Answer / susie

Answer :

4 1

Explanation:

The boolean expression needs to be evaluated only till the
truth value of the expression is not known. j is not equal
to zero itself means that the expression’s truth value is 1.
Because it is followed by || and true || (anything) => true
where (anything) will not be evaluated. So the remaining
expression is not evaluated and so the value of i remains
the same.

Similarly when && operator is involved in an expression,
when any of the operands become false, the whole
expression’s truth value becomes false and hence the
remaining expression will not be evaluated.

false && (anything) => false where (anything) will
not be evaluated.

main() { if ((1||0) && (0||1)) { printf("OK I am done."); } else { printf("OK I am gone."); } } a. OK I am done b. OK I am gone c. compile error d. none of the above

5 Answers   HCL,

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

0 Answers  

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main() { float f=5,g=10; enum{i=10,j=20,k=50}; printf("%d\n",++k); printf("%f\n",f<<2); printf("%lf\n",f%g); printf("%lf\n",fmod(f,g)); }

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Write a C program that defines a 2-dimentional integer array called A [50][50]. Then the elements of this array should randomly be initialized either to 1 or 0. The program should then print out all the elements in the diagonal (i.e. a[0][0], a[1][1],a[2][2], a[3][3], ……..a[49][49]). Finally, print out how many zeros and ones in the diagonal.

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Write a function to find the depth of a binary tree.

13 Answers   Adobe, Amazon, EFI, Imagination Technologies,

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#include<stdio.h> main() { register i=5; char j[]= "hello"; printf("%s %d",j,i); }

2 Answers  

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

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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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how to test pierrot divisor

0 Answers  

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main( ) { static int a[ ] = {0,1,2,3,4}; int *p[ ] = {a,a+1,a+2,a+3,a+4}; int **ptr = p; ptr++; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); *ptr++; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); *++ptr; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); ++*ptr; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); }

2 Answers   Persistent,

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main() { float i=1.5; switch(i) { case 1: printf("1"); case 2: printf("2"); default : printf("0"); } }

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struct Foo { char *pName; }; main() { struct Foo *obj = malloc(sizeof(struct Foo)); clrscr(); strcpy(obj->pName,"Your Name"); printf("%s", obj->pName); } a. Your Name b. compile error c. Name d. Runtime error

3 Answers   HCL,

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