I am very new to C++ and I am trying to deconstruct someone's code, and I am not quite sure what to Google for, hence I am just going to ask here. This is a second attempt at a question I asked earlier which was poorly posed. Should this one not measure up, please let me know and I shall try to rectify.
Here is a structurally identical MWE, of the piece of code I am trying to understand.
#include <iostream>
using namespace std;
int square(int x){
// Function that squares without using *
int result = 0;
for (int counter = 0; counter < x; ++counter){
result += x;
}
return result;
}
int main()
{
int const D = 4;
int myArray[D] = {}; // all elements 0 in C++
char colour[D] = {'c','o','e','g'}; // Initialize String Array
int AEST = 5; // Initialise AEST
for (int d =0; d<D; d++){
if (colour[d]!='c' && colour[d]!='o'){
double aux= square (d);
if (aux!=0){
myArray[d]=aux;
}else{
return AEST;
}
}
}
// Lets see what we achieved.
for (int d =0; d<D; d++){
cout << myArray[d];
}
return 0;
}
Now then, lets crack on with some questions.
Precisely what I do not fully understand is this block:
}else{
return AEST;
}
Please not, AEST is not an error code, it is a numerical value that the code calculates. I have only initialized it here for the purpose of this MWE, but in actuality, it is calculated earlier on in the original code block.
My question is as follows:
The if statement is only true if the colours are not c or o and in which case we square d. In the MWE we square d twice. Hence, is the code then saying that we break out of the loop (with return AEST) IF we stumble upon a colour that is not c or o? But if we do break out of the loop under these conditions, why must we return AEST? It is already initialised AEST=5 earlier on, and nothing we do inside this loop will affect it (remember this block is structurally identical to what I am trying to understand, but obviously not fully identical). This is why I do not understand the else bit.
Again, if there is not enough information, please let me know.
The return AEST part in question exits the main() function. That means the program exits in state 5.
This is done to have some sort of error code detection. For example. If you have various things that can go wrong, you try to retun those with specific codes so you can look up and identify where the problem occured.
It is common to return 0 if everything is fine.
Related
Here is the chunk of code in question that I've pulled from my program:
#include <vector>
using namespace std;
vector<double> permittingConstructionCosts(56);
static const int PERMITTING_PERIODS = 0;
static const int CONSTRUCTION_PERIODS = 11;
static const double CONSTRUCTION_COSTS = 2169506;
static const double PERMITTING_COSTS = 142085;
static const int PERMITTING_CONSTRUCTION_PERIODS = PERMITTING_PERIODS + CONSTRUCTION_PERIODS;
void calcExpenses // Calculates permitting and construction expenses
(
vector<double>& expense,
double value1,
double value2
)
{
int i;
for (i=0; i<=PERMITTING_PERIODS + 1; i++)
{
expense[i] = value1;
}
for (i=PERMITTING_PERIODS + 2; i<expense.size(); i++)
{
if (i < PERMITTING_CONSTRUCTION_PERIODS + 2)
{
expense[i] = value2;
}
}
}
int main()
{
if (PERMITTING_PERIODS != 0)
{
calcExpenses(permittingConstructionCosts, -PERMITTING_COSTS/PERMITTING_PERIODS, -CONSTRUCTION_COSTS/CONSTRUCTION_PERIODS);
}
else
{
calcExpenses(permittingConstructionCosts, 0, -CONSTRUCTION_COSTS/CONSTRUCTION_PERIODS);
}
return 0;
}
According to ideone (http://ideone.com/LpzUny) the code has a runtime error that returns "time: 0 memory: 3456 signal:11".
I've tried to look for solutions on SO and found the following links:
How can I avoid a warning about division-by-zero in this template code?
How to eliminate "divide by 0" error in template code
However, I don't know how to use templates because I am new to c++ and I'm not sure I need to use them in this case so I have no clue how to adapt those solutions to my particular problem if it's even possible.
I'm pretty sure that the "-PERMITTING_COSTS/PERMITTING_PERIODS" is causing the problem but I thought that simply checking the divisor would solve the problem. This function seems to work for every other value other than 0 but I need to account for the case where PERMITTING_PERIODS = 0 somehow.
I would very much appreciate any help I can get. Thanks in advance!
Edit: I actually do initialize the vector in my program but I forgot to put that in because the size is decided elsewhere in the program. The chunk of code works once I fix that part by putting in a number but my program still has a runtime error when I set PERMITTING_PERIODS to 0 so I guess I have to go bug hunting elsewhere. Thanks for the help!
