Answer / asit mahato

The conditional test if(i=4) is true for evrey non zero
value of i.As here i=3 here execute the if statement.But
printf can't print nothing because we have not mation the
proper syntax of printf.if we replace
printf(i=4)by printf("i=4")
it will give the following output:
i=4

Answer / akbar shan

There Wont be any output for this program.It will show
error.Because 1:inside if "=" is used.2:"printf" cant print
the value

Answer / sathish

output is 3

Answer / jignesh patel

output of this program is 3

Answer / stuti

This program will give error. Becoz the format of if
statement and printf statement- both are wrong.
In if, instead of if(i==4), if(i=4) is written and in
printf statement, there should be printf("i==4"); and printf
("i=3");
So it will not be compiled.

Answer / raj

hi
frinds they are not compare with i it is error

Answer / 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;

Answer / ssssssssss

output
3

Answer / ruchi

outout will be 4 in case code is printf("i=4");
printf("i=3");
else this program will show two error

Answer / asdf

un predictable

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