解释给定代码的输出

Question

In the below code, variable i has been declared globally as well as locally in the for loop . Due to high precedence of local variable, i will be initialized with the value 10 . But in the first occurrence of the loop, due to i++in the for loop, value of i will become 11, so should it not exit the loop after the first instance only?

#include<stdio.h>
int main(){
int i;
for(i=0;i<5;i++){
     int i=10;
     printf(" %d",i);
     i++;
}
return 0;

}

PS: The answer is 10 10 10 10 10

Answer 1

Precedence is not just about initialization. The variable i inside the loop body is a different variable from the one outside. It "shadows" the outer i , making it inaccessible in the loop body.

The loop control statement is outside the loop body. Its i is the outer i , and nothing done to the inner i has any effect on it, so it starts at 0 and counts up to 5. At each loop iteration, the inner i is re-initialized to 10, and that's the one that gets printed.

Answer 2

The reason it always prints is because the inner i shadows the outer i . For the same reason the i++; inside the loop increments the inner i , not the i that's used as the for loop counter.

But the i that controls the loop gets incremented after the scope of the inner i is over and has no relation to i++; done inside the loop. It's completely independent of the inner i and thus the loop runs 5 times.

GCC provides an option to warn about such shadowing: -Wshadow .

Answer 3

Yet another answer because there is one aspect missing so far. If you change your code slightly:

#include<stdio.h>     
int main(){
int i;
for(i=0;i<5;i++){
     static int i=10;
     printf(" %d",i);
     i++;
}
return 0;
}

You will get a more "expected" result:

10 11 12 13 14

A variable inside a block stops to exist after the block (read: pair of curly braces) is left (and it is left here after every step to execute the control statements of for ) ... so with your original code, you get a new inner i each time. Declaring the inner i static still doesn't change the scope (it is only visible inside the loop body), but makes the variable survive and still be available when the same block is entered again.

Answer 4

Yes the answer is correct.

As loop iterates 5 times and for every iteration i inside loop is set to 10 and outer one is overshadowed by inner one. After i=10 its value is printed therefore 5 times 10 is output.