The problem lies inside the function, which is called by the else statement in the main function:
for (i=0; i<=PERMITTING_PERIODS + 1; i++)
{
expense[i] = value1;
}
Here, PERMITTING_PERIODS is 0, thus you loop from 0 to 2 (inclusive).
However, expense.size() is 0, since your vector is empty. As a result, you are trying to access an empty vector, which causes a segmentation fault.
With that said, print the value of i inside the loop, you should see that you try to access expense[0], but the vector is empty, so it has no first slot (basically it doesn't have any)!!
So replace that with:
expense.push_back(value1);
which will allocate enough space for your values to be pushed into the vector.
The answer given in the cited links, (i.e. "How to eliminate "divide by 0" error in template code") applies equally well here. The other answers were given in the context of templates, but this is completely irrelevant. The sample principle applies equally well with non-template code, too. The key principle is to compute a division, but if the denominator is zero, you want to compute the value of zero instead of the division.
So we want to compute -PERMITTING_COSTS/PERMITTING_PERIODS, but use the value of 0 instead of the division when PERMITTING_PERIODS is 0. Fine:
int main()
{
calcExpenses(permittingConstructionCosts,
(PERMITTING_PERIODS == 0 ? 0: -PERMITTING_COSTS)/
(PERMITTING_PERIODS == 0 ? 1: PERMITTING_PERIODS),
-CONSTRUCTION_COSTS/CONSTRUCTION_PERIODS);
return 0;
}
What I'm trying to do:
#include<iostream>
#include<string>
using namespace std;
int main(){
string st("HabibRehman");
string a="", b="";
int y=st.length();
int d=y/2;
int f;
f=d;
for(int i=1;i<st.length();i++){
for( int j=0;j<y;j++){
cout<<" ";
}
for(int k=0;k<1;k++)/*I'm not sure what this loop should be*/{
cout<<a+st.substr(d,i)<<endl;
/*int d=(st.length()/2);
static int f=d;//I was trying to keep the values static here
static int g=d;*/
//this doesn't work in here
}
--f;//if i put this outside the for loop it works wrong without the termination error.
a=st.substr(f,i);//I'm using this to get the value prior to the st.substr(d,i)
y--;
}
system("pause");
return 0;
}
I keep getting a termination error. Ive tried almost everything I knew like changing the loops, the positions of substrings and variables, when I get the output I want, the program gives me an termination error, otherwise it gives me a wrong output without the termination error.
I do not fully understand this code, it really writed very bad.
But after 1,5 minutes of debugging I found that exception is thrown because f in a=st.substr(f,i); is lower than 0.
In for(int i=1;i<st.length();i++){ you decrement the variable f st.length() times, but f is half of st.length() and will get after st.length()/2 iterations -1 value.
I added after --f;
this statement:
if(f < 0 ) return 0;
and it seems that programm working well.
Please, learn how to write beatiful and readable code.
I am trying to find the 1D peak through Divide and Conquer technique in this particular question,
my program even though it runs,
but at the time of giving the final output it says that there has been some problem with the execution,
I have got the answer from a different method, but I would like to know where am I at fault here.
#include<iostream>
using namespace std;
int a[8];
class pg1
{
public:
int func(int n)
{
if(a[n] <= a[n+1])
{
func(n++);
}
else if(a[n] <=a [n-1])
{
func(n--);
}
else
{
return n;
}
}
};
int main()
{
pg1 ob;
for(int i=0;i<8;i++)
{
cin >> a[i];
}
int x = ob.func(4);
cout << endl << x;
return 0;
}
Input-
5
6
8
5
4
3
6
4
Errors are-
1D Peak.exe has stopped working.
A problem caused the program to stop working correctly.Windows will close the program and notify you aif a solution is available.
End Result-
Process Exited with return value 3221225725
Don't use postincrement and similar in function calls.
Here's the problem condensed down to a really simple piece of code
#include <iostream>
int test(int n){
if(n == 1){
std::cout << "Function called!";
return test(n++);
}else{
return 0;
}
}
int main() {
test(1);
return 0;
}
Before you run this, ask yourself what you expect to happen here. Did it do what you thought?
When you run this you'll see that the code doesn't terminate properly. The output shows the function gets called infinitely many times, eventually the stack runs out of space and the program crashes.
