#define f(g,g2) g##g2

main()
{
   int var12=100;
   printf("%d",f(var,12));
 }

100

 #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);
}

M

p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11

 Is the following code legal?

typedef struct a
{
   int x;
   aType *b;
}aType

No

The typename aType is not known at the point of declaring the structure (forward references are not made for typedefs).

 main( ){
 static int  a[ ]   = {0,1,2,3,4};
 int  *p[ ] = {a,a+1,a+2,a+3,a+4};
 int  **ptr =  p;
 ptr++;
 printf(�\n %d  %d  %d�, ptr-p, *ptr-a, **ptr);
 *ptr++;
 printf(�\n %d  %d  %d�, ptr-p

 main()

{
      extern int i;
      i=20;
      printf("%d",i);
}

give buld error as unresolved extern symbol i

20

it will produce the run time error because it search the external variable while program running and that is not found

Unresolved external symbol i ...extern keyword shows that the symbol i is not declared in current scope, and in the program i is not declared in any other scope.

Undefined variable i

20

error

Compiler time ErrorUndefined variable i extern int i  is only a declaration not a definition

here linker error will come because  extern will say it is defined some where else outside main. so to make it run do int i; outside main

Runtime error,because there is no external defnition of i

Runtime error,because there is no external defnition of i

linker error

error

The answer will b 20.....since the variable declared as a i just take d reference of variable declared globally

Linker error

20

20

20

20

it will show an error b'coz extern variable cann't be initialized it can initialize only through external linkage.

error, there is no definition for i.

 main()

{
   int i=5;
   printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}

45545

The arguments in function call are pushed into stack from left to right. The evaluation is from right to left.

5 6 6 5 5

54545

54545

45545GrabageValue

There are 6 %d...

Compiler dependent. function parameters are not evaluated in a defined order in C/C++
Can be.. 4 5 5 5 5 or 4 5 5 4 5

5 5 6 4 5

i++  i--  ++i  --i  i
 5    6     6   5    5

45545
because in printf() calculation will be evaluated right to left.
hence (i++,i--,++i,--i,i)
right to left: i=5,--i=4,(now i=4), ++i=5(now i=5), i--will be 5 this time and printed and (i--=4, b

 main()

{
   printf("\nab"); 
   printf("\bsi");
   printf("\rha");
}

hai

\n  - newline
\b  - backspace
\r  - linefeed 

hai

hai

ha

ha

hai

 main()
     {
      main();
      }

Runtime error : Stack overflow.

main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate

 main()
   {
      char *cptr,c;
      void *vptr,v;
      c=10;  v=0;
      cptr=&c; vptr=&v;
      printf("%c%v",c,v);
   }

Compiler error (at line number 4): size of v is Unknown.

You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vpt

10,0

 main()
{
   char *p = "ayqm";
   char c;
   c = ++*p++;
   printf("%c",c);
}

b

There is no difference between the expression ++*(p++) and ++*p++. Parenthesis just works as a visual clue for the reader to see which expression is first evaluated.