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You can't do that. Macros are expanded early in the compilation process, and only then is the resulting C (or C++) code compiled. The a in your code is therefore treated as a text string "a", rather than the number 1.
In C++, you could do something similar:
template <int n, class T> T FUNC(T x) { return x + n; }
...
int const a = 1;
int result = FUNC<a>(100);
Note the const before the a.
Freedom is the freedom to say that two plus two make four. If that is granted, all else follows.
-- 6079 Smith W.
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Why would you want all those extra macros? Why not just:
#define FUNC_N(x, y) ((x) + (y))
int result2 = a + 100;
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if you insist on using macros, then you will need to do something ugly, like:
#define FUNC_N(n,x) (n==1 ? FUNC_1(x) : n==2 ? FUNC_2(x) : n==3 ? FUNC_3(x) : n==4 ? FUNC_4(x) : 0)
it's much better to just make a function that can switch on n:
int FUNC_N(int n, int x)
{
switch (n)
{
case 1: return FUNC_1(x);
}
}
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Sorry, but it would take far too much time and space to explain that code in detail. If you do not understand the C language there are plenty of on-line tutorials, or you could find a book on Amazon. Alternatively, you could go back to StackOverflow and ask the person who wrote it.
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Actually i just asked to explanation code which green sign sembol answer from stackoverflow . I try to ask my question with using comment but need 50 reputation .
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You should ask the person who posted the code on Stackoverflow. This is CodeProject.
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I have a table with 2 columns; here’s two typical examples:
TABLE 1 TABLE 2
X Y X Y
-46.3 16.0 -50.3 71.2
-40.1 -28.1 -43.6 117.7
-34.0 -154.0 -36.9 165.7
-27.8 -171.8 -30.2 176.9
-21.6 178.0 -23.5 -179.2
-15.4 166.2 -16.8 -173.3
-9.3 120.1 -10.1 -149.3
-3.1 -2.0 -3.4 -86.2
3.1 -28.6 3.4 -67.0
9.3 -80.7 10.1 -72.8
15.4 -147.7 16.8 -93.5
21.6 -175.5 23.5 -151.7
27.8 162.6 30.2 112.9
34.0 120.2 36.9 80.4
40.1 49.4 43.6 67.7
46.3 15.4 50.3 71.4
In the table 1, the Y decreases, while in the table 2 the Y increases. But notice the ambiguity in the table 1 for Y= 150 and in the table 2 for Y= 70.
I generate one table at runtime (I plan to use from 20 to 50 rows), the column Y is an angle (I use radians from –pi to pi and double precision number, but here I used degrees for the sake of simplicity).
The program generates an angle from –pi to pi and I need to find the two X’s that bracket the angle. For example, if the angle is 150, for the table 1 the function should find [-15.4, -9.3] and [27.8, 34.0].
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If you wonder why no one answers your question, it may be because of the confusing way you are describing your problem:
Member 3648633 wrote: In the table 1, the Y decreases, while in the table 2 the Y increases.
No! In both tables, X increases, but Y does not strictly increase or decrease over the full range.
Member 3648633 wrote: But notice the ambiguity in the table 1 for Y= 150 and in the table 2 for Y= 70.
Ambiguity of what? What are you talking about?
Member 3648633 wrote: the column Y is an angle (I use radians from –pi to pi
That would be the interval [-3.1415, +3.1415]. Your values are much greater than that! If these are angles, your values are in degrees, not radians!
Member 3648633 wrote: I need to find the two X’s that bracket the angle. For example, if the angle is 150, for the table 1 the function should find [-15.4, -9.3] and [27.8, 34.0].
Well, finally! Why didn't you start your posting with that simple description of the task? It all makes more sense now. Moreover, there's no 'ambiguity' - there are simply multiple solutions to your problem.
Now there's only one thing left to do: pose a question. I didn't find any.
P.S.: Since you didn't pose a question, I have one for you: What should the program return for an angle input of 179 degrees? That is a possible input after all...
GOTOs are a bit like wire coat hangers: they tend to breed in the darkness, such that where there once were few, eventually there are many, and the program's architecture collapses beneath them. (Fran Poretto)
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Let us consider, 100n + 5 = O(n). For this function there are multiple n_0 and c possible.
Solution1: 100n+5 ≤100n+n=101n≤101n,for all n ≥5,n_0=5 and c=101 is a solution.
Solution2:100n+5≤100n+5n=105n≤105n,for all n≥1,n_0=1 and c=105 is also a solution
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Thank you for that bit of enlightenment. 
The difficult we do right away...
...the impossible takes slightly longer.
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The difficulty of understanding when and how it began.
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Thanks for replying!
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But how could time "begin"? Surely that implies a transition from one state to another, which requires the concept of "time" to already exist.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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The scientists tell us it started with an enormous bang.
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But again, "started" requires the concept of time to already exist.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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Hey, I didn't invent big bang. I'm only repeating what they say happened: space and time did not exist, then they did. Go figure (as the scientists probably don't say).
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Setting the date on my watch.
"(I) am amazed to see myself here rather than there ... now rather than then".
