Answer / susie

Answer :

1

Explanation:

Arrays cannot be passed to functions as arguments and only
the pointers can be passed. So the argument is equivalent to
int * array (this is one of the very few places where [] and
* usage are equivalent). The return statement becomes,
sizeof(int *)/ sizeof(int) that happens to be equal in this
case.

Answer / manoj kumar

Arrays cannot be passed to functions as arguments and only
the pointers can be passed.

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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writte a c-programm to display smill paces

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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

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main() { int i, n; char *x = “girl”; n = strlen(x); *x = x[n]; for(i=0; i<n; ++i) { printf(“%s\n”,x); x++; } }

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What is "far" and "near" pointers in "c"...?

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What is the output for the following program main() { int arr2D[3][3]; printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0])) ); }

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{ int *ptr=(int*)malloc(sizeof(int)); *ptr=4; printf("%d",(*ptr)+++*ptr++); }

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find simple interest & compund interest

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why do you use macros? Explain a situation where you had to incorporate macros in your proc report? use a simple instream data example with code ?

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find A^B using Recursive function

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main() { int i; printf("%d",scanf("%d",&i)); // value 10 is given as input here }

2 Answers   IBM,

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