Answer / susie

Answer :

Compiler Error. We cannot apply indirection on type void*.

Explanation:

Void pointer is a generic pointer type. No pointer
arithmetic can be done on it. Void pointers are normally
used for,

1. Passing generic pointers to functions and returning such
pointers.

2. As a intermediate pointer type.

3. Used when the exact pointer type will be known at a later
point of time.

void main() { if(~0 == (unsigned int)-1) printf(“You can answer this if you know how values are represented in memory”); }

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int i,j; for(i=0;i<=10;i++) { j+=5; assert(i<5); }

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Write out a function that prints out all the permutations of a string. For example, abc would give you abc, acb, bac, bca, cab, cba. You can assume that all the characters will be unique.

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#include<stdio.h> int main() { int x=2,y; y=++x*x++*++x; printf("%d",y); } Output for this program is 64. can you explain how this output is come??

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Set up procedure for generating a wire frame display of a polyhedron with the hidden edges of the object drawn with dashed lines

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void main () { int x = 10; printf ("x = %d, y = %d", x,--x++); } a. 10, 10 b. 10, 9 c. 10, 11 d. none of the above

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write a simple calculator c program to perform addition, subtraction, mul and div.

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main() { int i=-1,j=-1,k=0,l=2,m; m=i++&&j++&&k++||l++; printf("%d %d %d %d %d",i,j,k,l,m); }

1 Answers  

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const int perplexed = 2; #define perplexed 3 main() { #ifdef perplexed #undef perplexed #define perplexed 4 #endif printf("%d",perplexed); } a. 0 b. 2 c. 4 d. none of the above

1 Answers   emc2, HCL,

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main() { int i = 0xff ; printf("\n%d", i<<2); } a. 4 b. 512 c. 1020 d. 1024

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