You can see this code in action here: http://ideone.com/QL0jCP
In your program you have the same problem:
int func(int n)// say n = 4
{
if(a[n] <= a[n+1])//say this is true
{
func(n++); //this calls func(4) THEN increments n afterwards
}
This calls func with the same value over and over.
The solution is to not use postincrement or postdecrement in your function calls. These create hard to diagnose bugs as you have seen in this question. Just a simple func(n+1) is all you need. If you needed to use the variable later then just create an explicit variable to do that, it's much cleaner coding style (as this problem you ran into here shows).
After you fix this you'll need to fix your array bounds checking.
if(a[n] <= a[n+1])
If n is the last spot in the array you suddenly are trying to access one place past the end of the array, if you are lucky you get a segfault and a crash, if you are unlucky you get some bug that messes up your system that is hard to find. You want to check the values are valid.
I have the following code compiled in linux terminal (c++ in linux) and am using OpenCv 2.4.3.
However, am getting a segmentation fault in run time and I really have no clue as to why. I have placed differnt cout statements to know if the program processed to the particular stage but in vain. Could you please help me? Please explain me what exactly is this segmentation fault. Am stuck here for a long time.
#include "opencv2/imgproc/imgproc.hpp"
#include "opencv2/highgui/highgui.hpp"
#include <iostream>
#include <stdlib.h>
using namespace cv;
using namespace std;
int main()
{
cout<<"check"<<flush;
Mat src,src_gray,dst;
int kernel_size = 3;
int scale = 1;
int delta = 0;
int ddepth = CV_16S;
char* window_name = "sharpness estimate";
int freq,rows,cols =0;
double *estimate,*min = 0;
Point *minLoc,*maxLoc = 0;
src = imread("/home/siddarth/examplescv/erez images/image53.jpg");
if( !src.data )
{
return -1;
}
namedWindow(window_name,CV_WINDOW_AUTOSIZE);
Mat abs_dst;
cvtColor(src,src_gray,CV_RGB2GRAY);
Laplacian(src_gray,dst,ddepth,kernel_size,scale,delta,BORDER_DEFAULT);
convertScaleAbs(dst, abs_dst);
minMaxLoc(dst,min,estimate,minLoc,maxLoc,noArray());
Size s = dst.size();
rows = s.height;
cols = s.width;
cout<<rows<<endl<<cols<<endl;
for(int i=0;i<=rows;i++)
{
for(int j=0;j<=cols;j++)
{
if(dst.at<double>(i,j) >= *estimate-100
&& dst.at<double>(i,j) <= *estimate+100)
{
cout<<freq++;
}
}
}
cout<<"estimate :"<<*estimate<<endl;
cout<<"frequency :"<<freq<<endl;
imshow(window_name,abs_dst);
waitKey(1000);
return 0;
}
The code doesn't cross the first "check" print statement just after the main function declaration. That is the confusing issue. But once I flushed the first print statement, it got executed. I am still facing issues.
Make sure you insert std::endl into cout so that the buffer is flushed. This will probably be why you're not seeing any output.
One immediate issue is that your for loops check the condition with <=, meaning that you're probably going one past the end. But since you're using at, you should have an exception thrown (assuming this Mat type acts like a standard container).
Also, you're creating lots of pointers to pass as some function arguments (for example, double* estimate). This doesn't actually give you a double object though, just a pointer. Unless the function you're passing them to is allocating a double for you (which I hope it's not), you're doing it wrong. You should be doing:
double estimate;
minMaxLoc(/* ... */, &estimate, /* ... */);
You'll need to do that with all of the values you're getting through output parameters.
Another thing to note: Doing int i, j = 0; only initialises j to 0, but not i. You need to do int i = 0, j = 0;.
Okay, I'm going to explain why fixing the initialisers works. I had to look up the definition of minMaxLoc to see what happens. Basically, the function is something like the following:
void setToFive(int* x)
{
if (x) {
*x = 5;
}
}
This function will take a pointer to an int, and then set that int to the value 5. However, if the pointer passed is a null pointer, the value will not be set (otherwise there'll be undefined behaviour because you're derefencing a null pointer). Basically, passing a null pointer says "I don't care about this value so don't give it to me".
Now when you were initialising your pointers, you were doing:
double *estimate, *min = 0;
This only sets min to the null pointer. Since estimate is left uninitialized, you can't rely on its value being null. You need to provide an initialiser for each declarator:
double *estimate = 0, *min = 0;
Thanks to #sftrabbit. The problem was the initialization. instead of
int freq,rows,cols=0;
The change was
int freq=0,rows=0,cols=0;
this removed the segmentation fault. Thanks a lot for your help :).
