I've seen instances where someone would use return statements with multiple values. For example: return 8, 10, 6; As far as I know, only one of these values will actually be returned. What is the benefit of using return this way?
Specifically, this question appears on my homework.
The statement: return 2 * 3 + 1, 1 + 5; returns the value ____.
I know it would always return 6, so why would I ever write it like this?
Forgive me if this is a simple question. I am still somewhat new to programming.
The statement return 2 * 3 + 1, 1 + 5; returns the value 6.
This the trick of comma operator in C++. You can read more about it here:
https://en.cppreference.com/w/cpp/language/operator_other
A comma operator is basically a list of expressions separated by commas, they will be evaluated from left to right, and the result of the last item will be treated as the result of the whole comma operator.
Here is a simple example demonstrating how comma operator works.
int foo() {
int i = 1;
return i += 2, i++, i + 5; // This is a comma operator with three items
// i += 2 will be evaluated first, then i == 3
// i++ will be evaluated second, then i == 4
// i + 5 will be evaluate last, and the result is 9
// the result of the last item is returned by the return statement
}
int main() {
std::cout << foo();
return 0;
}
This code prints 9.
When you do this you are avoiding using math in your function block and this make your function like working 3 time and result would be the one you want
I am asking the user to input an expression which will be evaluated in postfix notation. The beginning of the expression is the variable name where the answer of the evaluated expression will be stored. Ex: A 4 5 * 6 + 2 * 1 – 6 / 4 2 + 3 * * = where A is the variable name and the equal sign means the answer to the expression will be stored in the variable A. The OUT A statement means that the number stored in the variable A will be printed out.
What I need help with is that when I input the second expression, I do not get the right answer. For example, my first expression A 4 5 * 6 + 2 * 1 – 6 / 4 2 + 3 * * = will evaluate to 153 and then when I input my second expression B A 10 * 35.50 + =, it has to evaluate to 1565.5, but it doesn't. It evaluates to 35.5. I cannot figure out why I am getting the wrong answer. Also, I need help with the OUT statement.
else if (isalpha(expr1[i]))
{
stackIt.push(mapVars1[expr1[i]]);
}
Will place the variable, or zero if the variable has not been set, onto the stack.
else if (isalpha(expr1[i]))
{
map<char, double>::iterator found = mapVars1.find(expr1[i]);
if (found != mapVars1.end())
{
stackIt.push(found->second);
}
else
{
// error message and exit loop
}
}
Is probably better.
Other suggestions:
Compilers are pretty sharp these days, but you may get a bit out of char cur = expr1[i]; and then using cur (or suitably descriptive variable name) in place of the remaining expr1[i]s in the loop.
Consider using isdigit instead of expr1[i] >= '0' && expr1[i] <= '9'
Test your code for expressions with multiple spaces in a row or a space after an operator. It looks like you will re-add the last number you parsed.
Test for input like 123a456. You might not like the result.
If spaces after each token in the expression are specified in the expression protocol, placing your input string into a stringstream will allow you to remove a great deal of your parsing code.
stringstream in(expr1);
string token;
while (in >> token)
{
if (token == "+" || token == "-'" || ...)
{
// operator code
}
else if (token == "=")
{
// equals code
}
else if (mapVars1.find(token) != mapVars1.end())
{
// push variable
}
else if (token.length() > 0)
{
char * endp;
double val = strtod(token.c_str(), &endp);
if (*endp == '\0')
{
// push val
}
}
}
To use previous symbol names in subsequent expressions add this to the if statements in your parsing loop:
else if (expr1[i] >= 'A' && expr1[i] <= 'Z')
{
stackIt.push(mapVars1[expr[i]]);
}
Also you need to pass mapVars by reference to accumulate its contents across Eval calls:
void Eval(string expr1, map<char, double> & mapVars1)
For the output (or any) other command I would recommend parsing the command token that's at the front of the string first. Then call different evaluators based on the command string. You are trying to check for OUT right now after you have already tried to evaluate the string as an arithmetic assignment command. You need to make that choice first.
