C:递归调用返回一个值

Question

int find(char* a, trie_node* node, int result){//need to make sure a is not NULL at beginning
    int i,temp;
    if ((a != NULL && a[0] !='\n') && node->children[a[0] - 97] == NULL)
    {
        result = 0;//not found any children satisfied the requirement
    }
    else if ((a != NULL && a[0] !='\n') && node->children[a[0] - 97] != NULL){
        temp = a[0];
        a++;
        find(a, node->children[temp - 97], result);
    } else{//a == NULL, which means end of the find procedure, just return the num_children
        result = node->num_children; //most inner one
    }
    return result;
}

我正在尝试从该函数返回结果。由于它是一个严格的 C 程序，因此我需要在函数末尾添加一个 return 语句。

在 gdb 中跟踪它后，最内部的函数调用返回正确的数字结果。然而，在返回到外部函数的过程中，结果的值会丢失。因此，该函数将返回 0，这是错误的。我如何返回并保留最内层调用的值？

Answer 1

您只需要添加 return 语句即可。您也根本不需要结果参数。尝试重写；

int find(char* a, trie_node* node) {
    //need to make sure a is not NULL at beginning
    int i,temp;
    if ((a != NULL && a[0] !='\n') && node->children[a[0] - 97] == NULL)
    {
        return 0;//not found any children satisfied the requirement
    }
    else if ((a != NULL && a[0] !='\n') && 
             node->children[a[0] - 97] != NULL)
    {
        temp = a[0];
        a++;
        return find(a, node->children[temp - 97]);
    } 
    //a == NULL, which means end of the find procedure, just return the num_children
    return node->num_children; //most inner one
}