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- Question
What will be the output of the program?
#include<stdio.h> int main() { int i = 5; while(i-- >= 0) printf("%d,", i); i = 5; printf("\n"); while(i-- >= 0) printf("%i,", i); while(i-- >= 0) printf("%d,", i); return 0; }
Options- A. 4, 3, 2, 1, 0, -1
4, 3, 2, 1, 0, -1 - B. 5, 4, 3, 2, 1, 0
5, 4, 3, 2, 1, 0 - C. Error
- D. 5, 4, 3, 2, 1, 0
5, 4, 3, 2, 1, 0
5, 4, 3, 2, 1, 0 - Correct Answer
- 4, 3, 2, 1, 0, -1
4, 3, 2, 1, 0, -1
ExplanationStep 1: Initially the value of variable i is '5'.
Loop 1: while(i-- >= 0) here i = 5, this statement becomes while(5-- >= 0) Hence the while condition is satisfied and it prints '4'. (variable 'i' is decremented by '1'(one) in previous while condition)
Loop 2: while(i-- >= 0) here i = 4, this statement becomes while(4-- >= 0) Hence the while condition is satisfied and it prints '3'. (variable 'i' is decremented by '1'(one) in previous while condition)
Loop 3: while(i-- >= 0) here i = 3, this statement becomes while(3-- >= 0) Hence the while condition is satisfied and it prints '2'. (variable 'i' is decremented by '1'(one) in previous while condition)
Loop 4: while(i-- >= 0) here i = 2, this statement becomes while(2-- >= 0) Hence the while condition is satisfied and it prints '1'. (variable 'i' is decremented by '1'(one) in previous while condition)
Loop 5: while(i-- >= 0) here i = 1, this statement becomes while(1-- >= 0) Hence the while condition is satisfied and it prints '0'. (variable 'i' is decremented by '1'(one) in previous while condition)
Loop 6: while(i-- >= 0) here i = 0, this statement becomes while(0-- >= 0) Hence the while condition is satisfied and it prints '-1'. (variable 'i' is decremented by '1'(one) in previous while condition)
Loop 7: while(i-- >= 0) here i = -1, this statement becomes while(-1-- >= 0) Hence the while condition is not satisfied and loop exits.
The output of first while loop is 4,3,2,1,0,-1Step 2: Then the value of variable i is initialized to '5' Then it prints a new line character(\n).
See the above Loop 1 to Loop 7 .
The output of second while loop is 4,3,2,1,0,-1
Step 3: The third while loop, while(i-- >= 0) here i = -1(because the variable 'i' is decremented to '-1' by previous while loop and it never initialized.). This statement becomes while(-1-- >= 0) Hence the while condition is not satisfied and loop exits.
Hence the output of the program is
4,3,2,1,0,-1
4,3,2,1,0,-1
Control Instructions problems
Search Results
- 1. What will be the output of the program?
#include<stdio.h> int main() { int i=0; for(; i<=5; i++); printf("%d", i); return 0; }
Options- A. 0, 1, 2, 3, 4, 5
- B. 5
- C. 1, 2, 3, 4
- D. 6 Discuss
Correct Answer: 6
Explanation:
Step 1: int i = 0; here variable i is an integer type and initialized to '0'.
Step 2: for(; i<=5; i++); variable i=0 is already assigned in previous step. The semi-colon at the end of this for loop tells, "there is no more statement is inside the loop".
Loop 1: here i=0, the condition in for(; 0<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 2: here i=1, the condition in for(; 1<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 3: here i=2, the condition in for(; 2<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 4: here i=3, the condition in for(; 3<=5; i++) loop satisfies and then i is increemented by '1'(one)
Loop 5: here i=4, the condition in for(; 4<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 6: here i=5, the condition in for(; 5<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 7: here i=6, the condition in for(; 6<=5; i++) loop fails and then i is not incremented.
Step 3: printf("%d", i); here the value of i is 6. Hence the output is '6'.
- 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; }
Options- A. Hi
- B. Hello
- C. Hi Hello
- D. None of above Discuss
Correct Answer: Hi
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'
Example:#include<stdio.h> int main() { float a=0.7; printf("%.10f %.10f\n",0.7, a); return 0; }Output:
0.7000000000 0.6999999881- 3. 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; }
Options- A. Infinite loop
- B. 0 1 2 ... 65535
- C. 0 1 2 ... 32767 - 32766 -32765 -1 0
- D. No output Discuss
Correct Answer: Infinite loop
Explanation:
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'.
- 4. What will be the output of the program?
