jorgechiquinv My solution without using vector:
#include <iostream> int main() { int N,i=0; cin>>N; int *A = new int[N]; while(std::cin>>A[i++]); while(std::cout<<A[--N]<<' ' && N); delete[] A; return 0; }
glennlopez wow. thats pretty smart can you tell me where you learned it or walk me through your though proccess for
int *A = new int[N];
SudhanshuMohan thats for dynamically allocating memory ie it adjusts its size according to the input during runtime
akshaybusa67 what is it mean by *A=new int[N]? why new keyword used? and what will happen by this expression?
meowmix2 not sure why people just downvote an honest question instead of responding with an answer - everyone is here to learn.
"As a rule of thumb, C++ pointers should only refer to objects allocated wth new." you can read here for more explanation: Section A3.8. Pointers http://www.horstmann.com/ccj2/ccjapp3.html
Rohith13 Can you explain me this step: " while(cout<
dalownerx3 It means that while the statement in the parenthesis is true, keep executing the statements inside the while loop.
arpitarora866 what
AND N
do?
imachin " ...AND N); is used to stop when N arrives at value 0, he decreases the N value for each iteration, so basically is the stop condition for the while loop cause the AND operator with a 0 value returns 0, it was pretty ingenious. Also he deletes the pointer to avoid memory leaks, very important.
alaniane [deleted]
shihaabmeerun Heres another solution:
int main() { int n; cin >> n; const int a = n; int arr[a]; return 0; }MoWaNoob_101 In your solution You havent accepted input to you array arr[] And neither have you printed the output (reverse order) All you have done is declared the size of the array.
p_rzetelski_note (jorgechiquinv) Yep! Same here (excpet without the whiles, that looks clean, I like it). I threw a couple of runtime exceptions to handle the constraints on N and Ai. I also had N as a reference to a variable named arraySize just for fun :)
shaikaslam186 I have a doubt .why you use the CONST???????
abhikapoor2000_1 to not able the program change value of n further
ness_eg Why did you add '&& N' in 'while(cout<
jorgechiquinv That's because only with
coutthe loop will never end, so I add other constraints, whenNwill be 0 the loop will end (N is like an inverse iterator).Sorry my english
ness_eg thank you ^^
danielgcwik thanks, Jorge, that worked for me
929315787cenmin1 what 's "and" in while mean?
mo_fa Its the termination condition for the whilr loop. For every number greater than or equal to 1 "and" treats its as 1, so when it reaches 0 the "and" will give the output 0 which will terminate the while loop.
vishalshukla044 while(cout<
AbdulQuadir delete[] A; what's this statement doing?
mo_fa It is used to free the memory assigned to A because its no longer needed.
alaniane @slanger125 You also want to make sure use delete[] A and not just delete A delete[] A frees the entire array from memory delete A will just free the pointer and leave the array in memory creating a memory leak.
sourabhitm First of all thanks for sharing your solution. I really appreciate it. But i found a bug. There is no exit condition for the first while loop. It will keep writing in the unauthorised memory and crash. It should be like this.
int main() { int N,i=0; cin>>N; int *A = new int[N]; while(cin>>A[i++] && (i < N)); while(cout<<A[--N]<<' ' && N); delete[] A; return 0; }
jorgechiquinv The firts while loop finishes when
cinreads EOF, i.e. whencinfinishes to read the whole input stream.Another way is using
scanffrom C, it returns the number of input data assigned or the definedmacro EOF(-1) when there is nothing less to read, so you can do the next one:while(scanf("%i",&A[i++])>0); //or while(~scanf("%i",&A[i++]));
Those are the way of reading in c++ the whole input, there is challenges where the input size is not given
Sorry my english.
radico there should be one more condition in the first while loop
while(cin>>A[i++] and i<N);
otherwise, will be an infinit loop. thumps up, nice approach
jorgechiquinv See the comment above, it already has been explained
MrNik can u pls explain me both the while statements, i am not getting it.
nittinsant_seth1 Can you explain, how did the array get reversed?
das_the_don999 Hi, the snippet is not working in codeblocks. Is there any particular reason behind it?
P.s - I am new to this.
grinning_moon_c1 Truly elegant. Shouldn't we free A or set it to NULL as well?
jorgechiquinv You have to free A (
delete[] A), if only set it to NULL you'll lose the block of memory and cause a leak memory.Sorry my english.
anbuibz66 why we must cout N
scott_spillmann Beware coders, g++ (likely the compiler being used here) allows variable length arrays. However, if you compile using g++ -pedantic, it will not compile because ISO C++ forbids variable length arrays. Typically array length is specified at compile time. This is a g++ extension to the language vs standard c++.
There is a solution using std::vector (std::vector allows run-time dynamic array length)
simonalsn I did the variable array length solution, and tried compiling it with clang++, that worked fine as well.
Axesilo Thanks for clarifying. I had thought that specifying the size of an array at runtime was not allowed in C++, but then I was second-guessing myself because the code compiled and ran fine.
mattvenezia Thank you for explaining this. I was very confused as to why the compiler let me get away with declaring an array size via a variable, instead of forcing me to make a call to new.
mpatriak My C++ Solution
int numElements; cin >> numElements; int arr[numElements]; for (int i = 0; i < numElements; i++) cin >> arr[i]; for (int i = numElements - 1; i >= 0; i--) cout << arr[i] << " "; return 0;
alan_kan Hey I'm just a bit confused. Using the example input, 4 and 1 4 3 2, when you say int numElements, numElements = 4 right? And doing int arr[numElemnts] that creates an array that includes 0 through 3?
mpatriak Right, 4 elements in the array, so that can be used when initializing the array. It will have 4 indices, which are then populated by the second line of input.
