INTERVIEW QUESTIONS C BRANCHES & LOOPS IN C DETAILS

Question: How to print 1 to 100 without using any conditional loop?

Answer: By using recursive function.

By using recursive function. Source: CoolInterview.com

Recursion is solution of this problem.

DisplayInfo(Int val = 1)
{
Printf("%d ", val);
if(val == 100)
return;
else
DisplayInfo(++val);
} Source: CoolInterview.com

for(i=1;i<=100;i++)
{
printf(/n"%d",i);
} Source: CoolInterview.com

DisplayInfo(Int val = 1)
{
Printf("%d ", val);
if(val == 100)
return;
else
DisplayInfo(++val);
} Source: CoolInterview.com

int a=1;
l:if(a<=100)
{
printf("%d",a);
goto a;
} Source: CoolInterview.com

U all have used condition in one way or another.
To solve this problem we have to
use 'asm' keyword.
google 'asm' to find out more details about asm.
thanks Source: CoolInterview.com

Reply amitabh u cant tell answers without observing the questions ok don't do again in future otherwise it will light bad on freshers. Source: CoolInterview.com

Display(Int val = 1)
{
Printf("%d ", val);
while(val == 100)
{
return;
}
Display(val++);
} Source: CoolInterview.com

Display(Int val = 1)
{
Printf("%d ", val);
while(val == 100)
{
return;
}
Display(val++);
} Source: CoolInterview.com

#include<stdio.h>
#include<conio.h>
void main()
{
int a=1;
clrscr();
SHOW:
if(a>100)
{
}
else
{
printf("%d
",a);
a+=2;
goto SHOW;
}
getch();
} Source: CoolInterview.com

void main()
{
unsigned int i=100;
back:printf("%d
",i);
i--;
getch();
} Source: CoolInterview.com

main()
{
static int i=1;
if(i<=100)
{
printf("%d",i++);
main();
}
} Source: CoolInterview.com

main()
{
static int i=1;
if(i<=100)
{
printf("%d",i++);
main();
}
}
Source: CoolInterview.com

fun(int a)
{
printf(a);
a++;
}

main()
{

a=(a>100)?func(a):1;
}
Source: CoolInterview.com

#include<stdio.h>

int print(int i)
{
printf(" %d ",i^1?print(i-1):i);
return i+1;
}
int main()
{
print(100);
return 0;
}
Source: CoolInterview.com

//in java;
k=1;
void disp(int k)
{
System.out.println(k);
if(k==100)
return void;
else
disp(++k);
} Source: CoolInterview.com

simple!
int main()
{
printf("1 to 100");
return 0;
}

Source: CoolInterview.com

void main()
{
int a=1,b;
b=sum(a);
printf("%d",a);
printf("%d",b);
getch();
}
int sum(int a1)
{
if(a1<=100)
sum(a1+1);
}
Source: CoolInterview.com

#include "stdafx.h"
#include "stdio.h"

void printnumber(int num)
{
num++;
printf("%d
",num);
if(num < 100)
printnumber(num);
}

void main()
{
int var = 0;
printnumber(var);
}

execute this with vc++ compiler. Source: CoolInterview.com

This is an open question with no bounds defined. However, the following one of the ways of doing it:

int foo(void)
{
int i = 1;

loop:
if(i <= 100)
{
printf("i
");
}
else
{
return;
}
goto loop;
}
Source: CoolInterview.com

# include <assert.h>
# include <stdio.h>
#include<iostream.h>
#include<conio.h>
main()
{
int n=1,r;

label:
printf("%d ", n);
assert(n!=100);
n++;
goto label;
} Source: CoolInterview.com

main()
{
int i =100;

back:
printf ("%d", i);
i --;
if(i == 0)
goto end;
goto back;

end:
exit()
} Source: CoolInterview.com

#include<stdio.h>
#include<conio.h>
int i=1;
void main()
{
if(i<101)
{
printf("%d ",i);
main();
}
else
{
printf(" bye");
}
getch();

} Source: CoolInterview.com

int disp(int i)
{
printf(" i = %d
", i);
return (i ^ 100) && disp(++i);
}

int main(void)
{
disp(1);
} Source: CoolInterview.com

int getsum(int n)
{
int sum=0;
if(n==0)
return(sum);
else
sum=sum+n+getsum(n-1);
return(sum);
} Source: CoolInterview.com

void main()
{
int i;
if(i<=100)
i++
cout<<i:} Source: CoolInterview.com

#include<conio.h>
#include<stdio.h>
void main()
{ int a;
clrscr();
for(;;){
if(a==100){break;}
else{
printf("
%d",a);}
}getch();
} Source: CoolInterview.com

#include<stdio.h>
#include<cono.h>
main()
{
int i;
i=1;
clrscr();
while(i<=100)
{
printf("%d
",i);
i++;
}
getch(); Source: CoolInterview.com

void main()
{
int i;
clrscr()
for(i=1;i<=100;i++)
printf("%d",i);
}
Source: CoolInterview.com

This can be solved by using TERENARY operators.
void main()
{
int i=1;
clrscr();
n:
i<=100?printf("%d",i):(exit(0));
i++;
goto n;
getch();
}
Source: CoolInterview.com

main()
{
static int a=1;
printf("%d ",a++);
if(a>100)
{
getch();
exit(0);
}
else
main()
} Source: CoolInterview.com

#include<stdio.h>
#include<conio.h>
void main()
{
int i=1;
top:
printf("%3d",i);
if(i<100)
{
i++;
goto top;
}
getch();
} Source: CoolInterview.com

#include<stdio.h>
#include<conio.h>
void main()
{
int i=1;
clrscr();
top:
prinf("%d",i);
if(i<100)
{
i++;
goto top:
}
getch();
}
Source: CoolInterview.com

#include<stdio.h>
#include<conio.h>
int n=1;
int func(int n)
{
if(n <= 100)
{
printf("%5d",n);
func(n+1);
}
else
return 1;
}
void main()
{
func(n);
getch();
} Source: CoolInterview.com

See Guys all are wrong.
U guys are using loops again.
Here is the correct answer


#include<stdio.h>
#include<conio.h>
int i=1;
void main()
{
i<=100 ? printf("%d
",i) : getch(); //conditional operator
i++;
main(); //recursive calling of main() function
} Source: CoolInterview.com

#-----------
#-----------

func(int x)
{
printf("%d",x++);
x<100?func(x):printf("print complete");
}

void main()
{
int x=0;
func(x);
} Source: CoolInterview.com

Hello folks,

I guess this code snippet is compact and understandable:-

void display(int i){
if( i==101 ) //Not 100, to print 100 as value
return 1;
printf("%d ", i);
display(i+1);
} Source: CoolInterview.com

#include<stdio.h>
#include<conio.h>
int main(){
void show(int);
show(100);
getch();
return 0;
}
void show(int i)
{
if(i==0)
return;
printf("%d",i);
show(i-1);
} Source: CoolInterview.com

If you have the better answer, then send it to us. We will display your answer after the approval.

Rules to Post Answers in CoolInterview.com:-

• There should not be any Spelling Mistakes.
• There should not be any Gramatical Errors.
• Answers must not contain any bad words.
• Answers should not be the repeat of same answer, already approved.
• Answer should be complete in itself.

Name :*
Email Id :*
Answer :*
Verification Code	Not readable? Load New Code
Enter the above shown code: *
	Inform me about updated answers to this question