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What is the output of the program given below
#include<stdio.h>
main()
{
char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
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Answer / sharan
#include<stdio.h>
main()
{
char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Here the CATCH is semicolon after for
so maximum positive value for the signed char is 127.
Hence it loops 127 times after that value of i wraps to
negative value that is -128.
Thus it prints -128.
| Is This Answer Correct ? | 11 Yes | 1 No |
Answer / karan
it will display the garbage value bcoz there is semicolon
at end of the for loop which will be
-128
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / gajanandon
after for there is semicolon...means empty statement. So no
effect of printf.
so for runs till i (char value) increments in positive
direction and terminates once it exceeds 127 (char
limitation).
Hence finally printf will execute and then prints -128.
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / prasanna kumar [cse dept. kln
i think the program will give the output as 0 or null....
because i=0 is in int datatype but in this program it is
declared as character datatype so it will give the output as
0 or null and it will goes for only one time after wards it
will incremented and goes infinitely....
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / akash dhal
initialized with 0
in for loop ist time condition satisfied so print 0,like
this 127 will be printed .as it is a signed no. so 127+1 is
-128 so condition false come out of the loop.
| Is This Answer Correct ? | 0 Yes | 0 No |
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C 
