Outputs or errors of given simple program with explanation (171 to 179)

171)      void main()

{

char ch;

for(ch=0;ch<=127;ch++)

printf(“%c   %d \n“, ch, ch);

}

Answer:

            Implementaion dependent

Explanation:

The char type may be signed or unsigned by default. If it is signed then ch++ is executed after ch reaches 127 and rotates back to -128. Thus ch is always smaller than 127.

172)      Is this code legal?

int *ptr;

ptr = (int *) 0x400;

Answer:

                        Yes

Explanation:

The pointer ptr will point at the integer in the memory location 0x400.

173)      main()

{

char a[4]="HELLO";

printf("%s",a);

}          

Answer:

                        Compiler error: Too many initializers

Explanation:

The array a is of size 4 but the string constant requires 6 bytes to get stored.

174)      main()

{          

char a[4]="HELL";

printf("%s",a);

}

Answer:

                        HELL%@!~@!@???@~~!

Explanation:

The character array has the memory just enough to hold the string “HELL” and doesnt have enough space to store the terminating null character. So it prints the HELL correctly and continues to print garbage values till it             accidentally comes across a NULL character.

175)      main()

{

                        int a=10,*j;

            void *k;

                        j=k=&a;

            j++; 

                        k++;

            printf("\n %u %u ",j,k);

}

Answer:

                        Compiler error: Cannot increment a void pointer

Explanation:

Void pointers are generic pointers and they can be used only when the type is not known and as an intermediate address storage type. No pointer arithmetic can be done on it and you cannot apply indirection operator (*) on void pointers.

176)      main()

                        {

                                    extern int i;

                        {           int i=20;

                         {         

                           const volatile unsigned i=30; printf("%d",i);

                         }

                        printf("%d",i);

                        }

                          printf("%d",i);

            }          

            int i;

177)      Printf can be implemented by using  __________ list.

Answer:

                        Variable length argument lists

178) char *someFun()

            {

            char *temp = “string constant";

            return temp;

            }

            int main()

            {

            puts(someFun());

            }

Answer:

            string constant

Explanation:

            The program suffers no problem and gives the output correctly because the character constants are stored in code/data area and not allocated in stack, so this doesn’t lead to dangling pointers.

179)      char *someFun1()

            {

            char temp[ ] = “string";

            return temp;

            }

            char *someFun2()

            {

            char temp[ ] = {‘s’, ‘t’,’r’,’i’,’n’,’g’};

            return temp;

            }

            int main()

            {

            puts(someFun1());

            puts(someFun2());

            }

Answer:

            Garbage values.

Explanation:

            Both the functions suffer from the problem of dangling pointers. In someFun1() temp is a character array and so the space for it is allocated in heap and is initialized with character string “string”. This is created dynamically as the function is called, so is also deleted dynamically on exiting the function so the string data is not available in the calling function main() leading to print some garbage values. The function someFun2() also suffers from the same problem but the problem can be easily identified in this case.