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It is 66, 0

Static variables will be declared once only and though g is declared globally, it is also declared inside main again, so while printing g, the one which declared locally will be preferred and not the one which is declared globally.

66 0
coz while printing 'g' it would take the local variable..and static and global variables retain values.

do you remember ..one question was from recursion..and we have to tell the output..i gues f(1)..?

66,0

Any 1 got any email confirmation abt FTE written?

The output would be add function was not declared.
C is case sensitive ;)

It is 66, 0

Static variables will be declared once only and though g is declared globally, it is also declared inside main again, so while printing g, the one which declared locally will be preferred and not the one which is declared globally.

It's 66, 0, static varaible s will be declared and initialized once.

The output will be 66,11 cause every time you pass i and that value is get added with g in function add

So, in last iteration of for loop i will be 11 and so after completion of add the value of g will be 11

priority of local variable is greater than global variable so value of g inside main function is what is initialize but s which is static variable it will initialize one times and s will 66

priority of local variable is greater than global variable so value of g inside main function is what is initialize but s which is static variable it will initialize one times and s will 66

answer should be 66, 0

value of g should be 0 because is scope is only inside main
value of j is 66(1+2+3+...+11) as its value depends upon static int variable which keeps on adding contiguous integers from 1 to 11 .and in the last iteration returned value will become value of j

Error..!!! come
undefined reference to 'add'.
it's a tricky question
otherwise answer will be 66,0

{66 0 is the real answer}

#include<stdio.h>
int g = 0;
int Add(int i)
{
static int s=0;
s =s+i;
g=g+i;
return s;
}
int main()
{
int ,j;
for(int i=1;i<=11;i++)
{ int g=0;s=0;
j = add(i);
}
printf("%d,%d",j,g);
return 0;
}


in this g declared in for loop then scop of g inside for loop. and printf outside for loop so it will print globa g i.e 66 coz global ge updated in add function.

ans ids 66,66

initially there was compile error as add has undefined reference.Add was used in function definition and add was used in calling function.Finally it is 66 as s is declared to be static and g=0 as g is initialised to 0 globally;

I remember that in the original question, the variable g was declared inside the for loop not in the main. So the output will be 66, 66

i hope it is 11 ,0

i hope u right ,its 11,0