I want to be able to generate a boolean condition during the runtime based on the states of some variables. My task looks simple at first. I have a large if () else if () statement what needs to determine if the number is in a certain range. It then does something depending on whether that number is inside that range.
Here's pseudocode to demonstrate what I want:
void fun(int num, int offset = 0) {
if (0...60) {
// do something
} else if (60...180) {
// do something else
} else if (180...240) {
} else if (240...360) {
}
}
The first if statement should work like this:
if (0 >= num && num <= 20) {
// do something
}
The caveat here is that in addition to int num, there is another parameter passed in, which I call the offset. The structure of the code here, including the do something inside the { } is the same. The only things that need to change are are ranges, based on the value of the offset. By the way, this is not a default parameter here, it is just pseudocode demonstrating what the value of int offset was passed in.
void fun(int num, int offset = 120) {
if (120...180) {
// do something
} else if (180...300) {
// do something else
} else if (300...360) {
} else if (360...120) {
}
}
That last else if () statement has been giving me some trouble.
} else if (360...120) {
}
What I'm actually trying to write here is:
} else if (num >= 360 || num <= 120) {
}
The reason for this is that my int num may have a value > 360. However, in that case for the purpose of my application it has to "wrap around" and be treated as a value 0...120.
This is for a mathematical application here. Whenever you have int num > 360, you go around the full circle and you end back at 0 where you started. So that is the effect which I want to achieve.
I don't want to write extra functions. I want my code to be generic because many different values for int num and int offset may be passed into my function. I want to generate the necessary conditions during the runtime based on the value of int offset.
The main problem here is that in the first situations, when int offset = 0 my condition is
} else if (240 >= num && num <= 360) {
}
However, for a different offset we wrap around and so I have to change the format of the entire condition! For example, when int offset = 120, as shown above:
} else if (num >= 360 || num <= 120) {
}
The problem is that in the first situation I had the && in the last else if (), but now I have the || to convey the same meaning. What I'm looking for is a way to be able to manipulate the operators inside the conditional statements as mere chars in a string, then "paste" the completed condition into the if () statements during the runtime!
What's even worse is that this "wrapping around" can occur inside any one of the if () statements, not just the last one. It is based on the value of the offset.
I can't use preprocessor tricks, because I want this to work during the runtime. Maybe it is possible to use function pointers or something for this, but I don't know how to do that. Please note that the ... above is not real C++ code, it is pseudocode! I'm aware that there is a "range-based" switch statement in C++, but I can't use that because of the "wrapping around" property mentioned above.
Related
If I put a break point on this function it will step through until the base case is met but when it hits the return 1 it doesn't actually exit the loop. Instead it goes to the bottom bracket and then bounces to int t = expo(m, n / 2) and steps downs to return t * t*m. It then goes back to the bottom bracket and repeats this process before eventually stopping. Can someone explain what is going on?
int expo(const int m, const unsigned int n) {
funcCallCounter++; //counts how many times the function is called
if (n == 0) {
return 1;
}
else {
if (n % 2 == 0) {
int t = expo(m, n / 2);
return t * t;
}
else {
int t = expo(m, n / 2);
return t * t*m;
}
}
}
When the flow of your program in a function encounters a return statement, control is returned to place that called that function. That is true even if it was the same function. This is completely normal and expected.
In your case, each call still has work to do after the recursive step (because your expo() call is never quite the last thing before return) and you're seeing the program get on with that work.
Keep an eye on your stack frame while debugging; you'll see that it's not really jumping around in the way that you think it is; you're returning to previous call contexts.
I don't quite understand the meaning of else if statements.
why not just to continue with the if statements in case one is false?
it works the same.
example with if only that will work the same with else if:
function testSize(num) {
if (num < 5){
return "Tiny";
}
if (num < 10){
return "small";
}
return "Change Me";
}
testSize(7);
In your actual code you specify a return statement in the code associated to the if statement.
Suppose you don't specify a return statement or suppose today you specify a return statement but tomorrow you remove it to do a common return at the end of the method.
This code will test each condition even if the first one is true :
if (num < 5){
// do something
}
if (num < 10){
// do something
}
This code will not test the next condition if the first one is true :
if (num < 5){
// do something
}
else if (num < 10){
// do something
}
These two ways of doing have two distinct meanings.
When you have a series of if statements, you expect that more than one condition may be true.
When you have a series of if-else-if statements, you expect to have not more than one condition true.
Using the first form (a series of if) while functionally you expect to have not more than one condition true is misleading.
