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Sample Technical Questions

C Sample Question

111. 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.

112. 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.

113. 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.

114. 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.

115. 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.

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