Question

with fear that I may overstep another question of mine (although this is a new problem alltogether) I still ask this question.

我有这一法典:

int blob_count(int y, int x, int gridCopy[][5], int sum){

    //Local vars
    int posX, posY;

    //Find the position 1 behind and 1 above the starting point, and start the loop there
    for(posX = -1;posX <=1; posX++){
        for(posY = -1; posY <= 1; posY++){
            if((y + posY) >= 0 && (x + posX) >= 0){
                if((y + posY) <= 5 && (x + posX) <= 5){
                    if(gridCopy[posY+y][posX+x] == 1){
                        //Set the starting point to 0 (so it wont get calculated again)
                        gridCopy[posY+y][posX+x] = 0;



                        y = posY+y;
                        x = posX+x;

                        sum++;
                        blob_count(y, x, gridCopy, sum);
                    }
                }
            }
        }
    }

    return sum;
}

问题是,每起休养费中,金额是1美元,是错失价值。通过为每次休养手术提供印刷结果:

sum = 1  
sum = 2  
sum = ...  
sum = n

然而,通过在休息室外(返回前的权利)印刷出钱,那是巨大的。

sum = n  
sum = ...  
sum = 2  
sum = 1  

return sum; // = 1

这一点显然是错误的,因为我想到的是全部数字,而不是最低数字。返回是否对错位有价值? 我试图在休养呼吁(在休息期间)之后将其付诸行动。

Answer 1

Okay let s get rid of the extra bits and simplify your problem down to the essentials. You have:

int blob_count(int sum)
{
    sum++;

    if (sum < 10)
        blob_count(sum);

    return sum;
}

If you add printf("sum==%d ", sum) right before the return then it will be called first at the innermost recursion (where sum == 10), then it will return to the next level out where sum == 9, print that, return to sum == 8 and so on.

如果你在回程电话上打到blob_count(sum),那么,在你重新开标之前,你将重印这些数值,以便从sum=0, amount = 1开始。

如果您希望<代码>sum成为您的再次入侵最深的层次,那么你要么通过收益价值回馈:

int blob_count(int sum)
{
    sum++;

    if (sum < 10)
        sum = blob_count(sum);

    return sum;
}

或您可以通过点子通过,以便修改原有变量:

void blob_count(int* sum)
{
    *sum++;

    if (*sum < 10)
        blob_count(sum);

    return;
}

第一个可能是你寻求的解决办法。

Answer 2

说了什么。每个休养电话的现值都经过复印,复制件被转至休养电话。如果你想改变功能中的物体,你必须将这些物体而不是物体本身当作点。

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