int i =10
main()
{
int i =20,n;
for(n=0;n<=i;)
{
int i=10
i++;
}
printf("%d", i);
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Answer / jai
printf() will never be executed since for() is an infinite
loop.
| Is This Answer Correct ? | 30 Yes | 2 No |
Answer / vignesh1988i
it will not print i value at all , since the loop is always
true for all cases. since the termination condition is
depending upon n&i,always n<=i, what ever possibe positive
value greater than zero ,it may be.
| Is This Answer Correct ? | 5 Yes | 0 No |
Answer / dips
it doesnt have any terminating condition so it will be
infinite loop ,no output
| Is This Answer Correct ? | 4 Yes | 0 No |
Answer / guest
the print would be 20.
the problem is about scope. the first i=10 is global scope.
But inside main() comes function scope. So i=20. The i
inside the for loop is of block scope and does not affect
the i outside it.
| Is This Answer Correct ? | 5 Yes | 9 No |
find the output? void r(int a[],int c, int n) { if(c>n) { a[c]=a[c]+c; r(a,++c,n); r(a,++c,n); } } int main() { int i,a[5]={0}; r(a,0,5); for(i=0;i<5;i++) printf("\n %d",a[i]); getch(); }
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Just came across this question, felt worth sharing, so here it is I want you to make a C/C++ program that for any positive integer n will print all the positive integers from 1 up to it and then back again! Let's say n=5 I want the program to print: 1 2 3 4 5 4 3 2 1. Too easy you say? Okay then... You can ONLY USE: 1 for loop 1 printf/cout statement 2 integers( i and n) and as many operations you want. NO if statements, NO ternary operators, NO tables, NO pointers, NO functions!
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