#include<stdio.h> int main() { int x = 10, y = 20; if(!(!x) && x) printf("x = %d\n", x); else printf("y = %d\n", y); return 0; }
Options- A. y =20
- B. x = 0
- C. x = 10
- D. x = 1 Discuss
Correct Answer: x = 10
Explanation:
The logical not operator takes expression and evaluates to true if the expression is false and evaluates to false if the expression is true. In other words it reverses the value of the expression.Step 1: if(!(!x) && x)
Step 2: if(!(!10) && 10)
Step 3: if(!(0) && 10)
Step 3: if(1 && 10)
Step 4: if(TRUE) here the if condition is satisfied. Hence it prints x = 10.- 5. What will be the output of the program?
#include<stdio.h> int main() { unsigned int i = 65536; /* Assume 2 byte integer*/ while(i != 0) printf("%d",++i); printf("\n"); return 0; }
Options- A. Infinite loop
- B. 0 1 2 ... 65535
- C. 0 1 2 ... 32767 - 32766 -32765 -1 0
- D. No output Discuss
Correct Answer: No output
Explanation:
Here unsigned int size is 2 bytes. It varies from 0,1,2,3, ... to 65535.Step 1:unsigned int i = 65536; here variable i becomes '0'(zero). because unsigned int varies from 0 to 65535.
Step 2: while(i != 0) this statement becomes while(0 != 0). Hence the while(FALSE) condition is not satisfied. So, the inside the statements of while loop will not get executed.
Hence there is no output.
Note: Don't forget that the size of int should be 2 bytes. If you run the above program in GCC it may run infinite loop, because in Linux platform the size of the integer is 4 bytes.
- 6. What will be the output of the program?
#include<stdio.h> int main() { int i = 1; switch(i) { printf("Hello\n"); case 1: printf("Hi\n"); break; case 2: printf("\nBye\n"); break; } return 0; }
Options- A. Hello
Hi - B. Hello
Bye - C. Hi
- D. Bye Discuss
Correct Answer: Hi
Explanation:
switch(i) has the variable i it has the value '1'(one).Then case 1: statements got executed. so, it prints "Hi". The break; statement make the program to be exited from switch-case statement.
switch-case do not execute any statements outside these blocks case and default
Hence the output is "Hi".
- 7. What will be the output of the program?
#include<stdio.h> int main() { int k, num = 30; k = (num < 10)? 100 : 200; printf("%d\n", num); return 0; }
Options- A. 200
- B. 30
- C. 100
- D. 500 Discuss
Correct Answer: 30
- 8. What will be the output of the program?
#include<stdio.h> int main() { int a=0, b=1, c=3; *((a)? &b : &a) = a? b : c; printf("%d, %d, %d\n", a, b, c); return 0; }
Options- A. 0, 1, 3
- B. 1, 2, 3
- C. 3, 1, 3
- D. 1, 3, 1 Discuss
Correct Answer: 3, 1, 3
Explanation:
Step 1: int a=0, b=1, c=3; here variable a, b, and c are declared as integer type and initialized to 0, 1, 3 respectively.Step 2: *((a) ? &b : &a) = a ? b : c; The right side of the expression(a?b:c) becomes (0?1:3). Hence it return the value '3'.
The left side of the expression *((a) ? &b : &a) becomes *((0) ? &b : &a). Hence this contains the address of the variable a *(&a).
Step 3: *((a) ? &b : &a) = a ? b : c; Finally this statement becomes *(&a)=3. Hence the variable a has the value '3'.
Step 4: printf("%d, %d, %d\n", a, b, c); It prints "3, 1, 3".
- 9. 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; }
Options- A. x and y are equal
- B. x and y are not equal
- C. Unpredictable
- D. No output Discuss
Correct Answer: x and y are equal
Explanation:
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".- 10. What will be the output of the program?
#include<stdio.h> int main() { int i=4; switch(i) { default: printf("This is default\n"); case 1: printf("This is case 1\n"); break; case 2: printf("This is case 2\n"); break; case 3: printf("This is case 3\n"); } return 0; }
Options- A. This is default
This is case 1 - B. This is case 3
This is default - C. This is case 1
This is case 3 - D. This is default Discuss
Correct Answer: This is default
This is case 1
Explanation:
In the very begining of switch-case statement default statement is encountered. So, it prints "This is default".In default statement there is no break; statement is included. So it prints the case 1 statements. "This is case 1".
Then the break; statement is encountered. Hence the program exits from the switch-case block.
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- 1. What will be the output of the program?