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- Question
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; }
Options- A. C-program
- B. Ps
- C. Error
- D. None of above
- Correct Answer
- C-program
ExplanationStep 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".
Control Instructions problems
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- 1. What will be the output of the program?
#include<stdio.h> int main() { int a = 300, b, c; if(a >= 400) b = 300; c = 200; printf("%d, %d, %d\n", a, b, c); return 0; }
Options- A. 300, 300, 200
- B. Garbage, 300, 200
- C. 300, Garbage, 200
- D. 300, 300, Garbage Discuss
Correct Answer: 300, Garbage, 200
Explanation:
Step 1: int a = 300, b, c; here variable a is initialized to '300', variable b and c are declared, but not initialized.
Step 2: if(a >= 400) means if(300 >= 400). Hence this condition will be failed.
Step 3: c = 200; here variable c is initialized to '200'.
Step 4: printf("%d, %d, %d\n", a, b, c); It prints "300, garbage value, 200". because variable b is not initialized.- 2. 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; }
Options- A. b = 300 c = 200
- B. b = 100 c = garbage
- C. b = 300 c = garbage
- D. b = 100 c = 200 Discuss
Correct Answer: b = 100 c = 200
Explanation:
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"- 3. 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.
- 4. 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".- 5. 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".
- 6. What will be the output of the program?
#include<stdio.h> int main() { char j=1; while(j < 5) { printf("%d, ", j); j = j+1; } printf("\n"); return 0; }
Options- A. 1 2 3 ... 127
- B. 1 2 3 ... 255
- C. 1 2 3 ... 127 128 0 1 2 3 ... infinite times
- D. 1, 2, 3, 4 Discuss
Correct Answer: 1, 2, 3, 4
- 7. What will be the output of the program?
#include<stdio.h> int main() { int i=3; switch(i) { case 1: printf("Hello\n"); case 2: printf("Hi\n"); case 3: continue; default: printf("Bye\n"); } return 0; }
Options- A. Error: Misplaced continue
- B. Bye
- C. No output
- D. Hello Hi Discuss
Correct Answer: Error: Misplaced continue
Explanation:
The keyword continue cannot be used in switch case. It must be used in for or while or do while loop. If there is any looping statement in switch case then we can use continue.- 8. What will be the output of the program?
#include<stdio.h> int main() { int x=1, y=1; for(; y; printf("%d %d\n", x, y)) { y = x++ <= 5; } printf("\n"); return 0; }
Options- A. 2 1
3 1
4 1
5 1
6 1
7 0 - B. 2 1
3 1
4 1
5 1
6 1 - C. 2 1
3 1
4 1
5 1 - D. 2 2
3 3
4 4
5 5
Discuss
Correct Answer: 2 1
3 1
4 1
5 1
6 1
7 0
- 9. Point out the error, if any in the for loop.
#include<stdio.h> int main() { int i=1; for(;;) { printf("%d\n", i++); if(i>10) break; } return 0; }
Options- A. There should be a condition in the for loop
- B. The two semicolons should be dropped
- C. The for loop should be replaced with while loop.
- D. No error Discuss
Correct Answer: No error
Explanation:
Step 1: for(;;) this statement will genereate infinite loop.
Step 2: printf("%d\n", i++); this statement will print the value of variable i and increement i by 1(one).
Step 3: if(i>10) here, if the variable i value is greater than 10, then the for loop breaks.
Hence the output of the program is
1
2
3
4
5
6
7
8
9
10
- 10. Point out the error, if any in the program.
#include<stdio.h> int main() { int P = 10; switch(P) { case 10: printf("Case 1"); case 20: printf("Case 2"); break; case P: printf("Case 2"); break; } return 0; }
Options- A. Error: No default value is specified
- B. Error: Constant expression required at line case P:
- C. Error: There is no break statement in each case.
- D. No error will be reported. Discuss
Correct Answer: Error: Constant expression required at line case P:
Explanation:
The compiler will report the error "Constant expression required" in the line case P: . Because, variable names cannot be used with case statements.The case statements will accept only constant expression.
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- 1. What will be the output of the program?