Besides, if the code is modified and you add a condition that is both true for two if statements while you don't want have this case, it would create an issue.
Your code is only showing your belief. What would happen in the example below?
function testSize(num) {
if (num < 5){
x = 1;
}
if (num < 10){
x = 2;
}
result = complex calculations;
}
function testSize2(num) {
if (num < 5){
x = 1;
} else if (num < 10){
x = 2;
}
return x * 2;
}
testSize(4); // result is 4
testSize2(4); // result is 2
x may also be involved in more calculations
if(condition) {...}
if(condition) {...}
if(condition) {...}
In above code, even if the first or second condition is true, third condition have to be checked.
if(condition) {}
else if(condition){}
else if(condition){}
Here if first condition is true, next two will not be checked. So, it saves time and is more readable logically.
A one way if statement takes an action if the specified condition is true.If the condition is false, nothing is done. But what if you want to take alternative actions when the conditions is false ? You can use a two-way if-else statement. The action that a two-way if-else statements specifies differ based on whether the condition is true or false.
Well, there is a bit different from this two statement.Consider the follow samples
if(a > 0) {
...
}
if( a == 0) {
...
}
if(a < 0) {
...
}
and
if(a > 0) {
...
}
else if( a == 0) {
...
}
else if(a < 0) {
...
}
when a is zero the last else if statement will not be execute while if need to compare third time.If a equals to 10, else if could be execute once while if is third.From this else if statement could be execute less and let your program a bit fast.
else if should be used when there are multiple conditions and you want only one of them to be executed, for instance:
if(num<3){ console.log('less than 3') }
else if(num<2){ console.log('less than 2')
If you use multiple if statements instead of using else if, it will be true for both the conditions if num = 1, and therefore it will print both the statements.
Multiple if statements are used when you want to run the code inside all those if statements whose conditions are true.
In your case it doesn't make a difference because the function will stop and return at the very first return statement it encounters. But let's say, the blocks' orders are interchanged, then your function will never return 'tiny' even if num = (anything less than 5).
I hope that helps!
If all your if branches terminate the function (e.g., but returning a value of throwing an exception), you're right, and you really don't need an else statement (although some coding standards might recommend it).
However, if the if branches don't terminate the function, you'd have to use an else or else if clause to prevent multiple blocks from being executed. Assume, e.g., you want to log a message to the console instead of returning it:
if (num < 5) {
console.log("Tiny");
} else if (num < 10) {
console.log("small");
} else {
console.log("Change Me");
}
So, i have a class called Vuelo, it has a method in which i can add a passenger to an airplane flight, i must check that the passenger id is not already in the array (the array is at first with all zeros), i must also check that there is enough space for another passenger to be added (max 10)
bool Vuelo :: agregarPasajero(int id)
{
int i = 0;
for(int iC = 0; iC < 10; iC++)
{
if (listaPasajeros[iC] == id)
{
i++;
return false;
}
}
if(i == 0)
{
if(cantidadPasajeros >= 10)
{
return false;
}
else
{
return true;
cantidadPasajeros++;
}
}
}
If i is not zero, you get to the end of the function without any kind of return statement. Since you declared the function to always return a bool, you should provide one for that case.
Now, you may know that i will never be zero at that spot, but the logic for that is fairly complex (I missed it on the first reading), and a compiler cannot be expected to realize that there is in fact no chance of control flow ever getting to the end of the function without encountering a return. In this case it's best to add a dummy return.
You can probably get away with not having a dummy return if you remove the bogus i == 0 test. i will necessarily always be zero at that point, since if it were ever increased, the function immediately returns false.
The statement cantidadPasajeros++; will never be executed since it is located after a return statement. Any halfway decent compiler also warns on that.
Here's a c++ program i tried to write for the above question.Our teacher told us to use a for loop.
void main()
int A[30],B[30],m,n,i,j,x,z;
cout<< "enter two numbers";
cin>>m>>n;
for(i=1,j=0;i<=m,j<30;i++,j++)
{
if(m%i==0)
{ A[j]=i;
z=j;
}
}
for(i=1,j=0;i<=n,j<30;i++,j++)
{
if(n%i==0)
{ B[j]=i;
x=j;
}
}
for(i=z;i>=0;--i)
{
for(j=x;j>=0;--j)
{
if(A[i]==B[j])
{ cout<<"gcd="<<A[i];
}
}
}
}
The output displays " Enter two numbers:" and when i entered 15 and 3, the result i got was a blinking cursor. Working through the program, I realised that the divisors for each number when stored in the arrays of A and B were not stored continuously or had gaps in between. If there isn't anything in the memory for say A[11], what happens when you check it against another variable with a number? Can somebody please modify this to make it work and tell me what's wrong? I am new to programming, so excuse my program if it is clumsy.
