what will be the output of this program?
void main()
{
int a[]={5,10,15};
int i=0,num;
num=a[++i] + ++i +(++i);
printf("%d",num);
}
Answers were Sorted based on User's Feedback
Answer / viswanath
Ans: 15.
Exp: a[++i] means i = 1; so 10 + 2 +3
| Is This Answer Correct ? | 7 Yes | 1 No |
Answer / nandkishor
Ans: 19.
Explanation: First expression (++i) execute. Hence i=1;
Than a[++i]=a[2]=15.
And than ++i=3.
Hence evaluation will be 15+3+1=19.
Note that expression closed in braces have high precedence.
| Is This Answer Correct ? | 0 Yes | 0 No |
What is the hidden bug with the following statement? assert(val++ != 0);
main() { int i=3; switch(i) { default:printf("zero"); case 1: printf("one"); break; case 2:printf("two"); break; case 3: printf("three"); break; } }
could you please send the program code for multiplying sparse matrix in c????
why array index always strats wuth zero?
PROG. TO PRODUCE 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1
main() { char *p = “ayqm”; printf(“%c”,++*(p++)); }
29 Answers IBM, TCS, UGC NET, Wipro,
main(){ char a[100]; a[0]='a';a[1]]='b';a[2]='c';a[4]='d'; abc(a); } abc(char a[]){ a++; printf("%c",*a); a++; printf("%c",*a); }
struct aaa{ struct aaa *prev; int i; struct aaa *next; }; main() { struct aaa abc,def,ghi,jkl; int x=100; abc.i=0;abc.prev=&jkl; abc.next=&def; def.i=1;def.prev=&abc;def.next=&ghi; ghi.i=2;ghi.prev=&def; ghi.next=&jkl; jkl.i=3;jkl.prev=&ghi;jkl.next=&abc; x=abc.next->next->prev->next->i; printf("%d",x); }
Is it possible to print a name without using commas, double quotes,semi-colons?
main() { main(); }
Write a complete program that consists of a function that can receive two numbers from a user (M and N) as a parameter. Then print all the numbers between the two numbers including the number itself. If the value of M is smaller than N, print the numbers in ascending flow. If the value of M is bigger than N, print the numbers in descending flow. may i know how the coding look like?
main( ) { int a[ ] = {10,20,30,40,50},j,*p; for(j=0; j<5; j++) { printf(“%d” ,*a); a++; } p = a; for(j=0; j<5; j++) { printf(“%d ” ,*p); p++; } }