― Blaise Pascal
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Can you please elaborate?
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Time to rehash Stroustrup's computer/phone comment?
Software rusts. Simon Stephenson, ca 1994. So does this signature. me, 2012
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Hello everyone,
I have been struggling since previous month about some question that I found on some university,
and I just cant think about a good way to solve it.
Please, help me to think about a way to solve this question.
The most important thing is that the solution has to work as fast as it can with the lowest complexity
that possible.
The question :
Quote: Vik the puffin is planning a long road trip around the circle road in Iceland, during which he wants to visit
all the landmarks along a path of length L. The tank of Vik's car can take up to F units of fuel and for every
unit of distance covered, his car consumes a unit of fuel. Using Google maps, Vik knows how far each of
the N gas stations are from the beginning of the path and the price per fuel unit each station offers. At the
starting point he has T units of fuel in his car.
1. Task
Write a program that will accept the above information and will calculate the minimum amount of money
Vik needs to spend on gas. If the journey is impossible to make, it should print -1.
2. Input
The first line contains four space separated integers:
N (0 < N < 50001): The total number of gas stations
F (0 < F < 1000001): The units of fuel Vik's car can take
T (0 <= T <= F): The units of fuel Vik's car has at the beginning of the trip
L (0 < L < 1000000001): The path length of the landmarks he plans to visit
Each of the following N lines will contain two integers: the first one, D_i (0 <= D_i <= L) corresponds
to the distance of the station from the starting point, and the second one, C_i (1 <= C_i <= 1,000,000)
represents the cost per fuel unit for that station.
Note: You may assume that the trip will be on a straight line where all gas stations are
spread on this line at the positions specified by their D_i values.
3. Output
The minimum amount of money to be spent or with -1 in case the trip is not feasible.
Note: There is a newline character at the end of the last line of the output.
4. Sample input
4 20 6 34
4 40
18 15
10 7
20 12
2
5. Sample output
348
6. Explanation
The rst line of the input is 4 20 6 34 which means that:
a. There are in total N=4 gas stations on the route
b. The (max) fuel capacity of Vik's car is F=20 liters
c. The tank currently has T=6 liters of gas
d. Vik wants to travel L=34 kms in total
Then the details for the 4 gas stations are provided in the form Di Ci, where Di is the distance of this
gas station from the starting point and Ci is the cost per liter of gas:
4 40
18 15
10 7
20 12
For simplicity assume that the whole trip is done in a straight line as depicted below:
link to show the trip from the example -> Vik The Puffin — imgbb.com[^]
Obviously Vik does not have enough fuel for all 34 kms, so he needs to refuel. The cheapest gas station is
the one labeled (B) above, however Vik does not (initially) have enough fuel in his tank to reach (B), since
B-S = 10 and he has T=6. So he needs to add an extra 4 liters from gas station A, so that he can the make
it until gas station B to get as much (cheap) as he can in order to make his 34 km journey. Thus he pays
(i) 4lt * 40$/lt = 160¿ and now he can make it until (B). Since until this moment he has only traveled
10 kms, he needs gas for another 34-10=24kms. Normally he would want to refuel his car with 24 liters
(since B is the cheapest gas station) but since his (max) fuel capacity is F=20 liters he will only take 20
liters and thus pay (ii) 20 lt * 7 $/lt = 140$. He knows however that up to point (B) he has only traveled
10kms and he needs to travel another 24kms to reach his goal, whereas he has gas for 20kms. So he would
have to stop at a later gas station (after he has traveled at least 4kms) to refuel another 4 liters of gas so
that he could complete the whole 34 kms journey. Since he now has quite some gas, he may decide whether
he wants to refuel at (C) or at (D) and since (D) is cheaper, it is more than 4kms away from (B) and is
within reach (based on his gas in the tank) he will choose to refuel another 4 liters at (D) and thus pay (iii)
4lt*12$/lt=48$). After that he can successfully reach the end point of his trip.
PLEASE HELP ME WITH THIS!
Thanks in advance!
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Quote: The most important thing is that the solution has to work as fast as it can with the lowest complexity
that possible.
I think the most important thing is that the solution "works"; worry about performance later.
Try a flow chart.
"(I) am amazed to see myself here rather than there ... now rather than then".
― Blaise Pascal
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Member 13870192 wrote: I have been struggling since previous month about some question that I found on some university,
..but mysteriously cannot link to..
It's a straightforward problem. Create a class called Vic, and have him go on each possible route, calculating the costs on the way. Select the cheapest, and done. It is a naive brute-force solution that I called "travelling Orca's". Since it resembles having an Orca in a Taxi take the actual route and phone back the results. Is as quick and efficient as the name suggests, but it works, and is a very good starting point to explain possible optimizations.
Member 13870192 wrote: The most important thing is that the solution has to work as fast as it can with the lowest complexity
that possible.
The TSP thingy has been done to death. If you skipped class, that's your problem. You are fishing for free labor.
Bastard Programmer from Hell 
If you can't read my code, try converting it here[^]
"If you just follow the bacon Eddy, wherever it leads you, then you won't have to think about politics." -- Some Bell.
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