我被困于尝试从算法书中实现伪代码。我的代码会编译并打印出正确的答案，只是我要打印的某些信息未正确显示。控制台输出如下所示：
Correct Solution Tested:
max_left= 7
max_right= 10
sum= 43
Failing Outputs:
curr_cross_low = 17; curr_cross_high = -1; curr_cross_sum = 38
curr_cross_low = -1; curr_cross_high = -1; curr_cross_sum = 18
curr_cross_low = 32766; curr_cross_high = -272632720; curr_cross_sum = 43
max_left_full= 32766
max_right_full= -272632512
sum_full= 43
Program ended with exit code: 0
打印出来的前三个值是正确算法的结果，该结果是通过蛮力实现算法的一部分而得出的。在代码中，这就是函数“ findMaxCrossingSubarray”本身。输出的第二部分是当我执行完整算法“ findMaximumSubarray”时。我认为应该打印出表明正在接近解决方案的结果。变量“ sum_full”给出的最终答案似乎是正确的，因为它与书中所说的蛮力解决方案相匹配。
我一直在尝试找到如何打印正确的max_left_full和max_right_full值，而不是我认为是内存地址的值。我的意思是，如果我在一个地方更改指针，它会使解决方案不正确或打印出内存地址。
有没有一种简单的方法可以找到我可能会丢球的地方？
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
//returns a pointer to a value equal to the set of changes
int * returnPriceChanges(int sz, int A[]){
int MAX_SIZE = 256;
static int* C;
C = malloc(MAX_SIZE *sizeof(int));
int i;
for(i=0;i<sz-1;i++){
C[i]=A[i+1]-A[i];
}
return C;
}
int findMaxCrossingSubarray(int A[], int low, int mid, int high, int* max_left, int* max_right){
double left_sum = -INFINITY;
int sum = 0;
for(int i=mid;i>=low;i--){
sum=sum+A[i];
if(sum > left_sum){
left_sum = sum;
*max_left = i;
}
}
double right_sum = -INFINITY;
sum = 0;
for(int j=mid+1; j<=high;j++){
sum=sum+A[j];
if(sum > right_sum){
right_sum = sum;
*max_right = j;
}
}
return (*max_left, *max_right, left_sum+right_sum);
}
int findMaximumSubarray(int A[], int low, int high){
int curr_left_low, curr_left_high, curr_left_sum;
int curr_right_low, curr_right_high, curr_right_sum;
int curr_cross_low, curr_cross_high, curr_cross_sum;
int mid = 0;
int* temp_max_left, temp_max_right;
if(high==low){
return(low, high, A[low]);
}
else{
mid =floor((high+low)/2);
curr_left_low, curr_left_high, curr_left_sum = findMaximumSubarray(A, low, mid);
curr_right_low, curr_right_high, curr_right_sum = findMaximumSubarray(A, mid+1,high);
curr_cross_low, curr_cross_high, curr_cross_sum = findMaxCrossingSubarray(A, low, mid, high, &temp_max_left, &temp_max_right);
if(curr_left_sum>=curr_right_sum && curr_left_sum>=curr_cross_sum){
return (curr_left_low, curr_left_high, curr_left_sum);
}
else if(curr_right_sum>= curr_left_sum && curr_right_sum>=curr_cross_sum){
return (curr_right_low, curr_right_high, curr_right_sum);
}
else{
printf("curr_cross_low = %d; curr_cross_high = %d; curr_cross_sum = %d\n", curr_cross_low, curr_cross_high, curr_cross_sum);
return (curr_cross_low, curr_cross_high, curr_cross_sum);
}
}
}
int main(){
int prices[] = {100,113,110,85,105,102,86,63,81,101,94,106,101,79,94,90,97};
int szPrices = sizeof(prices)/sizeof(prices[0]);
int changes[szPrices-1];
int *P;
P = returnPriceChanges(szPrices,prices);
//set C = to list of changes
for(int i=0; i<szPrices-1; i++){
changes[i]=*(P+i);
}
int max_left, max_right, sum;
max_left, &max_right, sum = findMaxCrossingSubarray(changes, 0, 8, 16, &max_left, &max_right);
printf("\nCorrect Solution Tested: \nmax_left= %d \nmax_right= %d \nsum= %d\n\n", max_left, max_right, sum);
printf("\nFailing Outputs:\n");
int max_left_full, max_right_full, sum_full;
max_left_full, &max_right_full, sum_full = findMaximumSubarray(changes, 0, 16);
printf("\nmax_left_full= %d \nmax_right_full= %d\nsum_full= %d\n\n", max_left_full, max_right_full, sum_full);
return 0;
}
答案 0 :(得分：1)
您不能从C中的函数返回元组。当您在C中使用逗号分隔值时，整个表达式将仅求值最后一个成员。
所以当您写：
a, b, c = some_function();
这真的意味着：
/* do nothing */, /* do nothing */, c = some_function();
如果要返回复合数据结构，请使用struct，即
struct subarray
{
int low;
int high;
int sum;
};
void findMaximumSubarray(int A[], int low, int high, struct subarray * result);
如果结构很小，并且您使用的是现代编译器，并且未在嵌入式系统上运行，那么您还可以按值返回该结构：
struct subarray findMaximumSubarray(int A[], int low, int high);
后一种语法简化了用法，但是如果您以这种方式启动returning huge structs，则可能成为问题。