即使参数是函数调用，C也会使用短路评估吗? [英] Does C use short circuit evaluation even when arguments are function calls?

问题描述

我知道逻辑运算符会进行短路检查.也就是说，如果存在类似A && B && C的语句，则如果A为false，则不会评估B和C.但这在B和C是函数调用的情况下也是如此吗?

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?

例如，此代码中的return语句:

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) );  
}

推荐答案

是的，如果root1->data == root2->data是false，则不会调用这些函数.

Yes, the functions are not called if root1->data == root2->data is false.

简单的检查就是这样做

#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);
}

这篇关于即使参数是函数调用，C也会使用短路评估吗?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！