Is there a better and shorter way of my for loop?

Question

I'm a newbie here and this my first question. I learn basic c++ on Visual Studio 2019. I study functions and loops.

I solved it with my code of 30 lines. But I just wonder is a there a shorter and better way of this?

using namespace std; int main() { for (int j = 0; j < 5; j++) { if (j == 1) break; cout << "*"; cout << endl; if (j == 2) break; cout << "**"; cout << endl; if (j == 3) break; cout << "***"; cout << endl; if (j == 4) break; cout << "****"; cout << endl; if (j == 5) break; cout << "*****"; cout << endl; } return 0; }

This is not a loop at all but some cout statements executed sequentially. In the second iteration j == 1 and the loop will be exited — Sebastian Hoffmann
– Sebastian Hoffmann, Commented Mar 1, 2021 at 10:29
@Caleth It doesn't even hit if (j == 2), because it will always break in the second iteration. — Lukas-T
– Lukas-T, Commented Mar 1, 2021 at 10:31
You have asked a very basic question, yet with a powerful concept. — Abhishek Dutt
– Abhishek Dutt, Commented Mar 1, 2021 at 10:41

IWonderWhatThisAPIDoes · Accepted Answer · 2021-03-01 10:56:51Z

You have made several mistakes in your code.

break statements should actually be continue statements. break terminates the entire loop altogether (ie. absolutely nothing will be printed in the code you provide).
i == 5 can never be true, as the loop terminates before thet happens (see the condition i < 5).

To answer your question, you can use a nested loop.

for (int i = 0; i < 5; i++) { for (int j = 0; j <= i; j++) { std::cout << '*'; } std::cout << std::endl; }

Or, if you don't need to be flexible, do it manually.

std::cout << "*\n**\n***\n****\n*****" << std::endl;

Oh, endl does always insert newline. My bad for misreading the documentation.
@IWonderWhatThisAPIDoes Nested loop has been my solve. However my code run on the visual studio 2019 but I understood my code is unnecessary work. Thanks a lot for your explanation.

Konrad Rudolph · Accepted Answer · 2021-03-01 10:52:52Z

Why a loop? Here’s the same logic in seven lines of code:

int main() { std::cout << "*\n"; std::cout << "**\n"; std::cout << "***\n"; std::cout << "****\n"; std::cout << "*****\n"; }

To be honest, this is probably as short as it gets (apart from cramming all the output into a single statement).

But the issue with your loop isn’t primarily its length, it’s that your loop isn’t variable: it fails to work as soon as you change the number of iterations, and that defeats the purpose. All the logic is hard-coded, which makes the loop useless.

But if you want to use loops, then make the logic of printing stars configurable:

void print_stars(int n) { for (int i = 0; i < n; ++i) { std::cout << "*"; } std::cout << "\n"; } int main() { for (int i = 0; i < 5; ++i) { print_stars(i + 1); } }

@Caleth That’d be the next level. I didn’t want to introduce any new concepts into the solution.
@Konrad Rudolph I like your void function and for loop. I'll study on this. Thanks a lot for your explanation.

Abhishek Dutt · Accepted Answer · 2021-03-01 10:39:32Z

#include<iostream> using namespace std; int main(){ for(int i = 1;i <= 5;i++){ for(int j = 1;j <= i;j++){ cout<<"*"; } cout<<"\n"; } }

Marek R · Accepted Answer · 2021-03-01 10:44:33Z

void printTrignleL(size_t n) { for (size_t i = 0; i < n; ++i) { std::cout << std::string(i + 1, '*') << '\n'; } }

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