Answer / narasimha

"near" and "far" pointers are actually non-standard
qualifiers that you'll find only on x86 systems. They
reflect the odd segmentation architecture of Intel
processors. In short, a near pointer is an offset only,
which refers to an address in a known segment. A far pointer
is a compound value, containing both a segment number and an
offset into that segment.

Segmentation still exists on Intel processors, but it is not
used in any of the mainstream 32-bit operating systems
developed for them, so you'll generally only find the "near"
and "far" keywords in source code developed for Windows 3.x,
MS-DOS, Xenix/80286, etc.

Answer / rag

It is used to access the higher memory addresses.

Answer / peter

I think it's kind of pointer for different computer
architecture such as X86, IBM power server and so on because
various model of memory model is different.

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()); }

2 Answers  

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write a program in c language to get the value of arroy keys pressed and display the message which arrow key is pressed?

1 Answers  

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void main() { int x,y=2,z; z=(z*=2)+(x=y=z); printf("%d",z); }

4 Answers  

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

3 Answers  

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Is the following code legal? typedef struct a { int x; aType *b; }aType

1 Answers  

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Predict the Output: int main() { int *p=(int *)2000; scanf("%d",2000); printf("%d",*p); return 0; } if input is 20 ,what will be print

2 Answers  

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prog. to produce 1 2 3 4 5 6 7 8 9 10

4 Answers   TCS,

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Sir... please give some important coding questions asked by product companies..

0 Answers  

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main() { int c[ ]={2.8,3.4,4,6.7,5}; int j,*p=c,*q=c; for(j=0;j<5;j++) { printf(" %d ",*c); ++q; } for(j=0;j<5;j++){ printf(" %d ",*p); ++p; } }

2 Answers   CSS, Wipro,

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void main() { int i; char a[]="\0"; if(printf("%s\n",a)) printf("Ok here \n"); else printf("Forget it\n"); }

3 Answers   Accenture,

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main() { int i=-1; +i; printf("i = %d, +i = %d \n",i,+i); }

1 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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