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what is a pointer
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Answer / chaitali anand tendulkar
A pointer is a variable which contains a address in memory
of another variable. in other words it is a address of
variable.
| Is This Answer Correct ? | 27 Yes | 2 No |
Answer / roshan
In a simple word.....
A pointer is a variable which is store the address of another variable.
| Is This Answer Correct ? | 11 Yes | 1 No |
Answer / anil yadav
A pointer is a variable which contains the address in memory of another variable. We can have a pointer to any variable type....
| Is This Answer Correct ? | 4 Yes | 0 No |
Answer / joshi
Pointer is a special variable which hold the address of the
next variable.
| Is This Answer Correct ? | 3 Yes | 2 No |
what is a far pointer
12 Answers ABB, DRDO, ITI, Maruti Suzuki, Steel Plant, TCS, Toyota, Vivo Mobiles,
HOW DO YOU HANDLE EXCEPTIONS IN C?
Tell me about low level programming languages.
Study the code: void show() main() { show(); } void show (char *s) { printf("%sn",s); } What will happen if it is compiled & run on an ANSI C Compiler? A)It will compile & nothing will be printed when it is executed B)it will compile but not link C)the compiler will generate an error D)the compiler will generate a warning
what will be the output of this program? #include<stdio.h> #define cube(x) x*x*x void main() { int i,j=5; i=cube(j+3); printf("i=%d",i); }
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explain memory layout of a C program
18)struct base {int a,b; base(); int virtual function1(); } struct derv1:base{ int b,c,d; derv1() int virtual function1(); } struct derv2 : base {int a,e; } base::base() { a=2;b=3; } derv1::derv1(){ b=5; c=10;d=11;} base::function1() {return(100); } derv1::function1() { return(200); } main() base ba; derv1 d1,d2; printf("%d %d",d1.a,d1.b) o/p is a)a=2;b=3; b)a=3; b=2; c)a=5; b=10; d)none 19) for the above program answer the following q's main() base da; derv1 d1; derv2 d2; printf("%d %d %d",da.function1(),d1.function1(),d2.function1 ()); o/p is a)100,200,200; b)200,100,200; c)200,200,100; d)none 20)struct { int x; int y; }abc; you can not access x by the following 1)abc-->x; 2)abc[0]-->x; abc.x; (abc)-->x; a)1,2,3 b)2&3 c)1&2 d)1,3,4
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