JChiquin My solution without using vector:
#include <iostream> int main() { int N,i=0; std::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
igor_c_palha how different is it from the "malloc"?
2210181043_Ekky malloc() is a function in C, and New is for C++
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.
pardumangarg7 array size is also wrong
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<
JChiquin 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 ^^
narutoitachi329 int main() { int n, i = 0; cin >> n; int a[n]; for (; i < n; i++) cin >> a[i]; for (; i >0; i--) cout << a[i- 1] <<" "; return 0; }
JChiquin You shouldn't use arrays like that. In fact,
ais initialized with the first value ofnwhat is garbage, something like39281233, because is initialized in compile time not execution time, for dinamic size you must use pointers and the keywordneworvectorfromstd.narutoitachi329 [deleted]JChiquin Why does my comment have negative votes?
priyankavithani1 a[i-1]
can you tell me how it is working?
anshup how this line is working i am not able to understand for (; i >0; i--) cout << a[i- 1] <<" ";
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; }
JChiquin 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.
balavemulakonda yes!!helped a lot
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?
Umar_Mukhtar actually Array didn't get reversed we are printing in reverse order (i.e from last index to first).
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?
JChiquin 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
Umar_Mukhtar For those of you who are wondering how the 1st while loop will terminate , it will terminate only if you type 'false' or 'eof' . It is a fancy way of solving this question but it is not optimised way because it takes a hell lot of machine cycles and instructions . take a look
while (cin >> A[i++] && (i < N)); 00B324BD mov eax,dword ptr [i]
00B324C0 mov ecx,dword ptr [A]
00B324C3 lea edx,[ecx+eax*4]
00B324C6 mov dword ptr [ebp-11Ch],edx
00B324CC mov eax,dword ptr [i]
00B324CF add eax,1
00B324D2 mov dword ptr [i],eax
00B324D5 mov esi,esp
00B324D7 mov ecx,dword ptr [ebp-11Ch]
00B324DD push ecx
00B324DE mov ecx,dword ptr [imp?cin@std@@3V?char_traits@D@std@@@1@A (0B3D098h)]
00B324E4 call dword ptr [__imp_std::basic_istream >::operator>> (0B3D09Ch)]
00B324EA cmp esi,esp
00B324EC call __RTC_CheckEsp (0B312DFh)
00B324F1 mov dword ptr [ebp-0FCh],eax
00B324F7 mov edx,dword ptr [ebp-0FCh]
00B324FD mov eax,dword ptr [edx]
00B324FF mov ecx,dword ptr [ebp-0FCh]
00B32505 add ecx,dword ptr [eax+4]
00B32508 mov esi,esp
00B3250A call dword ptr [__imp_std::ios_base::operator bool (0B3D0D0h)]
00B32510 cmp esi,esp
00B32512 call __RTC_CheckEsp (0B312DFh)
00B32517 movzx ecx,al
00B3251A test ecx,ecx
00B3251C je main+0E8h (0B32528h)
00B3251E mov eax,dword ptr [i]
00B32521 cmp eax,dword ptr [N]
00B32524 jge main+0E8h (0B32528h)
00B32526 jmp main+7Dh (0B324BDh)
while (cout << A[--N] << ' ' && N); 00B32528 mov eax,dword ptr [N]
00B3252B sub eax,1
00B3252E mov dword ptr [N],eax
00B32531 push 20h
00B32533 mov esi,esp
00B32535 mov ecx,dword ptr [N]
00B32538 mov edx,dword ptr [A]
00B3253B mov eax,dword ptr [edx+ecx*4]
00B3253E push eax
00B3253F mov ecx,dword ptr [imp?cout@std@@3V?char_traits@D@std@@@1@A (0B3D0D4h)]
00B32545 call dword ptr [__imp_std::basic_ostream >::operator<< (0B3D0A4h)]
00B3254B cmp esi,esp
00B3254D call __RTC_CheckEsp (0B312DFh)
00B32552 push eax
00B32553 call std::operator<< > (0B314A6h)
00B32558 add esp,8
while (cout << A[--N] << ' ' && N); 00B3255B mov dword ptr [ebp-108h],eax
00B32561 mov ecx,dword ptr [ebp-108h]
00B32567 mov edx,dword ptr [ecx]
00B32569 mov ecx,dword ptr [ebp-108h]
00B3256F add ecx,dword ptr [edx+4]
00B32572 mov esi,esp
00B32574 call dword ptr [__imp_std::ios_base::operator bool (0B3D0D0h)]
00B3257A cmp esi,esp
00B3257C call __RTC_CheckEsp (0B312DFh)
00B32581 movzx eax,al
00B32584 test eax,eax
00B32586 je main+150h (0B32590h)
00B32588 cmp dword ptr [N],0
00B3258C je main+150h (0B32590h)
00B3258E jmp main+0E8h (0B32528h)dafnis this is nice but allocating memory havent been tocuched in previous lessons...
