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+3

How to draw X shape using *

I tried this code but only works when num=5: void draw_X_Shape(int n) { int i,j; for(i=0;i<5;i++) { for(j=0;j<5;j++) { if((i==j) || (i+j)==(n-1)) { printf("*"); } else printf(" "); } printf("\n"); } } int main() { int num=5; draw_X_Shape(num); return 0; }

10/16/2018 8:32:43 AM

Aisha Ali

5 Answers

New Answer

+6

Try it like this: void draw_X_Shape(int n) { int i,j; for(i=0;i<n;i++) { for(j=0;j<n;j++) { if((i==j) || (i+j)==(n-1)) { printf("*"); } else printf(" "); } printf("\n"); } } int main() { int num=7; draw_X_Shape(num); return 0; } I have changed the '5' in the lines that start with for into 'n'.

+1

Thanks Paul Jacobs

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but if n is even then x is not look like x

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iqtiza hasan yes u r right, did you try how to solve this?

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public static void Main(string[] args) { int n=5; for (int i =0;i<n;i++){ for(int j=0;j<n;j++){ if (j==i||(n-j)==i){ Console.Writeline("*"); } else { Console.Writeline(" "); } } } Console.Writeline(""); } } Probably the simplest way i could have think of ..