Trivia: Test On Output Of C Programs! Quiz
Are you a computer programmer looking for a way to refresh what you know when it comes to the output of C programs? If you said yes, then you are in luck as the quiz is prepared to help you do just that. Give it a try and see if beyond might need to spend more hours studying for your finals.
Questions and Answers
- 1.
What is the final value of x when the code int x; for(x=0; x<10; x++) {} is run?
- A.
10
- B.
9
- C.
0
- D.
1
Correct Answer
A. 10Explanation
The final value of x when the code is run is 10. This is because the code initializes x as 0 and then increments it by 1 each time the loop runs. The loop continues to run as long as x is less than 10. Once x reaches 10, the loop condition is no longer true and the loop terminates. Therefore, the final value of x is 10.Rate this question:
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- 2.
What will be output if you will compile and execute the following c code? #include<stdio.h> void call(int,int,int); int main(){ int a=10; call(a,a++,++a); return 0; } void call(int x,int y,int z){ printf("%d %d %d",x,y,z); }
- A.
10 10 12
- B.
12 11 11
- C.
12 12 12
- D.
10 11 12
- E.
Compiler Error
Correct Answer
B. 12 11 11Explanation
The code snippet calls the function "call" with three arguments: a, a++, and ++a. The value of "a" is initially 10. In the function call, the value of "a" is passed as 10, the value of "a++" is passed as 10 (but the value of "a" is incremented to 11 after the evaluation), and the value of "++a" is passed as 11 (since "a" is incremented to 11 before the evaluation). Therefore, the output will be "12 11 11".Rate this question:
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- 3.
What will be output if you will compile and execute the following c code? #include<stdio.h> int main(){ register int i,x; scanf("%d",&i); x=++i + ++i + ++i; printf("%d",x); return 0; }
- A.
17
- B.
18
- C.
21
- D.
22
- E.
Compiler Error
Correct Answer
E. Compiler ErrorExplanation
The code will result in a compiler error. The reason for this is the use of the "register" keyword before the variable declaration. In modern C programming, the "register" keyword is considered obsolete and is no longer necessary. Some compilers may still support it, but in this case, it causes a conflict with the use of the "++" operator multiple times on the same variable in the expression "++i + ++i + ++i". This results in undefined behavior and a compiler error.Rate this question:
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- 4.
What will be output if you will compile and execute the following c code? #include<stdio.h> int main(){ int a=5; int b=10; { int a=2; a++; b++; } printf("%d %d",a,b); return 0; }
- A.
5 10
- B.
6 11
- C.
5 11
- D.
6 10
- E.
Compiler error
Correct Answer
C. 5 11Explanation
The code declares two variables, a and b, and initializes them with the values 5 and 10 respectively. Then, inside a block of code, a new variable a is declared and initialized with the value 2. The value of this inner a is then incremented by 1, but it does not affect the value of the outer a. Similarly, the value of b is incremented by 1. Finally, the values of the outer a and b are printed, which are 5 and 11 respectively.Rate this question:
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- 5.
What will be output if you will compile and execute the following c code? #include<stdio.h> int main(){ float f=3.4e39; printf("%f",f); return 0; }
- A.
3.4e39
- B.
3.40000…
- C.
+INF
- D.
Compiler error
- E.
Run time error
Correct Answer
C. +INFExplanation
The code declares a variable 'f' of type float and assigns it the value 3.4e39, which is a very large number. The printf function is then used to print the value of 'f'. Since the value of 'f' exceeds the maximum representable value for a float, it will be represented as positive infinity (+INF) when printed.Rate this question:
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- 6.
What will be output if you will compile and execute the following c code? #include<stdio.h> int main(){ enum color{ RED,GREEN=-20,BLUE,YELLOW }; enum color x; x=YELLOW; printf("%d",x); return 0; }
- A.
-22
- B.
-18
- C.
1
- D.
Compiler error
- E.
None of the above
Correct Answer
B. -18Explanation
The code defines an enumeration called "color" with four possible values: RED, GREEN, BLUE, and YELLOW. The values are assigned implicitly, with RED being 0, GREEN being -20, BLUE being -19, and YELLOW being -18. The variable "x" is declared as an enum color and is assigned the value YELLOW. The printf statement then prints the value of "x", which is -18.Rate this question:
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- 7.
What will be output if you will compile and execute the following c code? #include<stdio.h> int main(){ asm{ mov bx,8; mov cx,10 add bx,cx; } printf("%d",_BX); return 0; }
- A.
18
- B.
8
- C.
0
- D.
Compiler error
- E.
None of above
Correct Answer
A. 18Explanation
The given C code uses inline assembly to perform addition of the values stored in the registers bx and cx. The result, 18, is then printed using the printf function. Therefore, the output of the code will be 18.Rate this question:
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- 8.
What will be output if you will compile and execute the following c code? #include<stdio.h> int main(){ enum xxx{ a,b,c=32767,d,e }; printf("%d",b); return 0; }
- A.
0
- B.
1
- C.
32766
- D.
Compiler error
- E.
None of above
Correct Answer
D. Compiler errorExplanation
The code will not compile because the enum value "b" is not assigned a value.Rate this question:
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- 9.
What will be output if you will compile and execute the following c code? #include<stdio.h> int main(){ char *url="c:\tc\bin\rw.c"; printf("%s",url); return 0; }
- A.
C:\tc\bin\rw.c
- B.
C:/tc/bin/rw.c
- C.
C: c inw.c
- D.
C:cinw.c
- E.
W.c in
Correct Answer
E. W.c inExplanation
The given code declares a character pointer variable named "url" and assigns it the value "c:\tc\bin\rw.c". The printf statement then prints the value of "url", which is "c:\tc\bin\rw.c". Therefore, the output of the code will be "c:\tc\bin\rw.c".Rate this question:
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- 10.
What will be output if you will compile and execute the following c code? #include<stdio.h> int main(){ goto abc; printf("main"); return 0; } void dispaly(){ abc: printf("display"); }
- A.
Main
- B.
Display
- C.
Maindisplay
- D.
Displaymain
- E.
Compiler error
Correct Answer
E. Compiler errorExplanation
The code will result in a compiler error because the label "abc" is defined within the function "dispaly()" but it is being used outside of that function in the "main()" function. Labels can only be used within the function they are defined in.Rate this question:
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