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# Loop Control Questions

NA
SHSTTON
21
Solv. Corr.
59
Solv. In. Corr.
80
Attempted
0 M:0 S
Avg. Time

1 / 61

What would be the Output

int i;
for(i=1 ;i<5;i++ );
printf("%d",i);

A1,2,3,4,5

B6

C2,3,4,5,6

D5

ENone of these

Explanation:

Here is no explanation for this answer

Workspace

NA
SHSTTON
29
Solv. Corr.
82
Solv. In. Corr.
111
Attempted
0 M:0 S
Avg. Time

2 / 61

What is the final value of i and final value of LOOPS ?
#include<stdio.h>
int main()
{
int i,j,k,l,lc=0;

printf("Enter the number  string:<1234 567>\n");
scanf("%2d%d%1d",&i,&j,&k);
for(;k;k--,i++)
for(l=0;printf("%d %d\n",i,l);)
printf("LOOPS= %d\n", lc-1);
}

AI = 16 and LOOPS=169

BI = 0 and LOOPS=16

CCompilation Error

DNo output

Explanation:

How?

Workspace

NA
SHSTTON
20
Solv. Corr.
51
Solv. In. Corr.
71
Attempted
0 M:0 S
Avg. Time

3 / 61

What is the output of the following C Program?
#include<stdio.h>
int main()
{
int i;
for (i=9;i<13; i++)
printf("%d %0x ",i,i);
return 0;
}

A9 9 10 10 11 11 12 12

B9 10 11 12

C9 9 10 b 11 b 12 c

DCompilation Error

E9 9 10 a 11 b 12 c

Explanation:

Here is no explanation for this answer

Workspace

NA
SHSTTON
10
Solv. Corr.
46
Solv. In. Corr.
56
Attempted
0 M:0 S
Avg. Time

4 / 61

What is the output of the following C Program?
#include<stdio.h>
int main(){
for(printf("1");!printf("0");printf("2"))
return 0;
}

D10

ECompilation error

Explanation:

Here is no explanation for this answer

Workspace

NA
SHSTTON
26
Solv. Corr.
69
Solv. In. Corr.
95
Attempted
0 M:0 S
Avg. Time

5 / 61

What is the output of the following C Program?
#include<stdio.h> void main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q;
}
for(j=0;j<5;j++){
printf(" %d ",*p);
++p;
}
}

A2 3 4 6 5 2 3 4 6 5

B2 2 2 2 2 2 3 4 6 5

C2.8 3.4 4 6.7 5 2.8 3.4 4 6.7 5

DNone of these

ECompiler Error

Explanation:

Here pointer c is assigned to both p and q.
In first loop, only q is incremented and not c , the value 2 will be printed 5 times.
In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed

Workspace

NA
SHSTTON
21
Solv. Corr.
23
Solv. In. Corr.
44
Attempted
0 M:0 S
Avg. Time

6 / 61

What is the output of the following C Program?
#include<stdio.h>
int main()
{
int i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
return 0;
}

A0

B1

C2

DNone of these

Explanation:

before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

Workspace

NA
SHSTTON
13
Solv. Corr.
90
Solv. In. Corr.
103
Attempted
0 M:0 S
Avg. Time

7 / 61

What is the output of the following C Program?
#include<stdio.h>
int main()
{
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}

Ac aptitude

Bc aptitude three time

Cno output No Error

DCompilation Error

ENone of these

Explanation:

Here "I" is an unsigned integer. It is compared with a signed value. Since the both types doesn't match, signed is promoted to unsigned value.
The unsigned equivalent of -2 is a huge value so condition becomes false and control comes out of the loop.

Workspace

NA
SHSTTON
18
Solv. Corr.
28
Solv. In. Corr.
46
Attempted
0 M:0 S
Avg. Time

8 / 61

What is the output of the following C Program?
#include<stdio.h>
void  main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}

AGarbage values

BCompilation Error

Cno output No Error

Dstack Overflow

Explanation:

The inner printf executes first to print some garbage value. The printf returns no of characters printed and this value also cannot be predicted. Still the outer printf prints something and so returns a non-zero value. So it encounters the break statement and comes out of the while statement.

Workspace

NA
SHSTTON
18
Solv. Corr.
21
Solv. In. Corr.
39
Attempted
0 M:0 S
Avg. Time

9 / 61

What is the output of the following C Program?
#include<stdio.h>
void  main()
{
signed char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}

A256

B128

C-256

D-128

ECompilation Error

Explanation:

Notice the semicolon at the end of the for loop. Tee initial value of the i is set to 0.
The inner loop executes to increment the value from
0 to 127 and then it rotates to the negative value of -128.
The condition in the for loop fails and so comes out of the for loop. It prints the current value of i that is -128.

Workspace

NA
SHSTTON
26
Solv. Corr.
14
Solv. In. Corr.
40
Attempted
0 M:0 S
Avg. Time

10 / 61

What will be output of the following "c" code?
#include<stdio.h>
int main() {
int i;
for(i=0;i<5;i++){
int i=10;
printf(" %d",i);
i++;
}
return 0;
}

A10 11 12 13 14

B10 10 10 10 10

C0 1 2 3 4

DInfinite loop

Explanation:

Here is no explanation for this answer

Workspace

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