c - 递归函数 C?

Question

为什么这个递归函数有效？我无法理解为什么它不会一直调用自己直到 n=1 然后停止，而不执行低于它调用自己的任何内容。

int main(){
    int t[] = {7,9,6,4,2};
    int min, max;
    search_extremes_rec(t, sizeof(t)/sizeof(t[0]), &min, &max);
    printf("min: %d, max: %d", min, max);
    return 0;
}
void search_extremes_rec(const int t[], int n, int *min, int *max){
    if(n<=1){
        *min = t[0];
        *max = t[0];
    }else{
        search_extremes_rec(t, n-1, min, max);
        if(*min > t[n-1]){
            *min = t[n-1];
        }   
        else if(*max < t[n-1]){
            *max = t[n-1];
        }
    }
}

Answer 1

每当您想了解递归函数的作用时，插入几个打印输出。

void search_extremes_rec(const int t[], int n, int *min, int *max){
        printf("Entering search_extremes_rec. n:%d\n", n);
        if(n<=1){
                printf("Base case. Exiting\n");
                *min = t[0];
                *max = t[0];
        }else{
                printf("Calling recursively...\n");
                search_extremes_rec(t, n-1, min, max);
                printf("Recursive call done. n:%d min:%d max:%d\n", n, *min, *max);

                if(*min > t[n-1]){
                        *min = t[n-1];
                }
                else if(*max < t[n-1]){
                        *max = t[n-1];
                }
        }
}

输出:

$ ./a.out 
Entering search_extremes_rec. n:5
Calling recursively...
Entering search_extremes_rec. n:4
Calling recursively...
Entering search_extremes_rec. n:3
Calling recursively...
Entering search_extremes_rec. n:2
Calling recursively...
Entering search_extremes_rec. n:1
Base case. Exiting
Recursive call done. n:2 min:7 max:7
Recursive call done. n:3 min:7 max:9
Recursive call done. n:4 min:6 max:9
Recursive call done. n:5 min:4 max:9
min: 2, max: 9

如您所见，它确实会一直持续到 n=1。