: --x> (note those sharp barbs on the arrow tip). It adds +1 to armor piercing, so it finishes the loop 1 iteration faster than regular arrow operator. Try it yourself:
int x = 10;
while( --x> 0 )
printf("%d ", x);
Why all the complication?
The simple answer to the original question is just:
#include <stdio.h>
int main()
{
int x = 10;
while (x > 0)
{
printf("%d ", x);
x = x-1;
}
}
It does the same thing. I am not saying you should do it like this, but it does the same thing and would have answered the question in one post.
The x-- is just shorthand for the above, and > is just a normal greater-than operator. No big mystery!
There are too many people making simple things complicated nowadays ;)
Conventional way we define condition in while loop parenthesis"()" and terminating condition inside the braces"{}", but this -- & > is a way one defines all at once.
For example:
int abc(){
int a = 5
while((a--) > 0){ // Decrement and comparison both at once
// Code
}
}
It says, decrement a and run the loop till the time a is greater than 0
Other way it should have been like:
int abc() {
int a = 5;
while(a > 0) {
a = a -1 // Decrement inside loop
// Code
}
}
Both ways, we do the same thing and achieve the same goals.
(x --> 0) means (x-- > 0).
You can use (x -->)
Output: 9 8 7 6 5 4 3 2 1 0
You can use (-- x > 0) It's mean (--x > 0)
Output: 9 8 7 6 5 4 3 2 1
You can use
(--\
\
x > 0)
Output: 9 8 7 6 5 4 3 2 1
You can use
(\
\
x --> 0)
Output: 9 8 7 6 5 4 3 2 1 0
You can use
(\
\
x --> 0
\
\
)
Output: 9 8 7 6 5 4 3 2 1 0
You can use also
(
x
-->
0
)
Output: 9 8 7 6 5 4 3 2 1 0
Likewise, you can try lot of methods to execute this command successfully.
This --> is not an operator at all. We have an operator like ->, but not like -->. It is just a wrong interpretation of while(x-- >0) which simply means x has the post decrement operator and this loop will run till it is greater than zero.
Another simple way of writing this code would be while(x--). The while loop will stop whenever it gets a false condition and here there is only one case, i.e., 0. So it will stop when the x value is decremented to zero.
Here -- is the unary post decrement operator.
while (x-- > 0) // x goes to 0
{
printf("%d ", x);
}
In the beginning, the condition will evaluate as
(x > 0) // 10 > 0
Now because the condition is true, it will go into the loop with a decremented value
x-- // x = 9
That's why the first printed value is 9
And so on. In the last loop x=1, so the condition is true. As per the unary operator, the value changed to x = 0 at the time of print.
Now, x = 0, which evaluates the condition (x > 0 ) as false and the while loop exits.
--> is not an operator, it is the juxtaposition of -- (post-decrement) and > (greater than comparison).
The loop will look more familiar as:
#include <stdio.h>
int main() {
int x = 10;
while (x-- > 0) { // x goes to 0
printf("%d ", x);
}
}
This loop is a classic idiom to enumerate values between 10 (the excluded upper bound) and 0 the included lower bound, useful to iterate over the elements of an array from the last to the first.
The initial value 10 is the total number of iterations (for example the length of the array), and one plus the first value used inside the loop. The 0 is the last value of x inside the loop, hence the comment x goes to 0.
Note that the value of x after the loop completes is -1.
Note also that this loop will operate the same way if x has an unsigned type such as size_t, which is a strong advantage over the naive alternative for (i = length-1; i >= 0; i--).
For this reason, I am actually a fan of this surprising syntax: while (x --> 0). I find this idiom eye-catching and elegant, just like for (;;) vs: while (1) (which looks confusingly similar to while (l)). It also works in other languages whose syntax is inspired by C: C++, Objective-C, java, javascript, C# to name a few.
That's what you mean.
while((x--) > 0)
We heard in childhood,
Stop don't, Let Go (روکو مت، جانے دو)
Where a Comma makes confusion
Stop, don't let go. (روکو، مت جانے دو)
Same Happens in Programming now, a SPACE makes confusion. :D
The operator you use is called "decrement-and-then-test". It is defined in the C99 standard, which is the latest version of the C programming language standard. The C99 standard added a number of new operators, including the "decrement-and-then-test" operator, to the C language. Many C++ compilers have adopted these new operators as extensions to the C++ language.
Here is how the code without using the "decrement-and-then-test" operator:
#include <stdio.h>
int main()
{
int x = 10;
while (x > 0)
{
printf("%d ", x);
x--;
}
}
In this version of the code, the while loop uses the > operator to test whether x is greater than 0. The x-- statement is used to decrement x by 1 at the end of each iteration of the loop.