C Programming and Technical

Q.
main()
{i
nt i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
}

A. 1

Explanation: before entering into the for loop the checking condition is "evaluated". Here it
evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

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C Other Question

output?
#define f(g,g2) g##g2
main()
{i
nt var12=100;
printf("%d",f(var,12));
}
Answer: 100 output??
Q.
#include
main()
{
char s[]={'a','b','c','n','c',''};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}

A. M

Explanation: p is pointing to character 'n'.str1 is pointing to character 'a' ++*p "p is pointing to
'n' and that is incremented by one." the ASCII value of 'n' is 10. then it is incremented to 11.
the value of ++*p is 11. ++*str1 "str1 is pointing to 'a' that is incremented by 1 and it becomes
'b'. ASCII value of 'b' is 98. Both 11 and 98 is added and result is subtracted from 32. i.e.
(11+98-32)=77("M");