\Code:#include<stdio.h> void main(){ float a; int arra[100]; int i=0,j; scanf("%f",&a); while(a>0){ if((int)a%2!=0) arra[i]=1; else arra[i]=0; i++; a=a/2; if(a!=(int)a) a=a-0.5; } j=i; for(i=j;i=0;i--) printf("%d",arra[i]); }
nothing is printed.please help
Because of this:Originally Posted by tommytmh
Few other things:Code:for(i=j;i=0;i--)
- The return type of main() is int, not void.
- There is no point at all in using a float here, since you cast to int when it counts anyway -- you even round down. Also, floating point numbers are represented differently. You might want to start here if you want to figure out the binary:
Floating point - Wikipedia, the free encyclopedia
You are doing the straightforward power of two unsigned int representation, so your input should be an unsigned int.
C programming resources:
GNU C Function and Macro Index -- glibc reference manual
The C Book -- nice online learner guide
Current ISO draft standard
CCAN -- new CPAN like open source library repository
3 (different) GNU debugger tutorials: #1 -- #2 -- #3
cpwiki -- our wiki on sourceforge
simply u can keep on %2 the decimal number till it is >0 and store each remainder in an array..
later on reverse print the array ...u get the binary easily...
" I failed in some subjects in exam , but my friend passed in all . Now he is an engineer in Microsoft and I am the owner of Microsoft !! "
- Bill Gates .
You could use itoa to obtain directly the number in binary as a char array.
Check this tutorial out:
Dystopian Code: Radix Conversion in C
See here itoa reference:
itoa - C++ Reference
