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C programming

• retagged by
3,947 views
1 1 vote 
void main()
{   unsigned char var=0;
    for(var=0;var<=255;var++)
    {
        printf("%d ",var);
    }

}

What is output of this code?

3 Answers

Best answer
5 5 votes
Infinite loop 0,1,......255,0,1,................................................
Reason is you are declaring var   as a unsigned char which gets 1B = 8 bits of memory  and max unsigned value possible is 255
Rangle 0 to 255
whenver var = 255 = 1111 1111   then var + 1  is  
1111 1111
           +1
________
0000 0000      carry flag = 1
________
value is wrapped around :D
• selected by
0 0 votes
The range of unsigned char is 255.so it will start from 0 to 255,then again starts from 0 and so on and will get stuck into infinite loop.
0 0 votes
infinite loop as range of char  in signed -128 to +127 but in unsigned

it will be 0 to 255 which means according to this question no value in between this no.s will be less then 255

hence infinite loop
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