Answer / manoj ku. dalai

Explaining With the examples.........

1) const char *a="xyz" (string is constant, pointer is not)
*a='x' (error)
a="hi" (legal)

2) char* const a="xyz" (pointer is constant, String is not)
*a='x' (legal)
a="hi" (error)

3) char const *a="xz" (string is constant, pointer is not)
a*='x' (error)
a="hi" (legal)

Answer / susie

Answer :

1. 'const' applies to char * rather than 'a' ( pointer to a
constant char )

*a='F' : illegal

a="Hi" : legal

2. 'const' applies to 'a' rather than to the value of
a (constant pointer to char )

*a='F' : legal

a="Hi" : illegal

3. Same as 1.

Answer / srinivas

The answers for first and third case is fine,but in 2nd
case we cannot assign *a='F',becoz *a points to starting
address of the array you cannot change the value at that
address where the reference to that pointer is lost.

main() { int i; clrscr(); for(i=0;i<5;i++) { printf("%d\n", 1L << i); } } a. 5, 4, 3, 2, 1 b. 0, 1, 2, 3, 4 c. 0, 1, 2, 4, 8 d. 1, 2, 4, 8, 16

4 Answers   HCL,

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int a=1; printf("%d %d %d",a++,a++,a); need o/p in 'c' and what explanation too

1 Answers  

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I need your help, i need a Turbo C code for this problem.. hope u'll help me guys.? Your program will have a 3x3 array. The user will input the sum of each row and each column. Then the user will input 3 values and store them anywhere, or any location or index, temporarily in the array. Your program will supply the remaining six (6) values and determine the exact location of each value in the array. Example: Input: Sum of row 1: 6 Sum of row 2: 15 Sum of row 3: 24 Sum of column 1: 12 Sum of column 2: 15 Sum of column 3: 18 Value 1: 3 Value 2: 5 Value 3: 6 Output: Sum of Row 1 2 3 6 4 5 6 15 7 8 9 24 Sum of Column 12 15 18 Note: Your program will not necessary sort the walues in the array Thanks..

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/*what is the output for*/ void main() { int r; printf("Naveen"); r=printf(); getch(); }

4 Answers  

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main() { int x=5; clrscr(); for(;x==0;x--) { printf("x=%d\n”", x--); } } a. 4, 3, 2, 1, 0 b. 1, 2, 3, 4, 5 c. 0, 1, 2, 3, 4 d. none of the above

3 Answers   HCL,

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main() { int a=2,*f1,*f2; f1=f2=&a; *f2+=*f2+=a+=2.5; printf("\n%d %d %d",a,*f1,*f2); }

6 Answers  

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void main() { char ch; for(ch=0;ch<=127;ch++) printf(“%c %d \n“, ch, ch); }

1 Answers  

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#include<stdio.h> main() { struct xx { int x=3; char name[]="hello"; }; struct xx *s; printf("%d",s->x); printf("%s",s->name); }

3 Answers   Hexaware,

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#if something == 0 int some=0; #endif main() { int thing = 0; printf("%d %d\n", some ,thing); }

1 Answers  

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main() { int i=5,j=6,z; printf("%d",i+++j); }

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3) Int Matrix of certain size was given, We had few valu= es in it like this. =97=97=97=97=97=97=97=97=97=97=97 1 = | 4 | | 5 | &= nbsp; | 45 =97=97=97=97=97=97=97=97=97=97=97 &n= bsp; | 3 | 3 | 5 | = | 4 =97=97=97=97=97=97=97=97=97=97=97 34 |&nbs= p; 3 | 3 | | 12 | &= nbsp; =97=97=97=97=97=97=97=97=97=97=97 3 | &nbs= p; | 3 | 4 | = | 3 =97=97=97=97=97=97=97=97=97=97=97 3 | = ; | | | = ; 3 | =97=97=97=97=97=97=97=97=97=97=97 &= nbsp; | | 4 | = ; | 4 | 3 We w= ere supposed to move back all the spaces in it at the end. Note: = If implemented this prog using recursion, would get higher preference.

0 Answers   RoboSoft,

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main() { int x=5; for(;x!=0;x--) { printf("x=%d\n", x--); } } a. 5, 4, 3, 2,1 b. 4, 3, 2, 1, 0 c. 5, 3, 1 d. none of the above

2 Answers   HCL,

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