Andreas has pointed out that there are other ways to achieve the goal of finding the gcd, but the point of the exercise is to get a better handle on some basic programming constructs. So lets go with your approach.
Your idea is to compute the two lists of divisors and then compare them. As you say, having a list with gaps in makes this harder.
So adapt your loop, only increment the storage index when you've stored something
for(i=1,j=0;i<=m && j<30;i++) // need the && here; a comma means something different
{
if(m%i==0)
{ A[j++]=i;
z=j;
}
}
Second, you have a typo you're not storing in B, so fix that
for(i=1,j=0;i<=n && j<30;i++)
{
if(n%i==0)
{ B[j++]=i; //B here not A
x=j;
}
}
That should help.
Try this:
int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
As taken from here: https://codereview.stackexchange.com/questions/66711/greatest-common-divisor
My code is:
#include<stdio.h>
void main(void)
{
float timeLeavingTP;
int transitNumber;
float transitTime;
printf("Please enter the time leaving TP.\n");
scanf_s("%f",&timeLeavingTP);
printf("Please enter bus number.\n");
scanf_s("%d",&transitNumber);
if(timeLeavingTP==1.00)
{
if(transitNumber==27)
{
transitTime=1.56;
}
else if(transitNumber==8);
{
transitTime=1.39;
}
}
if(timeLeavingTP==6.30)
{
if(transitNumber==27)
{
transitTime=7.32;
}
else if(transitNumber==8)
{
transitTime=7.29;
}
printf("The time reached home is %f\n",transitTime);
}
}
After debugging i got
Please enter the time leaving TP
1.00
Please enter bus number
27
Please enter to continue...
My question is How do i adjust the program to make it look like the one below instead. What kind of error did i commit?
Please enter the time leaving TP
1.00
Please enter bus number
27
The time reached home is 1.56
Thanks for the help in advance!
Hi guys after including == i still got the same for my debugging? Is there something else that i did wrong?
Part 1: = vs ==
Note that:
if(timeLeavingTP=1.00)
Does not do what you expect. It assigns timeLeavingTP with 1.00.
You probably want:
if(timeLeavingTP==1.00)
Additionally, note that this error occurs 6 times in your program.
Part 2: comparing floating point numbers
Your code might work in this case, but I'm not 100% sure if it will or not. It's often difficult to directly compare 2 floating point numbers, because of the inaccuracy of storing them (for example, 0.1 is usually not representable in floating point).
Most people solve this problem in one of a few ways:
Test a range around the number.
Convert to some fixes width format. Perhaps you could store the number as an integer, knowing that it's representation is actually 0.01 * the stored number.
In this case, you could actually just store the information as strings, and compare those.
Part 3: conditionals
To write a proper conditional, it should look like:
if (condition) {
...
} else if (condition) {
...
} else if (condition) {
...
} else {
...
}
You can certainly nest conditionals as well:
if (condition) {
if (condition) {
...
} else {
...
}
} else if (condition) {
...
}
Your code, for example, messes this up when you do:
}
else(transitNumber=8);
{
transitTime=1.39;
}
Note that the else statement does not accept a conditional after it.
Part 4: excessive semicolons
Additionally, note that after the else and if statements there are no semicolons. The semicolons only appear within the braces. So this statement:
if(timeLeavingTP=6.30);
While semantically valid, does not do what you expect. You actually want to remove that semicolon.
if(timeLeavingTP == 1.00)
{
if(transitNumber == 27)
{
transitTime=1.56;
}
else if(transitNumber == 8)
{
transitTime=1.39;
}
}
else if(timeLeavingTP == 6.30)
{
if(transitNumber == 27)
{
transitTime == 7.32;
}
if(transitNumber ==8)
{
transitTime=7.29;
}
}
printf("The time reached home is %f\n",transitTime);
}
if(transitNumber=27)
{
transitTime=1.56;
}
else(transitNumber=8);
{
transitTime=1.39; //this line is executed all the time
}
This code is completly invalid!
First, you do not compare anything... transitNumber = 27 is an assignment.
Second else(transitNumber=8); again this is an assignment and it should be else if(...). Also ; at the and means that transitTime = 1.39(inside bracket) will always happen, even if transitNumber != 8
Change
if(timeLeavingTP=1.00)
to
if(timeLeavingTP==1.00)
so that you can compare timeLeavingTP correctly.