Since you are in a Linux environment, you can use valgrind to find out exactly where the segmentation fault is happening. Just type valgrind before the name of the program, or the way you execute your program. For example, if you execute your program with the following command:
hello -print
issue the following command instead:
valgrind hello -print
I see you already solved this one, but this may be helpful in the future!
as the title explains this is a program to find lcm of numbers between 1 to 20. i found an algorithm to do this, here's the link
http://www.cut-the-knot.org/Curriculum/Arithmetic/LCM.shtml
there is a java applet on the webpage that might explain the algorithm better
Problem: i wrote the code compiler shows no error but when i run the code the program goes berserk, i guess may be some infinite loopig but i can't figure it out for the life of me. i use turbo c++ 4.5 so basically if anyone can look at the code and help me out it would be great . thanks in advance
Algorithm:
say we need to find lcm of 2,6,8
first we find the least of the series and add to it the number above it, i.e the series become
4,6,8
now we find the least value again and add to it the intitial value in the column i.e 2
6,6,8
so the next iteration becomes
8,6,8
8,12,8
10,12,8
10,12,16
12,12,16
14,12,16
14,18,16
16,18,16
18,18,16
18,18,24
20,18,24
20,24,24
22,24,24
24,24,24
as you can see at one point all numbers become equal which is our lcm
#include<iostream.h>
/*function to check if all the elements of an array are equal*/
int equl(int a[20], int n)
{
int i=0;
while(n==1&&i<20)
{
if (a[i]==a[i+1])
n=1;
else
n=0;
i++;
}
return n;
}
/*function to calculate lcm and return that value to main function*/
int lcm()
{
int i,k,j,check=1,a[20],b[20];
/*loading both arrays with numbers from 1 to 20*/
for(i=0;i<20;i++)
{
a[i]=i+1;
b[i]=i+1;
}
check= equl(a,1);
/*actual implementation of the algorith*/
while(check==0)
{
k=a[0]; /*looks for the least value in the array*/
for(i=0;i<20;i++)
{
if(a[i+1]<k)
{
k=a[i+1]; /*find the least value*/
j=i+1; /*mark the position in array */
}
else
continue;
}
a[j]=k+b[j]; /*adding the least value with its corresponding number*/
check= equl(a,1);
}
return (a[0]);
/*at this point all numbers in the array must be same thus any value gives us the lcm*/
}
void main()
{
int l;
l=lcm();
cout<<l;
}
In this line:
a[j]=k+b[j];
You use j but it is unitialized so it's some huge value and you are outside of the array bounds and thus you get a segmentation fault.
You also have some weird things going on in your code. void main() and you use cout without either saying std::cout or using namespace std; or something similar. An odd practice.
Also don't you think you should pass the arrays as arguments if you're going to make lcm() a function? That is int lcm(int a[], int b[]);.
You might look into using a debugger also and improving your coding practices. I found this error within 30 seconds of pasting your code into the compiler with the help of the debugger.
Your loop condition is:
while(n==1&&i<20)
So your equl function will never return 1 because if n happens to be 1 then the loop will just keep going and never return a 1.
However, your program still does not appear to return the correct result. You can split the piece of your code that finds the minimum element and replace it with this for cleanliness:
int least(int a[], int size){
int minPos = 0;
for(int i=0; i<size ;i++){
if (a[i] < a[minPos] ){
minPos = i;
}
}
return minPos;
}
Then you can call it by saying j = least(a, 20);. I will leave further work on your program to you. Consider calling your variables something meaningful instead of i,j,k,a,b.
Your equl function is using array indices from 0-20, but the arrays only have 1-19
j in lcm() is uninitialized if the first element is the smallest. It should be set to 0 at the top of the while loop
In the following code, when i=19, you are accessing a[20], which is out of the bounds of the array. Should be for(i=0;i<19;i++)
for(i=0;i<20;i++) {
if(a[i+1]<k)
You are not actually using the std namespace for the cout. this should be std::cout<<l
Your are including iostream.h. The standard is iostream without the .h, this may not work on such an old compiler tho
instead of hard-coding 20 everywhere, you should use a #define. This is not an error, just a style thing.
The following code does nothing. This is the default behavior
else
continue;