Answer / sanjay bhosale

Segments are nothing but sections for e.g code segment(also called text region),data segment, stack segment and heap segment.
They represent different parts of the process or different aspects of the process.

Offsets represents address where next read of write should begin.

Answer / mayank

When paging technique is performed ,The page will breaks in
to the segments and its sequence is said to be segments and
is width or can be said as length is called is offset.
In sort...
segment is physical address..
and offset is logical address..

main() { int c = 5; printf("%d", main||c); } a. 1 b. 5 c. 0 d. none of the above

2 Answers   HCL,

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# include<stdio.h> aaa() { printf("hi"); } bbb(){ printf("hello"); } ccc(){ printf("bye"); } main() { int (*ptr[3])(); ptr[0]=aaa; ptr[1]=bbb; ptr[2]=ccc; ptr[2](); }

1 Answers  

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write a program to find out roots of quadratic equation "x=-b+-(b^2-4ac0^-1/2/2a"

2 Answers  

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How will u find whether a linked list has a loop or not?

8 Answers   Microsoft,

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main() { char s[ ]="man"; int i; for(i=0;s[ i ];i++) printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]); }

1 Answers   DCE,

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What are the following notations of defining functions known as? i. int abc(int a,float b) { /* some code */ } ii. int abc(a,b) int a; float b; { /* some code*/ }

1 Answers  

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How do you sort a Linked List (singly connected) in O(n) please mail to pawan.10k@gmail.com if u can find an anser...i m desperate to knw...

6 Answers   Microsoft, MSD, Oracle,

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find simple interest & compund interest

2 Answers  

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main() { char * strA; char * strB = I am OK; memcpy( strA, strB, 6); } a. Runtime error. b. I am OK c. Compile error d. I am O

4 Answers   HCL,

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main() { static char names[5][20]={"pascal","ada","cobol","fortran","perl"}; int i; char *t; t=names[3]; names[3]=names[4]; names[4]=t; for (i=0;i<=4;i++) printf("%s",names[i]); }

2 Answers  

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main() { char *p = "hello world"; p[0] = 'H'; printf("%s", p); } a. Runtime error. b. “Hello world” c. Compile error d. “hello world”

5 Answers   HCL,

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struct Foo { char *pName; char *pAddress; }; main() { struct Foo *obj = malloc(sizeof(struct Foo)); clrscr(); obj->pName = malloc(100); obj->pAddress = malloc(100); strcpy(obj->pName,"Your Name"); strcpy(obj->pAddress, "Your Address"); free(obj); printf("%s", obj->pName); printf("%s", obj->pAddress); } a. Your Name, Your Address b. Your Address, Your Address c. Your Name Your Name d. None of the above

2 Answers   HCL,

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