C Programming Ultimate Quiz: Trivia Test!
20 Questions
| Attempts: 250
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Questions and Answers
- 1.
void main() { int const * p=5; printf("%d",++(*p)); }
- A.
5
- B.
6
- C.
Some address
- D.
Compiler error: Cannot modify a constant value.
- E.
Linker error
Correct Answer
D. Compiler error: Cannot modify a constant value.Explanation
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".Rate this question:
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- 2.
Main() { char s[ ]="man"; int i; for(i=0;s[ i ];i++) printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]); }
- A.
Man0 man1 man2
- B.
Mann mann mann
- C.
Mmmm aaaa nnnn
- D.
Compiler error
- E.
Linker error
Correct Answer
C. Mmmm aaaa nnnnExplanation
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].Rate this question:
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- 3.
Main() { float me = 1.1; double you = 1.1; if(me==you) printf("I love U"); else printf("I hate U"); }
- A.
I love U
- B.
I hate U
- C.
No Out put
- D.
Compiler error
- E.
Linker error
Correct Answer
B. I hate UExplanation
For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.Rate this question:
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- 4.
Main() { static int var = 5; printf("%d ",var--); if(var) main(); }
- A.
5
- B.
4
- C.
5 4 3 2 1
- D.
Compiler error
- E.
Linker error
Correct Answer
C. 5 4 3 2 1Explanation
When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.Rate this question:
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- 5.
Main() { int c[ ]={2.8,3.4,4,6.7,5}; int j,*p=c,*q=c; for(j=0;j<5;j++) { printf(" %d ",*c); ++q; } for(j=0;j<5;j++){ printf(" %d ",*p); ++p; } }
- A.
2.8,3.4,4,6.7,5 2.8,3.4,4,6.7,5
- B.
2.8,3.4,4,6.7,5 2 2 2 2 2
- C.
2 3 4 6 5 2 2 2 2 2
- D.
2 2 2 2 2 2 3 4 6 5
- E.
Compiler error
Correct Answer
D. 2 2 2 2 2 2 3 4 6 5Explanation
Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.Rate this question:
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- 6.
Main() { extern int i; i=20; printf("%d",i); }
- A.
20
- B.
0
- C.
No out put
- D.
Compiler error
- E.
Linker Error : Undefined symbol '_i'
Correct Answer
E. Linker Error : Undefined symbol '_i'Explanation
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurredRate this question:
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- 7.
Main() { int i=-1,j=-1,k=0,l=2,m; m=i++&&j++&&k++||l++; printf("%d %d %d %d %d",i,j,k,l,m); }
- A.
0 0 1 3 1
- B.
0 0 0 3 0
- C.
0 0 1 3 0
- D.
Compiler error
- E.
Linker Error : Undefined symbol '_m'
Correct Answer
A. 0 0 1 3 1Explanation
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‗i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‗0 || 0‘ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.Rate this question:
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- 8.
Main() { char *p; printf("%d %d ",sizeof(*p),sizeof(p)); }
- A.
1 0
- B.
1 1
- C.
1 2
- D.
Compiler error
- E.
Linker Error : Undefined symbol '_sizeof'
Correct Answer
C. 1 2Explanation
The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.Rate this question:
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- 9.
Main() { int i=3; switch(i) { default:printf("zero"); case 1: printf("one"); break; case 2:printf("two"); break; case 3: printf("three"); break; } }
- A.
Zero
- B.
One
- C.
Two
- D.
Three
- E.
Zero one
Correct Answer
D. ThreeExplanation
The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.Rate this question:
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- 10.
Main() { printf("%x",-1<<4); }
- A.
0
- B.
1
- C.
Fff0
- D.
Compiler error
- E.
Linker Error : Undefined symbol '_-1
Correct Answer
C. Fff0Explanation
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.Rate this question:
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- 11.
Main() { char string[]="Hello World"; display(string); } void display(char *string) { printf("%s",string); }
- A.
Hello world
- B.
Hello
- C.
World
- D.
Compiler Error : Type mismatch in redeclaration of function display
- E.
Linker Error : Undefined symbol '_string'
Correct Answer
D. Compiler Error : Type mismatch in redeclaration of function displayExplanation
In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.Rate this question:
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- 12.
Main() { int c= - -2; printf("%d",c); }
- A.
-2
- B.
2
- C.
0
- D.
Compiler Error
- E.
Linker Error : Undefined symbol '_c'
Correct Answer
B. 2Explanation
Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.Rate this question:
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- 13.
#define int char main() { int i=65; printf("%d",sizeof(i)); }
- A.
65
- B.
1
- C.
2
- D.
4
- E.
Compiler Error
Correct Answer
B. 1Explanation
Since the #define replaces the string int by the macro charRate this question:
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- 14.
Main() { int i=10; i=!i>14; printf("%d",i); }
- A.
10
- B.
14
- C.
1
- D.
0
- E.
Compiler Error
Correct Answer
D. 0Explanation
In the expression !i>14 , NOT (!) operator has more precedence than ‗ >‘ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).Rate this question:
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- 15.
#include<stdio.h> main() { char s[]={'a','b','c','\n','c','\0'}; char *p,*str,*str1; p=&s[3]; str=p; str1=s; printf("%d",++*p + ++*str1-32); }
- A.
0
- B.
1
- C.
77
- D.
-77
- E.
Compiler Error
Correct Answer
C. 77Explanation
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77("M");
So we get the output 77 :: "M" (Ascii is 77).Rate this question:
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- 16.
#include<stdio.h> main() { int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} }; int *p,*q; p=&a[2][2][2]; *q=***a; printf("%d----%d",*p,*q); }
- A.
10 2 3 4 5 6 7 8
- B.
5 6 7 8 10 2 3 4
- C.
SomeGarbageValue---1
- D.
Compiler Error
- E.
Linker Error
Correct Answer
C. SomeGarbageValue---1Explanation
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.Rate this question:
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- 17.
#include<stdio.h> main() { struct xx { int x=3; char name[]="hello"; }; struct xx *s; printf("%d",s->x); printf("%s",s->name); }
- A.
3 hello
- B.
3 h
- C.
No out put
- D.
Compiler Error
- E.
Linker Error
Correct Answer
D. Compiler ErrorExplanation
You should not initialize variables in declarationRate this question:
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- 18.
Is the declaration is correct #include<stdio.h> main() { struct xx { int x; struct yy { char s; struct xx *p; }; struct yy *q; }; }
- A.
True
- B.
Don't Know
- C.
May be correct
- D.
Compiler Error
- E.
Linker Error
Correct Answer
D. Compiler ErrorExplanation
The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.Rate this question:
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- 19.
Main() { printf("\nab"); printf("\bsi"); printf("\rha"); }
- A.
Ab si ha
- B.
Absiha
- C.
Hai
- D.
Compiler Error
- E.
Linker Error
Correct Answer
C. HaiExplanation
\n - newline
\b - backspace
\r - linefeedRate this question:
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- 20.
Main() { int i=5; printf("%d%d%d%d%d%d",i++,i--,++i,--i,i); }
- A.
5 5 6 4 5
- B.
5 6 5 7 6 5
- C.
4 5 5 5 4
- D.
4 5 5 4 5
- E.
5 4 5 4 5
Correct Answer
D. 4 5 5 4 5Explanation
The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.Rate this question:
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Current Version
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Mar 21, 2022Quiz Edited by
ProProfs Editorial Team -
Jun 24, 2010Quiz Created by
Srinivasdareddy
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