Question

有没有人知道如何解决警告：当我返回new_cost时，return从整数中生成指针而没有强制转换[默认启用]？

int *cost(int num_nodes, edge new_solution[][10])
{
    int new_cost = 0;
    int num_edges = 1; //set number of edges back to default
    int x, y, weight;

    for (x = 1; x <= num_nodes; x++) //print out new_solution
    {
        weight =0;//find the largest tx on each node
        for (y = 1; y <= num_nodes; y++)
            if (new_solution[x][y].label == 1)
            {
                printf("\n Edge %d:(%d %d) energy:%d",
                       num_edges++, x, y, new_solution[x][y].weight);
                if (weight < new_solution[x][y].weight) //find highest energy used per node
                {
                    weight = new_solution[x][y].weight;
                    //printf("\n weight:%d accum:%d", weight, new_cost);
                }
            }
        new_cost += weight; //find total weight
    }
    printf("\n Total cost is %d\n\n", new_cost);
    return new_cost;
}

是因为我无法将权重变量分配给数组吗？对于返回成本，我收到以下函数的相同警告。有谁知道如何解决这个问题？

int *acceptance_prob(float T, int old_cost, int new_cost, edge new_solution[][10], 
                     edge current_SA[][10], FILE *fp, int num_nodes)
{
    int delta = 0, x, y;
    float ap = 0.0;
    int cost;

    delta = (old_cost - new_cost);
    ap = (exp(delta/T)); //this is the typical equation used

    if (new_cost < old_cost)//if new_solution has less energy select it
    {
        fprintf(fp, "NO");
        for (x = 1; x <= num_nodes; x++)
            for (y = 1; y <= num_nodes; y++)
                current_SA[x][y] = new_solution[x][y];
        old_cost = new_cost;
    }
    else //if new_solution uses more energy maybe select it
    {
        if (ap > rand_float())
        {
            fprintf(fp,"YES");
            for (x = 1; x <= num_nodes; x++)
                for (y = 1; y <= num_nodes; y++)
                    current_SA[x][y] = new_solution[x][y];
            old_cost = new_cost;
        }
    }
    cost = old_cost;

    return cost;
}

Answer 1

有没有人知道如何解决警告：当我返回new_cost时，返回从整数中生成指针而没有强制转换[默认启用]？

将函数的返回类型从int*更改为int。

从该函数返回int*无论如何都没有意义。

return从整数中生成指针而不进行强制转换[默认情况下启用]

1 个答案: