Exercise: Control Instructions

Questions for: Control Instructions

What will be the output of the program, if a short int is 2 bytes wide? 
#include<stdio.h>
int main()
{
    short int i = 0;
    for(i<=5 && i>=-1; ++i; i>0)
        printf("%u,", i);
    return 0;
}
A:
1 ... 65535
B:
Expression syntax error
C:
No output
D:
0, 1, 2, 3, 4, 5
Answer: A

for(i<=5 && i>=-1; ++i; i>0) so expression i<=5 && i>=-1 initializes for loop. expression ++i is the loop condition. expression i>0 is the increment expression.

In for( i <= 5 && i >= -1; ++i; i>0) expression i<=5 && i>=-1 evaluates to one.

Loop condition always get evaluated to true. Also at this point it increases i by one.

An increment_expression i>0 has no effect on value of i.so for loop get executed till the limit of integer (ie. 65535)

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What will be the output of the program? 
#include<stdio.h>
int main()
{
    int x = 3;
    float y = 3.0;
    if(x == y)
        printf("x and y are equal");
    else
        printf("x and y are not equal");
    return 0;
}
A:
x and y are equal
B:
x and y are not equal
C:
Unpredictable
D:
No output
Answer: A

Step 1: int x = 3; here variable x is an integer type and initialized to '3'.
Step 2: float y = 3.0; here variable y is an float type and initialized to '3.0'
Step 3: if(x == y) here we are comparing if(3 == 3.0) hence this condition is satisfied.
Hence it prints "x and y are equal".

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What will be the output of the program? 
#include<stdio.h>
int main()
{
    unsigned int i = 65535; /* Assume 2 byte integer*/
    while(i++ != 0)
        printf("%d",++i);
    printf("\n");
    return 0;
}
A:
Infinite loop
B:
0 1 2 ... 65535
C:
0 1 2 ... 32767 - 32766 -32765 -1 0
D:
No output
Answer: A

Here unsigned int size is 2 bytes. It varies from 0,1,2,3, ... to 65535.

Step 1:unsigned int i = 65535;

Step 2:
Loop 1: while(i++ != 0) this statement becomes while(65535 != 0). Hence the while(TRUE) condition is satisfied. Then the printf("%d", ++i); prints '1'(variable 'i' is already incremented by '1' in while statement and now incremented by '1' in printf statement) Loop 2: while(i++ != 0) this statement becomes while(1 != 0). Hence the while(TRUE) condition is satisfied. Then the printf("%d", ++i); prints '3'(variable 'i' is already incremented by '1' in while statement and now incremented by '1' in printf statement)
....
....

The while loop will never stops executing, because variable i will never become '0'(zero). Hence it is an 'Infinite loop'.

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What will be the output of the program? 
#include<stdio.h>
int main()
{
    int a = 500, b = 100, c;
    if(!a >= 400)
        b = 300;
    c = 200;
    printf("b = %d c = %d\n", b, c);
    return 0;
}
A:
b = 300 c = 200
B:
b = 100 c = garbage
C:
b = 300 c = garbage
D:
b = 100 c = 200
Answer: D

Initially variables a = 500, b = 100 and c is not assigned.

Step 1: if(!a >= 400)
Step 2: if(!500 >= 400)
Step 3: if(0 >= 400)
Step 4: if(FALSE) Hence the if condition is failed.
Step 5: So, variable c is assigned to a value '200'.
Step 6: printf("b = %d c = %d\n", b, c); It prints value of b and c.
Hence the output is "b = 100 c = 200"

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What will be the output of the program? 
#include<stdio.h>
int main()
{
    char str[]="C-program";
    int a = 5;
    printf(a >10?"Ps\n":"%s\n", str);
    return 0;
}
A:
C-program
B:
Ps
C:
Error
D:
None of above
Answer: A

Step 1: char str[]="C-program"; here variable str contains "C-program".
Step 2: int a = 5; here variable a contains "5".
Step 3: printf(a >10?"Ps\n":"%s\n", str); this statement can be written as


if(a > 10)
{
    printf("Ps\n");
}
else
{
    printf("%s\n", str);
}

Here we are checking a > 10 means 5 > 10. Hence this condition will be failed. So it prints variable str.

Hence the output is "C-program".

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