Reputation: 416
I know that logical operators do short-circuit checking. That is, if there is a statement like A && B && C
, then if A
is false, B
and C
are not evaluated. But is this also true in cases where B
and C
are function calls?
For example, the return statement in this code:
bool areIdentical(struct node * root1, struct node *root2)
{
/* base cases */
if(root1 == NULL && root2 == NULL)
return true;
if(root1 == NULL || root2 == NULL)
return false;
/* Check if the data of both roots is same and data of left and right
subtrees are also same */
return (root1->data == root2->data && //I am talking about this statement
areIdentical(root1->left, root2->left) &&
areIdentical(root1->right, root2->right) );
}
Upvotes: 2
Views: 982
Reputation: 657
yes, it is true in function calls also.
#include<stdio.h>
void main()
{
if(0&&printf("hello"))
{
printf("true");
}
else
printf("false");
}
for example consider the above code, it will give output as false. However replacing 0 by 1 in "if condition' will give output as "hellotrue."
Upvotes: 0
Reputation: 4366
Yes, the functions are not called if root1->data == root2->data
is false
.
Simple check is to do this:
#include <unistd.h>
#include <stdlib.h>
int main(void)
{
write(1, "z", 1);
if ((1 == 0) && write(1, "a", 1) && write(1, "b", 1))
{
write(1, "c", 1);
}
write(1, "d", 1);
return (EXIT_SUCCESS);
}
Upvotes: 3
Reputation: 158479
The Logical and operator will short circuit regardless of what the operands are, if we look at the draft C99 standard section 6.5.13
Logical AND operator paragraph 4 says(emphasis mine):
Unlike the bitwise binary & operator, the && operator guarantees left-to-right evaluation; there is a sequence point after the evaluation of the first operand. If the first operand compares equal to 0, the second operand is not evaluated.
Note, the second operand will not be evaluated only if the first is false
. Also note it guarantees right to left evaluation and a sequence point after the first evaluation.
Upvotes: 3