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C Aptitude Questions and Answers



Predict the output or error(s) for the following:
33.       main()
{
            int i=0;
            while(+(+i--)!=0)
                        i-=i++;
            printf("%d",i);
}

Answer:
-1

Explanation:
Unary + is the only dummy operator in C. So it has no effect on the expression and now the while loop is,             while(i--!=0) which is false and so breaks out of while loop. The value –1 is printed due to the post-decrement operator.

34.       main()
{
            float f=5,g=10;
            enum{i=10,j=20,k=50};
            printf("%d\n",++k);
            printf("%f\n",f<<2);
            printf("%lf\n",f%g);
            printf("%lf\n",fmod(f,g));
}

Answer:
Line no 5: Error: Lvalue required
Line no 6: Cannot apply leftshift to float
Line no 7: Cannot apply mod to float

Explanation:
                        Enumeration constants cannot be modified, so you cannot apply ++.
                        Bit-wise operators and % operators cannot be applied on float values.
                        fmod() is to find the modulus values for floats as % operator is for ints. 

35.       main()
{
            int i=10;
            void pascal f(int,int,int);
f(i++,i++,i++);
            printf(" %d",i);
}
void pascal f(integer :i,integer:j,integer :k)
{
write(i,j,k);
}

Answer:
Compiler error:  unknown type integer
Compiler error:  undeclared function write

Explanation:
Pascal keyword doesn’t mean that pascal code can be used. It means that the function follows Pascal argument passing mechanism in calling the functions.

36.    void pascal f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
void cdecl f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
main()
{
            int i=10;
f(i++,i++,i++);
            printf(" %d\n",i);
i=10;
f(i++,i++,i++);
printf(" %d",i);
}

Answer:
10 11 12 13
12 11 10 13

Explanation:
Pascal argument passing mechanism forces the arguments to be called from left to right. cdecl is the normal C argument passing mechanism where the arguments are passed from right to left.

37. What is the output of the program given below
 
main()
    {
       signed char i=0;
       for(;i>=0;i++) ;
       printf("%d\n",i);
    }

Answer
                        -128

Explanation
Notice the semicolon at the end of the for loop. THe initial value of the i is set to 0. The inner loop executes to increment the value from 0 to 127 (the positive range of char) and then it rotates to the negative value of -128. The condition in the for loop fails and so comes out of the for loop. It prints the current value of i that is -128.

38.  main()
    {
       unsigned char i=0;
       for(;i>=0;i++) ;
       printf("%d\n",i);
    }

Answer
            infinite loop

Explanation
The difference between the previous question and this one is that the char is declared to be unsigned. So the i++ can never yield negative value and i>=0 never becomes false so that it can come out of the for loop.

39.   main()
            {
       char i=0;
       for(;i>=0;i++) ;
       printf("%d\n",i);
       
 }

Answer:
                        Behavior is implementation dependent.

Explanation:
The detail if the char is signed/unsigned by default is implementation dependent. If the implementation treats the char to be signed by default the program will print –128 and terminate. On the other hand if it considers char to be unsigned by default, it goes to infinite loop.
Rule:
You can write programs that have implementation dependent behavior. But dont write programs that depend on such behavior.

40. Is the following statement a declaration/definition. Find what does it mean?
int (*x)[10];

Answer
                        Definition.
            x is a pointer to array of(size 10) integers.
 
                        Apply clock-wise rule to find the meaning of this definition.

41. What is the output for the program given below
 
     typedef enum errorType{warning, error, exception,}error;
     main()
    {
        error g1;
        g1=1;
        printf("%d",g1);
     }

Answer
                        Compiler error: Multiple declaration for error

Explanation
The name error is used in the two meanings. One means that it is a enumerator constant with value 1. The another use is that it is a type name (due to typedef) for enum errorType. Given a situation the compiler cannot distinguish the meaning of error to know in what sense the error is used:
            error g1;
g1=error;
            // which error it refers in each case?
When the compiler can distinguish between usages then it will not issue error (in pure technical terms, names can only be overloaded in different namespaces).
Note: the extra comma in the declaration,
enum errorType{warning, error, exception,}
is not an error. An extra comma is valid and is provided just for programmer’s convenience.

42.  typedef struct error{int warning, error, exception;}error;
     main()
    {
        error g1;
        g1.error =1;
        printf("%d",g1.error);
     }

 
Answer
                        1

Explanation
The three usages of name errors can be distinguishable by the compiler at any instance, so valid (they are in different namespaces).
Typedef struct error{int warning, error, exception;}error;
This error can be used only by preceding the error by struct kayword as in:
struct error someError;
typedef struct error{int warning, error, exception;}error;
This can be used only after . (dot) or -> (arrow) operator preceded by the variable

name as in :
g1.error =1;
            printf("%d",g1.error);
                        typedef struct error{int warning, error, exception;}error;
This can be used to define variables without using the preceding struct keyword as in:
error g1;
Since the compiler can perfectly distinguish between these three usages, it is perfectly legal and valid.
 
Note
This code is given here to just explain the concept behind. In real programming don’t use such overloading of names. It reduces the readability of the code. Possible doesn’t mean that we should use it!

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