Result of the following program is
main()
{
int i=0;
for(i=0;i<20;i++)
{
switch(i)
case 0:i+=5;
case 1:i+=2;
case 5:i+=5;
default i+=4;
break;}
printf("%d,",i);
}
}
a)0,5,9,13,17
b)5,9,13,17
c)12,17,22
d)16,21
e)syntax error
Re: Result of the following program is
main()
{
int i=0;
for(i=0;i<20;i++)
{
switch(i)
case 0:i+=5;
case 1:i+=2;
case 5:i+=5;
default i+=4;
break;}
printf("%d,",i);
}
}
a)0,5,9,13,17
b)5,9,13,17
c)12,17,22
d)16,21
e)syntax error
Re: Result of the following program is
main()
{
int i=0;
for(i=0;i<20;i++)
{
switch(i)
case 0:i+=5;
case 1:i+=2;
case 5:i+=5;
default i+=4;
break;}
printf("%d,",i);
}
}
a)0,5,9,13,17
b)5,9,13,17
c)12,17,22
d)16,21
e)syntax error
Answer is e), since opening and closing flower braces do
not match in numbers and default do not have colon
following it. Assuming switch(i) has an opening flower
brace and default has colon after it "switch(i) {, ...
default: i+= 4;"answer would be d).
All cases will be fall-through including default:
Re: Result of the following program is
main()
{
int i=0;
for(i=0;i<20;i++)
{
switch(i)
case 0:i+=5;
case 1:i+=2;
case 5:i+=5;
default i+=4;
break;}
printf("%d,",i);
}
}
a)0,5,9,13,17
b)5,9,13,17
c)12,17,22
d)16,21
e)syntax error
a memory of 20 bytes is allocated to a string declared as
char *s then the following two statements are executed:
s="Etrance"
l=strlen(s);
what is the value of l ?
a.20
b.8
c.9
d.21
What is the output for the following program
#include<stdio.h>
main()
{
char a[5][5],flag;
a[0][0]='A';
flag=((a==*a)&&(*a==a[0]));
printf("%d\n",flag);
}
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