I wrote the program as follows :
#include<cstdio>
#define max(a,b) a>b?a:b
using namespace std;
int main()
{
int sum=0,i,k;
for(i=0;i<5;i++)
{
sum=sum+max(i,3);
}
printf("%d\n",sum);
return 0;
}
I got the output : 4
But when I stored max(i,3) in a variable k and then added to sum, I got the correct output:
#include<cstdio>
#define max(a,b) a>b?a:b
using namespace std;
int main()
{
int sum=0,i,k;
for(i=0;i<5;i++)
{
k=max(i,3);
sum=sum+k;
}
printf("%d\n",sum);
return 0;
}
Output : 16
Can somebody please explain why is it happening?
hash-define macros are a string expansion, not a "language" thing.
sum=sum+max(i,3);
expands to:
sum=sum+i>3?i:3;
And if you are writing that with no () round it you deserve to get the wrong answer. Try this:
#define max(a,b) (a>b?a:b)
but there are still many situations where it will fail. As others point out an even better macro is:
#define max(a,b) ((a)>(b)?(a):(b))
but it will still fail in too many situations, such as arguments with side effects getting evaluated twice. You are much much better off avoiding macros where possible and doing something like this:
template <typename T> T max(T a, T b) { return a>b?a:b; }
or, infact, using std::max and std::min which have already been written for you!
This line:
sum=sum+max(i,3);
expands to:
sum = sum + i > 3 ? i : 3;
Which, when set up with parens to make it clearer is:
sum = (sum + i) > 3 ? i : 3;
So on the 5-passes through the loop, the expressions are:
sum = (0 + 0) > 3 ? 0 : 3; // Result, sum = 3
sum = (3 + 1) > 3 ? 1 : 3; // Result: sum = 3
sum = (3 + 2) > 3 ? 2 : 3; // Result: sum = 3
sum = (3 + 3) > 3 ? 3 : 3; // Result: sum = 3
sum = (3 + 4) > 3 ? 4 : 3; // Result: sum = 4
And that's where your answer comes from.
The conventional way to solve this is to change the #define to:
#define max(a,b) (((a)>(b))?(a):(b))
But even this has some pitfalls.
I think you are having operator precedence issues, you have to remember that define will lead to a textual replacement in your source code. You should change your define to
#define max(a,b) ((a) > (b) ? (a) : (b))
The output of the prepocessor (view it with the -E flag) will be:
sum = sum+i>3?i:3;
which is the same as
sum = (sum+i)>3?i:3;
which is not what you meant because + has a higher precedence than >. You should use:
#define max(a,b) (a>b?a:b)
instead.
Replacing your macro in the line sum=sum+max(i,3); gives the following form :
sum=sum+i>3?i:3 ;
which is asking that if sum + i is greater than 3 than assign sum's value accordingly. Hence, you have 4 because each time a new assignment happens inside the loop. Use the template method suggested by Andrew.
(The loop evaluates the condition (sum + i) > 3 ? i : 3 every time. There is no cumulative addition here.)
I have a sample midterm question that I am not too sure about. Here it is:
#include <iostream.h>
void f( int i )
{
if( i = 4 || i = 5 ) return;
cout << "hello world\n" ;
}
int main()
{
f( 3 );
f( 4 );
f( 5 );
return 0;
}
So I understand that the logical OR operator has a higher precedence and that it is read left to right. I also understand that what's being used is an assignment operator instead of the relational operator. I just dont get how to make sense of it all. The first thing the compiler would check would be 4 || i? How is that evaluated and what happens after that?
Let's add all the implied parentheses (remembering that || has higher precedence than = and that = is right-associative):
i = ((4 || i) = 5)
So, it first evaluates 4 || i, which evaluates to true (actually, it even ignores i, since 4 is true and || short-circuits). It then tries to assign 5 to this, which errors out.
As written, the code doesn't compile, since operator precedence means it's i = ((4 || i) = 5) or something, and you can't assign to a temporary value like (4 || i).
If the operations are supposed to be assignment = rather than comparison == for some reason, and the assignment expressions are supposed to be the operands of ||, then you'd need parentheses
(i = 4) || (i = 5)
As you say, the result of i=4 is 4 (or, more exactly, an lvalue referring to i, which now has the value 4). That's used in a boolean context, so it's converted to bool by comparing it with zero: zero would become false, and any other value becomes true.
Since the first operand of || is true, the second isn't evaluated, and the overall result is true. So i is left with the value 4, then the function returns. The program won't print anything, whatever values you pass to the function.
It would make rather more sense using comparison operations
i == 4 || i == 5
meaning the function would only print something when the argument is neither 4 nor 5; so it would just print once in your example, for f(3).
Note that <iostream.h> hasn't been a standard header for decades. You're being taught an obsolete version of the language, using some extremely dubious code. You should get yourself a good book and stop wasting time on this course.
The compiler shall isuue an error because expression 4 || i is not a lvalue and may not be assigned.
As for the expression itself then the value of it is always equal to true because 4 is not equal to zero.
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.