devanshudg909 in input section when i run in c++ it isnt leaving space
nevets_oomra I get the concept but for some reason I don't understand it very well (this question I mean). Can you please put comments in your code to explain properly what's happening here? I would really appreciate it... thanks.
bence_meszaros [deleted]hmt905 if i start loop i=o then compiler display error.when i=1 then ok but again show compiler error because quetion demand i/o using stdin and stdout so how i face difficulty ,please give me solution
nathanlct Hey I am using visual studio and apparently int arr[numElements] wouldn't work (error of "expression must have a constant value")
naveen010gupta because array needs an numeric value whenever you write an array and want to scan some values through condition , assign array to be '0' and size can be managed under loop..
denis_dzinic In first for loop if you replace cin >> arr[i] with following: int x; cin >> x; arr[i] = x;
That will work otherwise it will fail if you leave your code as it is and try following values it will fail:
8 6676 3216 4063 8373 423 586 8850 6762
179tanmay why is it not taking directly input in the array?
naveen010gupta mine is also same but its showing some test case failed and compiler error occure as segmentation error.........
kaushake I did the same thing, but some of the testcases get abort called! Is this because some of the numbers are too big?
dc7024781447 i too figured out th same logic but i can not see the output on screen ???
upasanadeka thanks. it worked. but I still am not able to find why my program showed "Segmentation fault" when I wrote the same exact code.
fatimazafarse for (int i = numElements - 1; i >= 0; i--) cout << arr[i] << " "; can you plesae explain this loop's conditions
Ethan13310 We don't want to use C-style array in C++. We use STL containers instead, which have a lot of advantages such as being compatible with STL algorithms, or providing useful member functions such as 'std::vector::size()' (to get the size of the array :D). In this case, we can use 'std::vector', which is a dynamic array. Here's the solution in true C++:
#include <vector> #include <iostream> #include <iterator> int main() { std::size_t size{}; // Here we use 'std::size_t' instead of 'int' // because an array size cannot be negative std::cin >> size; std::vector<int> vect(size); for (std::size_t i{}; i < size; ++i) std::cin >> vect[i]; // One of the many advantages to use C++ STL containers: // we can use reverse iterators for (auto rit = vect.rbegin(); rit != vect.rend(); ++rit) std::cout << *rit << ' '; std::cout << std::endl; return 0; }
Darkling Dunno why those but in fact vector is much more heavier than plain array. So there is a good point in using it instead vector. Imho.
Ethan13310 If you don't want to use std::vector here, you can use std::array instead, not C-style arrays.
And no, std::vector is not 'much more heavier'. It depends on how you use it (i.e. if you do a lot of reallocations). In my code, there is only one allocation since I provide the size at the initialization.
jorgechiquinv Why did you use "auto"? What's it means?
I thought it was used to declare local variables, is it another use?
xxyxxyz std::array can not be used here because the size of an std::array must be provided at compile time.
zzmmss Does
size_t size{}==size_t size? what does the{}means here please?
thebartman I'm trying to get better with the STL. Here's what I came up with. I just ignored the first line of input.
#include <vector> #include <iostream> #include <algorithm> #include <iterator> int main() { std::cin.ignore(256, '\n'); std::vector<int> v; std::copy(std::istream_iterator<int> (std::cin), std::istream_iterator<int> (), back_inserter(v)); std::copy(v.rbegin(), v.rend(), std::ostream_iterator<int> (std::cout, " ")); return 0; }
ronnie_c995 Since you are practising getting better, it's better style in c++14 to use copy(rbegin(v),rend(v),ostream_iterator(std::cout," "));
thebartman Oops, yup, always forget to do that. Thank you!
SmittyMcSmith why is this better style? rbegin/rend are methods in the vector's class... why wouldn't you call them as he originally wrote?
ronnie_c995 Because begin(), end() functions (not methods) are standard and work for all containers and even arrays like char[]. They are specialized to work with any container you might write too as long as it follows certain guidelines.
vdeshmuk This is so concise. Love it!
SmittyMcSmith this will only terminate when the stream ends which doesn't happen with a carrage return... how would you modify this such that it itertively reads in till it hits a cr?
hmt905 why we use vector concept here.pelase explain
anamsiddiqui099 what is segmentation fault? how to correct it?
abhishreesharma1 include
using namespace std;
int main() { int n,rev=0,d; cin>>n; while (n>0) { d=n%10; rev=rev*10+d; n=n/10; } cout<
}
zumrudu_anka The solution which i hope u'll like :)#include <iostream> using namespace std; int main(){ int n,temp; cin>>n; int k[n]; for(int i=0;i<n;i++){ cin>>k[i]; } for(int i=0;i<n/2;i++){ temp=k[i]; k[i]=k[n-1-i]; k[n-1-i]=temp; } for(int i=0;i<n;i++){ cout<<k[i]<<" "; } return 0; }
LardBoi Here's my pretty simple solution. I don't think this is too complicated. Unless you've never used vectors before. You could clean this up. A lot.
#include <iostream> #include <vector> using namespace std; int main() { vector<int> numArray; int maxNums; int nums; cin >> maxNums; for(int i = 0; i < maxNums; i++) { cin >> nums; numArray.push_back(nums); } int loop = maxNums; for(int i = 0; i < maxNums; i++) { cout << numArray[loop-1]; cout << " "; loop--; } return 0; }
rohitroy37100 my solution for beginners:-
int main() { /* Enter your code here. Read input from STDIN. Print output to STDOUT */
int len; cin>>len; int *arr = new int[len]; for(int i=0;i<len;i++) cin>>arr[i]; for(int i=(len-1);i>=0;i--) cout<<arr[i]<<" "; return 0;}
alridam98 [deleted]
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