Write a C program to print 1 2 3 ... 100 without using
loops?
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Answer / sutanu
void main ()
{
static int i;
if (i <= 100)
{
printf ("%d\n", i);
i++;
main ();
}
}
| Is This Answer Correct ? | 53 Yes | 15 No |
Answer / gunabalan
void main()
{
int i=1;
e:
printf("%d\t",i);
i++;
if(i<=100)
goto e;
}
| Is This Answer Correct ? | 36 Yes | 12 No |
Answer / lakshmipraba
#include<stdio.h>
int i;
void main()
{
if(i<=100)
{
printf("%d ",i);
i++;
main();
}
if(i>100)
exit(0);
}
| Is This Answer Correct ? | 14 Yes | 3 No |
Answer / vivek
void main()
{
int i=1;
if(i<=100)
printf("%d",i);
continue;
getch();
}
| Is This Answer Correct ? | 6 Yes | 2 No |
Answer / neha
int i;
void main(void)
{
if(i<=100)
printf("%d\n", i);
i++;
main();
getch();
}
| Is This Answer Correct ? | 29 Yes | 26 No |
Answer / uma maheswari
#include<stdio.h>
#include<conio.h>
int i=1;
void main()
{
i<=100 ? printf("%d\n",i) : getch(); //conditional operator
i++;
main(); //recursive calling of main() function
}
| Is This Answer Correct ? | 10 Yes | 7 No |
Answer / sagar
we can also create such a program without using loops and if
statement too ...
void main()
{
int i,n;
clrscr();
A:
printf("%d",i);
n=i++;
switch(n)
{
case 100: break;
default : goto A;
}
getch();
}
| Is This Answer Correct ? | 9 Yes | 7 No |
Answer / harsha
#include<stdio.h>
int i=0;
void main()
{
if(i==0)
clrscr();
if(i<100) {
printf("%d \t",++i);
main(); }
else {
getch();
exit(0); }
}
| Is This Answer Correct ? | 2 Yes | 1 No |
Answer / maxerp
In general, by using recursion, since it is the only
alternative to looping constructs
| Is This Answer Correct ? | 11 Yes | 12 No |
Answer / vivek
#include<stdio>
void main()
{
int i=1;
if(i<=100)
printf("%d",i);
i++;
continue;
getch;
}
| Is This Answer Correct ? | 2 Yes | 3 No |
Why is c so important?
18)struct base {int a,b; base(); int virtual function1(); } struct derv1:base{ int b,c,d; derv1() int virtual function1(); } struct derv2 : base {int a,e; } base::base() { a=2;b=3; } derv1::derv1(){ b=5; c=10;d=11;} base::function1() {return(100); } derv1::function1() { return(200); } main() base ba; derv1 d1,d2; printf("%d %d",d1.a,d1.b) o/p is a)a=2;b=3; b)a=3; b=2; c)a=5; b=10; d)none 19) for the above program answer the following q's main() base da; derv1 d1; derv2 d2; printf("%d %d %d",da.function1(),d1.function1(),d2.function1 ()); o/p is a)100,200,200; b)200,100,200; c)200,200,100; d)none 20)struct { int x; int y; }abc; you can not access x by the following 1)abc-->x; 2)abc[0]-->x; abc.x; (abc)-->x; a)1,2,3 b)2&3 c)1&2 d)1,3,4
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