int i;
i=2;
i++;
if(i=4)
{
printf(i=4);
}
else
{
printf(i=3);
}
output of the program ?
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Answer Posted / manishsoni
it show errors,bcoz
in c language,to print anyone we want to enclose that in " ".
so both the printf statement are wrong,so it is not show
proper result.
To show proper result we show that program and describe line
to line....
#include<stdio.h>
#include<conio.h>
int main()
{
int i;
i=2;
i++;
if(i=4)
{
printf("4");
}
else
{
printf("3");
}
getch();
}
In this prg i is declare as int type,after that 2 is store
into the i after i increased by not store so there is no
affect of the i's value.
at if statement
if(i=4)
here 4 is assigned into the i variable so
if statement is look like as;
if(4)which is treat is as ;
the if statement is thought that any non zero value is true
so if statement is true..
and execute first printf statement so the value is
printf("i=4");
so the final answer or result is:
i=4;
| Is This Answer Correct ? | 1 Yes | 0 No |
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