goto and junping loop by loop c++

Question

im a new programmer and user here. I don't want to use goto in my code because I don't want to be called " a bad programmer". I need your experience to get rid of that.

#include <iostream> #include <time.h> using namespace std; int main() { srand(time(NULL)); int control = 1; char a[50], harf[5][10]; a[0] = 65 + rand() % 26; for (int i = 1; i < 50; i++) { if (i % 2 == 0) a[i] = 65 + rand() % 26; else a[i] = 97 + rand() % 26; come: for (int j = 0; j < i; j++) { if (a[i] == a[j]) { if (i % 2 == 0) { a[i] = 65 + rand() % 26; goto come; } else { a[i] = 97 + rand() % 26; goto come; } } else continue; } cout << a[i]<<" "; } system("pause"); }

screenshot

StackOverflow may not be the best venue for this. Perhaps SoftwareEngineering is. Anyway, I recommend pulling the inner for loop into its own function. — AndyG
– AndyG, Commented Nov 16, 2018 at 20:30
It would help if you told us what the code is supposed to do. — axblount
– axblount, Commented Nov 16, 2018 at 21:01
Prefer to use character literals instead of their decimal ASCII encodings. For example, replace 65 with 'A'. Makes the code easier to read also. — Thomas Matthews
– Thomas Matthews, Commented Nov 16, 2018 at 21:02

L.C. · Accepted Answer · 2018-11-16 20:36:39Z

1

No need to create a function, replace your "goto" with "j = -1;"

answered Nov 16, 2018 at 20:36

L.C.

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Martin Heralecký · Accepted Answer · 2018-11-16 20:32:29Z

Simply replace the come block (the inner for-cycle) with a function. Then, instead of goto come, you call come() (assuming the name of the function is "come").

Tim Randall · Accepted Answer · 2018-11-16 20:50:26Z

Replace

come: for (int j = 0; j < i; j++) { if (a[i] == a[j]) { if (i % 2 == 0) { a[i] = 65 + rand() % 26; goto come; } else { a[i] = 97 + rand() % 26; goto come; } } else continue; }

with

int loop=0; do { for (int j = 0; j < i; j++) { if (a[i] == a[j]) { if (i % 2 == 0) { a[i] = 65 + rand() % 26; } else { a[i] = 97 + rand() % 26; } loop=1; break; } else loop=0; } }while(loop);

This is a more general sort of solution than some other answers. It's based on realizing that when you find a match (a[i] == a[j]) you just want to run the main for loop again, from the top. (You can achive the same result by setting j to -1, but that won't work in all situations, and certainly not if j needs to be unsigned)

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