param3108 thanks bro, it was helpful.
damian_beles [deleted]damian_beles [deleted]EdinHajdarevic your code requires to enter char/string so it can print the value
ashutosh_kaushi1 [deleted]
guilhermenicchio [deleted]guilhermenicchio [deleted]
sujeetpoudel_np Despite the code posted by JChiquin passed all of the test cases, the Visual Studio console did not give the same output until I modified the first while loop as:
while (std::cin >> A[i++] && i<N);Did anyone stop by the same issue?
sumudu_lansakara You may reuse
Nand ditchiJChiquin Hi, I had not think it, but can be done like this:
#include <iostream> int main() { int N; std::cin>>N; int *A = new int[N]; while(std::cin>>A[--N]); while(A[++N] && std::cout<<A[N]<<' '); delete[] A; return 0; }
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
iam_nevets 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
dharani18199 but when i do like this ..im getting runtime errors :(
svarnypetr Nice, learned that I can define the initial i in the for loop.
gomezrobledoosv1 Once my teachers said me that declaring a pointer like int numElements; cin >> numElements; int arr[numElements]; would lead me to static memory issues. They suggest me to implement dinamic memory. Is there a special fact to ignore my teachers in this case? Help me, because i don't see anyone complained with this. Sorry for my english.
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.
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.
zzmmss Does
size_t size{}==size_t size? what does the{}means here please?
abinayavasan int main() {
cin>>n; int a[n]; for(i=0;i<n;i++) { cin>>a[i]; } for(i=n-1;i>=0;i--) { cout<<a[i]<<" "; }}
abinayajuin easy solution superb
vaishnavijayam [deleted]
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
Madophs Cool C++ Solution
#include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> using namespace std; int main() { /* Enter your code here. Read input from STDIN. Print output to STDOUT */ int n; cin>>n; vector<int> vec(n,0); while(n--)cin>>vec[n]; for(int x=0; x<vec.size(); x++)cout<<vec[x]<<" "; return 0; }
tritchie329 #include <iostream> using namespace std; int main() { int size; int min; cin >> size; int *myArray = new int[size]; for (int j = 0; j < size; j++) { cin >> myArray[j]; } for (int i = size - 1; i >= 0; i--) { cout << myArray[i] << " "; } return 0; }
Aminson_Jets Are the int size and int min always available? Can you explain me what is a int j and int i
ramey_steven Great solution by 'jorgechiquinv', adding some comments for posterity if you're rusty with C++ like me :-)
#include <iostream> using namespace std; // Every C++ program starts with the main method int main() { // Declare a variable, N for the number of // values entered in from stdin int N; // Declare a variable i for incrementing int i=0; // Read in the initial value from cin into N, // for the number of elements in the array cin>>N; // Declare a pointer of type int, called A, // and allocated new memory for it of size N int *A = new int[N]; // Setup a while loop for as long as cin // is reading in variables, all into the array, A while(std::cin>>A[i++]); // Setup another while loop to output (cout) every // value in A in reverse, using --N to decrement the // the value of N so long as N is true (not equal to zero) while(std::cout<<A[--N]<<' ' && N); // Un-allocate the memory given to A delete[] A; // Return zero to indicate success return 0; }
tudorian95 This is the solution using std vector:
int main() { int n; cin>>n; int nr; vector v; for (int i =0; i>nr; v.push_back(nr); }
for (int i = v.size() -1 ; i>=0; i--){ cout<
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