Answer / harsha

Going by the answer 7..

max=0;
sum=0;

for i in 0 .. n
sum = sum + numbers[i];
if( sum > max ) {
max = sum;
}
else if (sum<0) {
sum=0;
}

#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??

1 Answers  

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#include<stdio.h> void fun(int); int main() { int a; a=3; fun(a); printf("\n"); return 0; } void fun(int i) { if(n>0) { fun(--n); printf("%d",n); fun(--n); } } the answer is 0 1 2 0..someone explain how the code is executed..?

1 Answers   Wipro,

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In a gymnastic competition, scoring is based on the average of all scores given by the judges excluding the maximum and minimum scores. Let the user input the number of judges, after that, input the scores from the judges. Output the average score. Note: In case, more than two judges give the same score and it happens that score is the maximum or minimum then just eliminate two scores. For example, if the number of judges is 5 and all of them give 10 points each. Then the maximum and minimum score is 10. So the computation would be 10+10+10, this time. The output should be 10 because 30/3 is 10.

0 Answers   TCS,

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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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What is the problem with the following code segment? while ((fgets(receiving array,50,file_ptr)) != EOF) ;

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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Write a single line c expression to delete a,b,c from aabbcc

2 Answers   Microsoft,

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main() { char *a = "Hello "; char *b = "World"; clrscr(); printf("%s", strcpy(a,b)); } a. “Hello” b. “Hello World” c. “HelloWorld” d. None of the above

4 Answers   Corporate Society, HCL,

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void ( * abc( int, void ( *def) () ) ) ();

1 Answers  

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Give a oneline C expression to test whether a number is a power of 2?

25 Answers   EA Electronic Arts, Google, Motorola,

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void main() { int i=10, j=2; int *ip= &i, *jp = &j; int k = *ip/*jp; printf(“%d”,k); }

1 Answers  

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main( ) { int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}}; printf(“%u %u %u %d \n”,a,*a,**a,***a); printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1); }

2 Answers  

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