Test 6

1. 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;
}

• 1 … 65535
• Expression syntax error
• No output
• 0, 1, 2, 3, 4, 5

Explanation 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)

---

2. What will be the output of the program?

#include<stdio.h>
int main()
{
    float a = 0.7;
    if(0.7 > a)
        printf("Hi\n");
    else
        printf("Hello\n");
    return 0;
}

• Hi
• Hello
• Hi Hello
• None of above

Explanation if(0.7 > a) here a is a float variable and 0.7 is a double constant. The double constant 0.7 is greater than the float variable a. Hence the if condition is satisfied and it prints ‘Hi’

---

12. What will be the output of the program ?

#include<stdio.h>
 
int main()
{
    char str1[] = "Hello";
    char str2[10];
    char *t, *s;
    s = str1;
    t = str2;
    while(*t=*s)
        *t++ = *s++;
    printf("%s\n", str2);
    return 0;
}

• Hello
• HelloHello
• No output
• ello

Explanation

• Iteration 1:
  • Condition: Copies ‘H’ from str1[0] to str2[0]. (Pointers don’t move).
  • Body: Copies ‘H’ from str1[0] to str2[0] again.
  • Increment: s moves to ‘e’, t moves to the next slot.
• Iteration 2:
  • Condition: Copies ‘e’ from str1[1] to str2[1].
  • Body: Copies ‘e’ again.
  • Increment: s moves to ‘l’, t moves to the next slot.

---

18. Point out the error in the program.

#include<stdio.h>
const char *fun();
 
int main()
{
    char *ptr = fun();
    return 0;
}
const char *fun()
{
    return "Hello";
}

• Error: Lvalue required
• Error: cannot convert ‘const char *’ to ‘char *’.
• No error and No output
• None of above

---

20. What will be the output of the program?

#include<stdio.h>
 
int main()
{
    int i;
    char c;
    for(i=1; i<=5; i++)
    {
        scanf("%c", &c); /* given input is 'a' */
        printf("%c", c);
        ungetc(c, stdin);
    }
    return 0;
}

• aaaa
• aaaaa
• Garbage value.
• Error in ungetc statement.

Explanation for(i=1; i⇐5; i++) Here the for loop runs 5 times.

Loop 1:
scanf(“%c”, &c); Here we give ‘a’ as input.
printf(“%c”, c); prints the character ‘a’ which is given in the previous “scanf()” statement.
ungetc(c, stdin); “ungetc()” function pushes character ‘a’ back into input stream.

Loop 2:
Here the scanf(“%c”, &c); get the input from “stdin” because of “ungetc” function.
printf(“%c”, c); Now variable c = ‘a’. So it prints the character ‘a’.
ungetc(c, stdin); “ungetc()” function pushes character ‘a’ back into input stream.

This above process will be repeated in Loop 3, Loop 4, Loop 5.